solved problems in geophysics (2012)
TRANSCRIPT
Solved Problems in Geophysics
Solving problems is an indispensable exercise for mastering the theory underlying the
various branches of geophysics. Without this practice, students often find it hard to
understand and relate theoretical concepts to their application in real-world situations.
This book is a collection of nearly 200 problems in geophysics, which are solved in
detail showing each step of their solution, the equations used and the assumptions made.
Simple figures are also included to help students understand how to reduce a problem to its
key elements. The book begins with an introduction to the equations most commonly used
in solving geophysical problems. The subsequent four chapters then present a series of
exercises for each of the main, classical areas of geophysics – gravity, geomagnetism,
seismology and heat flow and geochronology. For each topic there are problems with
different degrees of difficulty, from simple exercises that can be used in the most elemen-
tary courses, to more complex problems suitable for graduate-level students.
This handy book is the ideal adjunct to core course textbooks on geophysical theory. It is
a convenient source of additional homework and exam questions for instructors, and
provides students with step-by-step examples that can be used as a practice or revision aid.
Elisa Buforn is a Professor of geophysics at the Universidad Complutense de Madrid (UCM)
where she teaches courses on geophysics, seismology, physics, and numerical methods.
Professor Buforn’s research focuses on source fracture processes, seismicity, and seismo-
tectonics, and she is Editor in Chief of Física de la Tierra and on the Editorial Board of the
Journal of Seismology.
Carmen Pro is an Associate Professor at the University of Extremadura, Spain, where she has
taught geophysics and astronomy for over 20 years. She has participated in several
geophysical research projects and is involved in college management.
Agustín Udías is an Emeritus Professor at UCM and is the author of a large number of papers
about seismicity, seismotectonics, and the physics of seismic sources, as well as the
textbook Principles of Seismology (Cambridge University Press, 1999). He has held
positions as Editor in Chief of Física de la Tierra and the Journal of Seismology and as
Vice President of the European Seismological Commission.
Solved Problems in Geophysics
ELISA BUFORNUniversidad Complutense, Madrid
CARMEN PROUniversidad de Extremadura, Spain
AGUSTÍN UDÍASUniversidad Complutense, Madrid
cambridge university press
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Cambridge University Press
The Edinburgh Building, Cambridge CB2 8RU, UK
Published in the United States of America by
Cambridge University Press, New York
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Information on this title: www.cambridge.org/9781107602717
# Elisa Buforn, Carmen Pro and Agustín Udías 2012
This publication is in copyright. Subject to statutory exception
and to the provisions of relevant collective licensing agreements,
no reproduction of any part may take place without
the written permission of Cambridge University Press.
First published 2012
Printed in the United Kingdom at the University Press, Cambridge
A catalogue record for this publication is available from the British Library
Library of Congress Cataloging-in-Publication Data
Buforn, E.
Solved problems in geophysics / Elisa Buforn, Carmen Pro, Agustín Udías.
p. cm.
Includes bibliographical references.
ISBN 978-1-107-60271-7 (Paperback)
1. Geophysics–Problems, exercises, etc. I. Pro, Carmen. II. Udías Vallina, Agustín. III. Title.
QC807.52.B84 2012
550.78–dc23
2011046101
ISBN 978-1-107-60271-7 Paperback
Cambridge University Press has no responsibility for the persistence or
accuracy of URLs for external or third-party internet websites referred to
in this publication, and does not guarantee that any content on such
websites is, or will remain, accurate or appropriate.
Contents
Preface page vii
1 Introduction 1
Gravity 1
Geomagnetism 4
Seismology 6
Heat flow 10
Geochronology 11
2 Gravity 13
Terrestrial geoid and ellipsoid 13
Earth’s gravity field and potential 25
Gravity anomalies. Isostasy 53
Tides 95
Gravity observations 116
3 Geomagnetism 121
Main field 121
Magnetic anomalies 142
External magnetic field 156
Main (internal), external, and anomalous magnetic fields 174
Paleomagnetism 201
4 Seismology 208
Elasticity 208
Wave propagation. Potentials and displacements 211
Reflection and refraction 224
Ray theory. Constant and variable velocity 243
Ray theory. Spherical media 277
Surface waves 307
Focal parameters 324
5 Heat flow and geochronology 335
Heat flow 335
Geochronology 345
Bibliography 352
v
Preface
This book presents a collection of 197 solved problems in geophysics. Our teaching
experience has shown us that there was a need for a work of this kind. Solving problems
is an indispensable exercise for understanding the theory contained in the various branches
of geophysics. Without this exercise, the student often finds it hard to understand and relate
the theoretical concepts with their application to practical cases. Although most teachers
present exercises and problems for their students during the course, the hours allotted to the
subject significantly limit how many exercises can be worked through in class. Although
the students may try to solve other problems outside of class time, if there are no solutions
available this significantly reduces the effectiveness of this type of study. It helps, there-
fore, both for the student and for the teacher who is explaining the subject if they have
problems whose solutions are given and whose steps can be followed in detail. Some
geophysics textbooks, for example, F.D. Stacey, Physics of the Earth; G.D. Garland,
Introduction to Geophysics; C.M. Fowler, The Solid Earth: An Introduction to Global
Geophysics; and W. Lowrie, Fundamentals of Geophysics, contain example problems, and,
in the case of Stacey’s, Fowler’s, and Lowrie’s textbooks, their solutions are provided on
the website of Cambridge University Press. The main difference in the present text is the
type of problems and the detail with which the solutions are given, and in the much greater
number.
All the problems proposed in the book are solved in detail, showing each step of their
solution, the equations used, and the assumptions made, so that their solution can be
followed without consulting any other book. When necessary, and indeed quite often, we
also include figures that allow the problems to be more clearly understood. For a given
topic, there are problems with different degrees of difficulty, from simple exercises that can
be used in the most elementary courses, to more complex problems with greater difficulty
and more suitable for teaching at a more advanced level.
The problems cover all parts of geophysics. The book begins with an Introduction
(Chapter 1) that includes the equations most used in solving the problems. The idea of
this chapter is not to develop the theory, but rather to simply give a list of the equations
most commonly used in solving the problems, at the same time as introducing the reader to
the nomenclature. The next four chapters correspond to the division of the problems into
the four thematic blocks that are classic in geophysics: gravity, geomagnetism, seismology,
and heat flow and geochronology. We have not included problems in geodynamics, since
this would depart too much from the approach we have taken, which is to facilitate
comprehension of the theory through its application to specific cases, sometimes cases
which are far from the real situation on Earth. Indeed, some of the problems may seem a bit
artificial, but their function is to help the student practise with what has been seen in the
vii
theory. Neither did we want to include specific problems of geophysical prospecting as this
would have considerably increased the length of the text, and moreover some of the topics
that would be covered in prospecting, such as gravimetric and geomagnetic anomalies, are
already included in other sections of this work.
Chapter 2 contains 68 problems in gravity divided into five sections. The first section is
dedicated to the terrestrial geoid and ellipsoid, proposing calculations of the parameters
that define them in order to help better understand these reference surfaces. The second
corresponds to calculating the gravitational field and potential for various models of the
Earth, including the existence of internal structures. Gravity anomalies are dealt with in the
third section, with a variety of problems to allow students to familiarize themselves with
the corrections to the observed gravity, with the concept of isostasy, and with the Airy and
Pratt hypotheses. The fourth section studies the phenomenon of the Earth’s tides and their
influence on the gravitational field. The last section is devoted to the observations of
gravity from measurements made with different types of gravimeters and the corrections
necessary in each case. We also include the application of these observations to the
accurate determination of different types of height.
Chapter 3 contains 42 problems in geomagnetism divided into five sections. The first is
devoted to the main (internal) field generated by a tilted dipole at the centre of the Earth. It
includes straightforward problems that correspond to the calculation of the geomagnetic
coordinates of a point and the theoretical components of the magnetic field. This section
also introduces the student to the use of the principal units used in geomagnetism. The
second considers the magnetic anomalies generated by different magnetized bodies and
their influence on the internal field. The third section is devoted to the external field and its
variation with time. In the fourth section, we propose problems of greater complexity
involving the internal field, the external field, and anomalous magnetized bodies at the
same time. The last section is devoted to problems in paleomagnetism.
Chapter 4 contains 69 problems in seismology divided into seven sections. The first
presents some simple exercises on the theory of elasticity. The second addresses the problem
of the propagation of seismic energy in the form of elastic waves, resolving the problems on
the basis of potentials, and calculating the components of their displacements. We study the
reflection and refraction of seismic waves in the third section. The fourth is devoted to the
problem of wave propagation using the theory of ray paths in a planemedium of constant and
variable velocity of propagation. The fifth studies the problem of the propagation of rays in a
spherical medium of either constant or variable propagation velocity, with the calculation of
the travel-time curves for both plane and spherical media. The sixth section contains
problems in the propagation of surface waves in layered media. The seventh section is
devoted to problems of calculating the focal parameters and the mechanism of earthquakes.
Chapter 5 includes 11 problems in heat flow with the propagation of heat in plane and
spherical media, and seven problems in geochronology involving the use of radioactive
elements for dating rocks.
Finally, we provide a bibliography of general textbooks on geophysics and of specific
textbooks for the topics of gravity, geomagnetism, and seismology. We have tried to
include only those most recent and commonly used textbooks which are likely to be found
in university libraries.
viii Preface
In sum, the book is a university text for students of physics, geology, geophysics,
planetary sciences, and engineering at the undergraduate or Master’s degree levels. It is
intended to be an aid to teaching the subjects of general geophysics, as well as the specific
topics of gravity, geomagnetism, seismology, and heat flow and geochronology contained
in university curricula.
The teaching experience of the authors in the universities of Barcelona, Extremadura,
and the Complutense of Madrid highlighted the need for a work of this kind. This text is
the result of the teaching work of its authors for over 20 years. Thanks are due to
the generations of students over those years who, with their comments, questions, and
suggestions, have really allowed this work to see the light. We are also especially grateful
to Prof. Greg McIntosh who provided us with some problems on paleomagnetism, to
Prof. Ana Negredo for her comments on heat flow and geochronology problems, and to
Dr R.A. Chatwin who worked on translating our text into English.
The text is an extension of the Spanish edition published by Pearson (Madrid, 2010).
E. BUFORN, C. PRO AND A. UDÍAS
ix Preface
1 Introduction
Gravity
As a first approximation the Earth’s gravity is given by that of a rotating sphere. The
gravitational potential of a sphere of mass M is:
V ¼GM
r
where r is the position vector (Fig. A) and G the universal gravitational constant.
If the sphere is rotating with angular velocity o the centrifugal potential at a point on the
surface is given by
F ¼1
2o2
r2sin2y
where y is the angle that r forms with the axis of rotation.
The gravity potential is their sum U ¼ V þ F.
The value of the acceleration due to gravity (the gravity ‘force’) is given by the gradient
of the potential:
g ¼ rU
The radial component of the gravity force is given by
gr ¼ GM
r2
þ ro2sin2y
The potential of the Earth to a first-order approximation corresponds to that of a rotating
ellipsoid, and is given by
U ¼GM
a
a
rJ2
2
a
r
3
3sin2’ 1
þm
2
r
a
2
cos2’
where ’ ¼ 90º y is the geocentric latitude and a the equatorial radius.
The coefficient m is the ratio between the centrifugal and gravitational forces on the
sphere of radius a at the equator:
m ¼a3o2
GM
1
The dynamic form factor J2 is defined as
J2 ¼C A
a2M
where C and A are the moments of inertia about the axis of rotation and an equatorial axis.
The flattening of the ellipsoid (the shape of the Earth to a first-order approximation) of
equatorial and polar radius a and c is:
a ¼a c
aIn terms of J2 and m,
a ¼3
2J2 þ
m
2
The dynamic ellipticity is
H ¼C A
C
The gravity flattening is
b ¼gp ge
ge
where gp and ge are the normal values of gravity at the pole and the equator, respectively.
w
q
North Pole
Equator
P
a
r
l
Fig. A
2 Introduction
The gravity at a point of geocentric latitude ’ = 90º y is
g ¼ ge 1þ bsin2’
The geocentric latitude of a point is the angle between the equator and the radius vector of
the point. The geodetic latitude is defined as the angle between the equatorial plane and the
normal to the ellipsoid surface at a point. Astronomical latitude is the angle between the
equatorial plane and the observed vertical at a point.
The normal or theoretical gravity at a point of geocentric latitude ’ referred to the
GRS1980 reference ellipsoid is
g ¼ 9:780327 ð1þ 0:0053024 sin2’ 0:0000059 sin22’Þms2
The effect of the Sun and Moon on the Earth is to produce the phenomenon of the tides.
If one considers more generally the tidal effect due to an astronomical body of mass M at a
distance R from the centre of the Earth, one must add the corresponding potential, which, in
the first-order approximation, is given by
c ¼GMr
2
2R33cos2# 1
where r is the geocentric radius vector of the point, and # is the angle the position vector r
forms with the distance vector R.
Gravity anomalies, defined as Dg ¼ g –g, are the effects of the existence of anomalous
masses inside the Earth. The gravity anomaly along the Z (vertical) axis at a point distance
x along the horizontal axis produced by a sphere of radius R, density contrast Dr, and
buried at a depth d, is given by
g x; zð Þ ¼@Va
@z¼
GMðzþ dÞ
x2 þ ðzþ dÞ2h i3=2
where Va is the potential produced by the anomalous spherical mass DM ¼ 4/3pR3 Dr.
For problems in two dimensions, one uses the anomaly produced by an infinite horizon-
tal cylinder at depth d, perpendicular to the plane under consideration. The anomalous
potential is given by
Va ¼ 2pGra2ln1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ ðzþ dÞ2q
0
B
@
1
C
A
and the anomaly by
gðx; zÞ ¼ @Va
@z¼
2pGra2ðzþ dÞ
x2 þ ðzþ dÞ2
To correct for the height above sea level at which measurements are made, one uses the
concepts of the free-air and Bouguer anomalies. The free-air anomaly is
gFA ¼ g gþ 3:086h
3 Gravity
where g is the observed gravity, h the height in metres, and the anomaly is obtained in
gu (gravity units) mms2.
The Bouguer anomaly is
gB ¼ g gþ ð3:086 0:419rÞh
with r being the density of the plate of thickness h.
To account for isostatic compensation at height in mountainous areas, one adds an
isostatic correction which can be calculated assuming either the Airy or Pratt hypotheses.
With the Airy hypothesis, the root t of a mountain is given by
t ¼rc
rM rch
where rc and rM are the densities of the crust and mantle, and h is the height of the
mountain. For an ocean zone, with water density ra, the anti-root is
t0 ¼
rc rarM ra
h0
With the Pratt hypothesis, the density contrast in a mountainous area is
r ¼ r r0 ¼h
Dþ hr0
where D is the level of compensation, h the height of the mountain, and r0 the density at
sea level. For an oceanic zone of depth h0:
r0 ¼r0D rah
0
D h0
r ¼ r0 r0
The isostatic correction can be calculated using a cylinder of radius a and height b, whose
base is located at a distance c beneath the point, and with density contrast Dr:
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
For mountainous zones, with the Airy hypothesis: b ¼ t, c ¼ h þ H þ t (H ¼ crustal
thickness, h ¼ height of the point); and with the Pratt hypothesis: b ¼ D, c ¼ D þ h.
Geomagnetism
To a first approximation, the internal magnetic field of the Earth can be approximated by a
centred dipole inclined at 11.5 to the axis of rotation. The potential created by a magnetic
dipole at a point distant r from its centre and forming an angle ywith the axis of the dipole is
F ¼Cm cos y
r2
4 Introduction
where C ¼ m0/4p with m0 ¼ 4p 107 Hm1, and m is the dipole moment in units
of Am². The product Cm is given in Tm3.
The components of the magnetic dipole field B are:
Br ¼ @F
@r¼
2Cm cos y
r3
By ¼ 1
r
@F
@y¼
2Cm sin y
r3
In the centred dipole approximation for the Earth’s magnetic field, the geomagnetic
coordinates (f*, l*) of a point (y ¼ 90º f*) in terms of its geographic coordinates
(f, l) and those of the Geomagnetic North Pole (GMNP) (fB, lB) can be calculated using
the expressions of spherical trigonometry (Fig. B):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
sin l ¼sinðl lBÞ cosf
cosf
The vertical and horizontal components of the field, the geomagnetic constant B0, and the
total field are given by:
Z ¼ 2B0 sinf
H ¼ B0 cosf
B0 ¼Cm
a3
F ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
H2 þ Z2p
¼ B0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ 3sin2fq
The units used for the components of the magnetic field are the tesla T and the nanotesla
nT ¼ 109 T. The NS (X*) and EW (Y*) components are
X ¼ H cosD
Y ¼ H sinD
and the declination and inclination are given by
90º – fB
90º – f
GMNP
GNP
180º – l∗
q = 90º – f∗ D∗
l – lB
P
Fig. B
5 Geomagnetism
sinD ¼ cosfB sinðl lBÞ
cosf
tan I ¼ 2 tanf
The radius vector at each point of the line of force is:
r ¼ r0cos2f ¼ r0sin
2y
where r0 is the radius vector of the point of the line of force located at the geomagnetic
equator.
Magnetic anomalies are produced by magnetic materials within the Earth. The anomal-
ous potential due to a vertical dipole buried at depth d is
FA ¼Cmcosy
r2
¼Cmðzþ dÞ
x2 þ ðzþ dÞ2h i3=2
The vertical (z) and the horizontal (x) components of the magnetic anomaly at the surface
(z ¼ 0) produced by a vertical magnetic dipole at depth d are:
Z ¼Cmð2d2 x2Þ
ðx2 þ d2Þ5=2
X ¼3Cmxd
ðx2 þ d2Þ5=2
The Earth is affected by an external magnetic field produced mainly by the activity of the
Sun. This field is variable in time, with distinct periods of variation. The most noticeable is
the diurnal variation (Sq) with a maximum at 12 noon local time. The most important non-
periodic variations are the so-called magnetic storms.
Seismology
Earthquakes produce elastic waves which propagate through the interior and along the
surface of the Earth. Using the plane-wave approximation, the displacements of the
internal P- and S-waves (uiP and ui
S) can be obtained from a scalar potential and a vector
potential:
ui ¼ uiP þ ui
S ¼ r’ð Þi þ r cj
i
’ ¼ A exp ika gjxj at
cj ¼ Bj exp ikb gjxj bt
where A and Bj are the amplitudes, xj the coordinates of the observation point, ka and kb the
wavenumbers, gj are the direction cosines defined from the azimuth az and angle of
incidence i of the ray as:
6 Introduction
g1 ¼ sin i cos az
g2 ¼ sin i sin az
g3 ¼ cos i
and a and b are the P- and S-wave velocities of propagation, respectively, defined from
the Lamé coefficients (l and shear modulus m) and the density r:
vP ¼ a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
s
vS ¼ b ¼
ffiffiffi
m
r
r
Units used are: displacement amplitudes (u) in µm; potential amplitudes (A, Bi) in 103 m2;
wavenumber (k) in km1; and wave velocity (a, b) in km s1.
Poisson’s ratio is defined in terms of the Lamé coefficients as
s ¼l
2 lþ mð Þ
The angle of polarization of S-wave e is defined as
e ¼ tan1 uSH
uSV
where uSH is the amplitude of the SH component, and uSV that of the SV component. SH
and SV are the horizontal and vertical components of the S-wave on the wavefront plane.
The coefficients of reflection V and transmission W are given by the respective
ratios between the amplitudes of the reflected or transmitted potentials and the incident
potential:
V ¼A
A0
W ¼A0
A0
where A0 is the amplitude of the incident wave potential, A that of the reflected potential,
and A0 of the transmitted potential.
Snell’s law for plane media is expressed as
p ¼sin i
v
and for spherical media
p ¼r sin i
v
where p is the ray parameter, i the angle of incidence, v the propagation velocity of the
medium, and r the position vector along the ray.
7 Seismology
In the case of plane media with propagation velocity varying with depth v(z), the
epicentral distance and the travel time of a ray for a surface focus are given by
x ¼ 2
ðh
0
pdzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 p2p
t ¼ 2
ðh
0
2dzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 p2p
where ¼ v1 and h is the depth of maximum penetration of the ray. The variation of the
epicentral distance x with the ray parameter p is given by
dx
dp¼
2
B0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
20 p2p þ 2
ðB
0
dB
dzdz
B2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 p2p
where
B ¼1
v
dv
dz
In spherical media with velocity varying with depth v(r), the epicentral distance,
trajectory along the ray, and travel time are given by
¼ 2
ð
r0
rp
p
r
drffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 r2p
s ¼ 2
ð
r0
rp
drffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 r2
p
t ¼ 2
ð
r0
rp
dr
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 r2
p
where ¼ rv1, r0 is the radius at the surface of the Earth, and rp is the radius at the point
of maximum penetration of the ray.
The variation of the distance from the epicentre D with the ray parameter p in a spherical
medium is
d
dp¼
2
ð1 B0Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
20 p2p þ 2
ðB
0
dB
drdr
ð1 B2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 p2p
where
B ¼r
v
dv
dr
The radial and vertical components (u1 and u3) of surface waves can be obtained from
the potentials ’ and c. The transverse component (u2) is kept apart
u1 ¼@’
@x1
@c
@x3¼ ’;1 c;3
u2 ¼ C exp iksx3 þ ikðx1 ctÞ½
u3 ¼@’
@x3þ
@c
@x1¼ ’;3 þ c;1
8 Introduction
where c is the wave propagation velocity and
’ ¼ A exp ikrx3 þ ikðx1 ctÞ½
c ¼ B exp iksx3 þ ikðx1 ctÞ½
r ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
a2 1
r
s ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b2 1
s
For surface waves, c < b < a, and hence r and s are imaginary.
For dispersive waves, the relationship between the phase velocity c and the group
velocity U is
U ¼ cþ kdc
dk
where k is the wavenumber.
The position of the seismic focus is given by the coordinates of the epicentre (’0, l0)
and the depth h. The time is that of the origin of the earthquake t0. The size is given by
the magnitude which is proportional to the logarithm of the amplitude of the recorded
waves. For surface waves this is:
Ms ¼ logA
Tþ 1:66 logþ 3:3
where A is the amplitude of ground motion in microns, T is the period in seconds, and ∆ the
epicentral distance in degrees.
The magnitude of the moment is given by
Mw ¼2
3logM0 6:1
whereM0 is the seismic moment in Nm (newtonmetres). The seismic moment is related to
the displacement of the fault ∆u and its area S:
M0 ¼ muS
The mechanism of earthquakes is given by the orientation of the fracture plane (fault)
defined by the angles ’ (azimuth), d (dip), and l (slip angle or rake), or by the vectors n
(the normal to the fault plane) and l (the direction of slip).
The elastic displacement of the waves produced by a point shear fault is
uk xs; tð Þ ¼ mu tð ÞS linj þ ljni @Gki
@xj
where Gki is the medium’s Green’s function which, for an isotropic, homogeneous, infinite
medium, and P-waves in the far-field regime, is given by
GPki ¼
1
4pra2rgigkd t
r
a
9 Seismology
The P-wave displacements are given by:
uPk xs; tð Þ ¼_u tð ÞS
4pra3rm linj þ ljni
gigjgk
This equation can be expressed also in terms of the moment tensor Mij
uPk xs; tð Þ ¼_M ij tð Þ
4pra3rgigjgk
Mij is a more general representation of a point source.
Heat flow
The Fourier law of heat transfer by diffusion states that the heat flux _q is proportional to the
gradient of the temperature T:
_q ¼ KrT
where K is the thermal conductivity coefficient. The units of heat flow are Wm2.
The heat diffusion equation, assuming that K is constant, is given by
kr2T þe
rCv
¼@T
@t
where Cv is the specific heat, r the density, e the heat generated per unit volume and unit
time (heat sources), and k the thermal diffusivity:
k ¼K
rCv
If there are no heat sources, the diffusion equation is
kr2T ¼@T
@t
In the case of one-dimensional flow with periodic variation of temperature over time,
one has:
T z; tð Þ ¼ T0 exp
ffiffiffiffiffiffi
o
2k
r
zþ i
ffiffiffiffiffiffiffiffi
o
2kz
r
þ ot
where z is the vertical direction (positive towards the nadir) and o the angular frequency.
In the case of stationary one-dimensional solutions (T constant in time) one obtains from
the diffusion equation:
T ¼ e
2Kz2 þ
_q0Kzþ T0
10 Introduction
where T0 and _q0 are the temperature and flow at the surface (z ¼ 0).
For a spherical Earth, assuming that the thermal conductivity is constant, and that
the amount of heat per unit volume depends only on time, the diffusion equation takes the
form:
K@2T
@r2þ2
r
@T
@r
þ eðtÞ ¼ rCv
@T
@t
where r is the radial direction.
For the stationary case, the above equation reduces to
1
r2
d
drr2 dT
dr
¼ e
K
Integrating twice, one has
T ¼ T0 þe
6KR2 r
2
where T0 is the temperature at the surface (r ¼ R).
Geochronology
Geochronology is based on determining the age of a rock by measuring the decay of its
radioactive elements. In a sample of radioactive material, the number of atoms that have
yet to disintegrate after time t is given by
nt ¼ n0elt
where n0 is the initial number of atoms, and l the decay constant. The rate of decay dn/dt is
the activity R, so that
R ¼ R0elt
where R0 is the initial activity (at t ¼ 0).
The half-life (or period) of the sample is the time it takes for the activity R to fall to half
its initial value. It is given by:
T1=2¼
0:693
l
The mean life-time t of one of the atoms that existed at the start is given by:
t ¼1
l
If a sample consists of NR radioactive nuclei and NE stable nuclei, the time to arrive at the
propotion NE/NR is given by
11 Geochronology
t ¼1
l1þ
NE
NR
If the rubidium–strontium (Rb-Sr) method is used to date a sample, a correction must be
made for the contamination of the stable 86Sr isotope relative to the radioisotope 87Sr:
87Sr86Sr
total ¼87Sr86Sr
initial þ87Rb86Sr
elt 1
This expression corresponds to a straight line (isochrone) of slope elt 1
and intercept
corresponding to the initial content87Sr86Sr
initial.
12 Introduction
2 Gravity
Terrestrial geoid and ellipsoid
1. Calculate the geodetic and geocentric latitudes of a point P on the ellipsoid
whose radius vector is 6370.031 km, given that m ¼ 3.4425 103, GM ¼ 39.86005
1013 m3 s2, and 6356.742 km is the polar radius. Determine J2 and b.
The major semi-axis (equatorial radius) is a and the minor (polar radius) is c, the geocentric
latitude is ’, and the geodetic latitude is ’d (Fig. 1).
The coefficient m is given by the equation
m ¼o2a3
GM
where G ¼ 6.671011 m3 kg1 s2 is the gravitational constant, M the Earth’s mass, and
the angular velocity is o = 2p/T, where T is the rotation period (T ¼ 24 h). We obtain for
the semi-axis a the value
a ¼mGMT 2
4p2
1=3
¼ 6378:127 km
The Earth’s flattening a can be obtained directly since we already know a and c so
a ¼a c
c¼
6378:127 6356:742
6378:127¼ 3:3529 103
The radius vector to the point P is given by the equation r ¼ a(1 a sin2 ’ )
From this equation we can calculate the geocentric latitude ’:
6370:031 ¼ 6378:127 1 3:3539 103 sin2 ’
’ ¼ 37 580 2200
The relation between the geocentric ’ and geodetic ’d latitudes is given by
tan’d ¼1
1 að Þ2tan’
Substituting the already obtained values for a and ’
’d ¼ 38 090 3500
13
The dynamic form factor J2 can be obtained from the equation
a ¼3
2J2 þ
m
2
Then
J2 ¼2
3a
m
2
¼ 1:0878 103
From this value we can determine the gravity flattening b using the equation
b ¼5
2m a ¼ 5:2533 103
2. Taking the first-order approximation, let two points of the ellipsoid at 45 N and 30 S
be situated at distances of 6367.444 km and 6372.790 km from the centre, respectively.
If the normal gravity values are 9.806193 m s2 for the first and 9.793242 m s2 for
the second, calculate: the flattening, gravity flattening, coefficient m, equatorial radius,
polar radius, dynamic form factor, and the Earth’s mass.
Data
r1 ¼ 6367.444 km w1 = 45 N g1 = 9.806193 m s2
r2 ¼ 6372.790 km w2 = 30 S g2 = 9.793242 m s2
The normal or theoretical gravity at a point can be expressed in terms of the normal gravity
at the equator ge, the gravity flattening b, and the latitude of the point ’:
g1 ¼ ge 1þ b sin2 ’1
g2 ¼ ge 1þ b sin2 ’2
If we divide both expressions we obtain:
g1g2
¼1þ b sin2 ’1
1þ b sin2 ’2
P
a
j jd
rC
Fig. 1
14 Gravity
From this expression we can obtain the gravity flattening, since we already know g1, g2,’1,’2:
b ¼g1 g2
g2 sin2 ’1 g1 sin
2 ’2
¼ 5:297 103
The distance r from the centre of the ellipsoid to points on its surface can be given as a
function of the flattening a, the equatorial radius a, and the latitude ’:
r1 ¼ að1 a sin2 ’1Þ
r2 ¼ að1 a sin2 ’2Þ
If we divide both expressions
r1
r2
¼1 asin2’1
1 asin2’2
Thus we obtain the value of the flattening,
a ¼ 3:353 103
From this value we find the equatorial radius,
a ¼r1
1 asin2’1
¼ 6378:137 km
The polar radius c can be found from this value and the flattening:
a ¼a c
aand c ¼ að1 aÞ ¼ 6356:751 km
The coefficient m is obtained from a and b:
aþ b ¼5
2m m ¼
2
5ðaþ bÞ ¼ 3:460 103
From this value we can obtain the value of the Earth’s mass M from
m ¼o2a3
GM
with o ¼2p
Tp where T ¼ 24 hours.
Therefore
M ¼4p2a3
T2Gm¼ 5:946 1024kg
3. Obtain the value of the terrestrial flattening in the first-order approximation, given
that the normal gravity values for two points of the ellipsoid are:
Point 1: w1 ¼ 42º 200 g1 ¼ 980.389 063 Gal
Point 2: w2 ¼ 47º 300 g2 ¼ 980.854 830 Gal
Take the equatorial radius to be 6378.388 km.
15 Terrestrial geoid and ellipsoid
For this problem we use the equations of Problems 1 and 2:
g1 ¼ geð1þ bsin2’1Þ
g2 ¼ geð1þ bsin2’2Þ
If we divide these expressions:
g1g2
¼1þ bsin2’1
1þ bsin2’2
and we can solve for the gravity flattening, b:
b ¼ 5:288 2675 103
Using the gravity flattening b we can determine the value of gravity at the equator, ge:
ge ¼g1
1þ bsin2’1
¼ 978:043 614Gal
Using the following equations
aþ b ¼5
2m
a ¼3
2J2 þ
m
2
m ¼o2a3
GM
we derive the expression
ge ¼GM
a2ð1 bÞ þ o2a
and substituting the values we obtain GM ¼ 3.986 5415 1014 m3 s2
From this value, taking T ¼ 24 hours, we obtain
m ¼o2a3
GM¼
4p2a3
T2GM¼ 3:442 5698 103
And finally the Earth’s flattening is
a ¼5
2m b ¼ 3:318 1575 103
4. P is a point of the terrestrial ellipsoid at latitude 60 ºS and distance to the centre of
6362.121 km. The Earth’s mass is 5.9761 1024 kg and the ratio between the polar
and equatorial semi-axes is 0.9966. Taking the first-order approximation, calculate:
(a) The flattening and the coefficient J2.
(b) The value of normal gravity in mGal at P.
As in the previous problems we use the equations given in Problems 1 and 2.
16 Gravity
(a) From the equation for the flattening (Problem 2)
a ¼ 1c
a¼ 3:4 103
The value of the equatorial radius a is obtained from the equation
a ¼r
1 asin2’¼ 6 378 386m
The coefficient m is given by
m ¼o2a3
GM¼
4p2a3
T2GM¼ 3:4429 103
The gravity flattening b is given by
b ¼5
2m a ¼ 5:2072 103
The dynamic form factor J2 is found from the relation
J2 ¼2a m
3¼ 1:1190 103
(b) The normal gravity at that point is given by
g ¼ geð1þ bsin2’Þ ¼ 981 856:3mGal
5. At a point P on the ellipsoid at latitude 50 ºS, the value of normal gravity is 9.810
752 m s2 and the distance to the centre of the Earth is 6365.587 km. Given that
the mass of the Earth is 5.976 1024 kg and the ratio between the minor and major
semi-axes is c/a ¼ 0.996 6509, calculate:
(a) The flattening, equatorial radius, gravity flattening, dynamic form factor, and
coefficient m.
(b) The normal gravity at the equator.
(c) The centrifugal force at P.
Data
’ ¼ 50 S r ¼ 6365.587 km g ¼ 9.810 752 m s2.
(a) According to Problem 1, the flattening is given by
a ¼ 1c
a¼ 3:349 103
and the equatorial radius a is
a ¼r
1 asin2’¼ 6378:122 km
17 Terrestrial geoid and ellipsoid
Taking T ¼ 24 hours, the coefficient m is then given by
m ¼4p2a3
T2GM¼ 3:4425 103
b and J2 can be obtained from the equations
b ¼5
2m a ¼ 5:2571 103 J2 ¼
2
3a
m
2
¼ 1:0852 103
(b) The normal gravity ge at the equator is given by
ge ¼g
1þ bsin2’¼ 9:780 579m s2
(c) The centrifugal force at point P is given by its radial and transverse components
f ¼ frer þ fyey
where, since 90 y = ’,
fr ¼ o2r sin2 y ¼ o2
r cos2 ’ ¼ 0:013 909m s2
fy ¼ o2r sin y cos y ¼ o2
r cos’ sin’ ¼ 0:016 576m s2
6. Taking the first-order approximation, calculate the Earth’s flattening a, gravity
flattening b, dynamic form factor J2, and polar radius c, given that:
ge ¼ 978.032 Gal (normal gravity at the equator)
a ¼ 6378.136 km (equatorial radius)
GM ¼ 39:8603 1013 m3 s2
and that for a point on the ellipsoid at latitude 60 ºN the normal gravity value is
981 921 mGal.
Calculate also the radius vector of this point and the gravitational potential.
Assuming a first-order approximation, the expression for the normal gravity is
g ¼ geð1þ b sin2 ’Þ
The value of b is given by
b ¼g ge
gesin2’
¼ 5:302 103
The coefficient m, taking T ¼ 24 hours, is
m ¼4p2a3
T2GM¼ 3:442 103
a and J2 are determined from the equations
a ¼5
2m b ¼ 3:303 103 J2 ¼
2
3a
m
2
¼ 1:055 103
18 Gravity
The polar radius c is determined from the flattening
a ¼a c
a) c ¼ að1 aÞ ¼ 6357:069 km
The radius vector at the point of latitude 60 ºN is given by
r ¼ að1 asin2’Þ ¼ 6362:335 km
The gravity potential at the same point, in the first-order approximation, is found using
Mac Cullagh’s formula,
U ¼GM
r1
J2
2
a
r
2
3 sin2 ’ 1
þr
a
3m
2cos2 ’
¼ 6:263 57 107 m2 s2
7. Assuming that the Moon is an ellipsoid of equatorial radius 1738 km and polar
radius 1737 km, with J2 = 3.8195 104 and a mass of 7.3483 1022 kg, calculate its
period of rotation.
First, we calculate the lunar flattening a,
a ¼a c
a¼ 5:7537 104
The coefficient m is found from the values of a and J2:
a ¼3
2J2 þ
m
2) m ¼ 2a 3J2 ¼ 7:5900 106
From the value of m we find the period of rotation T:
m ¼4p2a3
T2GM) T ¼
4p2a3
mGM
12
¼ 27:32 days
8. Calculate, in the first-order approximation, the latitude and radius vector of a point
P of the terrestrial ellipsoid for which the value of normal gravity is 979.992 Gal, given
that the Earth’s mass is 5.976 1024 kg and that normal gravity for another point Q at
50º S latitude is 981.067 Gal, for the equator is 978.032 Gal, and that J2 is 1.083 103.
The gravity flattening b is found from the normal gravity at Q with latitude ’1 = 50:
b ¼
g1ge
1
sin2 ’Q
¼ 5:288 103
Using the same expression and the normal gravity at P we calculate its latitude
sin2’P ¼
gAge
1
b; ’P ¼ 37 590 46:5700
The Earth’s flattening is given by
a ¼15
8J2 þ
b
4¼ 3:353 103
19 Terrestrial geoid and ellipsoid
and the value of the coefficient m by
m ¼ 2a 3J2 ¼ 3:457 103
From the value of m we obtain the equatorial radius a:
a ¼mGM
o2
13
¼ 6 387 062:758m
where we have substituted o = 2p/T, taking T ¼ 24 hours.
Finally, we find the radius vector of point P
r ¼ að1 asin2’Þ ¼ 6 378 946:678m
9. At a point P on the terrestrial ellipsoid of latitude 70 ºS and radius vector 6359.253
km, the value of normal gravity is 982.609 Gal. If the mass of the Earth is 5.9769 1024
kg and the equatorial radius is 6378.136 km, calculate the value of normal gravity at
the Pole, the dynamic form factor, and the centrifugal force at the Pole and the equator.
The flattening a is given by
a ¼1
sin2’1
r
a
¼ 3:3528 103
Putting T ¼ 24 hours, the coefficient m is obtained from the equation
m ¼o2a3
GM¼ 3:4425 103
From a and m we find the gravity flattening b:
b ¼5
2m a ¼ 5:2535 103
Normal gravity at the equator is found from the value of the gravity at point P:
ge ¼g
1þ b sin2 ’¼ 9:780 72m s2
From this value we find the normal gravity at the Pole:
gp ¼ ge 1þ bð Þ ¼ 9:832 10m s2
The dynamic form factor is found from the values of a and m:
J2 ¼2
3a
m
2
¼ 1:0877 103
The centrifugal force is given by the expression:
f ¼ o2r sin2 yer þ o2
r sin y cos yey
At the pole, y = 90 ’ = 0 ! f = 0.
At the equator, y ¼ 90º ! f = o2 a er = 0.033 73 er m s2
20 Gravity
10. Let two points of the ellipsoid be of latitudes w1 and w2, with radius vectors
6372.819 km and 6362.121 km, respectively. The ratio of the normal gravities is
0.997 37, the flattening 3.3529 103, and the gravity flattening 5.2884 103.
Calculate:
(a) The Earth’s mass.
(b) The latitude of each point and the dynamic form factor.
(a) The equatorial radius a can be obtained from the ratio of the two normal gravities
g1g2
¼1þ bsin2’1
1þ bsin2’2
¼1þ b
a r1
aa
1þ ba r2
aa
¼ l
where l ¼ 0.997 37. We solve for a and obtain
a ¼b lr2 r1ð Þ
aþ bð Þ l 1ð Þ¼ 6382:94 km
We calculate the mass of the Earth from the flattening and gravity flattening as in Problem 2:
m ¼2
5aþ bð Þ ¼ 3:4565 103 M ¼
4p2a3
T2Gm¼ 5:9653 1024 kg
(b) The latitudes at each point are calculated from the radius vectors
sin2’1 ¼a r1
aa! ’1 ¼ 43 270 5800
sin2’2 ¼a r2
aa! ’2 ¼ 80 330 4600
The dynamic form factor is obtained from the values of a, b, and m:
b ¼ 2aþ m9
2J2
J2 ¼ 2aþ m bð Þ2
9¼ 1:0831 103
11. Let a point A have a value of gravity of 9793 626.8 gu and a geopotential number
of 32.614 gpu. Calculate the gravity at a point B, knowing that the increments in
dynamic and Helmert height over point A are 271.116 m and 271.456 m, respectively.
Take g45 ¼ 9.806 2940 m s2. Give the units for each parameter.
The dynamic heights at points A and B are given by:
HAD ¼
CA
g45
HBD ¼
CB
g45
where C is the value of the geopotential at each pointP
N
j¼1
gjdhj
!
and g45 the normal
gravity for a point on the ellipsoid at 45º latitude.
21 Terrestrial geoid and ellipsoid
Subtracting both equations,
HBD HA
D ¼CB CAð Þ
g45
Solving for CB,
CB ¼ CA þ g45 HBD HA
D
¼ 298:478 gpu
If heights are given in kmand normal gravity inGal, geopotentials are in gpu (geopotential units)
1 gpu ¼ 1 kGal m ¼ 1 Gal km
The Helmert orthometric height H is given by
H ¼C
g þ 0:0424Hð11:1Þ
where C is in gpu, g in Gal, and H in km.
Solving for H:
H ¼g
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
g2 þ 4 0:0424Cp
2 0:0424
Since the point A is above the geoid (CA > 0), we take the positive solution,
HA ¼ 33:301m
Then, the Helmert height at point B is
HB ¼ HA þHBA ¼ 304:757m
The gravity at point B is calculated using Equation (11.1)
gB ¼CB
HB
0:0424HB ¼ 979:382 75Gal
12. Calculate the value of gravity in gravimetric units and mGal of a point on the
Earth’s surface whose orthometric (Helmert) and dynamic heights are 678.612 m and
679.919 m, respectively, taking g45 ¼ 9.806 294 m s2.
The geopotential is calculated from the dynamic height,
HD ¼C
g45) C ¼ HDg45 ¼ 666:748 gpu
where HD is given in km and g45 in Gal.
Knowing the geopotential, we calculate the gravity from the orthometric (Helmert)
height H, using its definition,
H ¼C
g þ 0:0424H) g ¼
C
H 0:0424H ¼ 982:489 34Gal ¼ 9 824 893:4 gu
13. If at a point on the surface of the Earth of Helmert height 1000 m one observes
a value of gravity of 9.796 235 m s2, calculate the average value of gravity
22 Gravity
between that point and the geoid along the direction of the plumb-line, and
the point’s geopotential number.
The mean value of gravity between a height H and the surface of the geoid is given by
g ¼1
H
ðH
0
gðzÞdz
where g(z) is the value of gravity at a distance z from the geoid along the vertical path to a
point of height H. This value can be obtained using the Poincaré and Prey reduction from
the value of g observed at the Earth’s surface at a point of height H,
gðzÞ ¼ g þ 0:0848ðH zÞ
Then
g ¼1
H
ðH
0
gðzÞdz ¼1
H
ðH
0
g þ 0:0848 H zð Þ½ dz
¼1
Hgzþ 0:0848 Hz 0:0424z2 H
0
g ¼ g þ 0:0424 H ¼ 979:6659Gal
where g is given in Gal and H in km.
The geopotential C can be obtained from the formula for the Helmert height,
H ¼C
g þ 0:0424H) C ¼ ðg þ 0:0424 HÞ H ¼ 979:666 gpu
14. For two points A and B belonging to a gravity measurement levelling line, one
obtained:
gA = 9.801 137 6 m s2
CA = 933.316 gpu
Gross increment elevation: DhBA ¼ 20:340
Increment in dynamic height: HAD HB
D ¼ 20:340m.
Given that the normal gravity at 45 º latitude is 9806 294 gu, calculate the Helmert
heigth of point B.
As in Problem 11, the dynamic heights at A and B are given by
HAD ¼
CA
g45
HBD ¼
CB
g45Subtracting both equations:
HBD HA
D ¼CB CAð Þ
g45
Solving for CB:
CB ¼ CA þ g45 HBD HA
D
¼ 913:371 gpu
23 Terrestrial geoid and ellipsoid
The geopotential at B can be obtained from the gross increment in elevation betweenA and B,
CB ¼ CA þgA þ gB
2
hBA
and, solving for gB,
gB ¼2ðCB CAÞ
hBA gA ¼ 980:103 08Gal
Finally we calculate the orthometric Helmert height at point B,
H ¼C
g þ 0:0424H
Substituting the values for point B, and solving for H we obtain
H ¼gB
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
gB2 þ 4 0:0424CB
p
2 0:0424¼ 931:875m
15. A, B, and C are points connected by a geometric levelling line. Given that the
normal gravity at a latitude of 45º is 980.6294 Gal, complete the following table:
Station A
Dynamic height:
HAD ¼
CA
g45¼ 678:118m
Helmert height:
H ¼C
g þ 0:0424H
H ¼g
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
g2 þ 4 0:0424 Cp
2 0:0424¼ 678:611m
ð15:1Þ
Station B
The geopotential number is found from the dynamic height:
CB ¼ g45HBD ¼ 664:452 gpu
Station
Gravity
(Gal)
Height
Increment (m)
Geopotential
Number (gpu)
Dynamic
Height (m)
Helmert
Height (m)
A 979.88696 – 664.982 ? ?
B ? 0.541 ? 677.577 ?
C 979.88665 ? ? ? 657.134
24 Gravity
From this value and the difference in height with respect to station Awe find the gravity at B:
CB ¼ CA þgA þ gB
2
hAB
from which we get gB = 979.877 84 Gal
The Helmert height of B is found as in station A:
HB ¼ 678:077m
Station C
From the known values of gravity and Helmert height we find the geopotential number
(Equation 15.1)
CC ¼ gHC þ 0:0424H2C ¼ 661:574 gpu
To calculate the difference in height of C with respect to B we begin with the expression
CC ¼ CB þgB þ gC
2
hCB
from which
hCB ¼2 CC CBð Þ
gB þ gC¼ 2:937m
The dynamic height is found directly from the geopotential number:
HCD ¼
CC
g45¼ 674:642m
The complete table is:
Earth’s gravity field and potential
16. Suppose an Earth is formed by a sphere of radius a and density r, and within it
there are two spheres of radius a/2 with centres located on the axis of rotation. The
density of that of the northern hemisphere is 5r and of that of the southern hemi-
sphere is r /5. The value of the rotation is such that m ¼ 0.1. Determine:
(a) The potential U in the r3 approximation.
(b) The values of gr and gu for a point on the equator in the r2
approximation.
Station
Gravity
(Gal)
Height
increment (m)
Geopotential
number (gpu)
Dynamic
height (m)
Helmert
height (m)
A 979.88696 – 664.982 678.118 678.611
B 979.87784 0.541 664.452 677.577 678.077
C 979.88665 2.937 661.574 674.642 657.134
25 Earth’s gravity field and potential
(c) The error made in (b) with respect to the exact solution.
(d) The deviation of the vertical from the radial at the equator.
(a) The gravitational potential is the sum of the potentials of the three spheres
V ¼ V1 þ V2 þ V3 ¼GM
rþGM1
qþGM2
q0ð16:1Þ
where r is the distance from a point P to the centre of the sphere of radius a and mass M,
where M is given by
M ¼4
3pra3
q and q0 are the distances to the centres of the two spheres in its interior in the northern and
southern hemispheres which have differential masses M1 and M2, respectively (Fig. 16).
The differential masses are those corresponding to the difference in density in each case
with respect to the large sphere:
M1 ¼4
3pð5r rÞ
a3
8¼
M
2
differential mass of the sphere in the northern hemisphere
M2 ¼4
3p
r
5 r
a3
8¼
M
10
differential mass of the sphere in the southern hemisphere
The distance q can be calculated using the cosine law
q ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r2 þa
2
2
2a
2r cos y
r
¼ r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þa
2r
2
2a
2r
cos y
r
Considering this expression, 1/q corresponds to one of the generating functions of the
Legendre polynomials. Then 1/q, in the first-order approximation, is given by
1
q¼
1
r1þ
a
2rcos yþ
1
2
a
2r
2
3cos2y 1
Since cos y0 = cos y, 1/q0 is given by
1
q0¼
1
r1
a
2rcos yþ
1
2
a
2r
2
3cos2y 1
P
θ
θ'a
r
r 2
r 3
a/2
α
q
q '
ρ
ρ/5
5ρ
Fig. 16
26 Gravity
If we substitute in Equation (16.1), the potentials for each sphere are given by
V1 ¼GM
rV2 ¼
GM
2
1
rþ
a
2r2cos yþ
a2
8r33cos2y 1
V3 ¼ GM
10
1
r
a
2r2cos yþ
a2
8r33cos2y 1
Then, the total gravity potential is the sum of the three gravitational potentials plus the
potential of the centrifugal force due to the rotation:
U ¼ GM 1þ1
2
1
10
1
rþ
1
4þ
1
20
a
r2cos y
þ1
16
1
80
a2
r3
3 cos2 y 1
þ1
2r2o2 sin2 y
In terms of the coefficient m, given here by m ¼o2a3
GM,
U ¼GM
a
7
5
a
rþ
3
10
a2
r2cos yþ
1
20
a3
r3
3cos2y 1
þ1
2m
r
a
2
sin2y
(b) Using this first-order approximation of the potential, the radial and tangential
components of gravity at the equator, r = a and y = 90, putting m ¼ 0.1, are
gr ¼@U
@r¼ 1:3
GM
a2
gy ¼1
r
@U
@y¼ 0:3
GM
a2
(c) To calculate exactly the value of gr at the equator we have to calculate
the exact contribution of each of the three spheres plus the centrifugal force (m¼ 0.1):
gr ¼ g1rþ g2
rþ g3
r m
GM
a2
g1r¼
GM
a2
g2r¼
GM
2r22cos a
g3r¼
GM
10r23cos a
where
r2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þa2
4
r
¼ r3
and a is the angle which forms r2 and r3 with the equator (Fig. 16)
27 Earth’s gravity field and potential
sin a ¼a=2
r2
¼1ffiffiffi
5p
Then
gr ¼ GM
a21þ
4
5ffiffiffi
5p
4
25ffiffiffi
5p m
¼ 1:19GM
a2
gy ¼ g1y þ g2y ¼ GM
25
4a2
1ffiffiffi
5p þ
GM
105
4a2
1ffiffiffi
5p
0
B
@
1
C
A
¼GM
a2
2
5ffiffiffi
5p þ
2
25ffiffiffi
5p
¼ 0:14GM
a2
The error made in the first-order approximation with respect to the exact solution is
gr ¼ 1:19þ 1:3ð ÞGM
a2¼ 0:11
GM
a2
gy ¼ ð0:3þ 0:14ÞGM
a2¼ 0:16
GM
a2
(d) The deviation of the vertical with respect to the radial direction is given by the angle i
which is determined from the gravity components gr and gy. At the equator this angle is:
• Using the first order approximation
tan i ¼gy
gr¼
0:3
1:3) i ¼ 13:0
• Using the exact values
tan i ¼0:16
1:19) i ¼ 7:6
17. A spherical planet is formed by a sphere of radius a and density r, and inside it a
sphere of radius a/2 and density 5r centred at the midpoint of the radius of the
northern hemisphere. There is no rotation.
(a) Determine J0, J1, and J2.
(b) What is the deviation of the vertical from the radial at the equator?
(a) The total gravitational potential is the sum of the potentials of the two spheres
(Fig. 17) where g, is the attraction due to the potential V1 and g2 that due to the
potential V2:
V ¼ V1 þ V2 ¼GM
rþGM 0
q
where r and q are the distances from a point P to the centres of the large and small spheres,
respectively.
28 Gravity
As we did in Problem 16, for the small sphere of radius a/2 we take the differential
mass M 0
M 0 ¼4
3pa3
85r rð Þ ¼
16
24pra3 ¼
M
2
where M is the mass of the sphere of radius a and density r.
For 1/q we take the first-order approximation of the Legendre polynomial, as we did in
Problem 16:
1
q¼
1
r1þ
a
2rcos yþ
1
2
a
2r
2
3cos2y 1
Then, the expression for the gravitational potential V is:
V ¼GM
rþGM
2
1
rþ
a
2r2cos yþ
1
4
a2
4r33cos2y 1
¼ GM3
2rþ
a
4r2cos yþ
a2
32r33cos2y 1
(b) We know that the potential can be expressed by an expansion in zonal
spherical harmonics (Legendre polynomials) given in the first-order approxima-
tion by
V ¼GM
aJ0
a
r
þa
r
2
J1 cos yþa
r
3
J21
23cos2y 1
Comparing the two expressions we obtain,
5r qP
r
a /2
g1
g2
a
r
q
i
Fig. 17
29 Earth’s gravity field and potential
J0 ¼3
2
J1 ¼1
4
J2 ¼1
16
The components of gravity at the surface of the large sphere (r ¼ a) are:
gr ¼@V
@r¼
GM
a23
21
2cos y
3
323cos2y 1
gy ¼1
r
@V
@y¼
GM
a21
4sin y
3
16cos y sin y
and at the equator, y ¼ 90:
gr ¼ 45GM
32a2
gy ¼ GM
4a2
At the equator the deviation of the vertical with respect to the radial direction is
tan i ¼gy
gr¼
8
45
i ¼ 10:08
18. Suppose an Earth is formed by a sphere of radius a and density r, and within it
there are two spheres of radius a/2 and density 2r with centres located on the axis of
rotation in each hemisphere. If M is the mass of the sphere of radius a, calculate:
(a) ThepotentialU(r,u) and the formof the equipotential surface passing through thePoles.
(b) The component gr of gravity in the first-order approximation for points on the surface.
(c) Calculate gr directly at the Pole and the equator, and compare with the first-order
approximation.
(a) This problem is similar to Problem 16, but now the density of the two spheres is the
same. The total gravity potential is the sum of the gravitational potentials of the
three spheres (V, V1 and V2) plus the potential due to the rotation F:
U ¼ V þ V1 þ V2 þ F ð18:1Þwhere
V ¼GM
r
V1 ¼GM 0
q1
V2 ¼GM 0
q2
F ¼1
2o2
r2sin2y
30 Gravity
and whereM is the mass of the large sphere of radius a andM 0 the differential mass of each
of the small spheres of radius a/2, r is the distance from a point P to the centre of the large
sphere, and q1 and q2 the distances from P to the centres of the small spheres (Fig. 18). As
in Problem 16 the differential mass is given by the difference in density between the large
and the small spheres:
M 0 ¼4
3p 2r rð Þ
a
2
3
¼M
8
The inverse of the distance 1/q can be approximated by
1
q1¼
1
r1þ
a
2rcos yþ
a
2r
2 1
23cos2y 1
and since cos y0 = cos y
1
q2¼
1
r1
a
2rcos yþ
a
2r
2 1
23cos2y 1
The potential of the rotation can be written in terms of the coefficient m = a3o2/GM,
F ¼GM
r
1
2
1
a3a3o2
GMr3
sin2y ¼GM
r
m
2
r
a
3
sin2y
Substituting in Equation (18.1)
U ¼GM
r
10
8þ1
8
a
2r
2
3cos2y 1
þr
a
3 m
2sin2y
ð18:2Þ
r
a
a/2
2r
2r
q
q’
q2
q1
a
r
P
q
Fig. 18
31 Earth’s gravity field and potential
At the Poles, r ¼a and y = 0, and the potential is
Upoles ¼GM
a
10
8þ
2
32
¼21
16
GM
a
The form of the equipotential surface which passes through the Poles (r ¼ a) is obtained
from Equation (18.2)
r ¼GM
Upoles
10
8þ1
8
a
2r
2
3cos2y 1
þr
a
3 m
2sin2y
Making the approximation r ¼ a in the right-hand side:
r ¼32
42a
10
8þm
2sin2yþ
1
323cos2y 1
and substituting cos2 y = 1 sin2 y, we obtain
r ¼ a 132
42
3
32m
2
sin2y
This is the equation of an ellipse with flattening a ¼ (32/42)(3/32m/2). Since there is
symmetry with respect to the axis of rotation, the equipotential surface is an ellipsoid of
revolution.
At the poles: y ¼ 0 ) rp ¼ a.
At the equator: y ¼ 90 ) re ¼ a 1þ32
42
m
2
3
32
Depending on the value of m, we have the following cases,
m
2¼
3
32) re ¼ a ) sphere
m
2<
3
32) re < a ) prolate ellipsoid
m
2>
3
32) re > a ) oblate ellipsoid
m ¼ 0 ) r ¼ a 13
42
< a ) prolate ellipsoid
(b) For the gravity at the Pole, in the first-order approximation, we take the derivative
of the potential (18.2) and substitute y ¼ 0 ) rp ¼ a:
gr ¼@U
@r¼
GM
a210
8
6
32
¼ 1:4375GM
a2
(c) The exact solution for the gravity at the pole is the sum of the attractions of the
three spheres:
gr ¼ GM
a2GM
2a2
GM
18a2¼ 1:5555
GM
a2
32 Gravity
At the equator we take the derivative of the potential and substitute r = a and
y = 90:
gr ¼@U
@r¼
GM
a210
8þ
3
32þ m
¼ GM
a237
32 m
¼ GM
a21:1562 m½
For the exact solution we write
gr ¼ GM
a22GM
8q2cos aþ o2a
From Fig. 18 the distance q is given by
q2 ¼a2
4þ a2 ¼
5
4a2
cos a ¼
ffiffiffi
4
5
r
Therefore
gr ¼ GM
a21þ
8
40
ffiffiffi
4
5
r
m
" #
¼ GM
a21:1789 m½
The approximated values are smaller than the exact solutions.
19. For the case of Problem 18, if GM = 4 103 m3 s2, a = 6 103 km, and v = 7
105 s1, calculate the values of J2, a, m, H, and b.
From the definition of m we obtain
m ¼a3o2
GM¼
216 1018 49 1010
4 1014¼ 2:6 103
The value of J2 is obtained by comparing the two expressions for the potential U
(Problem 18):
U ¼GM
r1þ
a
r
2
J21
23cos2y 1
þm
2
r
a
3
sin2y
U ¼5
4
GM
r1þ
1
40
a
r
2
3cos2y 1
þm
2
r
a
3
sin2y
Then, J2 = 0.05.
The flattening is obtained from the relation
a ¼3
2J2 þ
m
2
a ¼3
2 50 103 þ
2:3 103
2¼ 0:0765
33 Earth’s gravity field and potential
The gravity flattening is given by
b ¼gp ge
ge¼
1:555 1:175
1:175¼ 0:323
where we have used the values of gravity at the Pole and equator obtained in Problem 18,
and in the latter we have substituted the value obtained for the coefficient m.
The dynamic ellipticity H is defined as the ratio of the moments of inertia with respect to
the polar and equatorial radius (Fig. 19a):
H ¼C A
C
where A and C are the moments of inertia of a sphere respect to the polar and equatorial
radi: (axes x1 and x3). The moment of inertia of a sphere of radius R is
Isph ¼2
5MR2
We have to add to the moment of inertia of the sphere of radius a the moments of inertia of
the two internal spheres of radius a/2. For the C-axis (x3) we have
IC ¼ Isph a þ 2Isph a2¼
2
5Ma2 þ 2
2
5
M
8
a2
4¼ 0:425Ma2
For the A-axis (x1) the moment of inertia of each of the small spheres is given by
(Fig. 19b)
X3
X1
2r
2r
r
a/2
a
Fig. 19a
34 Gravity
I ¼ ICM þMh2
since in this case the A axis does not coincide with the centre of mass, where R ¼ a/2
and h ¼ a/2:
Isph a=2 ¼2
5
M
8
a2
4þM
8
a2
4¼ 0:044Ma2
IA ¼ Isph a þ 2Isph a=2 ¼2
5Ma2 þ 2 0:044Ma2 ¼ 0:488Ma2
Finally
H ¼C A
C¼
IC IA
IC¼
0:425 0:488
0:425¼ 0:147
20. Suppose an Earth is formed by a sphere of radius a and density r, and within it
there is a sphere of radius a/2 and density 5r centred at the midpoint of the northern-
hemisphere polar radius. If m ¼ 1/8 and M is the mass of the sphere of radius a,
determine:
(a) The form of the equipotential surface passing through the North Pole.
(b) For latitude 45º, the astronomical latitude and the deviation of the vertical from
the radial.
(a) The gravitational potential is the sum of the potentials for the sphere of radius a
and that of the sphere of radius a/2 (Fig. 20):
V ¼ V1 þ V2 ¼GM
rþGM1
q
As in the previous problems the potential of the small sphere is given in terms of
differential mass M1:
a
R
h
b
Fig. 19b
35 Earth’s gravity field and potential
M1 ¼4
3p 5r rð Þ
a
2
3
¼M
2
and for the inverse of the distance 1/q we use the approximation
1
q¼
1
r1þ
a
2rcos yþ
a
2r
2 1
23cos2y 1
Then, the total gravitational potential is
V ¼GM
r
3
2þ
a
4rcos yþ
1
16
a
r
2
3cos2y 1
The total potential U is the sum of the gravitational potential V plus the potential of rotation
F, where
F ¼1
2r2o2sin2y
and using the coefficient m = o2a3/GM = 1/8, we have
U ¼GM
r
3
2þ
a
4rcos yþ
1
16
a
r
2
3cos2y 1
þr
a
3 m
2sin2y
At the North Pole, y = 0 and r ¼ a, and the value of the potential is
Up ¼GM
a
3
2þ1
4þ1
8
¼15GM
8a
a/2
a
5r
r
q j
qg
gr
gq
fa
i
P
Fig. 20
36 Gravity
The form of the equipotential surface is found by putting U ¼ Up:
15GM
8a¼
GM
r
3
2þ
a
4rcos yþ
1
16
a
r
2
3cos2y 1
þr
a
3 m
2sin2y
Putting r ¼ a inside the square brackets and solving for r we find
r ¼ a4
51þ
1
6cos yþ
1
12cos2y
(b) The deviation of the vertical with respect to the radial direction is given by the angle i:
tan i ¼gy
gr
To find this value we have to calculate the two components of gravity
gr ¼@U
@r¼ GM
3
2
1
r22a
4
1
r3cos y
3a2
16
1
r43 cos2 y 1
þ2r
16a3sin2 y
gy ¼1
r
@U
@y¼
GM
r
a
4r2sin y
a2
16r36 cos y sin yþ
r2
8a3sin y cos y
For a point on the surface we put r ¼a:
gr ¼ GM
a219
16þ1
2cos yþ
11
16cos2y
gy ¼ GM
a21
4sin yþ
1
4sin y cos y
and for latitude 45º
gr ¼ GM
a21:88
gy ¼ GM
a20:30
Then the angle i is given by tan i ¼0:30
1:88) i ¼ 9:0.
The astronomical latitude is
fa ¼ 90 y i ¼ 36:0
21. If the internal sphere of Problem 20 is located on the equatorial radius at
longitude zero, find expressions for the components of gravity: gr , gu , gl.
As in the previous problem the differential mass of the small sphere M1 is (Fig. 21a):
M1 ¼M
2
The total potential U is the sum of the gravitational potentials V and V1, and the potential
due to rotation F. According to Fig. 21b, using the relations of spherical triangles, if ’ and
l are the coordinates of the point where the potential is evaluated, then
37 Earth’s gravity field and potential
cosc ¼ cos 90º cos(90º’)þ sin 90º sin(90º’) cosl
cosc ¼ cos’ cosl
Using the expression for 1/q as in Problem 16,
1
q¼
1
r1þ
a
2rcoscþ
a
2r
2 1
23cos2c 1
The gravitational potential of the small sphere is given by
V1 ¼GM
2r1þ
a
2rcos’ cos lþ
a2
8r23cos2’cos2l 1
The total potential U is given by:
U ¼GM
r
3
2þ
a
4rcos’ cos lþ
a2
16r23cos2’cos2l 1
þ1
2r2o2cos2’
l = 0° A
l
a/2
y
P
B
rq
j
Fig. 21a
90° 90°– j
y
l
B
P
A
Fig. 21b
38 Gravity
Using the coefficient m and sin y = cos ’, we obtain
U ¼ GM3
2rþ
a2
4r2sin y cos lþ
a2
16r33sin2ycos2l 1
þr2
a3m
2sin2y
The three components of gravity are found by differentiating U with respect to r, y, and l:
gr ¼@U
@r¼ GM
3
2r2
a2
2r3sin y cos l
3a2
16r43sin2ycos2l 1
þr
a3msin2y
gy ¼1
r
@U
@y¼ GM
a2
4r3cos y cos lþ
a2
16r46 cos l sin y cos yþ
r
a3m sin y cos y
gl ¼1
r sin y
@U
@l¼ GM
a2
4r3sin l
3a2
8r4sin y cos l sin l
22. A planet is formed by a sphere of radius a and density r, with a spherical core of
density 5r and radius a/2 centred on the axis of rotation in the northern hemisphere
and tangential to the equator. The planet rotates with m ¼ 1/4. For the point at
coordinates (45º N, 45º E), calculate:
(a) The astronomical latitude.
(b) The deviation of the vertical from the radial.
(c) The angular velocity of rotation that would be required for this deviation to be zero.
(a) The gravitational potential is the sum of the potentials of the two spheres (Fig. 22):
V ¼ V1 þ V2 ¼GM
rþGM 0
qð22:1Þ
5r
a
r
a/2
q
ggr
gq
r
j
P
i
Fig. 22
39 Earth’s gravity field and potential
As in Problem 16, the inverse of the distance from a point P to the centre of the small
sphere, 1/q, can be approximated by
1
q¼
1
r1þ
a
2rcos yþ
1
2
a
2r
2
3cos2y 1
As in previous problems we use the differential mass of the small sphere,
M 0 ¼4
3pa3
8ð5r rÞ ¼
2
3pra3
and since M ¼4
3pra3, then M 0 = M/2.
Substituting in Equation (22.1) we obtain
V ¼GM
rþGM
2
1
rþ
a
2r2cos yþ
1
2
a2
4r33cos2y 1
¼ GM3
2rþ
a
4r2cos yþ
a2
16r33cos2y 1
The total potential U is the sum of V plus the potential due to rotation F:
U ¼ V þ F ¼GM
r
3
2þ
a
4rcos yþ
a2
16r23cos2y 1
þ1
2r2o2sin2y
The components of gravity gr and gy are
gr ¼@U
@r¼ GM
3
2r2
a
2r3cos y
3a2
16r43cos2y 1
þ ro2sin2y ð22:2Þ
gy ¼1
r
@U
@y¼ GM
a
4r3sin y
a2
16r46 cos y sin y
þ ro2 sin y cos y ð22:3Þ
For a point on the surface of the large sphere and coordinates 45º N, 45º E, we have that
y = 45, r ¼ a, and
m ¼a3o2
GM¼
1
4
Putting these values in (22.2) and (22.3), we obtain
gr ¼ 1:82GM
a2and gy ¼ 0:24
GM
a2
To find the astronomical latitude we first have to find the deviation of the vertical with
respect to the radial:
tan i ¼gy
gr¼ 0:13 ; i ¼ 7:5
The astronomical latitude is, then, f = ’ i = 45 7.5 = 37.5.
(b) Thedeviation of the verticalwith respect to the radial direction, as already found, is i=7.5.
(c) If we want the deviation of the vertical to be null, i ¼ 0, this implies gy = 0.
40 Gravity
writing Equation (22.3) in terms of the coefficient m, where
m ¼a3o2
GMð22:4Þ
we have
gy ¼ 0 ¼GM
a22ffiffiffi
2p
þ 3
16m
2
and solving for m gives m ¼ 0.73.
Substituting in Equation (22.4) we obtain
o ¼
ffiffiffiffiffiffiffiffi
GM
a3
r
0:85
23. A planet consists of a very thin spherical shell of mass M and radius a, within
which is a solid sphere of radius a/2 and mass M 0 centred at the midpoint of the
equatorial radius of the zero meridian. The planet rotates with angular velocity v
about an axis normal to the equatorial plane. Calculate:
(a) The potential at points on the surface as a function of latitude and longitude.
(b) The components of the gravity vector.
(c) If M 0 ¼ 10 M, what is the ratio between the tangential and radial components of
gravity at the North Pole?
(a) The potentialU is the sumof the gravitational potentials due to the spherical shellV1, and
to the interior sphere V2, plus the potential due to the rotation of the planet F (Fig. 23):
U ¼ V1 þ V2 þ F
F ¼1
2o2
r2cos2’
V1 ¼GM
r
V2 ¼GM 0
q
ð23:1Þ
where r is the distance from a point P on the surface of the planet to its centre, q is the
distance from point P to the centre of the interior sphere, and ’ the latitude of point P.
Using the cosine law,
q ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r2 þ
a2
4 ar cosc
r
where c is the angle between r and the equatorial radius, and its inverse can be approxi-
mated by (Problem 16)
1
q¼
1
r1þ
a
2rcoscþ
a2
8r23cos2c 1
ð23:2Þ
Using the relation for spherical triangles
cos a ¼ cos b cos cþ sin b sin c cosA
41 Earth’s gravity field and potential
putting b ¼ 90º ’, c ¼90º, A ¼ l, and a ¼ c, where l is the longitude of P, then
cos c = cos ’ cos l
Substituting in (23.1), the potential due to the small sphere is
V2 ¼ GM 0
r1þ
a
2rcos’ cos lþ
a2
8r23cos2’cos2l 1
The total potential U is
U ¼GM
rþGM 0
r1þ
a
2rcos’ cos lþ
a2
8r23cos2’cos2l 1
þ1
2o2
r2cos2’
ð23:3Þ
(b) The components of the gravity vector are obtained from Equation (23.3):
gr ¼@U
@r
¼GM
r2
þ GM 0 1
r2
a
r3cos’ cos l
3a2
8r43cos2’cos2l 1
þ o2rcos2’
gy ¼1
r
@U
@y¼
1
r
@U
@’
¼GM 0
r
a
2r2sin’ cos lþ
a2
8r36 cos’cos2l sin’
þ o2r cos’ sin’
a
l
l
a/2
y
y
r q
j
P
90°–
j
x
Fig. 23
42 Gravity
gl ¼1
r cos’
@U
@l¼
GM 0
r cos’
a
2r2cos’ sin l
a2
8r36cos2’ cos l sin l
(c) At the North Pole, ’ = 90 and r ¼ a. Putting M 0 ¼10M and substituting in the
previous equations we obtain
gr ¼ GM
a2GM 0
a2þ3GM 0a2
8a4¼ 7:25
GM
a2
gy ¼GM 0
2a2¼
5GM
a2
The ratio between the radial and the tangential components of gravity at the North
Pole is
gr
gy¼ 1:45
24. An Earth consists of a sphere of radius a and density r, within which there are two
spheres of radius a/2 centred on the axis of rotation and tangent to each other. The
density of that of the northern hemisphere is 4r and that of the southern hemisphere
is r/4.
(a) Express the gravitational potential in terms ofM (the mass of the large sphere) up
to terms of 1/r3.
(b) What astronomical latitude corresponds to points on the equator (without
rotation)?
(c) What error is made by using the 1/r3 approximation in calculating the value of gr
at the equator?
(a) The total gravitational potential V is the sum of the potentials of the sphere of
radius a (V0) and of the two spheres of radius a/2 situated in the northern (V1) and
southern (V2) hemispheres (Fig. 24):
V ¼ V0 þ V1 þ V2
As in previous problems the large sphere is considered to have uniform density
r and the effect of the two interior spheres is calculated using their differential
masses
M ¼4
3pra3
M1 ¼4
3p 4r rð Þ
a3
8¼
3M
8
M2 ¼4
3p
r
4 1
a3
8¼
3M
32
The potentials are
43 Earth’s gravity field and potential
V0 ¼GM
r
V1 ¼GM1
r1
¼3GM
8
1
rþ
a
2r2cos yþ
a2
8r33cos2y 1
V2 ¼GM2
r2
¼ 3GM
32
1
r
a
2r2cos yþ
a2
8r33cos2y 1
where r1 and r2 have been calculated as in Problem 16. Then, the total gravitational
potential in the 1/r3 approximation is
V ¼ GM41
32rþ15
64
a
r2cos yþ
9
256
a2
r3
3cos2y 1
(b) The components of the gravity vector, taking into account that there is no rotation, are
gr ¼@V
@r¼ GM
41
32r2
15a
32r3cos y
27a2
256r43cos2y 1
gy ¼1
r
@V
@y¼ GM
15a
64r3sin y
27a2
128r3cos y sin y
ð24:1Þ
At the equator, r ¼ a and y = 90 and we obtain
r
r/4
4r
a/2
a
r2
r1
a
x
x
Fig. 24
44 Gravity
gr ¼ 1:175GM
a2
gy ¼ 0:243GM
a2
The astronomical latitude (’a) is the angle between the vertical and the equatorial plane. In
our case at the equator this is given by the deviation of the vertical from the radial
direction:
tan’a ¼gy
gr¼ 0:207
Then
’a ¼ 11:68 N
(c) If we want to calculate the exact value of gr at the equator, we calculate the exact
attractions of each sphere and add them:
g0r¼
GM
a2g1r¼
3GM
8r21cos a g2
r¼
3GM
32r22cos a ð24:2Þ
where r1 and r2 are the distances from the centre of each of the two interior
spheres (Fig. 24):
r1 ¼ r2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þa2
4
r
¼affiffiffi
5p
2
and a is the angle which r1 and r2 form with the equatorial plane:
sin a ¼a=2
r1
¼1ffiffiffi
5p
The radial component of gravity is given by
gTr¼ g0
rþ g1
rþ g2
r¼ 1:335
GM
a2
The error we make using the approximation is
gapprox gexact ¼ 0:160GM
a2; that is; 16%:
25. An Earth consists of a sphere of radius a and density r within which there are
two spheres of radius a/2 centred on the axis of rotation and tangent to each other.
The density of that of the northern hemisphere is 2r and that of the southern
hemisphere is r/2.
(a) Express the potential V in terms ofM (the mass of the large sphere), G, and r up to
terms in 1/r3.
45 Earth’s gravity field and potential
(b) According to the value of this potential V, which astronomical latitudes corres-
pond to the geocentric latitudes 45º N and 45º S?
(c) What must the rotation period be for the astronomical and geocentric latitudes to
coincide?
(d) What error is made by the 1/r3 approximation in calculating the value of gr at the
equator? And at the North Pole?
(a) As in previous problems the effect of the interior spheres is given in terms of their
differential masses (Fig. 25):
M1 ¼4
3p 2r rð Þ
a
2
3
¼M
8
M2 ¼4
3p
r
2 r
a
2
3
¼ M
16
The distances q1 and q2 from the centre of each sphere to an arbitrary point P are found
using the cosine law:
q1 ¼ r2 þ
a2
4 2
ar
2cos y
q2 ¼ r2 þ
a2
4þ 2
ar
2cos y
r
r
2r
a/2
a
ja
qj
b
gr
gq gr
r/2
q2q1
x
x
P
Fig. 25
46 Gravity
Using the approximation for 1/q (Problem 16), the total gravitational potential V is the sum
of the potentials of the three spheres:
V ¼GM
rþGM1
q1þGM2
q2
¼GM
r
17
16þ
3
32
a
rcos yþ
1
128
a
r
2
3cos2y 1
(b) The components of the gravity vector are given by
gr ¼@V
@r¼ GM
1
r2
17
16þ
6
32
a
r3cos yþ
3
128
a2
r43cos2y 1
gy ¼1
r
@V
@y¼
GM
r
3
32
a
r2sin yþ
1
128
a2
r36 cos y sin y
ð25:1Þ
If the point P is at the surface, r ¼ a, then
gr ¼ GM
a217
16þ
3
16cos yþ
3
1283cos2y 1
gy ¼ GM
a23
32sin yþ
6
128cos y sin y
At geocentric latitude 45º N, y ¼ 45º,
gr ¼ 1:21GM
a2
gy ¼ 0:09GM
a2
The deviation of the vertical with respect to the radial direction i is given by
tan i ¼gy
gr¼ 0:074 ) i45 ¼ 4:2
According to Fig. 25, the astronomical latitude ’a can be determined from the deviation of
the vertical i,
’a þ iþ 180 ’ð Þ ¼ 180 ) ’a ¼ ’ i ¼ 45 4:2 ¼ 40:8 N
In the same way, for geocentric latitude 45º S (y = 135)
gr ¼ 0:94GM
a2
gy ¼ 0:04GM
a2
Then tan i135 ¼gy
gr¼ 0:043 ) i135 ¼ 2:5
Then the astronomical latitude is
’a ¼ 45 2:5 ¼ 47:5 ¼ 47:5 S
47 Earth’s gravity field and potential
(c) If we want the astronomical and geocentric latitudes to coincide, then the
deviation of the vertical must be null, i ¼ 0º. This implies that gtotaly must be
zero. To do this by means of the rotation, we have to make the tangential
component of the centrifugal force gRy be equal and of opposite sign to that of
the gravitational potential gVy :
gytotal ¼ gy
V þ gyR ¼ 0 ) gy
V ¼ gyR
The tangential component due to rotation is
gRy ¼1
r
@F
@y
where F ¼1
2o2
r2sin2y. Then
gRy ¼ o2r cos y sin y
For a point on the surface at latitude 45º N, r = a and y = 45, so
gyV ¼ gy
R ) 0:09GM=a2 ¼ o2 a=2
From here we can calculate the period of rotation
T ¼2p
o¼
2pffiffiffiffiffiffiffiffiffi
0:18p
ffiffiffiffiffiffiffiffi
a3
GM
r
For a point at latitude 45º S, r = a and y = 135, so
gRy ¼ gVy ) 1
2o2a ¼ 0:04
GM
a2) T ¼
2pffiffiffiffiffiffiffiffiffi
0:08p
ffiffiffiffiffiffiffiffi
a3
GM
r
(d) The value of the radial component of gravity at the equator, r ¼ a, y = 90, by
substitution in (25.1), is
gr ¼ 1:04GM
a2
If we calculate the exact value by adding the contributions of the three spheres
(Fig. 25)
gexactr
¼ gMrþ g1
rþ g2
r
gMr
¼ GM
a2
g1r¼ g1 cos b
g2r¼ g2 cos b
where
48 Gravity
cos b ¼affiffiffi
5
4
r
a
¼2ffiffiffi
5p
g1 ¼ GM
8
4
5a2
g2 ¼GM
16
4
5a2
gexactr
¼ GM
a20:96
The error in the approximation is:
gerrorr
¼ 0:96GM
a2 1:04
GM
a2
¼ 0:08GM
a2
In a similar way, for a point at the North Pole, r ¼ a, y ¼ 0º:
gr ¼ 1:30GM=a2
gexactr
¼ gMrþ g1
rþ g2
r
gMr
¼ GM
a2
g1r¼
GM
8a
2
2¼
GM
2a2
g2r¼
GM
16 aþa
2
2¼
GM
36a2
gexactr
¼ GM
a21:47
The error in the approximation is
gerrorr
¼ 1:47GM
a2 1:30
GM
a2
¼ 0:17GM
a2
26. A sphericalEarth of radius ahas a core of radius a/2whose centre is displaced a/2 along
the axis of rotation towards the North Pole. The core density is twice that of the mantle.
(a) What should the period of rotation of the Earth be for the direction of the plumb-
line to coincide with the radius at a latitude of 45º S?
(b) What are the values of J0, J1, J2, and m?
(a) As in previous problems we calculate the gravitational potential by the sum of the
potentials of the two spheres, using for the core the differential mass (Fig. 26):
V ¼ V1 þ V2 ¼GM
rþGM 0
q
M 0 ¼4
3p 2r rð Þ
a2
8¼
M
8
49 Earth’s gravity field and potential
As we saw in Problem 16, we use for 1/q the first-order approximation
V ¼GM
rþGM
8
1
rþ
a
2r2cos yþ
1
2
a2
4r33cos2y 1
ð26:1Þ
The total potential U is the sum of the gravitational potential V and the potential due to
rotation
F ¼1
2r2o2sin2y
U ¼ GM9
8
1
rþ
a
16r2cos yþ
a2
64r33cos2y 1
þ1
2r2o2sin2y
In order that the direction of the plumb-line coincides with the radial direction, the
tangential component of gravity, gy, must be null:
gy ¼1
r
@U
@y¼GM
r
a
16
1
r2sin y
a2
64
1
r36 cos y sin y
þ ro2 sin y cos y
For a point on the surface at latitude 45º S, the tangential component of gravity is, with
r = a, y = 135,
gy ¼ 0:003GM
a2ao2
2
a
r
q j
r
gq
gr
i
g
q
x
45°
2r
a/2 P
Fig. 26
50 Gravity
Putting this component equal to zero, we find the value of the period of rotation T:
gy ¼ 0:003GM
a2ao2
2¼ 0 ) o2 ¼ 0:006
GM
a3) T ¼
2p
0:077
ffiffiffiffiffiffiffiffi
a3
GM
r
(b) The gravitational potential V of Equation (26.1) can be written as
V ¼GM
r
9
81þ
a
18rcos yþ
a2
72r23cos2y 1
We obtain the values of J1 and J2 by comparison with the equation
V ¼GM
rJ0 þ J1
a
rP1 þ J2
a
r
2
P2
Since the total mass is (9/8)M, we obtain
J0 ¼ 1
J1 ¼1
18
J2 ¼1
72
27. Within a spherical planet of radius a and density r there are two spherical cores of
radius a/2 and density r 0 with centres located on the axis of rotation at a/2 from the
planet’s centre, one in the northern hemisphere and the other in the southern hemisphere.
(a) Neglecting rotation of the planet, calculate what the ratio r0/r should be for the
gravity flattening to be 1/8.
(b) If the planet rotates so that m ¼ 1/16, and the ratio of the densities is that found in
part (a), calculate the astronomical latitude which corresponds to the geocentric
latitude 45º N.
(a) Since there is no rotation the total potential U is the sum of the gravitational
potentials of the three spheres (Fig. 27). As in previous problems we use the mass
M of the planet with uniform density r and for the two cores the differential
masses M 0. For 1/q we use the approximation as in Problem 16:
M 0 ¼4
3pa3
8r0 rð Þ ¼
4
3pa3
8r0 rð Þ
r
r¼
M
8
r0
r 1
ð27:1Þ
The potential U is
U ¼GM
rþGM 0
r2þ
a
2r
2
3cos2y 1
The radial components of gravity at the equator and the Pole are found by taking the
derivative of the potential U:
gr ¼@U
@r¼
GM
r2
þ GM 0 2
r23a2
4r43cos2y 1
51 Earth’s gravity field and potential
On the surface r ¼ a, and at the equator y = 90 and at the Pole y = 0, so
ger¼
GM
a2GM 0
a25
4
gpr¼
GM
a2GM 0
a214
4
The gravity flattening is given by
b ¼gp ge
ge¼
1
8
By substituting the values of gravity we find the relation between M and M 0:
1
8¼
M 7
2M 0 þM þ
5
4M 0
M 5
4M 0
) M ¼67
4M 0
Putting M 0 in terms of M from Equation (27.1) we find the ratio of the densities:
M 0 ¼4
67M ¼
M
8
r0
r 1
)r0
r¼ 1:48
(b) For a rotating planet we add to the potential U the rotational potential, F:
U ¼GM
rþGM 0
r2þ
a
2r
2
3cos2y 1
þGM
r
r
a
3 m
2sin2y
The radial and tangential components of gravity are now
gr ¼@U
@r¼
GM
r2
þ GM 0 2
r23a2
4r43cos2y 1
þ GMrm
a3sin2y
¼ 1:11GM
r2
gy ¼1
r
@U
@y¼GM 0
r2
a
2r
2
6 cos y sin y
þGM
r2
r
a
3
m sin y cos y
¼ 0:013GM
r2
From Fig. 27 we see that the relation between the geocentric and astronomical latitudes is
’a ¼ ’ i
where i is the deviation of the vertical with respect to the radial direction, which is given by
tan i ¼gy
gr¼ 0:012 ) i ¼ 0:7
52 Gravity
Then the astronomical latitude for geocentric latitude 45º is
’a ¼ 45 0:7 ¼ 44:3
Gravity anomalies. Isostasy
28. For two-dimensional problems, the gravitational potential of an infinite horizontal
cylinder of radius a is
V ¼ 2pGra2 ln1
r
where r is the distance measured perpendicular to the axis. Assume that a horizontal
cylinder is buried at depth d as measured from the surface to the cylinder’s axis.
(a) Calculate the anomaly along a line of zero elevation on the surface perpendicular
to the axis of the cylinder.
(b) At what point on this line is the anomaly greatest?
(c) What is the relationship between the distance at which the anomaly is half the
maximum and the depth at which the cylinder is buried?
(d) For a sphere of equivalent mass to produce the same anomaly, would it be at a
greater or lesser depth?
r
r
r’
a
x
x
a/2
ja
ggr
gq
j = 45°
i
Fig. 27
53 Gravity anomalies. Isostasy
(a) The gravity anomaly produced by an infinite horizontal cylinder buried at depth d, with
centre at x ¼ 0 (Fig. 28), is given by the derivative in the vertical direction (z-axis) of
the gravitational potential V:
V ¼ 2pGra2 ln1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ ðzþ dÞ2q
0
B
@
1
C
A
g ¼ gz ¼ @V
@z¼
2prGa2d
x2 þ ðzþ dÞ2ð28:1Þ
For points on the surface (z ¼ 0):
g ¼2prGa2d
x2 þ d2
(b) To find the point at which the anomaly has its maximum value, we take the
derivative with respect to x and put it equal to zero:
@g
@x¼ 0 ) 2prGa2d2x ¼ 0 ) x ¼ 0
Substituting x ¼ 0 in (28.1):
gmax ¼2pGra2
d
(c) The distance at which the anomaly has a value equal to half its maximum value
gives us the depth d at which the cylinder is buried:
x
x
d
a
r
gr
gz
P
r
∆g
x
∆gmax
x1/2
(1/2)∆gmax
Fig. 28
54 Gravity
gmax
2¼ g )
2pGra2
2d¼
2pGra2d
x2 þ d2) x1=2 ¼ d
(d) The gravitational potential produced by a sphere of differential mass DM buried at
depth d under x ¼ 0 is given by
V ¼GM
r¼
GM
x2 þ ðzþ dÞ2 1=2
The gravity anomaly is
gz ¼@V
@z¼
GMðzþ dÞ
x2 þ ðzþ dÞ2 3=2
and for a point on the surface z ¼ 0,
gz ¼GMd
x2 þ d2ð Þ3=2
The maximum value for x ¼ 0 is
gmax ¼GM
d2
The distance at which the anomaly has half its maximum value is
GMd
x21=2 þ d2
3=2¼
GM
2d2
x1=2 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
22=3 1p
d ¼ 0:766d
Therefore, the sphere is at a greater depth than the cylinder.
29. At a point at latitude 42º 290 19 00 and height 378.7 m the value of gravity is
observed to be 980 252.25 mGal. Calculate in gravimetric units (gu):
(a) The free-air anomaly.
(b) The Bouguer anomaly if the density of the crust is 2.65 g cm3.
(a) We first calculate the normal or theoretical value of gravity given by the expression
g ¼ 9:7803268 1þ 0:00530244sin2’ 0:0000058sin22’
ms2
We substitute for ’ its value 42º 290 1900 and obtain
g ¼ 9:803 9299m s2
The free-air anomaly, using the free-air correction, is
gFA ¼ g þ 3:086h g ¼ 238:7 gu
55 Gravity anomalies. Isostasy
(b) The Bouguer anomaly is calculated from the free-air anomaly using the Bouguer
correction with a crust density of 2.65 g cm3:
gB ¼ g þ 3:086h 2pGrh g ¼ gFA 2pGrh ¼ 659:3 gu
30. An anomalous mass is formed by two equal tangent spheres of radius R, with
centres at the same depth d ( d R ) and density contrast Dr.
(a) Calculate the Bouguer anomaly at the surface (z ¼ 0) produced by the mass
anomaly along a profile passing through the centres of the two spheres.
(b) Represent it graphically for x ¼ 0 (above the tangent point), 500, 1000, and 2000 m
taking R ¼ 1 km, d ¼ 3 km, and Dr ¼ 1 g cm3.
(a) For one sphere the anomaly for points on the surface (z ¼ 0) is (Problem 28)
g ¼GMd
x2 þ d2ð Þ3=2
For two spheres the anomaly is the sum of the attractions of the two spheres (Fig. 30a):
g ¼GMd
r31
þGMd
r32
where
r1 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x Rð Þ2 þ d2q
r2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
xþ Rð Þ2 þ d2q
Then
g ¼GMd
x Rð Þ2þd2h i3=2
þGMd
xþ Rð Þ2þd2h i3=2
R
d r2r1
xP(x,0)
x x
Fig. 30a
56 Gravity
(b) To represent graphically the curve of the anomaly (Fig. 30b), we first find the point
at which it is a maximum:
@g
@x¼
@
@x
GMd
x Rð Þ2þd2h i3=2
þGMd
xþ Rð Þ2þd2h i3=2
0
B
@
1
C
A¼ 0
xþ Rð Þ2 x Rð Þ2 þ d2h i5
¼ x Rð Þ2 xþ Rð Þ2 þ d2h i5
) x ¼ 0maximum
Using the data given in the problem, we find the values of the anomaly for the five points,
with Dr = 1 g cm3, R = 1 km, d = 3 km
M ¼4
3pR3
r ¼4
3p 109 103 ¼ 4:19 1012 kg
x (m) Dg (gu)
0 53.0
500 52.0
1000 48.9
1500 43.9
2000 37.5
55
50
45
Anom
aly
(g
µ)
40
–2000 –1000 0
Distance (m)
1000 2000
Fig. 30b
57 Gravity anomalies. Isostasy
31. At a point at geocentric latitude 45º N and height 2000 m the observed value of
gravity is g ¼ 6690 000 gu. Taking the approximation that the Earth is an ellipsoid of
equatorial radius a ¼ 6000 km, density ¼ 4 g cm3, J2 ¼ 103, and m ¼ 103,
calculate for that point:
(a) The free-air and the Bouguer anomalies.
(b) The distance from the free surface to that of the sphere of radius a (precision 1 gu).
(a) The volume of an ellipsoid is:
V ¼4
3pa3 1þ 2að Þ
The flattening is
a ¼3J2
2þm
2¼ 2 103
and the mass is
Me ¼ Vr ¼4
3pa3 1þ 2að Þr ¼ 3:624 1024 kg
Using G = 6.67 1011 m3 kg1 s2
GM
a2¼
6:67 1011 3:624 1024
36 1012¼ 6:732 994m s2
The value of gravity at the equator in the first-order approximation is given by
ge ¼GM
a21þ
3
2J2 m
¼ 6:736 361m s2
For a point at latitude 45ºN the radial component of gravity is
gr ¼ ge 1þ b sin2 ’
¼ 6:738 045m s2
where we have used the value of the gravity flattening b given by
b ¼5
2m a ¼ 0:5 103
The free-air correction is
CFA ¼ 2GM
a3h ¼ 2:24 106h m s2 ¼ 2:244h gu
Then, the free-air anomaly at that point is
gFA ¼ g gþ CFA
¼ 6 690 000 6 738 045þ 2:244 2000 ¼ 43 557 gu
In order to calculate the Bouguer anomaly, we first calculate the Bouguer correction
CB ¼ 2pGrh ¼ 1:676 106 hms2 ¼ 1:676h gu
58 Gravity
Then, the Bouguer anomaly is:
gB ¼ g gþ CAL 2pGrh ¼ gFA CB
¼ 45 557 1:676 2000 ¼ 46 909 gu
(b) If we call N the distance at the given point between the free surface and the surface
of the sphere of radius a (Fig. 31), this is given by:
N ¼ a r h
where r is the radius of the ellipsoid at latitude 45º N which to a first approximation is
r ¼ a 1 asin2’
¼ 6000 1 2 103 1
2
¼ 5994 km
Then,
N ¼ 6000 5994 2 ¼ 4 km
32. Beneath a point A at height 400 m there exists an anomalous spherical mass of
radius 200 m, density 3.5 g cm3, whose centre is 200 m below the reference level.
A point B is located at a height of 200 m and a horizontal distance of 400 m from A,
and a third point C is at a height of 0 m and at a horizontal distance of 800 m from
A. The density of the medium above the reference level is 2.6 g cm3, and below the
reference level it is 2.5 g cm3. The theoretical value of gravity is 980 000 mGal.
Calculate:
(a) The values of gravity at A, B, and C.
(b) The Bouguer anomalies at these points.
a
a
h
N
r
Fig. 31
59 Gravity anomalies. Isostasy
Precision 1 gu.
(a) The gravity at each point is given by
g ¼ g CFA þ CB þ Cam
where
Normal gravity: g ¼ 9800 000 gu
Free-air correction: CFA ¼ 3.086 h
Bouguer correction: CB ¼ 0.419 r1 h
r1 ¼ 2.6 g cm3 is the density of the material above the reference level
C am is the anomaly produced by the buried sphere at a point at height h and a horizontal
distance x from its centre:
Cam ¼GMðhþ dÞ
x2 þ ðhþ dÞ2 3=2
¼G4
3pR3ðrsph r2Þðhþ dÞ
x2 þ ðhþ dÞ2 3=2
ð32:1Þ
where d is the depth to the centre from the reference level; and rsph and r2 are the densities
of the sphere and of the medium where it is located, respectively. In our case: d ¼ 200 m,
rsph ¼ 3.5 g cm3, and r2 ¼ 2.5 g cm3.
For point A, x ¼ 0, we obtain
CFA ¼ 1234 gu CB ¼ 436 gu
Cam ¼GM
hþ dð Þ2¼
G4
3pR3ðrsph r2Þ
hþ dð Þ2¼ 6 gu
The value of gravity is gA ¼ 9 799 208 gu.
At point B:
CFA ¼ 617 gu
CB ¼ 218 gu
The anomaly produced by the sphere is calculated by Equation (32.1), substituting
x ¼ xB ¼ 400 m and h ¼ hB ¼ 200 m
C am ¼ 5 gu
We obtain gB ¼ 9 799 606 gu.
At point C:
The free-air and Bouguer corrections are null, because the point is at the reference level.
The anomaly due to the sphere, by substitution in Equation (32.1), x ¼ xC ¼ 800 m, and
h ¼ hC ¼ 0, is
C am ¼ 1 gu
The value of gravity is: gC ¼ 9800 001 gu.
(b) The Bouguer anomaly is given by
DgB ¼ g þ CFA– CB
– g
60 Gravity
By substitution of the values for each point we obtain that the anomalies correspond to
those produced by the sphere:
gBA ¼ 6 gu
gBB ¼ 5 gu
gBC ¼ 1 gu
33. For a series of points in a line and at zero height which are affected by the
gravitational attraction exerted by a buried sphere of density contrast 1.5 g cm3, the
anomaly versus horizontal distance curve has a maximum of 4.526 mGal and a point
of inflexion at 250 m from the maximum. Calculate:
(a) The depth, anomalous mass, and radius of the sphere.
(b) The horizontal distance to the centre of the sphere of the point at which the
anomaly is half the maximum.
(a) We know that the inflection point of the curve of the anomaly produced by a
sphere buried at depth d corresponds to the horizontal distance d/2. Then
xinf ¼d
2) d ¼ 2xinf ¼ 2 250 ¼ 500m
The maximum value of the anomaly at x ¼ 0 is
gmax ¼GM
d2¼ 4:526mGal ¼ 45:26 gu
and solving for DM
DM ¼45:26 106 m s2 5002 m2
6:67 1011 m3 s2 kg1¼ 1:6964 1011 kg
From this value we calculate the radius of the sphere:
M ¼4
3pR3
r ) R ¼3M
4pr
1=3
¼3 1:6964 1011 kg
4 3:14 1:5 103 kgm3
1=3
¼ 300m
(b) In order that g ¼ 12gmax with z ¼ 0 we write
GMd
x21=2 þ d2
3=2¼
1
2
GM
d2) x1=2 ¼ 383m
34. At a point at height 2000 m, the measured value of gravity is 9.794 815 m s2.
The reference value at sea level is 9.8 m s2. The crust is 10 km thick and of density
2 g cm3, and the mantle density is 3 g cm3. Calculate:
(a) The free-air, Bouguer, and isostatic anomalies. Use the Pratt hypothesis with a
cylinder of radius 10 km and a 40 km depth of compensation.
(b) If beneath this point there is a spherical anomalous mass of GDM ¼ 160 m3 s2 at
a 2000 m depth, what should the compensatory cylinder’s density be for the
compensation to be total?
61 Gravity anomalies. Isostasy
(a) The free-air anomaly is
gFA ¼ g gþ CFA ¼ 9 794 815 9 800 000þ 3:086 2000 ¼ 987 gu
The Bouguer anomaly is
gB ¼ g gþ CFA CB ¼ gFA 0:4191rh
¼ 987 0:4191 2 2000 ¼ 689 gu
To calculate the isostatic anomaly (Fig. 34a) we begin with the calculation of the isostatic
correction assuming Pratt’s hypothesis and using only a vertical cylinder of radius 10 km
under the point and the compensation level at 40 km. In this way, the correction consists
of the gravitational attraction of a cylinder of radius a and height b at a point at distance c
from the base of the cylinder, which is given by
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
ð34:1Þ
where Dr is the contrast of densities, which according to Pratt’s hypothesis is given by
r ¼hr0Dþ h
ð34:2Þ
where r0 is the density for a block at sea level, which in our case is formed by a crust
of density 2 g cm3 and thickness 10 km over a mantle of density 3 g cm3 and thickness
30 km. For the whole 40 km we use a mean value of density
r0 ¼2 10þ 3 30
40¼ 2:75 g cm3
P
C a
b
Fig. 34a
62 Gravity
By substitution in (34.2) we obtain
r ¼2 2:75
42¼ 0:13 g cm3
The isostatic correction (34.1) is
CI ¼ 2 3:1416 6:67 1011 0:13 103b40þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
100þ 4p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
100þ 1764p
c 103
¼ 382 gu
Finally the isostatic anomaly is given by
gI ¼ g gþ CFA þ CB þ CI ¼ gB þ CI ¼ 689þ 382 ¼ 307 gu
(b) If under the point considered there is an anomalous spherical mass (Fig. 34b) at
depth d ¼ 2 km, the anomaly it produces is
gam ¼GM
d2¼
160
2000ð Þ2¼ 40 gu
The total anomaly now is the Bouguer anomaly plus the anomaly due to the sphere:
g ¼ 689 40 ¼ 729 gu
If the isostatic compensation is total (isostatic anomaly equal to zero), this anomaly must
be compensated by the cylinder. Thus, the necessary contrast of densities Dr to do this can
be calculated using expression (34.1):
729 106 ¼ 2 3:1416 6:67 1011
40þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
100þ 4p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
100þ 1764p
103 r
10 km
D = 40 km
P
h
drc= 2 g cm–3
rM= 3 g cm–3
Fig. 34b
63 Gravity anomalies. Isostasy
so
r ¼ 0:25 g cm3
As the mean value (crust–mantle) of the density is 2.75 g cm3, the density of the cylinder
must now be
r0 r ¼ r ¼ 0:25 ¼ 2:75 r ) r ¼ 2:50 g cm3
35. At a point on the Earth at height 1000 m, the observed value of gravity is 979 700
mGal. The value at sea level is 980 000 mGal.
(a) Calculate the free-air and Bouguer anomalies.
(b) According to the Airy hypothesis, which is the state of compensation of that height?
(c) What should the depth of the root be for the compensation to be total?
To calculate the compensation, use cylinders of radius 40 km, crustal thickness
H ¼ 30 km, crust density 2.7 g cm3, and mantle density 3.3 g cm
3.
(a) The free-air anomaly is
gFA ¼ g gþ CFA ¼ g gþ 3:086h ¼ 86 gu
The Bouguer anomaly is
gB ¼ g gþ CFA CB ¼ gFA 2p G r h ¼ 1046 gu
(b) To calculate the isostatic anomaly according to the Airy hypothesis we first need to
obtain the value of the root given by the equation
t ¼rc
rM rch ¼ 4500m
The isostatic correction is given by
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
Substituting the values
Dr ¼ rM – rC ¼ 600 kg m3
b ¼ t ¼ 4500 m
c ¼ h þ H þ t ¼ 35 500 m
a ¼ 40 km
we obtain:
CI ¼ 409 gu
the isostatic anomaly is:
gI ¼ g gþ CFA þ CB þ CI ¼ gB þ CI ¼ 637 gu
The negative value of the anomaly indicates that the zone is overcompensated.
64 Gravity
(c) If we want the compensation to be total, the value of the isostatic correction
must be
DgI ¼ DgB þ CI ¼ 0 ¼> CI ¼ – DgB ¼ 1046 gu
Since the isostatic correction under the Airy hypothesis is
CI ¼ 2pGr t þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ hþ Hð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ ðhþ H þ tÞ2q
substituting and solving for t, we obtain
t ¼ 13 068m
For a total isostatic compensation the value of the root (13 068 m) must be much larger
than that corresponding to the 1000 m height, which is only 4500 m.
36. Gravity measurements are made at two points A and B of altitude 1000 m and
1000 m above the reference level, respectively, 2 km apart along a W-E profile at
latitude 38.80º N. Below a point C located in the direction AB and 1 km from A
is buried a sphere of radius 1 km and centre 3 km below the reference level, of density
r ¼ 1.76 g cm3. Calculate:
(a) The value of gravity at A and B.
(b) Using the Airy assumption and neglecting the sphere, calculate the root at A and B.
Crustal density rC ¼ 2.76 g cm3, mantle density rM ¼ 3.72 g cm3, a ¼ 10 km, and
H ¼ 30 km.
(a) The gravity observed at points A and B is given by
gA ¼ g CFA þ CB þ Cam
gB ¼ gþ CFA CB þ Cam
where g is the theoretical gravity, CFA the free-air correction, CB the Bouguer correction,
and C am the attraction due to the anomalous mass.
The theoretical gravity at the observation point at latitude 38.80º N is
g ¼ 9:780 32 1þ 0:005 3025sin2’ 0:000 0058sin22’
¼ 9:800714m s2
The free-air and Bouguer corrections are:
CFA ¼ 3:806h ¼ 3:806 1000 ¼ 3806 gu
CB ¼ 0:419rCh ¼ 0:419 2:76 1000 ¼ 1156 gu
The attraction due to the spherical anomalous mass (Fig. 36) is given by
Cam ¼GMðzþ dÞ
x2 þ ðzþ dÞ2 3=2
65 Gravity anomalies. Isostasy
For points A and B, by substitution of the values
M ¼4
3pR3
r
R ¼ 1000m; r ¼ 1000 kgm3
zA ¼ h ¼ 1000m
zB ¼ h ¼ 1000m
xA ¼ xB ¼ 1000m
d ¼ 3000m
we find
CamA ¼ 16 gu
CamB ¼ 50 gu
Then the values of gravity at both points are
gA ¼ 9800 627:9 3806þ 1156:4 16 ¼ 9:798 048m s2
gB ¼ 9800 627:9þ 3806 1156:4 50 ¼ 9:803 314m s2
(b) To calculate the value of the root under A and B according to the Airy hypothesis we
use the equation
t ¼rC
rM rCh
A
C
h
d
R
–h
B
XB
XA
ρ
r c = 2.76 g / cm3
x
Fig. 36
66 Gravity
where rC and rM are the crust and mantle densities. Then we find
tA ¼2:76
3:72 2:76 1000 ¼ 2875m
tB ¼2:76
3:72 2:76 ð1000Þ ¼ 2875m
37. At a point at latitude 43º N, the observed value of gravity is 9800 317 gu, and the
free-air anomaly is 1000 gu.
(a) Calculate the Bouguer anomaly. Take rC ¼ 2.67 g cm3.
(b) If the isostatic compensation is due to a cylinder of radius 10 km which is beneath
the point of measurement, what percentage of the Bouguer anomaly is compen-
sated by the classical models of Airy and Pratt?
(c) According to the Pratt hypothesis, what density should the cylinder have for the
compensation to be total?
(a) First we calculate the normal gravity at latitude 43ºN:
g ¼ 978:0320 1þ 0:005 3025sin2’ 0:000 0058sin22’
Gal
g ¼ 9804 385 gu
The height of the point is determined from the free-air anomaly,
gFA ¼ g gþ 3:086h ¼ 9 800 317 9 804 385þ 3:086h ¼ 1000 gu
and solving for h,
h ¼ 1642m
From this value we calculate the Bouguer anomaly
gB ¼ g gþ 1:967h ¼ 838 gu
(b) To apply the isostatic compensation using the Airy hypothesis we first calculate the
root corresponding to the height h ¼ 1642 m:
t ¼ 4:45h ¼ 7307m
The isostatic correction is determined using Equation (34.1) of Problem 34:
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
ð37:1Þ
where
a ¼ 10 km; b ¼ t ¼ 7307m; c ¼ t þ 30 000þ h ¼ 38 949m
r ¼ rM rC ¼ 3:27 2:67 ¼ 0:6 g cm3
which results in CI ¼ 70 gu. This represents 8% of the observed Bouguer anomaly.
67 Gravity anomalies. Isostasy
If we use the Pratt hypothesis, the contrast of densities corresponding to h ¼ 1642 m is
given by
r ¼hr0Dþ h
¼ 0:043 g cm3
where we have used r0 ¼ rC ¼ 2.67 g cm3 as the density of the crust. We now substitute
in Equation (37.1), b = D = 100 km, c = D þ h = 101.642 km and the obtained value of
Dr ¼ 0.043 g cm3, and obtain
CI ¼ 148 gu
We have to determine again the Bouguer anomaly using the density according to the Pratt
hypothesis
r ¼Dr0Dþ h
¼ 2:63 g cm3
gB ¼ gAL 2pGrh ¼ 810 gu
The isostatic correction corresponds now to 18% of the Bouguer anomaly.
(c) If the compensation is total the isostatic correction must be equal to the Bouguer
anomaly with changed sign:
CI ¼ gB
Using the Pratt hypothesis in order to calculate the density r of the cylinder under the
point, we have to take into account that this density must also be the density used in the
determination of the Bouguer anomaly. Then we write
CI ¼ gB ¼ gFA þ 2p G r h
2pGðr0 rÞ bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
¼ gFA þ 2pGrh
and putting
N ¼ bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
and solving for r, we obtain
r ¼gFA þ 2pGr0N
2pGðhþ NÞ¼ 2:46 g cm3
where we have used the values r0 ¼ 2.67 g cm3 and DgFA ¼ 1000 gu.
38. At a point on the Earth’s surface, a measurement of gravity gave a value of
9795 462 gu. The point is 2000 m above sea level. At sea level the crust is 20 km thick
and of density rC ¼ 2 g cm3. The density of the mantle is rM ¼ 4 g cm
3.
(a) Calculate the free-air and Bouguer anomalies.
68 Gravity
(b) Calculate the isostatic anomaly according to the Airy and Pratt assumptions.
Use cylinders of 10 km radius and compensation depth of 60 km.
(c) Beneath the point, there is an anomalous spherical mass of GDM ¼ 1200 m3 s2.
How deep is it?
Take g ¼ 9.8 m s2.
(a) The free-air anomaly is given by
gFA ¼ g gþ 3:086h ¼ 9 795 462 9 800 000þ 3:086h ¼ 1634 gu
For the Bouguer anomaly we first calculate the Bouguer correction
CB ¼ 0:419rh ¼ 0:419 2 2000 ¼ 1676 gu
Then we obtain
gB ¼ g gþ CFA CB ¼ gFA CB ¼ 42 gu
(b) To calculate the isostatic anomaly according to the Airy hypothesis we determine
first the value of the root corresponding to the height 2000 m:
t ¼rc
rM rch ¼
2
4 22000 ¼ 2000m
The isostatic correction, using a single cylinder under the point, is given by
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
ð38:1Þ
where (Fig. 38a)
P
C a
b
Fig. 38a
69 Gravity anomalies. Isostasy
a ¼ 10 km; b ¼ t ¼ 2 km; c ¼ H þ t þ h ¼ 20þ 2þ 2 ¼ 24 km
Calling A the term inside the brackets in Equation (38.1)
A ¼ 2þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
100þ 484p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
100þ 576p
¼ 0:166 km
The isostatic correction is, then, given by
CI ¼ 2 3:1416 6:67 108 cm3=gs2 2 g=cm3 166 102 cm ¼ 139 gu
Finally, the isostatic anomaly using the Airy hypothesis is
gI ¼ g gþ CFA CB þ CI ¼ 97 gu
According to the Pratt hypothesis, the regional density is given by
r ¼D
Dþ hr0 ¼
60
60þ 23:33 ¼ 3:22 g cm3
where D is the compensation depth (in this problem 60 km) and for r0 (Fig. 38b) we have
used the mean value of the density of the crust (2 g cm3) and of the mantle (4 g cm3)
along the compensation depth
r0 ¼1
32þ
2
34 ¼ 3:33 g cm3
The contrast of densities is
r ¼ 3:33 3:22 ¼ 0:11 g cm3
For the isostatic correction, using the Pratt hypothesis, the term A is now
A ¼ 60þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
102 þ 62 60ð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
102 þ 622p
¼ 7:4 km
and the correction
20 km
40 km
h
P
rM= 4 g cm–3
rC= 2 g cm–3
x
Fig. 38b
70 Gravity
CI ¼ 2 3:1416 6:67 1011 m3=kg1s2 0:11 103 kg=m3 7:4 103 m
¼ 341 gu
Since according to the Pratt hypothesis, the density of the compensating cylinder extends
to the surface of the height 2000 m, we have to calculate again the Bouguer anomaly using
this density (3.33 g cm3). We find for the Bouguer and isostatic anomalies the values
CB¼ 2 p G r h ¼ 2699 gu
DgB ¼ DgFA – CB ¼ 1065 gu
DgI ¼ DgB þ CI ¼ 724 gu
(c) If we assume that the isostatic anomaly is produced by a spherical anomalous mass
buried under the point at a depth d under sea level its gravitational effect is given by
gmax ¼GM
ðhþ dÞ2¼ gI ¼ 724 gu
Solving for d we obtain
hþ d ¼ 3517m ) d ¼ 1517m
39. At a point P at height 2000 m above sea level, a measurement is made of gravity.
The crust at sea level, where gravity is 9.8 m s2, is 20 km thick and of density
3 g cm3, and the density of the mantle is 4 g cm3. Below the point P, at 2000 m depth
under sea level, is an anomalous spherical mass of GDM ¼ 1200 m3 s2.
(a) Neglecting the isostatic compensation, what would be the value of gravity at the
point P?
(b) With isostatic compensation, what now is the value of gravity at that point?
Use the Airy and Pratt assumptions for the isostatic compensation (Pratt
depth of compensation, 100 km) with single cylinders of 20 km radius under the
point.
(a) Without isostatic compensation, the gravity observed at point P is equal to the sum of
the normal gravity plus the free-air and Bouguer corrections and the effect of the
anomalous mass. Remember that the free-air correction has negative sign:
gP ¼ g CFA þ CB þ Cam ð39:1Þ
where,
g ¼ 9800 000 gu
CFA ¼ 3:086 h ¼ 3:086 2000 ¼ 6172 gu
CB ¼ 0:419rh ¼ 0:419 3 2000 ¼ 2514 gu
The anomaly due to the anomalous mass is given by
Cam ¼ gmax ¼GM
hþ dð Þ2ð39:2Þ
71 Gravity anomalies. Isostasy
By substitution of the values,
Cam ¼GM
ðhþ dÞ2¼ 75 gu
The gravity observed a P is, then, given by
gP ¼ 9 800 000 6172þ 2514þ 75 ¼ 9 796 417 gu
(b) If there is isostatic compensation, according to the Airy hypothesis, we determine first
the depth of the root, using the density of the crust rC and of the mantle rM:
t ¼rC
rM rCh ¼
3
4 3 2000 ¼ 6000m
The isostatic correction is
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
ð39:3Þ
where a ¼ 20 km, b ¼ 6 km, c ¼ 2 þ 20 þ 6 ¼ 28 km, Dr ¼ 1 g cm3, so
CI¼ 23:146:67 1011 1103 6þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
202 þ 28 6ð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
202 þ 282p
103
¼ 553 gu
Then the observed gravity at point P is
gP ¼ 9796 417 553 ¼ 9795 864 gu
According to the Pratt hypothesis we first determine the contrast of densities
r ¼hr0Dþ h
where D is the level of compensation (100 km) and r0 is the mean density for the crust and
mantle down to depth 100 km:
r0 ¼20 3þ 80 4
100¼ 3:8 g cm3
Substituting we find
r ¼2 3:8
100þ 2¼ 0:074 g cm3
The isostatic correction is
CI ¼ 2 3:14 6:67 1011 0:074 103
100þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
202 þ 102 100ð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
202 þ 1022p
103
¼ 501 gu
72 Gravity
where we have used in Equation (39.3) the values
a ¼ 20 km; b ¼ 100 km; c ¼ 100þ 2 ¼ 102 km
The value of gravity at point P is now
gP ¼ 9796 417 501 ¼ 9795 916 gu
which is larger by 52 gu than using the Airy hypothesis
40. A point P is at altitude 1000 m above sea level. Beneath this point is a sphere
of 1 km radius and GDM ¼ 650 m3 s2, with its centre 4 km vertically below the
point P. Given that the density of the sphere is twice that of the crust and 3/2 that of
the mantle, calculate:
(a) The density of the sphere, crust, and mantle.
(b) The value of gravity that would be observed at P for the isostatic compensation to
be total including the sphere.
(c) The radius of the sphere for the root to be null. Comment on the result.
Use the Airy hypothesis for the isostatic compensation with H ¼ 30 km, 20 km radius
of the cylinder, and theoretical gravity g ¼ 980 Gal.
(a) If we know GDM we can calculate the contrast of densities between the anomalous
mass and the crust
GM ¼4
3prR3G ) r ¼
3GM
4pR3G¼
3 650
4 3:1416 109 6:67 1011
¼ 2:326 g cm3
Since the density of the sphere rsph is double that of the crust rc, the densities of the crust
and mantle are
rsph ¼ 2rC ) r ¼ rsph rC ¼ 2rC rC ¼ rC ¼ 2:326 g cm3
rsph ¼ 4:652 g cm3
rM ¼2
3rsph ¼ 3:101 g cm3
(b) If isostatic compensation is total we have
gI ¼ 0 ¼ gþ gP þ CFA þ CB þ CI þ Cam ð40:1Þ
where g is the normal gravity, gP the observed gravity at point P, CFA the free-air correction,
CB the Bouguer correction, CI the isostatic correction and C am the gravitational effect of
the anomalous mass.
The free-air and Bouguer corrections are given by
CFA ¼ 3:806h ¼ 3:806 1000 ¼ 3086 gu
CB ¼ 0:419r h ¼ 0:419 2:326 1000 ¼ 975 gu
73 Gravity anomalies. Isostasy
The effect of the spherical mass is
Cam ¼GM
hþ dð Þ2¼
650
4ð Þ2 106¼ 41 gu
We calculate the isostatic correction using the Airy hypothesis and taking into account the
presence of the spherical anomalous mass. Thus, according to Fig. 40, the equilibrium
between the gravity at P and at sea level far from P is given by
pa2rCH þ pa2rMt ¼ pa2rC hþ H þ tð Þ þ4
3pR3 rsph rC
and solving for t:
t ¼a2hrC þ
4
3R3 rsph rC
a2 rM rCð Þ¼ 3011m
As in previous problems we calculate the isostatic correction using a cylinder under point P
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
P
h
H
t
R
rM
rCrsph
a
d
x
Fig. 40
74 Gravity
where a ¼ 20 km, b ¼ t ¼ 3011 m, c ¼h þH þ t ¼ 34 011 m and Dr ¼ rM – rC ¼
0.775 g cm3, resulting in
CI ¼ 145 gu
If compensation is total, the isostatic anomaly must be null. This implies that the Bouguer
anomaly is equal, with opposite sign, to the isostatic correction
gI ¼ 0 ¼ gB CI ) gB ¼ 145gu
But the Bouguer correction can be obtained from Equation (40.1):
gI ¼ 0 ¼ gþ gP þ CFA þ CB þ CI þ Cam
Solving for CB we obtain
CB ¼ 975 gu
From the definition of the Bouguer anomaly we can find the value gP of gravity at P:
gB ¼ gP gþ CFA CB þ Cam
145 ¼ gP 9800 000þ 3086 975þ 41 ) gP ¼ 9797 703 gu
(c) Since the density of the sphere is greater than the density of the crust, there is an excess
of gravity at P with respect to other points at sea level far from P, which must be
compensated by a root of crustal material inside the mantle with negative gravitational
influence. In this situation the root can never be null.
41. At 10 km beneath sea level vertically under a point P of height 2000 m there exists
an anomalous spherical mass GDM ¼ 104 m3 s2. At sea level, gravity is 9800 000 gu
and the crustal thickness 20 km. The density of the crust is 2 g cm3, and of the
mantle 4 g cm3. Using the Airy assumption for the isostatic compensation with a
cylinder of 10 km radius, calculate for that point:
(a) The observed gravity.
(b) The free-air, Bouguer, and isostatic anomalies.
(a) For point P the Bouguer correction is
CB ¼ 0:419rh ¼ 0:419 2 h ¼ 0:838h ¼ 1676 gu
The gravity at point P, if there is no isostatic compensation and other effects, can be
obtained from the normal gravity and the free-air and Bouguer corrections
gP ¼ g CFA þ CB ¼ 9800 000 3:086 2000þ 0:838 2000
¼ 9795504 guð41:1Þ
Since there is an anomalous mass under point P we have to add its gravitational contribu-
tion to the gravity at P. For a spherical mass at depth h þ d under P the gravitational
attraction is
75 Gravity anomalies. Isostasy
Cam ¼GM
hþ dð Þ2¼
104 m3 s2
108 m2¼ 104 ms2 ¼ 100 gu
We calculate the root corresponding to the isostatic compensation, assuming the Airy
hypothesis, and taking into account the presence of the anomalous mass in the same way as
in Problem 40:
pa2rCH þ pa2rMt ¼ pa2rC hþ H þ tð Þ þ4
3pR3 rsph rC
¼ pa2rC hþ H þ tð Þ þMa
t ¼pa2rChþMa
pa2ðrM rCÞ¼ 2239m
The isostatic correction using a cylinder is given by
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
ð41:2Þ
where a¼ 10 km, b¼ t¼ 2.239 km, c¼ 20þ 2þ 2¼ 24.239 km, Dr¼ 4 2¼ 2 g cm3,
resulting in
CI ¼ 154 gu
The gravity at point P is the value obtained in (41.1) plus the contribution of the anomalous
mass and minus the isostatic correction:
gP ¼ 9795 504þ 100 154 ¼ 9795 450 gu
(b) The free-air anomaly is equal to this observed value minus the normal gravity plus
the free-air correction:
gFA ¼ gP gþ CAL
Substituting the values we obtain
gFA ¼ 9795 450 9800 000þ 3:086 2000 ¼ 1622 gu
The Bouguer anomaly is given by
gB ¼ gP gþ CAL CB ¼ 54 gu
Finally the isostatic anomaly is the Bouguer anomaly plus the isostatic correction:
gI ¼ 54þ 154 ¼ 100 gu
This value corresponds to the gravitational contribution of the anomalous mass.
42. At a point P of height 2000 m above sea level the measured value of gravity is
979.5717Gal. Beneath P is a sphere centred at a depth of 12 kmbelow sea level, 1 g cm3
density, and radius 5 km. Assuming the Airy hypothesis (H ¼ 30 km, rC ¼ 2.5 g cm3),
rM ¼ 3.0 g cm3), calculate the isostatic anomaly at the point in gu and mGal.
76 Gravity
For the compensation, assume cylinders of the same radius as the sphere. Normal
gravity g ¼ 9.8 m s2.
We first calculate the root t, assuming the Airy hypothesis, corresponding to the height
2000 m of point P. If the situation is of total isostatic equilibrium, we have to introduce the
effect produced by the sphere in the determination of the root (Fig. 40):
4
3p rsph rC
a3 þ pa2rChþ pa2rCH þ pa2rCt ¼ pa2HrC þ pa2rMt
so
t ¼rChþ
4
3a rsph rC
rM rC¼ 10 000m ð42:1Þ
The negative value of t (anti-root) is due to the deficit of mass produced by the presence of
the sphere (rsph < rC) under point P.
The isostatic correction, as in previous problems, is calculated taking a cylinder under the point:
CI ¼ 2pG rM rCð Þ bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
where b ¼ t ¼ 10 000 m, a ¼ 5000, and c ¼ H þ h ¼ 32 000 m.
We obtain CI ¼ 36 gu.
The isostatic anomaly at P is equal to the observed gravity minus the normal gravity and
the free-air, Bouguer, and isostatic corrections, and the attraction of the spherical mass:
gI ¼ gP gþ CFA CB CI Cam ð42:2Þ
The effect of the anomalous mass is given by
Cam ¼G4
3pa3 rsph rC
hþ dð Þ2¼ 267 gu
By substitution in (42.2)
gI ¼ 9 795 717 9 800 000þ 3:086 2000 0:419 2:5 2000þ 267 36
¼ 25 gu
43. A point A on the Earth’s surface is at an altitude of 2100 m above sea level.
Calculate:
(a) The value of gravity at A if the isostatic anomaly is 2.5 mGal. Assume the Airy
hypothesis (rC ¼ 2.6 g cm3, rM ¼ 3.3 g cm3, H ¼ 30 km).
(b) If the previous value had been measured with a Worden gravimeter of constant
0.301 82 mGal/division giving a reading of 630.6, calculate the value of gravity at
another point B at which the device reads 510.1 (both readings corrected for drift).
(c) At what depth is the centre of a sphere of density 4 g cm3 and radius 5 km which
is buried in the crust, given that the anomaly created at a point A, 12 km from the
77 Gravity anomalies. Isostasy
centre of the sphere, not in the same vertical, is 321 gu. Also calculate the
horizontal distance from the centre to point A.
Take, for compensation, cylinders of 10 km radius. g ¼ 9.8 m s2
(a) The isostatic anomaly is given by
gI ¼ gA gþ 3:086h 0:419rChþ CI ð43:1Þ
To calculate the isostatic correction CI, assuming the Airy hypothesis, we must first
calculate the root t that corresponds to the height h
t ¼rCh
rM rC¼ 7800m
As in other problems the isostatic correction is calculated using
CI ¼ 2pG rM rCð Þ bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
and substituting the values
b ¼ t ¼ 7800m; c ¼ hþ H þ t ¼ 39900m;
a ¼ 10 km; r ¼ rM rC ¼ 700 kgm3
we obtain
CI ¼ 84 gu
Solving for gA in Equation (43.1) we obtain
gA ¼ gI þ g 3:086hþ 0:419rCh CI ¼ 9795 748 gu
(b) For a Worden gravimeter the increment in gravity between two points (Dg) is propor-
tional to the increment in the values given by the instrument (DL) corrected by the
instrumental variations
g ¼ KL
gB gA ¼ KðLB LAÞ ) gB ¼ gA þ KðLB LAÞ
¼ gA 364 gu ¼ 9795 384 gu
where K is the constant of the gravimeter.
(c) The anomaly produced by a sphere buried at depth d under sea level at a point at height
h and at a horizontal distance x from the centre of the sphere is given by
Cam ¼G4
3pa3 rsph rC
ðhþ dÞ
x2 þ ðhþ dÞ2 3=2
ð43:2Þ
78 Gravity
For point A (Fig. 43) if r is the distance from the centre of the sphere to the point A,
x2 þ (h þd)2 ¼ r2 and solving for d in Equation (43.2) gives
d ¼Cam
r3
G4
3pa3 rsph rC
h
Substituting the values r ¼ 12 km, a ¼ 5 km, rsph – rC ¼ 1400 kg m3, h ¼ 2100 m,
C am ¼ 321 106 m s2, we obtain:
d ¼ 9245m
The horizontal distance is:
x ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r2 ðhþ dÞ2
q
¼ 3910m
44. In a gravity survey, two points A and B on the Earth’s surface gave the values 159
and 80 mGal for the free-air anomaly, and 51 and 25 mGal for the Bouguer
anomaly, respectively. Given that B is at an altitude 1000 m lower than A, and that the
density of the mantle is 25% greater than that of the crust, calculate:
(a) The value of gravity at A and B, and the densities of the crust and mantle.
(b) The isostatic anomaly according to the hypotheses of Airy (H ¼ 30 km) and Pratt
(D ¼ 100 km, r0 the value determined in the previous part) at point A. Take, for
compensation, cylinders of 10 km radius. g ¼ 980 Gal.
(a) The free-air anomaly at point A is given by
gFAA ¼ gA gþ CFAA ¼ gA gþ 3:086hA ð44:1Þ
r
d
x
a
rC
rsph
h
A
x
Fig. 43
79 Gravity anomalies. Isostasy
The Bouguer anomaly is
gBA ¼ gA gþ CFAA CB
A ¼ gFAA 0:419rChA
Changing values from mGal to gu, we write for the Bouguer anomalies at points A and B
gBA ¼ 510 ¼ 1590 0:419rChA ) 0:419rChA ¼ 2100 gu
gBB ¼ 250 ¼ 800 0:419rChB ) 0:419rChB ¼ 1050 guð44:2Þ
Dividing both equations, we find
hA
hB¼ 2
Knowing that the difference in height between A and B is 1000 m, we obtain for the
heights of both points,
hB ¼ hA 1000 ) hA ¼ hB þ 1000 ¼ 2hB ) hB ¼ 1000m ) hA ¼ 2000m
The density of the crust can be obtained from Equation (44.2):
0:419rChB ¼ 105 105 ms2 ¼ 0:419rC 1000 ) rC ¼ 2:505 g cm3
The density of the mantle is 25% more than that of the crust, so
rM ¼ rC 1þ 0:25ð Þ ¼ 1:2 2:505 ¼ 3:131g cm3
The gravity at A and B is obtained using Equation (44.1):
gA ¼ gFAA 3:086hA þ g ¼ 1590 3:086 2000þ 9800 000 ¼ 9 795 418 gu
gB ¼ 800 3:086 1000þ 9800 000 ¼ 9797 714 gu
(b) For the isostatic anomaly at point A, according to the Airy hypothesis, we first
calculate the value of the root t corresponding to its height:
t ¼rChA
1:25rC rC¼
hA
0:25¼ 8000m
For the isostatic correction we use, as in other problems, a cylinder
CI ¼ 2pG rM rCð Þ bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
Substituting the values
b ¼ t ¼ 8000m; c ¼ hþ H þ t ¼ 40 000m;
a ¼ 10 km; r ¼ rM rC ¼ 626 kgm3
we obtain
CI ¼ 77 gu
80 Gravity
The isostatic anomaly is
gIA ¼ gBA þ CI ¼ 510þ 77 ¼ 433 gu
If we use the Pratt hypothesis, we first calculate the density corresponding to the material
under point A:
r ¼Dr0Dþ h
¼100 2:505
100þ 2¼ 2:456 g cm3
and the contrast of density
r ¼ r0 r ¼ 2:505 2:456 ¼ 0:049 g cm3
The isostatic correction is determined using a cylinder,
CI ¼ 2pG rM rCð Þ bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
where Dr ¼ 0.049 g cm3, b ¼ D ¼ 100 km, a ¼ 10 km, c ¼ D þ h ¼ 102 km.
Then, we obtain
CI ¼ 158 gu
The isostatic anomaly will be the Bouguer anomaly plus the isostatic correction
gIA ¼ gBA þ CI ¼ 510þ 158 ¼ 352 gu
In both cases the anomaly is negative, but using the Airy model the value is greater than
using the Pratt model.
45. At a point P on the Earth’s surface, the observed value of gravity is 9.795
636 m s2, and the Bouguer anomaly is 26 mGal. Assuming the Airy hypothesis
(rC ¼ 2.7 g cm3, rM ¼ 3.3 g cm3, H ¼ 30 km), calculate:
(a) The height of the point.
(b) The isostatic anomaly.
(c) The value of gravity that would be observed at the point if beneath it were a
sphere at a depth of 10 km below sea level, with a density of 2.5 g cm3 and a
radius of 5 km, such that the compensation was total.
Compensation with cylinders of 5 km radius; g ¼ 9.8 m s2.
(a) We calculate the height of point P from the Bouguer anomaly:
gB ¼ gP gþ 3:086h 0:419rCh ) h ¼gB gP þ g
3:086 0:419rC
so
h ¼260 9 795 636þ 9 800 000
3:086 0:419 2:7¼ 2099:8m ffi 2100m
81 Gravity anomalies. Isostasy
(b) To calculate the isostatic anomaly, using the Airy hypothesis, we first calculate the
value of the root t corresponding to the height of the point:
t ¼rC
rM rCh ¼
2:7
3:3 2:7 2100 ¼ 9450m
Using a cylinder, the isostatic correction is given by
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
ð45:1Þ
where we substitute the values
a ¼ 5 km; b ¼ t ¼ 9450m;
c ¼ hþ H þ t ¼ 2100þ 30 000þ 9450 ¼ 41 550m;
r ¼ 3:3 2:7 ¼ 0:6 g cm3
and obtain
CI ¼ 22 gu
The isostatic anomaly is then
gI ¼ gB þ CI ¼ 260þ 22 ¼ 238 gu
(c) If the compensation is total then the isostatic anomaly must be zero. But now we have
to include the gravitational effect Cam produced by the presence of the anomalous mass
of the sphere.
gI ¼ 0 ¼ gP gþ 3:086h 0:419rCh Cam þ CI
Solving for gP:
gP ¼ g 3:086hþ 0:419rChþ Cam CI ð45:2Þ
where the effect of the sphere is given by
Cam ¼G4
3pa3 rsph rC
hþ dð Þ2¼ 48 gu
We calculate the isostatic correction according to the Airy hypothesis. First we calculate
the value of root t, but now we add the effect of the sphere on point P (Fig. 45):
rCpa2hþ rCpa
2H þ4
3pa3 rsph rC
þ rCpa2t ¼ rCpa
2H þ rMpa2t
Solving for t, we obtain
t ¼rChþ
4
3a rsph rC
rM rC¼ 7228m
82 Gravity
We substitute this value of t in Equation (45.1) together with the other values
a ¼ 5 km; b ¼ 7 ¼ 7228m;
c ¼ hþ H þ t ¼ 39 328m; r ¼ 3:3 2:7 ¼ 0:5 kgm3
and obtain
CI ¼ 18 gu
By substitution in (45.2) we find the value of the gravity at P under the given conditions:
gP ¼ 9800000 3:086 2100þ 0:419 2:7 2100 48 18 ¼ 9795830 gu
46. Consider a point on the surface of the Earth in an overcompensated region
at which the values of the free-air and the Bouguer anomalies are 1300 gu and
1200 gu, respectively.
(a) Is this a mountainous or an oceanic zone? Give reasons.
(b) Calculate the altitude and the value of gravity at the point given that the density
of the crust is 2.72 g cm3.
(c) If the isostatic anomaly is -1062 gu calculate, according to the Airy hypothesis
(rC ¼ 2.72 g cm3, rM ¼ 3.30 g cm3, H ¼ 30 km), the value of the root
responsible for this anomaly. Compare it with the value that it would have if
the region were in isostatic equilibrium.
H
t
h
d
P
rC
rM
rsph
ax
Fig. 45
83 Gravity anomalies. Isostasy
Compensation with cylinders of 10 km radius; g ¼ 9.8 m s2.
(a) Since the free-air anomaly is positive and the Bouguer anomaly is negative, this
indicates that this is a mountainous region.
(b) From the free-air and Bouguer anomalies we can easily calculate the height of the point:
gFA ¼ gP gþ 3:086h
gB ¼ gP gþ 3:086h 0:419rChð46:1Þ
and solving for h,
h ¼gFA gB
0:419rC¼ 2193m
The observed gravity at P can be obtained from either of the two equations (46.1):
gP ¼ g 3:086hþgFA ¼ 9794 532 gu
(c) The isostatic correction is found from the known Bouguer and isostatic anomalies:
gI ¼ gB þ CI ) CI ¼ 1062þ 1200 ¼ 138 gu
The isostatic correction, using the Airy hypothesis, is given, as in previous problems, as a
function of the root t, by
CI ¼ 138 ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
where we substitute
r ¼ 3:3 2:72 ¼ 0:58 g cm3; a ¼ 10 km;
b ¼ t; c ¼ hþ H þ t ¼ 2:193þ 30þ t km
and obtain for t,
t ¼ 19 984 m
If the region is in equilibrium the root due to the height h would be
t ¼rCh
rM rC¼
2:72 2193
3:3 2:72¼ 10 284m
Since we have already obtained a larger value (t ¼ 19 984 m), this indicates that the region
is overcompensated.
47. In an oceanic region, gravity is measured at a point on the surface of the sea,
obtaining a value of 979.7950 Gal. Calculate, using the Airy hypothesis (H ¼ 30 km,
rC ¼ 2.9 g cm3, rM ¼ 3.2 g cm3, rW ¼ 1.04 g cm3):
(a) The isostatic anomaly if the thickness of the crust is 8.4 km.
84 Gravity
(b) The thickness that the water layer would have to have if 15 km vertically below
the point there was centred a sphere of 10 km radius such that the anti-root is null
and the compensation total. Also calculate the density of the sphere.
For the compensation, take cylinders of 10 km radius. g ¼ 9.8 m s2.
(a) We are in an oceanic region, therefore in the calculation of the root for the isostatic
compensation according to the Airy hypothesis we have to consider the layer of water
of density rW. The value of the root is now given by
t0 ¼
rC rWrM rC
h0 ¼2:9 1:04
3:2 2:9h0 ¼ 6:2h0 ð47:1Þ
According to Fig. 47a, we have the following relation: H h0 t0 ¼ e) h0 ¼ H t
0 e,
where e is the thickness of the crust at the oceanic region, H the thickness of
the normal (sea level) continental crust, h0 the thickness of the water layer, and t0 the
negative root.
Substituting the values of t 0 from Equation (47.1) we obtain for h0
h0 ¼ 30 6:2h0 8:4 ) h0 ¼ 3 km ¼ 3000m
t0 ¼ 6:2 3000 ¼ 18 600m
From the value of the root we calculate the isostatic correction using a cylinder of 10 km
radius,
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
where Dr ¼ 3.2 2.9 ¼ 0.3 g cm3 ¼ 300 kg m3, b ¼ t0, c ¼ H ¼ 30 000 m, and so
CI ¼ 269 gu
To calculate the isostatic anomaly we first have to apply the Bouguer correction which in
this case consists of two terms: the first to eliminate the attraction of the water layer
rM
rC
rW
t
h
eH
P
Fig. 47a
85 Gravity anomalies. Isostasy
(2pGrWh0) and the second to replace this layer by one of density equal to the crustal
density (þ2pGrCh0). Since the point is at sea level the free-air correction is null:
DgI ¼ g gþ CB CI
CB ¼ 0:419ðrC rWÞh0 ¼ 2338 gu
DgI ¼ 19 gu
(b) If the anti-root is null and there is total compensation, then we have
t0 ¼ 0
gI ¼ 0 ¼ g g 2pGrWh0 þ 2pGrCh
0 Camð47:2Þ
Since the isostatic anomaly must be null, then the anomalous spherical mass and the water
layer must compensate each other. The attraction of the anomalous mass is Cam ¼ GDM/d2
where d is the depth of its centre below sea level. Then we can write
g gþ 2pG rC rWð Þh0 GM
d2¼ 0
where the mass of the sphere is
M ¼4
3pa3rsph
If the point P is totally isostatically compensated and the anti-root is null, then (Fig. 47b)
pa2h0rW þ pa2ðH h0ÞrC þ4
3pa3ðrsph rCÞ ¼ pa2HrC
Solving for h0 gives
h0 ¼4a rC rsph
3 rW rCð Þð47:3Þ
rCa
rsph
d
rW
P
h
e
H
x
Fig. 47b
86 Gravity
Substituting this value in (47.2) we obtain
rsph ¼g g
4
3pG 2a
a3
d2
þ rC ¼ 3372 kgm3
and putting this value in (47.3), h 0 ¼ 3369 m.
48. At a point on the Earth’s surface, 500 m below sea level, a gravity value is
measured of 980.0991 Gal. If the region is in isostatic equilibrium calculate, using
the Airy hypothesis (H ¼ 30 km, rC ¼ 2.7 g cm3, rM ¼ 3.2 g cm3):
(a) The thickness of the crust.
(b) The isostatic anomaly in gu, with reasons for the sign of each correction. Take
compensating cylinders of 5 km radius; g ¼ 9.8 m s2.
(a) We calculate first the root, according to the Airy hypothesis which corresponds to
the depth of the point, applying the condition of isostatic equilibrium (Fig. 48)
rCH ¼ rCðH h0 t0Þ þ rMt
0
so
t0 ¼
rCrM rC
h0 ¼ 2700m
The thickness of the crust e at that point is
e ¼ H h0 t0 ¼ 30 000 2700 500 ¼ 26 800m
(b) The isostatic anomaly is given by
gI ¼ gP g 3:086h0 þ 0:419rCh0 CI
The free-air correction (3.086h0) is negative because the point is below sea level. For the
same reason the Bouguer correction (0.419 rC h0) is positive. The isostatic correction,
using the Airy hypothesis, is calculated as in previous problems using a cylinder under the
point of 5 km radius and density contrast D r ¼ 3.2 2.7 ¼ 0.5 g cm3. The value of the
anti-root t 0 has already been calculated, so
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
trM
rC
Ph
eH
Fig. 48
87 Gravity anomalies. Isostasy
Substituting c ¼ H – h0 and b ¼ t0, we obtain
CI ¼ 9 gu
Then the isostatic anomaly is DgI ¼ 5 gu.
49. In an oceanic region where the density of the crust is 2.90 g cm3 and that of the
mantle 3.27 g cm3, the value of gravity measured at a point P on the sea floor at
depth 4000 m is 9.806 341 m s2.
Calculate, according to the Airy hypothesis:
(a) The thickness of the crust.
(b) The isostatic anomaly in gravimetric units.
Data: rw ¼ 1.04 g cm3, H ¼ 30 km, g ¼ 9.8 m s2. Take, for compensation, cylinders
of 10 km radius.
(a) First we calculate the value of the root according to the Airy hypothesis
t0 ¼
rC rWrM rC
h0 ¼ 20 108m
The thickness of the crust under the point is found by (Fig. 49)
e ¼ H h0 t0 ¼ 30000 40000 2018 ¼ 5892m
(b) Because the point is located at the bottom of the sea, to reduce the observed value of
gravity to the surface of the geoid (sea level) we eliminate first the attraction of the
water layer. Then we apply the free-air and the Bouguer corrections, to take into
account the attraction of a layer of crustal material which replaces the water. Finally we
apply the isostatic correction:
gI ¼ gP gþ 0:419rWh0 3:086h0 þ 0:419rCh
0 CI
The isostatic correction is calculated using a cylinder of 10 km radius,
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
h
e
t rM
rW P
rCH
Fig. 49
88 Gravity
where b ¼ t0 ¼ 20 108 m, c ¼ H h0 ¼ 26 000 m, Dr ¼ 3.27 2.9 ¼ 0.37 g cm3 ¼
370 kg m3
By substitution we obtain
CI ¼ 598 gu
The isostatic anomaly is:
gI ¼ 9 806 341 9 800 000 3:086 4000þ 0:419 1:04þ 2:9ð Þ 4000 598
¼ 4 gu
50. At a point with coordinates 42.78º N, 0.5º E and height 1572 m, the observed value
of gravity is 980.0317 Gal.
(a) Calculate the free-air and Bouguer anomalies.
(b) If cylinders of 10 km radius beneath that point are used for the isostatic compen-
sation, calculate the gravimetric attraction of the mass defect corresponding to
the altitude of the point according to the Airy and Pratt hypotheses. Take, for
the crust, H ¼ 30 km, rC ¼ 2.67 g cm3, for the mantle, rM ¼ 3.27 g cm3, and
D ¼ 100 km for the Pratt level of compensation.
(c) How deep should the root of the Airy model be for the compensation to be
total?
(a) We calculate first the normal gravity at the point where gravity has been observed
using the expression
g ¼ 9:780 32 1þ 0:005 3025 sin2 ’
¼ 9:804 243m s2
where ’ is the latitude.
The free-air anomaly is given by
gFA ¼ gP gþ CFA ¼ 9800 317 9804 243þ 3:086 1572 ¼ 925 gu
and the Bouguer anomaly by
gB ¼ gP gþ CFA þ CB ¼ 9800 317 9804 243þ 1:967 1572 ¼ 834 gu
(b) If we approximate the isostatic compensation by means of a cylinder of radius a
under the point, we use the expression
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
ð50:1Þ
where Dr is the density contrast, b the height of the cylinder, and c the distance from the
base of the cylinder to the observation point.
Airy: We calculate first the root given by the equation
t ¼ 4:45 h ¼ 4:45 1572 ¼ 6995 km
For the isostatic correction we substitute in (50.1) the values
89 Gravity anomalies. Isostasy
a ¼ 10 km; b ¼ t ¼ 6995 km; c ¼ t þ H þ h ¼ 38 567 km;
r ¼ rM rC ¼ 3:27 2:67 ¼ 0:6 g cm3
and obtain
CI ¼ 2 3:1416 6:67 1011 m3 s2 kg1 0:5 103 kgm3 A
where:
A ¼ 6:995þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
102 þ 38:567 6:995ð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
102 þ 38:5672p
103 ¼ 270m
so
CI ¼ 68 gu
Pratt: The contrast of densities is now given by
r ¼h
Dþ hr0 ¼
1575
100 000þ 1572 2:67 ¼ 0:04 g cm3
and substituting in Equation (50.1) with the values
a ¼ 10 km, b ¼ 100 km, c ¼ D + h ¼ 101 572 m, we have
CI ¼ 135 gu
(c) If the isostatic compensation is total (isostatic anomaly null) the isostatic correc-
tion, according to the Airy hypothesis, coincides with the value of the Bouguer
anomaly (834 gu):
gB ¼ 2pGr t þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 ðH þ hÞ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ t þ H þ hð Þ2q
so
t þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 ðH þ hÞ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ t þ H þ hð Þ2q
¼gB
2pGr
In this expression we solve for the value of the root t:
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ t þ H þ hð Þ2q
¼ t þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 ðH þ hÞ2q
gB
2pGr¼ t þ N
a2 þ t þ H þ hð Þ2 ¼ t2 þ N 2 þ 2tN
a2 þ t2 þ ðH þ hÞ2 þ 2tðH þ hÞ ¼ t
2 þ N2 þ 2tN
so
t ¼N 2 a2 ðH þ hÞ2
2ðH þ h NÞ
90 Gravity
By substitution of the values of N, a, H, and h we obtain
t ¼ 58 875m
Because this root has a negative value greater than the thickness of the crust, total
compensation is not possible.
51. Calculate the free-air anomaly observed on a mountain of height 2000 m which
is fully compensated by a root of depth t ¼ 10 km. The compensation is by a
cylinder of radius 20 km, the density of the crust is 2.67g cm3, and that of the
mantle is 3.27g cm3.
The free-air anomaly is given by
gFA ¼ g gþ 3:086h
Since the point is isostatically compensated, we calculate the isostatic correction using a
cylinder as in previous problems:
CI ¼ 2pGr bþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ c2p
where we substitute
b ¼ t ¼ 10 km, c ¼ h þH þ t ¼ 42 km, a ¼ 20 km, Dr ¼ 600 kg m3
and obtain
CI ¼ 306 gu
Since the point is totally compensated the isostatic anomaly must be zero:
gI ¼ g gþ CFA CB þ CI ¼ 0
The free-air anomaly can now be written as
gFA ¼ g gþ CFA ¼ CB CI ð51:1Þ
We can calculate the Bouguer correction:
CB ¼ 1:119h ¼ 1:119 2000 ¼ 2238 gu
and substituting in (51.1) we obtain, for the free-air anomaly,
gFA ¼ 2238 301:5 ¼ 1932 gu
52. Calculate for a point P at height 2100 m, latitude 40ºN, and observed gravity
979.7166 Gal the refined Bouguer anomaly in gravimetric units. Consider a surplus
mass compartment of 3000 m, mean height, 3520 m inner radius, 5240 m outer
radius, with n ¼ 16. The density of the crust is 2.5 g cm3.
The refined Bouguer anomaly is obtained using, besides the free-air and Bouguer correc-
tions, the correction for the topography or topographic or terrain correction (T):
gB ¼ g gþ CFA CB þ T
91 Gravity anomalies. Isostasy
The normal gravity at latitude 40º N is given by
g ¼ 9:780 32 1þ 0:0053025 sin2 ’
¼ 9:801 747m s2
the free-air correction by
CFA ¼ 3:086h ¼ 6481 gu
and the Bouguer correction by
CB ¼ 0:419rCh ¼ 2200 gu
The topographic correction is introduced in order to correct for the topographic masses not
included in the Bouguer correction, that is, in this case those above the height h (Fig. 52).
Remember that the Bouguer correction corrects for an infinite layer or plate of thickness h
and doesn’t consider the additional masses above h or the lack of masses below h. The
topographic correction is always positive because the masses above height h produce on
point P an attraction of negative sign which must be added and the lack of mass under h
must also be taken into account with a positive sign, since it has been subtracted in the
Bouguer correction.
To calculate the attraction of the mass above height h we use the attraction of
concentric cylinders (in our case two) with axis passing through point P and with
height equal to the difference between the height h of the point P and the height of
the mass of the topography hm above h. The cylinder is divided into n sectors with
radius a1 and a2 to approximate the topography (Fig. 52). Then the topographic
correction is given by
T ¼2pGrC
n
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a22 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a22 þ c2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a21 þ c bð Þ2q
þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a21 þ c2q
In our case we substitute the values
a2 ¼ 5240m; a1 ¼ 3520m; b ¼ hm h ¼ 900m; c ¼ 0; n ¼ 16
P
hm
rC
a1
a1
a2
a2
P x
h
Fig. 52
92 Gravity
and obtain
T ¼ 2 gu
The refined Bouguer anomaly is
gB ¼ g gþ CFA CB þ T ¼ 298 gu
53. Calculate the topographic correction for a terrestrial compartment of inner radius
a1 ¼ 5240 m, outer radius a2 ¼ 8440 m, n¼ 20, mean height 120 m, with 2000 m being
the height of the point P. Take r ¼ 2.65 g cm3.
In this problem we consider the topographic correction for the case of the lack of mass in
the topography at heights below that of the point P. Since in the Bouguer correction we
have subtracted an infinite layer of thickness h, we have to correct for the places where the
mass was not present (Fig. 53).
The topographic or terrain correction T in this case is calculated in the same way as in
the previous problem. Thus we take n sectors of cylinders with axis at point P and height
equal to the difference between h and hm (Fig. 52b). The correction is then given by
T ¼2pGrC
n
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a22 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a22 þ c2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a21 þ c bð Þ2q
þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a21 þ c2q
where we substitute a2 ¼ 8440 m, a1 ¼ 5240 m, b ¼ c ¼ h – hm, n ¼ 20 to obtain
T ¼ 0:67mGal
54. Calculate the topographic correction for an oceanic sector or compartment
of inner radius a1 ¼ 5240 m, outer radius a2 ¼ 8440 m, n ¼ 20, mean depth
525 m, with 600 m being the height of the point P. Take rC ¼ 2.67 g cm3, rW ¼
1.03 g cm3.
In this problem we have to correct for the lack of mass in the oceanic area near the point P,
between the sea level and height h (column 1 in Fig. 54). Also we have to take into account
the attraction produced by the water layer between sea level and the bottom of the sea
(column 2 in Fig. 54).
hmrC
a1
a2
P
h
Fig. 53
93 Gravity anomalies. Isostasy
To calculate the necessary topographic correction we proceed as in Problems 52 and 53,
using cylindrical sectors:
T ¼2pGr
n
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a22 þ c bð Þ2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a22 þ c2q
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a21 þ c bð Þ2q
þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a21 þ c2q
where
a2 ¼ 8440m; a1 ¼ 5240m; n ¼ 20
For the correction corresponding to the attraction of column 1, between height h and sea
level, we substitute the values:
b ¼ h ¼ c; r ¼ rC
and obtain
T1 ¼ 0:07mGal
For the correction of the attraction of column 2 between sea level and the bottom of the sea
we use the difference between the densities of the crust and of water:
b ¼ p ¼ 525m
c ¼ pþ hs ¼ 1125m
c b ¼ hs
r ¼ rC rW
T2 ¼ 0:11mGal
The total topographic correction is
T ¼ T1 þ T2 ¼ 0:18mGal
rw
a2
a1 P
P
h h1
2
Fig. 54
94 Gravity
Tides
55. Two spherical planets A and B of radii 2a and a and masses 3m and m are
separated by a distance (centre to centre) of 6a. The only forces acting are gravita-
tional, and the system formed by the two planets rotates in the equatorial plane.
(a) Calculate the value of the components of the acceleration of the tides at the Pole of
each planet directly and using the tidal potential. On which planet are they greater?
(b) If each planet spins on its axis with the same angular velocity as the system, what,
for each planet, is the ratio between the centrifugal force and the maximum of the
tidal force at the equator? On which planet is this ratio greater?
(a) From Fig. 55a we can deduce that at the Pole of planet A, the radial component of
the acceleration of the tides produced by planet B is
gTr¼
Gm
q2cos a
where q is the distance from the Pole of planet A to the centre of planet B and a the angle
formed by q and the radius at the Pole of planet A.
By substitution of the required values we obtain
q ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
6að Þ2 þ 2að Þ2q
¼ffiffiffiffiffi
40p
a
cos a ¼2affiffiffiffiffi
40p
a) a ¼ 71:6
gTr¼
Gm
40a2cos 71:6ð Þ ¼ 0:008
Gm
a2
The tangential component gTy is given by (Fig. 55a)
gTy ¼ Gm
36a2þGm
q2sin a ¼
Gm
36a2þ
Gm
40a2sin 71:6ð Þ
¼Gm
a20:028þ 0:024ð Þ ¼ 0:004
Gm
a2
2a
3m
6a
qa
A
q
b a
m
B
Fig. 55a
95 Tides
If we use the tidal potential,
c ¼GMr
2
2R33cos2# 1
where R is the distance between the centres of planets A and B (Fig. 55a), and # is the
angle the position vector r forms with the distance vector R (in this case it is equal to the
colatitude, # ¼ y)
c ¼GMr
2
2R33cos2y 1
The radial and tangential components of the acceleration are given by
gTr¼
@c
@r¼
@
@r
Gmr2
2 6að Þ33cos2y 1
¼Gmr
216a33cos2y 1
gTy ¼1
r
@c
@y¼
1
r
@
@y
Gmr2
2 6að Þ33cos2y 1
¼Gmr
216a33 cos y sin yð Þ
For planet A, at the Pole, r ¼ 2a and y ¼ 90º, so
gTr¼
Gm
108a2¼ 0:009
Gm
a2
gTy ¼ 0
For planet B, we proceed in a similar manner:
q02 ¼ 36a2 þ a2 ¼ 37a2
cos a ¼affiffiffiffiffi
37p
a) a ¼ 80:5
Therefore,
gTr¼
G3m
37a2cos 80:5ð Þ ¼ 0:013
Gm
a2
gTy ¼ G3m
36a2þG3m
37a2sin 80:5ð Þ ¼ 0:003
Gm
a2
Using the tidal potential, we obtain the acceleration components
c ¼G3mr2
2 6að Þ33cos2y 1
gTr¼
@c
@r¼
Gmr
72a33cos2y 1
gTy ¼ 1
r
@c
@y¼
9
216
Gmr
að Þ3cos y sin yð Þ
Substituting at the Pole of planet B, r ¼ a and y ¼ 90º, we have
gTr¼ 0:014
Gm
a2
gTy ¼ 0
96 Gravity
(b) First we calculate the centre of gravity of the system formed by the two planets
measured from the centre of planet A (Fig. 55b):
x ¼3m 0þ m 6a
3mþ m¼
3
2a
The rotation radius for planet A is 3/2a and for planet B
6a3
2a ¼
9
2a
In the rotating system the centrifugal force equals the force of gravitational attraction,
which at the equator (y ¼ 0º) is
fg ¼Gm3m
6að Þ2¼ fC ¼ o2
r ¼ mo2 9
2a
From this expression we obtain the value of the angular velocity o of the rotation system:
3Gm2
36a2¼ mo2 9
2a ) o2 ¼
Gm
54a3
Since the angular velocity of the spin of each planet is equal to that of the system, the spin
centrifugal force at the equator of planet A is
fC ¼ o2r ¼
Gm
54a32a ¼
Gm
27a2
The tidal force is
fT ¼ gTr¼
2Gmr
R3¼
2Gm2a
63a3
2a
3m
(3/2)a (9/2)a
6a
A
m
B
w
Fig. 55b
97 Tides
and their ratio
fC
fT¼
Gm
27a2
Gm
54a2
¼ 2
If we repeat these calculations for planet B, we obtain
fC ¼ o2r ¼
Gm
54a3a ¼
Gm
54a2
fT ¼2G3ma
63a3¼
Gm
36a2
and the ratio
fC
fT¼
Gm
54a2
Gm
36a2
¼ 0:666
The ratio is larger for planet A, as expected owing to its larger radius.
56. Two planets of mass M and radius a are separated by a centre-to-centre distance
of 8a. The planets spin on their own axes with an angular velocity such that the value
of the centrifugal force at the equator is equal to the maximum of the tidal force (the
equatorial plane is the plane in which the system formed by the two planets rotates
around an axis normal to that plane).
(a) Calculate the value of the components of the vector g as multiples of GM/a2 for a
point of l'¼ 60º and w ¼ 45º (with l¼ 0º, being the meridian in front of the other
planet) including all the forces that act.
(b) What is the relationship between the angular velocity of each planet and that of
the system?
(a) The tidal potential is given by
c ¼GMr
2
2R33cos2# 1
where R is the centre-to-centre distance between the planets, and # the angle formed by the
vector r to a point and R (Fig. 56a). From this potential we calculate the radial component
of the tidal force:
f Tr¼
@c
@r¼
@
@r
GMr2
2R33cos2# 1
¼GMr
R33cos2# 1
At the equator of one planet # ¼ 0º, r ¼ a, and R ¼ 8a, so
f Tr¼
GMa
8að Þ32 ¼
GM
256a2
98 Gravity
The spin centrifugal force is
fC ¼ o2a
Equating these two expressions we find the value of the spin angular velocity,
fC ¼ f Tr) o2a ¼
GM
256a2) o ¼
1
16
ffiffiffiffiffiffiffiffi
GM
a3
r
At a point on the surface of one of the planets the total potential is the sum of the
gravitational potential V, the spin potential F, and the tidal potential c:
U ¼ V þ Fþ c ¼GM
rþ1
2o2
r2cos2’þ
GMr2
2R33cos2# 1
For a point P at latitude ’ and longitude l (Fig. 56b)
cos# ¼ cos’ cos l
and the potential U is
U ¼GM
rþ1
2o2
r2cos2’þ
GMr2
2R33cos2’cos2l 1
The components of gravity including the three effects are,
gr ¼@U
@r¼
GM
r2
þ o2r cos2 ’þ
GMr
R33 cos2 ’ cos2 l 1
g’ ¼ 1
r
@U
@’¼ o2
r cos’ sin’þGMr
R33 cos2 l cos’ sin’
gl ¼1
r cos’
@U
@l¼
1
r cos’
GMr2
2R36 cos2 ’ cos l sin l
¼ GMr
R33 cos’ cos l sin l
ww
gr
gλ
gθ
θ
a
MMR
Ω
x
Fig. 56a
99 Tides
At the required point,
r ¼ a; ’ ¼ 45;
l ¼ 60 ) cos# ¼ cos 45 cos 60 ¼ 0:35 ) # ¼ 69:3
so
gr ¼ GM
a2þ
GM
256a3
a1
2þ
GMa
8að Þ331
2
1
4 1
¼ 0:9993GM
a2
g’ ¼GM
256a3a
1ffiffiffi
2p
1ffiffiffi
2p þ
GMa
8a3ð Þ31
4
1ffiffiffi
2p
1ffiffiffi
2p ¼ 0:0027
GM
a2
gl ¼ GMa
ð8aÞ33
ffiffiffi
2p
2
1
2
ffiffiffi
3p
2¼ 0:0018
GM
a2
(b) To obtain the angular velocity of the rotation of the system (O) we take into account
that the centrifugal force due to the rotation of the system at the equatorial plane is
equal to the gravitational attraction between the two planets:
M2p ¼ G
MM
ð8aÞ2
where p is the distance from the centre of one planet to the centre of gravity of the system.
Then we find
M24a ¼ G
MM
ð8aÞ2)
2 ¼GM
4 64a3¼
GM
256a3
and finally O ¼ o.
l
j
j
a
P
ϑ
ϑ
Fig. 56b
100 Gravity
57. Consider two planets of equal massm and radius a separated by a centre-to-centre
distance of 8a. Only gravitational forces act.
(a) Calculate the tidal force at the equator on one of the planets directly, using the
formula of the tidal potential (do so at l ¼ 0º, i.e. the point in front of the other
planet). Express the result in mGal given that Gm/a2 ¼ 980 000 mGal.
(b) Compare and comment on the reason for the difference between the results of the
direct calculation and using the tidal potential.
(c) What relationship must there be between the angular velocities of the planet’s
spin and of the system’s rotation for the centrifugal force due to the planet’s spin
to be equal to the tidal force at the equator and l ¼ 0º?
(a) The exact calculation of the tidal force at a point located at the equatorial plane in
front of the other planet is (Fig. 57)
fT ¼Gm
7að Þ2
Gm
8að Þ2¼
Gm
a21
49
1
64
¼ 46 875 gu
Using the tidal potential
c ¼Gmr2
2 Rð Þ33cos2# 1
where R is the centre-to-centre distance between the planets (8a), r the radius to the point
where the tide is evaluated (a), and # the angle between r and R (at the equator in front of
the other planet # ¼ 0º), the radial component of the tidal force can be derived from the
potential. At the equator this is the total tidal force
f Tr¼
@c
@r¼
@
@r
Gmr2
2 8að Þ33cos2# 1
!
¼2Gma
2 512a3ð Þ2 ¼ 38 281 gu
(b) The difference between the value obtained by the exact calculation and by using
the tidal potential is 8594 gu, that is, 18%. This is explained because the tidal
w w
a
a
m m
8a
Ω
x x
Fig. 57
101 Tides
potential is a first-order approximation corresponding to terms of the order of
(r/R)2. For a relatively large value of r/R (1/8) this approximation is not very good.
(c) Now we make the spin centrifugal force equal to the tidal force for a point at the
equator of one of the planets which is given by exact calculation
fT ¼Gm
a21
49
1
64
If the spin angular velocity is op, the spin centrifugal force at the equator is given by
fC ¼ o2pa
Equating these two expressions we obtain
fC ¼ fT ) o2pa ¼
Gm
a21
49
1
64
) o2p ¼
Gm
a31
49
1
64
ð57:1Þ
The centrifugal force due the rotation of the system with angular velocity os is equal to the
gravitational attraction between the two planets:
mo2s4a ¼
Gm
64a2) o2
s ¼Gm2
256a3ð57:2Þ
From (57.1) and (57.2) we obtain the relation between the two angular velocities:
o2p
o2s
¼
Gm
a31
49
1
64
Gm
256a3
¼ 1:22 )op
os
¼ 1:10
58. Two planets of equal mass m and radius a are separated by a distance R. The spin
angular velocity of each planet is such that the centrifugal force at the equator is
equal to the maximum of the tidal force. If the sum of the two forces at the equator
cancels the gravitational force, what is the distance R?
The tidal potential is given by
c ¼Gmr2
R3
1
23cos2# 1
where R is the centre-to-centre distance between the planets, r the radius to the point where
the tide is evaluated, and # the angle between r and R. The maximum value is at a point at
the equator in front of the other planet, # ¼ 0º and r ¼a. Then
fT ¼@c
@r¼
Gmr
R33cos2# 1
¼2Gma
R3ð58:1Þ
The spin centrifugal force for a point at the equator is
fC ¼ o2a ð58:2Þ
Equating (58.1) and (58.2) we obtain the spin angular velocity,
102 Gravity
2Gma
R3¼ o2a ) o2 ¼
2Gm
R3
If, at the point considered, the sum of the spin centrifugal force and the tidal force cancel
the gravitational force of the planet, then
F þ fC þ fT ¼ 0
where F ¼ Gm/a2.
The value of R must be
4Gma
R3¼
Gm
a2) R ¼
ffiffiffiffiffi
4a3p
59. Two spherical planets A and B of radii 2a and a and masses 5M and M spin on
their axes with equal angular velocities. They are separated by a centre-to-centre
distance of 8a, and form a system that rotates in the equatorial plane of both planets
with an angular velocity that is equal to that of the spin angular velocity of each one.
(a) Determine the total potential for points on planet A.
(b) Determine the expression for the three components of the total gravity, including
the tide, for a point on the surface of planet A at longitude 0º.
(c) If the Love number h on planet A is 0.5, determine the height of the terrestrial tide
as a multiple of a at the equator, at local noon with respect to planet B, in the case
that the system’s rotational angular velocity is the same as that of the spin of the
two planets about their axes.
(a) We calculate the centre of gravity of the system, measured from the centre of planet
A (Fig. 59a):
X ¼5M 0þM 8a
5M þM¼
4
3a
8a
A
(4/3)a
2a
5M
M
B
xx
Fig. 59a
103 Tides
The total potential U at a point on the surface of planet A at latitude ’ is given by the sum
of the gravitational potential V, plus the spin potential Ф, plus the tidal potential c
produced by planet B:
U ¼ V þ Fþ c
U ¼5GM
rþ1
2o2
r2cos2’þ
GMr2
8að Þ323cos2# 1 ð59:1Þ
where according to Fig. 59b
cos# ¼ cos’ cos t
where t is the local time of planet Bwith respect to planet A (hour-angle), at a point of l¼ 0º,
the geographical longitude at planet A. For t ¼ 0, the planet B is in front of the point, so
U ¼5GM
rþ1
2o2
r2cos2’þ
GMr2
2 8að Þ33cos2’cos2t 1
(b) The components of gravity are
gr ¼@U
@r¼
5GM
r2
þ o2rcos2’þ
GMr
8að Þ33cos2’cos2t 1
gy ¼ 1
r
@U
@’¼ o2
r cos’ sin’þGMr
8að Þ33 cos’ sin’cos2t
gl ¼1
r cos’
@U
@t¼
GMr
8að Þ33 cos’ cos t sin t
2a
l
j
ϑ
ϑ
j
P
Fig. 59b
104 Gravity
At a point on the surface of planet A, r ¼2a,
gr ¼ 5GM
4a2þ o22acos2’þ
GM2a
8að Þ33cos2’cos2t 1
gy ¼ o22a cos’ sin’þGM2
83a23 cos’ sin’cos2t
gl ¼GM2
83a23 cos’ cos t sin t
(c) At the equator ’ ¼ 0º, at 12 h with respect to B, t¼ 180º, h ¼ 1/2, and o ¼ O. The
height of the equilibrium terrestrial tide is given by
B ¼ hc
g
At the equator of planet A the tidal potential (59.1) is
c ¼2GM 2að Þ2
2 8að Þ3¼
GM
128a
If we approximate g by gr
gr ¼ 5GM
4a2þ o22aþ
GM2a
8að Þ32 ¼
5GM
4a2þ
GM
128a2þ o22a
¼ 159GM
128a2þ 2o2a
and the height of the equilibrium tide is
B ¼1
2
GM
128a159MG
128a2þ 2ao2
ð59:2Þ
We know that the angular velocity of the rotation of the system is equal to the spin angular
velocity of both planets, so the spin angular velocity is given by
5Mo2 4
3a ¼
G5M 2
8að Þ2) o2 ¼
3GM
256a2
By substitution in (59.2)
B ¼1
2
GM
128a159MG
128a2þ 2a
3GM
256a2
¼ a
312
60. The Earth is formed by a sphere of radius a and density r, and a core of radius a/2
and density 2r, in the northern hemisphere, centred on the axis of rotation and
105 Tides
tangent to the equatorial plane. The Moon has mass M/4 (whereM ¼ (4/3)pra3), is at
a distance (centre-to-centre) of 4a, and orbits in the equatorial plane. Determine:
(a) The total potential and the components of gravity including the tidal forces.
(b) The total deviation of the vertical from the radial at lunar noon, and the deviation
due to the tide at the same hour for latitude 45º N, with m ¼ 1/8.
(a) The total potential U is equal to the gravitational potential of the planet V1 with
uniform density plus that of the core V2 using the differential mass, the spin
potential F, and the tidal potential c. The gravitational potentials are given by
(Fig. 60):
V1 ¼GM
r
V2 ¼GM 0
q
where the differential mass of the core is
M 0 ¼4
3pð2r rÞ
a3
8¼
M
8) V2 ¼
GM
8q
and q is the distance to the centre of the core. Its inverse can be approximated by
1
q¼
1
r1þ
a
2rcos yþ
a
2r
2 1
23cos2y 1
The spin potential is
F ¼1
2o2
r2sin2y
and the total potential is
U ¼GM
r1þ
1
81þ
a
2rcos yþ
a2
2rð Þ21
23cos2y 1
!
þr
a
3 m
2sin2y
" #
þ c
r
2r
4a
a/2q P
ϑ
q r
j L
t – 180°
w
Fig. 60
106 Gravity
where m ¼ o2 a3/GM.
The tidal potential c due to the Moon is given by
c ¼GMLr
2
2R33cos2# 1
ð60:1Þ
According to Fig. 60
cos2# ¼ cos2’cos2 t 180ð Þ
By substitution in (60.1), since R (the centre-to-centre distance between the Earth and the
Moon) is 4a, ML ¼ M/4, and ’ ¼ 90º y, we obtain
c ¼GMr
2
512a33cos2’cos2 t 180ð Þ 1
The potential U is
U ¼ GM5
8rþ
a
16r2sin’þ
a2
64r33 sin2 ’ 1
þr2
a3m
2cos2 ’þ
r2
512a33 cos2 ’ cos2 t 1
The components of gravity are found by taking the derivatives of U with respect to r and ’:
gr ¼ GM 5
8r2
2a
16r3sin’
3a2
64r43 sin2 ’ 1
þr
a3m cos2 ’þ
2r
512a33 cos2 ’ cos2 t 1
gy ¼ 1
r
@U
@’
¼GM
r
a
16r2cos’
a2
64r36 cos’ sin’ð Þ
þr2
a3m
22 sin’ cos’þ
r2
512a36 sin’ cos’ cos2 t
By substituting r ¼ a, ’ ¼ 45º, m ¼ 1/8, and at 12 h lunar time, t ¼ 180º, we obtain
gr ¼ 1:128GM
a2
gy ¼ 0:023GM
a2
(b) The deviation of the vertical with respect to the radial direction at 12 h lunar time is
given by
tan i ¼gy
gr¼ 0:020 ) i ¼ 1:2
107 Tides
The part of the deviation due to the lunar tide is given by
tan i0 ¼gMygr
¼1
gr
1
r
@c
@’¼
1
gr
GMr2
r512a36 sin’ cos’cos2t
and by substitution of the same values
i0 ¼ 0:3
The greater part of the deviation is due to the core.
61. Two spherical planets, planet A of radius 2a and mass 3m and planet B of radius a
and mass m, are separated by a centre-to-centre distance of 6a. The system rotates in
the equatorial plane and each planet spins on its axis with the same angular velocity.
What, for each planet, is the ratio between the force of gravity and the maximum of
the tidal force at the equator in front of the other planet?
The centre of gravity of the system measured from the centre of planet A is (Fig. 61)
0 3mþ 6am
4m¼
3
2a
The angular velocity of the system is given by
3mGm
6að Þ2¼ 3mo2 3
2a ) o2 ¼
Gm
54a3
The tidal force (radial component) can be calculated from the tidal potential
c ¼GMr
2
2R33cos2# 1
) f Tr¼
@c
@r¼
GMr
R33cos2# 1
where R is the centre-to-centre distance between the planets. For a point on the equator in
front of the other planet, # ¼ 0º, and
f Tr¼
2GMr
6að Þ3
Planet A : f Tr¼
2Gm2a
216a3¼ 0:018
Gm
a2
Planet B : f Tr¼
2G3ma
216a3¼ 0:028
Gm
a2
Gravity without tides is the sum of the gravitational and centrifugal forces:
g ¼Gm
r2 o2
r
Planet A : g ¼G3m
4a2 o22a ¼
3Gm
4a2
Gm
54a32a ¼ 0:71
Gm
a2
Planet B : g ¼Gm
a2 o2a ¼
Gm
a2
Gm
54a3a ¼ 0:98
Gm
a2
108 Gravity
The ratios between the gravity and tidal forces are:
Planet A :g
f Tr
¼0:71
0:018¼ 39:44
Planet B :g
f Tr
¼0:98
0:028¼ 35:00
62. The Earth is of radius a and density r, with a core of radius a/2 and density 3r
on the axis of rotation in the southern hemisphere tangent to the equatorial
plane. The Moon has mass M/2 and its centre is at 4a from the centre of the Earth
(M ¼ 4/3pra3).
(a) Write down the total potential.
(b) What is the value of the angular velocity of the Earth if at the point 30º N, 30º E at
06:00 lunar time the radial component of gravity is equal to GM/a2?
(c) In this case, what is the ratio between the angular velocity of the Earth’s rotation
and that of the system?
(a) As in previous problems the total potential U is given by
U ¼ V1 þ V2 þ Fþ c
The differential mass of the core is:
M1 ¼4p
33r rð Þ
a3
8¼
M
4
A
2a
3m
(3/2)a (9/2)a
6am
a
B
xx
w
Fig. 61
109 Tides
Using the approximation for 1/q, where q is the distance from a point on the surface of the
Earth to the centre of the core, the gravitational potential due to the core is
V2 ¼GM
4q¼
GM
4
1
r
a
2r2cos yþ
a2
8r33cos2y 1
The potential due to the spin of the Earth is given by
F ¼1
2o2
r2sin2y ¼
GM
a3r2 m
2sin2y
where
m ¼o2a3
GM
and the tidal potential due to the Moon is
c ¼GM
2
r2
64a31
23cos2# 1
¼GMr
2
256a33sin2ycos2t 1
where t is the hour-angle of the Moon (Fig. 62)
Then, the total potential U is
U ¼ GM5
4r
a
8r2cos yþ
a2
32r33 cos2 y 1
þr3
a3m
2sin2 yþ
r2
256a33 sin2 y cos2 t 1
(b) The radial component of gravity is given by
gr ¼@U
@r
¼ GM 5
4r2þ
2a
8r3cos y
3a2
32r43 cos2 y 1
þrm sin2 y
a3þ
2r
256a33 sin2 y cos2 t 1
w
P
q
q
t
ϑ
r
a/2 x
3r
4a L90° – q
Fig. 62
110 Gravity
Substituting y ¼ 60º and at 6 h, t ¼ 90, we have
gr ¼GM
a2141
128þ m
3
4
1
128
¼GM
a2
Solving for m we obtain m ¼ 2.81, and the spin angular velocity is
o ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2:81GM
a3
r
rad s1
(c) The centre of gravity r0 of the Earth–Moon system, measured from the centre of the
Earth, is given by
r0 ¼
MTr1 þMLr2
MT þML
¼
M
24a
5
4þ1
2
M
¼8a
7
where the mass of the Earth MT includes that of the core,
MT ¼ M þM
4and ML ¼
M
2
and r1 ¼ 0 and r2 ¼ 4a.
To calculate the angular velocity of the system, O, we put the centripetal force at the
Moon equal to the gravitational force between the Earth and the Moon:
ML r2 r0ð Þ2 ¼ G
MTML
r2
Substituting and solving for O, we obtain
G
5
8M 2
16a2¼
M
2O2 4
8
7
a; so O ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
GM
a37
256
r
Then, the ratio between the angular velocities of the spin of the Earth and of the system is:
o
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2:81GM
a3
r
ffiffiffiffiffiffiffiffiffiffiffiffi
7GM
256a3
r ¼ 10:14
63. Consider two planets of equal mass m and radius a separated by a centre-to-
centre distance of 8a. The planets revolve around their centre of mass and
spin around their own axes. Their spin angular velocity is such that the value
of the centrifugal force is equal to the maximum tidal force of the two at the
equator.
111 Tides
(a) Calculate, for a point on the equator of one of the planets and longitude l¼ 90º, the
value of the vector g including all the forces acting at that point (l¼ 0º corresponds
to the point on the line joining the two centres of the planets) at t ¼ 0.
(b) What is the deviation of the vertical from the radial at the point w ¼ 45º, l ¼ 0º?
(a) If o is the spin angular velocity of the two planets, the centrifugal force at the
equator y ¼ 90, r ¼ a, only has radial component:
f Cr
¼ o2a ð63:1Þ
The radial component of the tidal force can be obtained from the tidal potential c which, in
the first-order approximation, is given by (Fig. 63)
c ¼Gmr2
2R33cos2# 1
where, if ’ is the latitude and l the longitude,
cos# ¼ cos’ cos t lð Þ
On the equator f ¼ 0º, so
c ¼Gmr2
2R33cos2ðt lÞ 1
The radial component of the tidal force is
f Tr¼
@c
@r¼
@
@r
Gmr2
2R33cos2ðt lÞ 1
¼Gmr
R33cos2ðt lÞ 1
Ω
w w
a a
90°
8ax x
m m
Fig. 63
112 Gravity
The maximum value is for t ¼ l and putting R ¼ 8a and r ¼ a, we obtain
f Tr¼
Gma
ð8aÞ32 ¼
Gm
256a2ð63:2Þ
Then, as the centrifugal force is equal to the tidal force, we put (63.1) equal to (63.2) and
solve for o:
o2a ¼Gm
256a2) o2 ¼
Gm
256a3ð63:3Þ
We know that m O2 4a ¼ Gm2/R2, so solving for O,
2 ¼
Gm
256a3
and then using (63.3), we obtain o/O ¼ 1.
The total potential U is the sum of the gravitational, spin, and tidal potentials, which for
t ¼ 0, is given by
U ¼ V þ Fþ c ¼Gm
rþ1
2o2
r2sin2yþ
Gmr2
2ð8aÞ33sin2ycos2l 1
The components of gravity including the tidal forces are
gr ¼@U
@r¼
Gm
r2þ o2
rsin2yþGmr
ð8aÞ33sin2ycos2l 1
gy ¼1
r
@U
@y¼ o2
r sin y cos yþGmr
ð8aÞ33 sin y cos ycos2l
gl ¼1
r sin y
@U
@l¼
Gmr
ð8aÞ33 sin y cos l sin l
ð63:4Þ
By substitution of r ¼ a, l ¼ 90º, and y ¼ 90º we have
gr ¼ Gm
a2þ o2a
Gm
512a2
gy ¼ 0
gl ¼ 0
(b) By substitution of r ¼ a, l ¼ 0, and y ¼ 45º in (63.4) we obtain
gr ¼ Gm
a2þ o2a
1
2þ
Gm
512a23
2 1
¼ Gm
a2þ
Gm
256a21
2þ
Gm
512a21
2¼
1021
1024
Gm
a2
gy ¼ o2a1
2þ
Gm
512a23
2¼
Gm
256a21
2þ
Gm
512a23
2¼
5
1024
Gm
a2
gl ¼ 0
113 Tides
The deviation of the vertical with respect to the radial direction is
tan i ¼gygr
¼ 0:28
64. Two spherical planets of radii 2a and a and masses 8M and M separated by a
centre-to-centre distance of 4a spin on their own axes and rotate in the equatorial
plane with the same angular velocity.
(a) Determine all the forces acting at a point on the smaller planet at geocentric
coordinates w ¼ 60º N, l ¼ 0º (00:00 h local time corresponds to passage of the
other planet through the zero meridian).
(b) For this same point, calculate the astronomical latitude and the tidal deviation of
the vertical.
(a) First we determine the centre of gravity of the system, putting the origin at the
centre of the small planet (Fig. 64):
x ¼0M þ 4a 8M
9M¼
32a
9
Because the spin angular velocity of each planet is equal to the angular velocity of the
system (o ¼ O), we can write, for the small planet, putting the gravitational attraction of
the two planets equal to the centripetal force:
G8MM
4að Þ2¼ Mo2 32a
9
and solving for o,
o2 ¼9GM
64a3
w
w
q ra
M
2a
4a
8M
q
x
Ω
x
Fig. 64
114 Gravity
On the small planet the gravitational force is
gr ¼ GM
r2
gy ¼ 0
and the force due to its spin is
fr ¼ o2rsin2y
fy ¼ o2r cos y sin y
For the point under consideration, r ¼ a, y ¼ 30º, we obtain
gGCr
¼ Gm
a2þ
9
64
GM
a21
4¼
247GM
256a2
gGCy ¼9
64
GM
a3a
ffiffiffi
3p
4¼
GM
a29ffiffiffi
3p
256
To add the tidal force we use the tidal potential in the first-order approximation,
c ¼G8Mr
2
2R33cos2# 1
where cos # ¼ sin y cos (t l).
The tidal force for the point considered, r ¼ a, y ¼ 30º, t ¼ l ¼ 0º, and R ¼ 4a, is
given by
f Tr¼
@c
@r¼
GM
a21
32
f Ty ¼1
r
@c
@y¼
GM
a23ffiffiffi
3p
32
The total force acting at the point is the sum of the three forces, gravitational, centrifugal,
and tidal:
gtotalr
¼ 255GM
256a2
gtotaly ¼GM
a212
ffiffiffi
3p
256
(b) The astronomical latitude is given by ’a ¼ ’ þ i, where i is the deviation of the
vertical without considering the tide:
tan i ¼gtotaly
gtotalr
¼9ffiffiffi
3p
247¼ 0:06 ) i ¼ 3:6 ) ’a ¼ 60þ 3:6 ¼ 63:6
The maximum deviation of the vertical due to the tide at the point considered is
i0 given by
115 Tides
tan i0 ¼f Tygtotalr
¼
3ffiffiffi
3p
32255
256
¼ 0:163 ) i0 ¼ 9:3
Gravity observations
65. Determine the values of gravity at the following series of points belonging to a
gravimetric survey with a Worden gravimeter, specifying the drift correction for each
of them.
The gravity at the base is 980.139 82 Gal, and the gravimeterconstant is 0.301 81 mGal/ru (ru: reading unit).
The instrument drift is given by
d ¼LAe LAb
tAe tAb
where LAb and LAe are the readings at the base A at the beginning and end of the
measurements taken at times tAb and tAe, respectively. By substitution we obtain
d ¼568:8 562:5
14:33 8:50¼ 1:08 ru=hour
The corrected reading for station j is given by
Ljc ¼ Lj – d (tj – tAb)
where Lj is the reading taken at time tj.
For a Worden gravimeter the increment in gravity between two points (D g)
is proportional to the increment in the readings corrected by the instrument drift
(D Lc):
D g ¼ K D Lc
where K is the instrument constant.
Thus, from the readings we obtain the following results.
Station Time Reading
A (base) 08:30 562.5
B 09:21 400.7
C 11:34 437.9
D 13:20 360.1
A 14:20 568.8
116 Gravity
66. Point A is at a geopotential level of 97.437 43 gpu. Point B is at a difference
of 15.213 m in height relative to A, and has a value of gravity of 9.712 611 m/s2.
Calculate:
(a) The value of gravity at point A, if the difference in readings of aWorden gravimeter
between B and A is 17.8 ru, and the gravimeter constant is 0.308 21 mGal/ru.
(b) The geopotential number, dynamic height, and Helmert height of the point B given
that the normal gravity at a point of latitude 45º on the ellipsoid is 980 629.40mGal.
(a) Using aWorden gravimeter the increment of gravity between pointsA andB is given by
DgBA ¼ KDLBA ¼ 5:5mGal
whereK is the instrument constant andDL is the difference between the readings at pointsAandB.
The gravity at point A is
gA ¼ gB þ DgBA ¼ 971 266:6mGal
(b) The geopotential number at B can be calculated from the value at A in the form
CB ¼ CA þgA þ gB
2
hBA ¼ 82:661 59 gpu
where gravity is given in Gal and increments in height in km, because the geopotential
units are, 1 gpu ¼ 1 kGal m ¼ 1 Gal km.
The dynamic height is given by
HBD ¼
CB
g45¼ 84:294m
The Helmert height can be calculated from the dynamic height by
H ¼C
g þ 0:0424H
where C is given in gpu, g in Gal, and H in km. Solving for H, we obtain
H ¼g
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
g2 þ 4 0:0424Cp
2 0:0424
Taking the positive solution because point B is above the geoid (CB > 0) we obtain
HB ¼ 85:107m
Station Corrected reading D g (mGal) g (mGal)
A (base) 562.5 980 139.82
B 399.8 49.10 980 090.72
C 434.6 10.50 980 101.22
D 354.9 24.05 980 077.17
A 562.5 62.66 980 139.83
117 Gravity observations
67. In a geometric survey with measurements of gravity using a Worden gravimeter,
the following values were obtained:
Calculate the gravimeter readings corrected for drift, and the gravimeter constant
The instrument drift is given by
d ¼LAe LAb
tAe tAb
where LAb and LAe are the readings at the base A at the beginning and end of the
measurements taken at times tAb and tAe, respectively. By substitution we obtain,
d ¼ 1:07 ru=hour
A reading corrected at station j is given by
Lcj ¼ Lj dðtj tAbÞ
where Lj is the reading at time tj.
For a Worden gravimeter the increment in gravity between two points (D g) is propor-
tional to the increment in the readings corrected by the instrument drift (D Lc):
D g ¼ K DLc
where K is the instrument constant. Thus, K can be calculated in the form
K ¼g
Lc
From each pair of observations we obtain a value of K. Finally we take the arithmetic mean
(Km) from all the values obtained. The results are given in the following table.
68. The following table is obtained from observations with a Lacoste–Romberg
gravimeter:
Station Gravimeter reading (ru) Time Gravity (gu) Height difference (m)
A (base) 1520.23 8 h 50 m 9 793 626.8
B 1759.15 9 h 15 m 9 794 363.9 30.410
C 1583.11 9 h 35 m 9 793 820.7 301.863
A 1521.30 9 h 50 m
Station Corrected reading (ru) Gravity (mGal) K (mGal/ru)
A (base) 1520.23 979 362.68
B 1758.70 979 436.39 0.3091
C 1582.31 979 382.07 0.3079
A 1520.23
Km ¼ 0.3085
118 Gravity
The gravimeter scale factor is 1.000 65, and the equivalence between reading units
and the relative value of gravity in mGal is given by
Given that the value of gravity at point A is 9.794 6312 m s2, calculate the values at
B and C.
First we correct the readings by the instrument drift:
d ¼LAe LAb
tAe tAb
where LAe and LAb are the readings at the base A at the end and the beginning of the survey
at times tAe and tAb. Then
d ¼ 0:0727 ru=hour
The corrected reading at each station j is given by
Lcj ¼ Lj dðtj tAbÞ
where Lj is the reading at time tj. The corrected readings are:
LcA ¼ 3614:351
LcB ¼ 3650:224
LcC ¼ 3610:600
These readings are converted into relative gravity values Rj using the conversion table.
The reading at station A is
LcA ¼ 3600þ 14:351
and the relative gravity value is
RA ¼ (3846.02 þ (14.351 1.07125)) 1.00065 ¼ 3863.90 mGal
For stations B and C,
Station Gravimeter reading Time
A 3614.351 10:10
B 3650.242 10:25
C 3610.633 10:37
A 3614.414 11:02
Reading Value in mGal Interval factor
3600 3846.02 1.071 25
3700 3953.15 1.071 40
119 Gravity observations
RB ¼ (3846.02 þ (50.224 1.07125)) 1.00065 ¼ 3902.36 mGal
RC ¼ (3846.02 þ (10.600 1.07125)) 1.00065 ¼ 3859.88 mGal
To convert the relative values into absolute values we need to know both values at one
station, in our case in station A:
gA ¼ 9.794 6312 m s2 ¼ 979 463.12 mGal
gB ¼ gA – RA þ RB ¼ 979 501.58 mGal
gC ¼ gA – RA þ RC ¼ 979 459.10 mGal
120 Gravity
3 Geomagnetism
Main field
69. Assume that the geomagnetic field of the Earth is a geocentric dipole with a North
Pole at 80 N, 45 E and a magnetic moment 8 1022 A m². Calculate for a point with
geographical coordinates 45 N, 30 W the components NS, EW, and Z of the Earth’s
magnetic field, the declination and inclination, and the geomagnetic longitude.
Earth’s radius: 6370 km and the constant C ¼ 107 H m1 (this value is used in all
problems).
We calculate first the geomagnetic latitude and longitude (f, l) from the geographical
coordinates (f, l) of the point and the geographical coordinates of the geomagnetic North
Pole (fB, lB) by the equation
sin’ ¼ sin’B sin’þ cos’B cos’ cosðl lBÞ
sin l ¼sinðl lBÞ cos’
cos’
Substituting the values
fB ¼ 80 N
lB ¼ 45 E
f ¼ 45 N
l ¼ 30 W ¼ 330
we obtain
f ¼ 46:70
l ¼ 84:82
In the geocentric magnetic dipole model, the vertical (Z ) and horizontal (H ) components
of the magnetic field can be obtained from
Z ¼ 2B0 sinf
H ¼ B0 cosf
B0 ¼Cm
a3
ð69:1Þ
121
In these equations B0 is the geomagnetic constant, m the magnetic moment of the dipole,
a the Earth’s radius, and the constant C ¼107 H m1.
In this case we are given that
m ¼ 8 1022 A m2
a ¼ 6370 km ¼ 6.37 106 m
By substitution in Equations (69.1) we obtain:
B0 ¼ 30 951 nT
Z ¼ 45 051 nT
H ¼ 21 227 nT
The geomagnetic declination is given by
sinD ¼ cosfB sinðl lBÞ
cosf
D ¼ 14:16
The NS (X ) and EW (Y ) components are
X ¼ H cosD ¼ 20 582 nT
Y ¼ H sinD ¼ 5193 nT
Finally, the geomagnetic inclination or dip (I ) at that point is given by
tan I ¼ 2 tanf ) I ¼ 64:77
70. Assume that the geomagnetic field is produced by a geocentric dipole of magnetic
moment 8 1022 Am², with North Pole at 80 N, 70 W, and that the Earth’s radius is
6370 km. Calculate for a point with geographical coordinates 60 N, 110 E:
(a) Its geomagnetic coordinates, the components of the Earth’s magnetic field (X ,
Y , Z , H ), the total field, the declination, and the inclination.
(b) The equation of the line of force passing through it.
(a) For this point the difference in longitude from the Geomagnetic North Pole
(GMNP) is 180º (Fig. 70), so both are on the same great circle. Then, the
geomagnetic coordinates are obtained from
f ¼ f ð90 fBÞ ¼ 50
l ¼ 180
The expressions for the geomagnetic vertical and horizontal components and for the total
geomagnetic field are
Z ¼ 2B0 sinf
H ¼ B0 cosf
F ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
H2 þ Z2p
B0 ¼Cm
a3
122 Geomagnetism
The values of the constants are
m ¼ 8 1022 A m2
a ¼ 6370 km ¼ 6.37 106 m
C ¼ 107 H m1
Substituting in the above equations we obtain
B0 ¼ 30 951 nT
Z ¼ 47 420 nT
H ¼ 19 895 nT
F ¼ 51 424 nT
For this point the geomagnetic declination D ¼ 0. So the NS (X ) and EW (Y )
components are
D ¼ 0
X ¼ H cosD ¼ H ¼ 19 895 nT
Y ¼ H sinD ¼ 0
The geomagnetic inclination (I ) is given by
tan I ¼ 2 tanf ) I ¼ 67:23
(b) The equation of the line of force passing through a point with geomagnetic
co-latitude y is
r ¼ r0 sin2 y
In this equation r0 is the distance from the Earth’s centre to a point on the line of force with
y ¼ 90º. The distance r0 is different for each line of force.
GMNP GNP
fB
f∗
l∗
l = 0
P
f
l
x
Fig. 70
123 Main field
For the point with geomagnetic latitude f ¼ 50º located in the Earth’s surface (r ¼ a),
y ¼ 90º – f ¼ 40º, so
r0 ¼a
sin2 y¼ 15 417km
71. Assume that the geomagnetic field is produced by a geocentric dipole of magnetic
moment 7.5 1022A m², with North Pole at 75 N, 65 W, and that the Earth’s radius
is 6372 km. Calculate:
(a) The NS and EW components for a point on the Earth’s surface at which the
inclination is 67 and the geomagnetic longitude is 120.
(b) The geographical coordinates of that point.
(c) The geomagnetic coordinates, field components, declination, and inclination of the
point on the geographical equator of zero geomagnetic longitude.
(a) The geomagnetic latitude f is obtained from
tan I ¼ 2 tanf ) f ¼ 49:7
The horizontal component, H , can be calculated from the geomagnetic constant, B0, and
the geomagnetic latitude:
B0 ¼Cm
a3¼ 28 989 nT
H ¼ B0 cosf ¼ 18 761 nT
To obtain the NS (X ) and EW (Y ) components it is necessary to calculate the declination (D )
from the spherical triangle with vertices at the Geographical North Pole (GNP), Geomagnetic
North Pole (GMNP), and the point P (Fig. 71a). But we need to calculate the geographic latitude
firstly by solving the spherical triangle. Applying the cosine law to the angle (90º – f):
cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ
þ sinð90 fBÞ sinð90 fÞ cosð180 lÞ
sinf ¼ sinfB sinf cosfB cosf
cos l ð71:1Þ
By substitution of the values,
f ¼ 55:1
To obtain the geomagnetic declination we apply the sine law in the spherical triangle of
Fig. 71a:
sinD
sinð90 fBÞ¼
sinð180 lÞ
sinð90 fÞ) sinD ¼
cosfB sin l
cosfð71:2Þ
and substituting the values we find
D ¼ 23:1
It is important to note that we have added a minus sign in the last equation in order for the
declination be positive toward the east.
124 Geomagnetism
The NS (X ) and EW (Y ) components are
X ¼ H cosD ¼ 17 262 nT
Y ¼ H sinD ¼ 7 349 nT
(b) The calculated geographical latitude is
f ¼ 55:1
The geographical longitude is obtained by applying the cosine law to the spherical triangle
of Fig. 71a:
cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ
þ sinð90 fBÞ sinð90 fÞ cosðl lBÞ
cos y ¼ sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
ð71:3Þ
so
cosðl lBÞ ¼sinf sinfB sinf
cosfB cosfð71:4Þ
Substituting the values, gives
l lB ¼ 101:6
To take the inverse cosine in the correct quadrant we bear in mind that l < 0 implies
that the point is to the west of the Geomagnetic North Pole, that is l – lB < 0. So we
obtain
l ¼ 166:6 W
(c) If this point is on the geographical equator (f ¼ 90) and has zero geomagnetic
longitude (l ¼ 0º), it is on the same geographical meridian as the Geomagnetic
North Pole. Then from Fig. 71b
90º– fB
GNP
GMNP
q = 90º– f∗
180º–l∗
l – lB
D∗
90º– f
P
Fig. 71a
125 Main field
f ¼ 90 fB ¼ 15:0
l ¼ 0
Z ¼ 2B0 sinf ¼ 15 006 nT
H ¼ B0 cosf ¼ 28 001 nT
D ¼ 0
X ¼ H
Y ¼ 0 nT
tan I ¼ 2 tanf ) I ¼ 28:2
72. Assume that the geomagnetic field is that of a dipole with North Pole at 75 N, 0 E.
What is the conjugate point of that of geographical coordinates 30 N, 30 E?
First, we calculate the geomagnetic coordinates (f , l ) (Problem 71; Fig. 71a):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
sin l ¼sinðl lBÞ cosf
cosf
ð72:1Þ
The values of the geographical coordinates are
fB ¼ 75 lB ¼ 0
f ¼ 30 l ¼ 30
f∗
P
fB
GMNPGNP
x
Fig. 71b
126 Geomagnetism
By substitution in Equations (72.1) we obtain:
f ¼ 42:6
l ¼ 36:0
A magnetic conjugate point is a point on the Earth’s surface that is located on the same line
of force and in the opposite hemisphere (Fig. 72, P and P1). Then, its geomagnetic
coordinates (f1, l1
) are
f1 ¼ f ¼ 42:6
l1 ¼ l ¼ 36:0
To calculate the geographical coordinates for this point (f1, l1) we use the spherical
triangle of Fig. 71a. We calculate the geographical latitude applying the cosine law:
cosð90 f1Þ ¼ cosð90 fBÞ cosð90 f
1Þ
þ sinð90 fBÞ sinð90 f
1Þ cosð180 l1Þ
sinf1 ¼ sinfB sinf1 cosfB cosf1
cos l1
f1 ¼ 53:9
To calculate the geographical longitude we apply the cosine law again:
cosð90 f1Þ ¼ cosð90 fBÞ cosð90
f1Þ
þ sinð90 fBÞ sinð90 f1Þ cosðl1 lBÞ
GMNP GNP
P
– f∗
f∗
x
P1
fB
x
Fig. 72
127 Main field
cosðl1 lBÞ ¼sinf
1 sinfB sinf
cosfB cosf
l1 lB ¼ 47:3
and taking the solution in the correct quadrant
l1 > 0 ) l1 lB > 0
l1 ¼ 47:3
73. Assume the centred dipole approximation, with the coordinates of the Geomag-
netic North Pole being 65 N, 0 E, and the magnetic moment of the dipole 8 1022
A m². Calculate, for a point on the Earth’s surface at geographical coordinates 30 N,
30 E:
(a) The geographical coordinates of the conjugate point.
(b) The declination, inclination, and vertical and horizontal components of the field
at both points. Compare and contrast the results.
Earth’s radius: 6370 km.
(a) First, we calculate the geomagnetic coordinates (f, l) for point P with geograph-
ical coordinates f ¼ 30 N, l ¼ 30 E using the equations (Problem 71; Fig. 71a)
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
sin l ¼sinðl lBÞ cosf
cosf
ð73:1Þ
The geographical coordinates of the Geomagnetic North Pole are
fB ¼ 65; lB ¼ 0
Substituting in Equations (73.1) results in
f ¼ 50:4
l ¼ 42:7
The geomagnetic coordinates (f1, l1
) of the magnetic conjugate point P1 satisfy
(Problem 72):
f1 ¼ f ¼ 50:4
l1 ¼ l ¼ 42:7
To calculate the geographical coordinates for this point (f1, l1) we use the spherical
triangle of Fig. 71a. We calculate the geographical latitude applying the cosine law
cosð90 f1Þ ¼ cosð90 fBÞ cosð90 f
1Þ
þ sinð90 fBÞ sinð90 f
1Þ cosð180 l1Þ
sinf1 ¼ sinfB sinf1 cosfB cosf
1 cos l
1
f1 ¼ 63:7
128 Geomagnetism
To calculate the geographical longitude we apply the cosine law again and, solving for l1,
cosðl lBÞ ¼sinf
1 sinfB sinf1
cosfB cosf1
l1 lB ¼ 77:1
l1 > 0 ) l1 lB > 0
l1 ¼ 77:1
(b) First we calculate the geomagnetic constant B0
B0 ¼Cm
a3¼
8 1022
ð6379Þ3 109¼ 30 951 nT
We calculate the declination D , inclination I , and vertical Z and horizontal H
components of the field at both points. The results are shown in the table, where we notice
that except for the declinations, which are very different, all other values are equal for both
points except in sign.
74. Assume the centred dipole approximation, with the coordinates of the Geomag-
netic North Pole 78.5 N, 70.0 W, and the magnetic dipole moment being 8.25 1022
A m². Calculate, for a point on the surface with coordinates 60.0 S, 170.0 W:
(a) Its geomagnetic coordinates, declination, inclination, and vertical and horizontal
components of the field.
(b) The potential at that point.
(c) The declination and inclination at the point diametrically opposite to it.
Earth’s radius: 6370 km.
(a) We calculate the geomagnetic coordinates (f, l) using the equations (Problem 71,
Fig. 71a)
sinf ¼ cos y ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
sin l ¼sinðl lBÞ cosf
cosf
ð74:1Þ
The values of the geographical coordinates are
fB ¼ 78:5 lB ¼ 70:0
f ¼ 60:0 l ¼ 170:0
P P1
sinD ¼ cosfB sinðl lBÞ
cosf sinD1 ¼
cosfB sinðl1 lBÞ
cosf1D ¼ 19.33 D
1 ¼ 40:2
tan I ¼ 2 tan f ) I ¼ 67.5 tan I1 ¼ 2 tanf1 ) I1 ¼ I1 ¼ 67:5
Z ¼ 2B0 sinf ¼ 47 662 nT Z
1 ¼ 2B0 sinf1 ¼ Z ¼ 47 662 nT
H ¼ B0 cosf ¼ 19 750 nT H
1 ¼ B0 cosf1 ¼ H ¼ 19 750 nT
129 Main field
Substitution in Equations (74.1) gives
f ¼ 60:0
l ¼ 80:0
The geomagnetic declination is given by (Problem 71, Fig. 71a)
sinD ¼ cosfB sinðl lBÞ
cosf
D ¼ 23:1
The inclination (I ) at that point is given by
tan I ¼ 2 tanf ) I ¼ 73:9
The vertical (Z ) and horizontal (H ) components of the magnetic field can be obtained
from
Z ¼ 2B0 sinf
H ¼ B0 cosf
B0 ¼Cm
a3
Substituting the values given we obtain:
B0 ¼ 31 918 nT
Z ¼ 2B0 sinf ¼ 55 279 nT
H ¼ B0 cosf ¼ 15 963 nT
(b) The potential at a point on the Earth’s surface (r ¼ a) at geomagnetic latitude f is
given by
F ¼Cm cos y
a2¼
Cm sinf
a2¼ 176 Tm
(c) We can observe in Fig. 74 that at the point diametrically opposite the geographical
and geomagnetic coordinates are
f1 ¼ f ¼ 60:0 N
l1 ¼ lþ 180 ¼ 10:0 E
f1 ¼ f ¼ 60:0
l1 ¼ l þ 180 ¼ 100:0
The geomagnetic declination at that point satisfies (Fig. 71a)
sinD1 ¼
cosfB sinðl1 lBÞ
cosf1
¼ cosfB sinðlþ 180 lBÞ
cosf
¼cosfB sinðl lBÞ
cosf ¼ sinD
130 Geomagnetism
Then, D1 ¼ D.
The geomagnetic inclination is given by
tan I1 ¼ 2 tanf1 ¼ 2 tanf ) I1 ¼ I
So, we can notice that the two points P and P1 (Fig. 74) that are diametrically opposite
are not magnetic conjugate points because the geomagnetic longitudes are different
by 180º.
75. Consider a point P on the Earth’s surface at coordinates 30 S, 10 W at which
the NS component of the geomagnetic field is 27 050 nT and the EW component is
5036 nT, with the geomagnetic inclination being negative. Assuming the centred
dipole hypothesis with magnetic moment 7.8 1022 Am², calculate:
(a) The geographical coordinates of the Geomagnetic North Pole.
(b) The geomagnetic coordinates of P’s conjugate point.
(a) We calculate first the geomagnetic constant B0:
B0 ¼Cm
a3¼ 30 177 nT
The geomagnetic declination D is obtained from the NS (X ) and EW (Y ) components
of the geomagnetic field:
tanD ¼Y
X ¼
5036
27 050) D ¼ 10:5
GMNPGNP
f1
fB
l∗
f∗f
P
P1
Fig. 74
131 Main field
The geomagnetic latitude f is calculated from the horizontal component H :
H ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þ Y 2p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
270502 þ ð5036Þ2q
¼ 27 515 nT
H ¼ B0 cosf ) f ¼
H
B0
¼ 24:3
We then have two solutions for the geomagnetic latitude. To choose the correct one we bear
in mind that a negative value of the geomagnetic inclination implies a negative value of the
geomagnetic latitude:
tan I ¼ 2 tanf
I < 0 ) f < 0
f ¼ 24:3
With these results we calculate the geographical coordinates of the Geomagnetic North
Pole (fB, lB) using the spherical triangle in Fig. 71a. Applying the cosine rule for the angle
90º – fB:
cosð90 fBÞ ¼ cosð90 fÞ cosð90 fÞ þ sinð90 fÞ sinð90 fÞ cosD
sinfB ¼ sinf sinfþ cosf cosf cosD
fB ¼ 79:0
To calculate the longitude lB of the Geomagnetic North Pole, we apply the cosine law for
the angle 90º – f :
cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ þ sinð90 fBÞ sinð90
fÞ cosðl lBÞ
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
cosðl lBÞ ¼sinf sinfB sinf
cosfB cosf) l lB ¼ 61:0
To choose the correct sign for the longitude we notice that the declination is negative and
then the point must be to the east of the Geomagnetic North Pole:
D < 0 ) l lB > 0
lB ¼ 61 10 ¼ 71:0 W
(b) The geomagnetic coordinates (f1, l1
) of P’s conjugate point verify that
f1 ¼ f ¼ 24:3
l1 ¼ l
We calculate the geomagnetic longitude l by
sin l ¼sinðl lBÞ cosf
cosf
l ¼ 56:2 ¼ l1
132 Geomagnetism
76. At a point P on the Earth’s surface with coordinates 45 N, 30 W, the value of the
total geomagnetic field is 49 801 nT, the horizontal component is 21 227 nT, and the
EW component is 5171 nT, with the magnetic inclination being positive. Calculate:
(a) The geographical coordinates of the Geomagnetic North Pole.
(b) The value of the geomagnetic potential at P.
(c) The distance from the Earth’s centre to the point at which the line of force passing
through P intersects the geomagnetic equator.
Earth’s radius: 6370 km.
(a) We calculate first the geomagnetic inclination, latitude, and declination by
cos I ¼H
F) I ¼ 64:8
tan I ¼ 2 tanf ) f ¼ 46:7
sinD ¼Y
H) D ¼ 14:1
With these results we calculate the geographical coordinates of the Geomagnetic North
Pole solving the spherical triangle (Fig. 71a) in the same way as in Problem 71:
sinfB ¼ sinf sinfþ cosf cosf cosD
fB ¼ 80:0
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
cosðl lBÞ ¼sinf sinfB sinf
cosfB cosf
l lB ¼ 75:2
D > 0 ) l lB < 0
lB ¼ 45:2 E
(b) The geomagnetic potential at point Pon theEarth’s surface (r¼ a¼ 6370km) is givenby
F ¼Cm cos y
a2¼ B0a sinf
ð76:1Þ
We calculate the geomagnetic constant B0 from the horizontal component H :
H ¼ B0 cosf ) B0 ¼
21 227
cosð46:7Þ¼ 30 951 nT
Substituting in the potential equation (76.1) we obtain
F ¼ 143 Tm
(c) The equation of the line of force passing through a point with geomagnetic co-
latitude y is
r ¼ r0 sin2 y
133 Main field
In this equation r0 is the distance from the Earth’s centre to the point at which the line of
force passing through P intersects the geomagnetic equator. Substituting r ¼ a ¼ 6370 km
gives
r0 ¼a
sin2 y¼
a
cos2 f ¼ 13 543 km
77. Assume the centred dipole approximation, with the coordinates of the Geo-
magnetic North Pole being 75 N, 65 W, and the magnetic moment of the dipole
7.5 1022 Am2. For a point on the Earth’s surface at which the inclination is 67 and
the geomagnetic longitude is 120, calculate:
(a) The NS and EW components.
(b) Its geographical coordinates.
The Earth’s radius: 6372 km.
(a) We calculate first the geomagnetic constant B0, latitude f , and horizontal H
component:
B0 ¼Cm
a3¼ 28 989 nT
tan I ¼ 2 tanf ) f ¼ 49:7
H ¼ B0 cosf ¼ 18 761 nT
To calculate the NS (X ) and EW (Y ) components it is necessary to obtain first the
geographic latitude (f ) and the geomagnetic declination (D ). Applying the cosine rule
for the angle 90º – f (Fig. 71a),
sinf ¼ sinfB sinf cosfB cosf
cos l
f ¼ 55:1
The geomagnetic declination is given by
sinD ¼ cosfB sin l
cosf
D ¼ 23:1
From this value we obtain the NS and EW components:
X ¼ H cosD ¼ 17 262 nT
Y ¼ H sinD ¼ 7349 nT
(b) The geographic latitude was already obtained,
f ¼ 55:1
134 Geomagnetism
To calculate the geographical longitude we apply the cosine law for the angle 90º – f
(Fig. 71a):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
cosðl lBÞ ¼sinf sinfB sinf
cosfB cosf
l lB ¼ 101:7
To choose the correct solution we notice that the declination is positive and then the point
must be to the west of the Geomagnetic North Pole
D > 0 ) l lB < 0
l ¼ 101:7 65 ¼ 166:7W
78. Assume a spherical Earth of radius 6370 km, with magnetic field produced by a
centred dipole whose northern magnetic pole is at 70 N, 60 W. Given that for a point
on the surface with coordinates 50 S, 80 W the horizontal component is 24 890 nT,
calculate:
(a) The magnetic dipole moment.
(b) The geographical coordinates of the conjugate point.
(a) We calculate first the geomagnetic latitude by (Fig. 71a)
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 30:9
To obtain the magnetic dipole moment m we need the geomagnetic constant B0, which is
related with the horizontal component H by
B0 ¼H
cosf ¼ 29 007 nT
B0 ¼Cm
a3) m ¼
B0a3
C¼ 7:5 1022 Am2
(b) Let us obtain first the geomagnetic longitude by (Problem 71, Fig. 71a)
sin l ¼sinðl lBÞ cosf
cosf
l ¼ 14:8
The geomagnetic coordinates of the conjugate point (f1, l1
) are (Fig. 78)
f1 ¼ f ¼ 30:9
l1 ¼ l ¼ 14:8
135 Main field
Solving again the spherical triangle (Fig. 71a) we calculate the geographical coordinates
(f1, l1):
sinf1 ¼ sinfB sinf1 cosfB cosf1
cos l1
f1 ¼ 11:4
cosðl lBÞ ¼sinf1
sinfB sinf1
cosfB cosf1
l1 lB ¼ 13:0
l < 0 ) l1 lB < 0
l1 ¼ 73:0
79. If the Earth’s geomagnetic field is produced by a centred dipole, tilted 15 away
from the axis of rotation, of magnetic moment 7.6 1022 Am2, and the Geomagnetic
North Pole is at longitude 65 W, calculate:
(a) The geomagnetic constant in nT.
(b) The geographical coordinates of a point on the Earth’s surface at which the
declination is D ¼ 14 15.50 and the inclination is I ¼ 65 23.50. Discuss the
possible solutions.
(c) The geographical and geomagnetic longitude of the agonic line.
Assume a spherical Earth of radius 6370 km.
GMNPGNP
fB
P1
f∗1
f∗
l∗
P
Fig. 78
136 Geomagnetism
(a) We calculate the geomagnetic constant from the magnetic moment m, the Earth’s
radius a, and the constant C¼107 Hm1, by
B0 ¼Cm
a3¼ 29 403 nT
(b) Let us obtain first the geomagnetic latitude from the inclination I by
tan I ¼ 2 tanf ) f ¼ 47:5
If the dipole is tilted 15 away from the axis of rotation the latitude of the Geomagnetic
North Pole will be
fB ¼ 90 15 ¼ 75:0
With these results we calculate the geographical latitude f solving the spherical triangle
(Fig. 71a). Applying the cosine rule,
cosð90 fBÞ ¼ cosð90 fÞ cosð90 fÞ þ sinð90 fÞ sinð90 fÞ cosD
sinfB ¼ sinf sinfþ cosf cosf cosD ð79:1Þ
To obtain the geographical latitude from this equation we can carry out a change of
variables, introducing two new variables (m, N ) such that
sinf ¼ m cosN
cosf cosD ¼ m sinNð79:2Þ
From these equations we can calculate P and N:
tanN ¼cosf cosD
sinf ¼cosD
tanf ) N ¼ 41:6
P ¼cosN
sinf ¼ 1:02
Substituting Equations (79.2) in Equations (79.1) we obtain
sinfB ¼ P cosN sinfþ P cosN cosf ¼ P sinðfþ NÞ
sinðfþ NÞ ¼sinfB
m¼
sinfB cosN
sinf
fþ N ¼ 78:4 ) f ¼ 36:8
But another solution is also possible:
fþ N ¼ 180 ð78:4Þ ¼ 258:4 ) f ¼ 60:0
The two solutions are correct and we don’t have any additional information to choose one
or the other.
(c) The agonic line is the line where the declination is zero and this implies that the
point is on the great circle that contains the Geographic North Pole and the
Geomagnetic North Pole. So the geomagnetic longitude l is zero or 180º:
l ¼ 0 ) l ¼ lB ¼ 65 W
l ¼ 180 ) l ¼ lB þ 180 ¼ 115 E
137 Main field
80. The Earth’s magnetic field is produced by two dipoles of equal moment
(M ¼ Cm ¼ 9.43 109 nT m3) and polarity, forming angles of 30 and 45 with the
axis of rotation, and contained in the plane corresponding to the 0 meridian. Find
the potential of the total field and the coordinates of the resulting magnetic North
pole, taking the Earth’s radius to be 6000 km.
The total potential at a point is the sum of the potentials of the two dipoles. If M is the
magnetic moment (M ¼ Cm), r is the distance from the dipole’s centre, and y1 and y2are the geomagnetic co-latitude relative to each dipole (Fig. 80), the total potential F is
given by
F ¼ F1 þ F2 ¼M cos y1
r2
M cos y2
r2
¼Mðcos y1 þ cos y2Þ
r2
ð80:1Þ
We calculate the angles y1 and y2 (Fig. 71a) by
cos y1 ¼ sinfB1 sinfþ cosfB1 cosf cosðl lB1Þ
cos y2 ¼ sinfB2 sinfþ cosfB2 cosf cosðl lB2Þð80:2Þ
The geographical coordinates of the two Geomagnetic North Poles are given by
fB1 ¼ 90 30 ¼ 60; lB1 ¼ 0
fB2 ¼ 90 45 ¼ 45; lB2 ¼ 180
GMNP1
GMNP2
GNP45° 30°
q2
q1
P
f
Fig. 80
138 Geomagnetism
Substituting these values in Equations (80.2):
cos y1 ¼ sinfB1 sinfþ cosfB1 cosf cos l
cos y2 ¼ sinfB2 sinf cosfB2 cosf cos l
Adding the two equations gives
cos y1 þ cos y2 ¼ ðsinfB1 þ sinfB2Þ sinfþ ðcosfB1 cosfB2Þ cosf cos l
and substituting in the equation of the potential (80.1)
F ¼M ½ðsinfB1 þ sinfB2Þ sinfþ ðcosfB1 cosfB2Þ cosf cos l
r2
F ¼M
ffiffiffi
3p
þffiffiffi
2p
sinfþ 1ffiffiffi
2p
cosf cos l
2r2
If we call # the geographic co-latitude, # ¼ 90 f, then
F ¼M
ffiffiffi
3p
þffiffiffi
2p
cos#þ 1ffiffiffi
2p
sin# cos l
2r2
The resulting magnetic North Pole, the point where the inclination I ¼ 90º, due to the
combined effect of the two dipoles is given by
tan I ¼Z
H
and therefore at the magnetic Pole, H ¼ 0.
We derive the component H by taking the gradient of the potential F
X ¼ B# ¼1
r
@F
@#¼
M ffiffiffi
3p
þffiffiffi
2p
sin#þ 1ffiffiffi
2p
cos# cos l
2r3
Y ¼ Bl ¼1
r sin#
@F
@l¼
M 1ffiffiffi
2p
sin l
2r3
H ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þ Y 2p
Since the magnetic North Pole is contained in the plane corresponding to the 0 geograph-
ical meridian, then its longitude is either 0º or 180º.
If the longitude is 0º
l ¼ 0 ) H ¼M
2r3
ffiffiffi
3p
þffiffiffi
2p
sin#þ 1ffiffiffi
2p
cos#h i
¼ 0
# ¼ 8 ¼ 172
But this result doesn’t correspond to the north hemisphere. Then we must take the
geographical longitude 180º:
l ¼ 180 ) H ¼M
2r3
ffiffiffi
3p
þffiffiffi
2p
sin# 1ffiffiffi
2p
cos#h i
¼ 0
# ¼ 8 ) fB ¼ 82
139 Main field
This is the correct result and the coordinates of the magnetic North Pole are
fB ¼ 82; lB ¼ 180
81. The Earth’s magnetic field is produced by one dipole in the direction of the axis
of rotation (negative pole in the northern hemisphere) and another with the same
moment in the equatorial plane which rotates with differential angular velocity v
with respect to the points on the surface of the Earth (consider that the Earth doesn’t
rotate). Its negative pole passes through the 45 E meridian at time t ¼ 0 and
completes a rotation with respect to that point in 24 hours. Consider a point of
geographical coordinates 45 N, 45 E.
(a) Calculate the magnetic field components (Br, Bu, Bl) at that point.
(b) Illustrate graphically how each of them varies with local time.
(a) The total potential at a point on the surface of the Earth is the sum of the potentials
of the two dipoles (Problem 80, Equation 80.1):
F ¼ F1 þ F2 ¼M cos y1
r2
M cos y2
r2
¼Mðcos y1 þ cos y2Þ
r2
Dipole 1 is in the direction of the axis of rotation and so the geomagnetic co-latitude of the
point with respect to this dipole (Fig. 81a) is equal to the geographical co-latitude,
y1 ¼ 90 f
cos y1 ¼ sinf
GMNP1
GMNP2
GNP
f
q1
q2
P
Fig. 81a
140 Geomagnetism
Dipole 2 is on the equatorial plane (fB2 ¼ 0) and rotates with respect to the points of
the surface. Owing to this rotation its geographical longitude lB2 changes with time t in
the form
lB2 ¼ ot þ 45
where o is the angular velocity, o ¼ 360/T, T being the rotation period of 24 h.
The co-latitude y2 is
cos y2 ¼ sinfB2 sinfþ cosfB2 cosf cosðl lB2Þ
Substituting the geographical coordinates of the negative geomagnetic equatorial Pole
(fB2, lB2):
cos y2 ¼ cos’ cosðl ot 45Þ
Substituting in the equation for the potential
F ¼ F1 þ F2 ¼M sinfþ cosf cosðl ot 45Þ½
r2
If we consider the geographical co-latitude # ¼ 90 f, the potential is given by
F ¼M cos#þ sin# cosðl ot 45Þ½
r2
We obtain the magnetic field components (Br, By, Bl) at the point (#, l) by taking the
gradient in spherical coordinates of the potential F:
Br ¼ @F
@r¼
2M cos#þ sin# cosðl ot 45Þ½
r3
B# ¼ 1
r
@F
@ #¼
M sin#þ cos# cosðl ot 45Þ½
r3
Bl ¼ 1
r sin#
@F
@l¼
M sinðl ot 45Þ
r3
Substituting the values # ¼ 45, l ¼ 45, B0 ¼ M/a3, gives
Br ¼ 2B0ffiffiffi
2p 1þ cosotð Þ
B# ¼ B0ffiffiffi
2p 1 cosotð Þ
Bl ¼ B0 sinot
(b) The variation of each component with local time is shown in Fig. 81b
141 Main field
Magnetic anomalies
82. Calculate the magnetic anomaly created by a magnetic dipole buried at depth d,
arbitrarily oriented, at an angle to the vertical of a. The negative pole is upwards.
Consider a point P with coordinates (x, z), where x is measured along the horizontal from
the projection of the centre of the dipole and z is the vertical from the reference level (the
Earth’s surface). The position vector r forms an angle b to the vertical (Fig. 82). The
anomalous magnetic potential created by the dipole for this point is
F ¼Cm cosðaþ bÞ
r2¼
Cmðsin b cos aþ cos b sin aÞ
r2ð82:1Þ
where (Fig. 82)
cos b ¼zþ d
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ ðzþ dÞ2q
sin b ¼x
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ ðzþ dÞ2q
0 5 10 15 20
t (h)
–2
–1
0
1
Bθ
Br
Bl
Br,
q,l
B0
Fig. 81b
142 Geomagnetism
Substituting in Equation (82.1) we obtain
F ¼Cm ðzþ dÞ cos a x sin a½
x2 þ ðzþ dÞ2h i3=2
To calculate the magnetic anomaly DB:
B ¼ rðFÞ
The vertical component of the magnetic field anomaly, taking the z-coordinate positive
downward, is
Z ¼@ðFÞ
@z¼
Cm x2 þ ðzþ dÞ2
cos a 3ðzþ dÞ ðzþ dÞ cos a x sin a½ h i
x2 þ ðzþ dÞ2h i5=2
For points on the Earth’s surface (z ¼ 0)
Z ¼Cm ðx2 2d2Þ cos aþ 3dx sin a½
x2 þ d2½ 5=2ð82:2Þ
The component of the magnetic anomaly in an arbitrary horizontal direction x for the
Earth’s surface points (z ¼ 0) is given by
X ¼ @ðFÞ
@x¼
Cm ð2x2 d2Þ sin a 3dx cos a½
x2 þ d2½ 5=2ð82:3Þ
z
Z
rd
b
a
Px
X
+
–
Fig. 82
143 Magnetic anomalies
83. Calculate the magnetic anomaly produced at a point with geographical
coordinates 38 N, 30 W by a horizontal dipole buried at a depth of 10 m with
Cm ¼ 5 105 Tm3 which is in the vertical plane of geographical east, and with the
negative pole to the west. Also calculate the total values of the field in the NS, EW, and
vertical directions, and total field F, as well as the variations in the magnetic declin-
ation and inclination due to the existence of the dipole. Consider the Earth’s magnetic
field to be produced by a centred dipole with North Pole at 72 N, 30 W, and with
B0 ¼ 32 000 nT.
We calculate the horizontal and vertical components of the magnetic anomaly from
Equations (82.2) and (82.3), taking a ¼ 90º because the dipole is horizontal and x ¼ 0
because the dipole’s centre is beneath the point (Fig. 83a). We call DX and DY the
horizontal components in the NS and EW directions, respectively. Because the dipole is
on the vertical east–west plane, the north–south component DX ¼ 0,
Z ¼ 0
Y ¼Cm ð2x2 d2Þ sin a 3dx cos a½
x2 þ d2½ 5=2¼
Cm
d3
Substituting
Cm ¼ 5 105 Tm3
d ¼ 10m
we obtain
Y ¼ 50 nT
jBj ¼ Y ¼ 50 nT
X ¼ 0
Z
+–
a
d
PE
Fig. 83a
144 Geomagnetism
To calculate the components of the magnetic anomaly in the direction of the Earth’s
magnetic field, F, and their horizontal component, H, we need to determine the magnetic
declination and inclination at the point:
H ¼ X cosD þY sinD
F ¼ H cos I þZ sin Ið83:1Þ
Because the point has the same longitude as the Geomagnetic North Pole (Fig. 83b),
f ¼ 90 ðfB fÞ ¼ 56:0
D ¼ 0
The inclination is given by
tan I ¼ 2 tanf ) I ¼ 71:4
Substituting these values in Equations (83.1) we obtain
H ¼ 0
F ¼ 0
The total value of the field in the NS, EW, and vertical directions, and total field F are
XT ¼ X þX
YT ¼ Y þY
ZT ¼ Z þZ
FT ¼ F þF
f∗
f
P
fB
GMNPGNP
Fig. 83b
145 Magnetic anomalies
The vertical Z and horizontal H components of the geomagnetic field are given by
Z ¼ 2B0 sinf ¼ 53 058 nT
H ¼ B0 cosf ¼ 17 894 nT
The NS (X ) and horizontal EW (Y ) components of the geomagnetic field and its
magnitude F are given by
X ¼ H cosD ¼ H ¼ 17 894 nT
Y ¼ H sinD ¼ 0
F ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
H2 þ Z2p
¼ 55 994 nT
We finally obtain that the total field components are
XT ¼ 17 894 nT
YT ¼ 50 nT
ZT ¼ 53 058 nT
FT ¼ 55 994 nT
The variations in magnetic declination and inclination due to the presence of the buried
dipole are
tanD0 ¼YT
XT
) D0 ¼ 0:02
D D0 ¼ 0:02
tan I 0 ¼ZT
HT
) I 0 71:4 ¼ I
84. Buried at a point with magnetic latitude 30 N and a depth of 50 m is a horizontal
magnetic dipole with Cm ¼ 107 nT m3 with the positive pole to the geographical
north.
(a) Calculate DF if B0 ¼ 30 000 nT and the declination at that point is 15. Find the
ratio DF /F.
(b) How far from the dipole’s centre along the north–south line will the dipole field
strength be in the same direction as that of the Earth (take D ¼ 0 ).
(a) The component of the magnetic anomaly DF in the direction of the Earth’s
magnetic field (the total field anomaly) is given by
F ¼ H cos I þZ sin I ð84:1Þ
We first calculate the components in the geographical directions of the magnetic anomaly
produced by the buried dipole using Equations (82.2) and (82.3) of Problem 82, substitut-
ing a ¼ 90º because the dipole is horizontal, and x ¼ 0 because the dipole’s centre is
beneath the point. In this problem DY ¼ 0 because the dipole is on the geographical north–
south vertical plane (Fig. 84). Then
146 Geomagnetism
X ¼ Cm
d3
Y ¼ 0
Z ¼ 0
Substituting the values
Cm ¼ 107 nTm3
d ¼ 50m
we obtain
X ¼ 80 nT
Substituting D ¼ 15º, the component of the magnetic anomaly in the direction of the
horizontal component H of the Earth’s magnetic field is
H ¼ X cosD ¼ 77 nT
At a point of magnetic latitude f ¼ 30 the magnetic inclination is
tan I ¼ 2 tanf ) I ¼ 49:1
Substituting in Equation (84.1), the total field anomaly is
F ¼ 50 nT
To calculate the geomagnetic field F we first obtain the components H and Z :
Z ¼ 2B0 sinf ¼ 30 000 nT
H ¼ B0 cosf ¼ 25 981 nT
F ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
H2 þ Z2p
¼ 39 686 nT
P
Z
N
a
d
– +
Fig. 84
147 Magnetic anomalies
The ratio of the total field anomaly and the Earth’s total magnetic field is
F
F¼ 1:26 103
(b) If the dipole field strength is in the same direction as that of the Earth then the
inclination I 0 due to the dipole is equal to that of the Earth’s field I , where
tan I 0 ¼Z
H
tan I ¼Z
H
Assuming D ¼ 0º, then
H ¼ E
If we substitute in Equations (82.2) and (82.3) of Problem 82, the angle a ¼ 90º because
the dipole is horizontal, we obtain
Z ¼3Cmdx
x2 þ d2 5=2
X ¼Cmð2x2 d2Þ
x2 þ d2 5=2
We have changed the sign of the vertical component because the negative pole is toward
the south.
Applying the condition, tan I 0 ¼ tan I , we obtain
Z
H¼
Z
X¼
Z
H
3Cmdx
x2 þ d25=2
Cm
2x2 d2Þ
x2 þ d25=2
¼3dx
2x2 d2¼
Z
H
2Zx2 3dHx Zd2 ¼ 0
Substituting the values
d ¼ 50 m
Z ¼ 30 000 nT
H ¼ 25 981 nT
and solving the equation, we obtain
x1 ¼ 80m
x2 ¼ 15m
We have two solutions: a point 80 m to the north from the surface projection of the dipole’s
centre and another 15 m to the south.
148 Geomagnetism
85. Located at a point with geocentric geographical coordinates 45 N, 30 W, at a
depth of 100 m, is a dipole of magnetic moment Cm¼ 1 T m3, tilted 45 from the
horizontal to true north, with the negative pole to the north and downwards. At this
point on the surface, the following magnetic field values were observed (in nT):
F ¼ 55 101; H ¼ 12 413; DF ¼ 1268; DH ¼ 547.
Determine:
(a) At the indicated point, the main field components X , Y , Z .
(b) At the indicated point, the deviation of the compass with respect to geomagnetic
north.
(c) The geocentric geographical coordinates of the North Pole of the Earth’s dipole.
Precision 1 nT.
(a) We calculate first the magnetic anomaly produced by the dipole, applying Equa-
tions (82.2) and (82.3) of Problem 82, substituting a ¼ 225º and x ¼ 0. The
horizontal component is in the NS direction (DX) (Fig. 85a)
Z ¼2Cm cos a
d3¼ 1414 nT
X ¼Cm sin a
d3¼ 707 nT
Y ¼ 0
ð85:1Þ
To calculate the declination we use the equation
H ¼ X cosD ) cosD ¼H
X
D ¼ 39:3
PN
Z
d
a
+
45°
–
Fig. 85a
149 Magnetic anomalies
To obtain the Earth’s main field we eliminate the buried dipole contribution from the
observed values:
F ¼ F F ¼ 56 369 nT
H ¼ H H ¼ 11 866 nT
Z ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðFÞ2 ðHÞ2q
¼ 55 106 nT
and the X and Y components
X ¼ H cosD ¼ 9182 nT
Y ¼ H sinD ¼ 7516 nT
(b) The observed declination is given by
tanD0 ¼Y
X¼
Y þY
X þX) D0 ¼ 37:2
The deviation of the compass due to the buried dipole with respect to geomagnetic north is
D0 D ¼ 2:1
(c) We calculate first the geomagnetic latitude of the point from the vertical and
horizontal components:
Z ¼ 2B0 sinf
H ¼ B0 cosf
tanf ¼Z
2H
f ¼ 66:7
With this value, the declination D and the geographical coordinates of the point (f, l), we
can solve the spherical triangle (Fig. 85b) and obtain the geographical coordinates of the
Geomagnetic North Pole:
GMNP
90º – f
q = 90º– f∗
90º – fB
l – lB
180º – l∗
D∗
GNP
P
Fig. 85b
150 Geomagnetism
cosð90 fBÞ ¼ cosð90 fÞ cosð90 fÞ þ sinð90 fÞ sinð90 fÞ cosD
sinfB ¼ sinf sinfþ cosf cosf cosD
fB ¼ 60:0
cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ þ sinð90 fBÞ sinð90 fÞ cosðl lBÞ
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
cosðl lBÞ ¼sinf sinfB sinf
cosfB cosf
l lB ¼ 30:0
The correct solution is the negative one because a positive value of the declination implies
that the point is to the west of the Geomagnetic North Pole:
D > 0 ) l lB < 0
lB ¼ 0
86. Located at a point with geographical coordinates 45 N, 30 W, at a depth of 100 m,
is a dipole of magnetic moment Cm ¼ 1 T m3, inclined 45 to the vertical towards the
south, with the positive pole upwards, and in the geographical north–south vertical
plane. The Earth’s dipole has its north pole at 60 N, 0 E and B0 ¼ 30 000 nT.
Calculate:
(a) The values of Z, H, F at the given point.
(b) Where does the compass point to at that same point?
(a) We calculate first the geomagnetic latitude corresponding to the point by
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 66:7
From this value we obtain the geomagnetic components Z , H and the total main field F :
Z ¼ 2B0 sinf ¼ 55 107 nT
H ¼ B0 cosf ¼ 11 866 nT
F ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðHÞ2 þ ðZÞ2q
¼ 56 370 nT
The geomagnetic declination D is given by
sinD ¼ cosfB sinðl lBÞ
cosf
D ¼ 39:2
and the geomagnetic inclination I by
tan I ¼ 2 tanf ) I ¼ 77:8
151 Magnetic anomalies
The magnetic anomaly created by the dipole buried at depth d is given by Equations (85.1)
of Problem 85. Substituting Cm ¼ 1 T m3, d ¼ 100 m, and a ¼ 45º, we obtain
Z ¼2Cm cos a
d3¼ 1414 nT
X ¼Cm sin a
d3¼ 707 nT
Y ¼ 0
The field anomalies DH and DF are given by
H ¼ X cosD ¼ 548 nT
F ¼ H cos I þZ sin I ¼ 1266 nT
Finally, the observed values are
Z ¼ Z þZ ¼ 53 693 nT
F ¼ F þF ¼ 55 104 nT
H ¼ H þH ¼ 12 414 nT
(b) To calculate in what direction the compass points we need the value of the observed
declination D’ including the effects of the geomagnetic field and the buried dipole:
tanD0 ¼Y
X¼
Y þY
X þX
Y ¼ H sinD ¼ 7500 nT
X ¼ H cosD ¼ 9195 nT
D0 ¼ 37:1
87. Located at a point on the Earth with geographical coordinates 45 N, 30 E, at a
depth of 100 m, is a dipole of magnetic moment Cm ¼ 107 nT m3, tilted 45 to the
vertical towards the south, with the positive pole downwards, and contained in the
plane of true north. The Earth’s field is produced by a centred dipole tilted 30 from
the axis of rotation in the plane of the 0 meridian, with B0 ¼ 30 000 nT. Calculate
the total values of F, Z, andH observed at the point of the surface above the centre of the
buried dipole.
We first calculate the geographical coordinates of the Geomagnetic North Pole and the
geomagnetic latitude
fB ¼ 90 30 ¼ 60
lB ¼ 0
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 66:7
152 Geomagnetism
The geomagnetic field components Z , H and the total main field F are given by
Z ¼ 2B0 sinf ¼ 55 107 nT
H ¼ B0 cosf ¼ 11 866 nT
F ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðHÞ2 þ ðZÞ2q
¼ 56 370 nT
We calculate the geomagnetic declination D by
sinD ¼ cosfB sinðl lBÞ
cosf
D ¼ 39:2
The inclination I is given by
tan I ¼ 2 tanf ) I ¼ 77:8
We obtain the magnetic anomaly produced by the buried dipole applying Equations (82.2)
and (82.3) of Problem 82, substituting a ¼ 45º and x ¼ 0. The horizontal component is in
the NS direction (DX):
Z ¼2Cm cos a
d3¼ 14 nT
X ¼Cm sin a
d3¼ 7 nT
Y ¼ 0
The field anomalies DH and DF are given by
H ¼ X cosD ¼ 5 nT
F ¼ H cos I þZ sin I ¼ 13 nT
Finally the observed values are
Z ¼ Z þZ ¼ 55 121 nT
F ¼ F þF ¼ 56 383 nT
H ¼ HþH ¼ 11 861 nT
88. Buried at a point with geographical latitude 20 N and the same longitude as the
geomagnetic pole, at a depth of 200 m, is a sphere of 50 m radius of material with
magnetic susceptibility 0.01. The Earth’s field is produced by a centred dipole tilted
10 from the axis of rotation and magnetic moment M ¼ 1030 g cm3 (Earth’s radius:
6000 km). Calculate:
(a) The anomaly produced by induced magnetization in the sphere at a point on the
Earth’s surface above the centre of the sphere. Give the vertical and horizontal
components in units of nT.
(b) The total anomaly for a point on the Earth’s surface 100 m south of the above
point.
153 Magnetic anomalies
(a) We first calculate the geomagnetic co-latitude (y) and latitude (f ) of the point,
knowing that it is in the same meridian as the Geomagnetic North Pole (Fig. 88a):
y ¼ 90 f ¼ 90 10 20 ¼ 60
f ¼ 30
The geomagnetic field is given by
B0 ¼M
a3¼ 4630 nT
Z ¼ 2B0 sinf ¼ 4630 nT
H ¼ B0 cosf ¼ 4009 nT
F ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðHÞ2 þ ðZÞ2q
¼ 6124 nT
The inclination is given by
tan I ¼ 2 tanf ) I ¼ 49
The magnetic anomaly created by a sphere is the same as the anomaly created by a
magnetic dipole oriented in the same direction as the geomagnetic field, that is, tilted
90 – I to the vertical and with the negative pole upwards (Fig. 88b). So we use Equations
(82.2) and (82.3) taking a ¼ 41º and x ¼ 0, but we change the sign of Equation (82.3)
because the negative pole is toward the north. The horizontal component is DX ¼ DH
because the dipole is in the magnetic north-vertical plane and DY ¼0:
Z ¼2Cm cos a
d3
H ¼Cm sin a
d3
GMNPGNP
10°
20°
q = 90° – f∗
P
Fig. 88a
154 Geomagnetism
To calculate Cm we use the magnetic susceptibility w and the volume V of the sphere:
Cm ¼ wFV ¼ 3:2 107 nT m3
Substituting this value in the equation of the components of the magnetic anomaly we
obtain
Z ¼2Cm cos a
d3¼
2 3:2 107 cos 41
8 106¼ 6 nT
X ¼Cm sin a
d3¼
3:2 107 sin 41
8 106¼ 3 nT
(b) The anomaly created by the sphere at a point at a distance x¼100 m to the south of
the above point is given by
Z ¼Cm
ðx2 2d2Þ cos aþ 3dx sin a
x2 þ d2 5=2
¼ 0:8 nT
H ¼Cm
ð2x2 d2Þ sin a 3dx cos a
x2 þ d2 5=2
¼ 3:3 nT
The total magnetic field anomaly is therefore
F ¼ H cos I þZ sin I ¼ 3:4 nT
P
Z
F∗
GN
MN
41°
49°
Fig. 88b
155 Magnetic anomalies
89. Buried at a point on the Earth at magnetic latitude 45 N, at a depth d, is a vertical
dipole of magnetic momentM (Cm) with negative pole upwards. IfM/d3 ¼ 10B0 (B0 is
the geomagnetic constant of the main field) calculate how far along the magnetic
meridian the direction of the buried dipole’s field will coincide with that of the Earth
(the terrestrial dipole field).
First we calculate the geomagnetic field components and the inclination by
Z ¼ 2B0 sinf ¼
ffiffiffi
2p
B0 nT
H ¼ B0 cosf ¼
ffiffiffi
2p
2B0 nT
tan I ¼Z
H¼ 2 ) I ¼ 63:4
The components of the magnetic field created by the dipole are given by Equations (82.2)
and (82.3) of Problem 82, putting a ¼ 0º:
Z ¼Cm
x2 2d2
x2 þ d25=2
H ¼Cm3dx
x2 þ d25=2
If the buried dipole’s field coincides with that of the Earth the magnetic inclinations due to
both have to be equal and so
Z
H¼ tan I ¼ 2
x2 þ 6xd 2d2 ¼ 0 ) x ¼ d 3ffiffiffiffiffi
28p
Of the two solutions, x ¼ 2.3d and x ¼ 8.3d, only the positive corresponds to the equal
direction of the two fields.
External magnetic field
90. The Earth’s magnetic field is produced by two dipoles of equal moment and
polarity that are at an angle of 60 to each other, with the bisector being the axis of
rotation. The dipoles are contained in the plane of the 0 geographical meridian.
(a) Calculate the potential on points of the Earth’s surface as a function of geograph-
ical coordinates f and l.
(b) At what points on the surface are the magnetic poles located?
(c) What form would the external field have in order to annul the internal field at the
magnetic equator?
156 Geomagnetism
(a) The total potential at a point is the sum of the potentials of the two dipoles (see
Equation 80.1 of Problem 80):
F ¼ F1 þ F2 ¼M cos y1
r2M cos y2
r2¼
Mðcos y1 þ cos y2Þ
r2
We calculate the geomagnetic co-latitudes by Equation (71.3) of Problem 71:
cos y1 ¼ sinfB1 sinfþ cosfB1 cosf cosðl lB1Þ
cos y2 ¼ sinfB2 sinfþ cosfB2 cosf cosðl lB2Þð90:1Þ
The geographical coordinates of the North Pole of each dipole are given by (Fig. 90)
fB1 ¼ fB2 ¼ 90 30 ¼ 60
lB1 ¼ 0
lB2 ¼ 180
Substituting these values in Equation (90.1):
cos y1 ¼
ffiffiffi
3p
2sinfþ
1
2cosf cos l
cos y2 ¼
ffiffiffi
3p
2sinf
1
2cosf cos l
Adding the two equations:
cos y1 þ cos y2 ¼ffiffiffi
3p
sinf
GMNP1GMNP2
GNP
P
30°
q1
q2
Fig. 90
157 External magnetic field
Substituting in the equation of the potential:
F ¼
ffiffiffi
3p
M sinf
r2
We note that this is the same potential as that produced by only a centred dipole with
magnetic momentffiffiffi
3p
M oriented in the direction of the rotation axis, because the geomag-
netic co-latitude is 90º f.
(b) Bearing in mind the last results, we calculate the inclination by
tan I ¼ 2 cotð90 fÞ ¼ 2 tanf ð90:2Þ
The magnetic poles are the points on the surface of the Earth where the value of the
inclination is equal to 90º:
I ¼ 90 )f ¼ 90
f ¼ 90
Therefore the magnetic poles coincide with the geographical poles.
(c) We derive themainfield by taking the gradient of the potentialF. The components are
Br ¼ @F
@r¼
2ffiffiffi
3p
M sinf
r3
Bf ¼1
r
@F
@f¼
ffiffiffi
3p
M cosf
r3
Bl ¼ 1
r cosf
@F
@l¼ 0
At the magnetic equator the inclination is null (I ¼ 0) and according to Equation (90.2) the
latitude is null too (f ¼ 0). Substituting in the last equations:
Br ¼ 0
Bf ¼
ffiffiffi
3p
M
r3
Bl ¼ 0
The external magnetic field to annul out the internal field is therefore
Be ¼ 0;
ffiffiffi
3p
M
r3
; 0
91. The Earth’s magnetic field is formed by a centred dipole with northern geomag-
netic pole at 60 N, 0 E and B0 ¼ 32 000 nTand a uniform external field from the Sun
of 10 000 nT parallel to the equatorial plane.
(a) For a point at coordinates 60 N, 60 W, calculate the components X, Y, Z of the
total field, and the values of D and I.
(b) How do D and Z of the total field vary throughout the day with local time t?
158 Geomagnetism
(a) We calculate the geomagnetic main field components Z , H obtaining the
geomagnetic latitude given by (71.3):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 61
Z ¼ 2B0 sinf ¼ 55 976 nT
H ¼ B0 cosf ¼ 15 514 nT
To calculate the NS (X ) and EW (Y ) components we need the geomagnetic declination
D (71.2):
sinD ¼ cosfB sinðl lBÞ
cosf
D ¼ 63
X ¼ H cosD ¼ 7043 nT
Y ¼ H sinD ¼ 13 823 nT
The external field is parallel to the equatorial plane and has a diurnal period (o ¼ 2p/24)
because it comes from the Sun. We assume that at local time t ¼ 0 the Sun is at the point’s
meridian. If we denote by N the modulus of the external field (N ¼ 10 000 nT) and bearing
in mind Fig. 91a (representation of the plane parallel to the equator that contains the point)
we have at time t 6¼ 0
Y ¼ Bel ¼ N sinot
The radial and tangential components (Fig. 91b) are
P
(a)
N
N coswt
Beλ
wt
Fig. 91a
159 External magnetic field
Z ¼ Br ¼ N cosot cosf
X ¼ Bf ¼ N cosot sinf
The total magnetic field is the sum of the two contributions:
ZT ¼ ð55 976þ 5000 cosotÞ nT
XT ¼ ð7043þ 8660 cosotÞ nT
YT ¼ ð13 823þ 10 000 sinotÞ nT
The declination D and inclination I are given by
tanD ¼YT
XT
¼13 823þ 10 000 sinot
7043þ 8660 cosot
tan I ¼ZT
HT
¼ZTffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2T þ Y 2
T
p
(b) To see how D and Z vary during the day with local time t we substitute several
values for t, obtaining the values in the table:
GNP
f
r
z = –Br
X = Bf
PNcosωt
Fig. 91b
t (h) Z (nT) D (º)
0 60976 41
6 55976 28
12 50976 83
18 55976 73
160 Geomagnetism
92. The Earth’s magnetic field is formed by a centred dipole of moment m and
north pole 60 N, 0 E, and a uniform external field of magnitude N ¼ B0/4 (B0 is
the geomagnetic constant of the internal field) parallel to the axis of rotation.
Determine:
(a) The total potential at any point.
(b) The coordinates of the boreal magnetic pole.
(c) The magnetic declination at the point 45 N, 45 E.
(d) The angle along the meridian between that point and the magnetic equator.
(a) The total potential (F) is the sum of two contributions: the main (internal) field
(Fi) and the external field (Fe):
F ¼ Fi þ F
e
The main field is formed by a centred dipole of moment m so the potential is given by
Fi ¼
Cm cos y
r2
We calculate the geomagnetic co-latitude y ¼ 90 – f at a point with geographical
coordinates (f, l) by (71.3):
cos y ¼ sinfB sinfþ cosfB cosf cosðl lBÞ ¼
ffiffiffi
3p
2sinfþ
1
2cosf cos l
GNP
P
f
ze
Xe
Fe
Fig. 92a
161 External magnetic field
The external field is parallel to the axis of rotation; therefore its components are in the
vertical and NS directions (Fig. 92a). If we call # ¼ 90 f the geographical co-latitude,
the components are given by
Ber¼ N cos# ¼ N sinf
Be# ¼ N sin# ¼ N cosf
Ze ¼ Ber¼ N sinf
X e ¼ Be# ¼ N cosf
Bearing in mind that Be ¼ ∇Fe the potential for the external field is
Fe ¼ Nr cos# ¼ Nr sinf
Therefore the total potential is given by
F ¼Cm
ffiffi
3p
2sinfþ 1
2cosf cos l
r2
þ Nr sinf
(b) At the magnetic boreal pole the inclination is I ¼ 90º and the horizontal field
components H, X, Y are given by
tan I ¼Z
H) H ¼ 0
H ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þ Y 2p
)X ¼ 0
Y ¼ 0
We obtain the components X and Y by taking the gradient of the potential
Be ¼ rFe
We bear in mind the relations@F
@y¼
@F
@f
X ¼ By
and obtain that the components are given by
X ¼ 1
r
@F
@f¼
Cmffiffi
3p
2cosf 1
2sinf cos l
r3
N cosf ¼ 0
Y ¼ 1
r cosf
@F
@l¼
Cm
2r3sin l ¼ 0 ) l ¼ 0
ð92:1Þ
Substituting in the equation of component X (92.1) the values
l ¼ 0
r ¼ a
B0 ¼Cm
a3
N ¼ B0=4
162 Geomagnetism
we obtain the geographical latitude of the boreal magnetic pole
tanfBM ¼2ffiffiffi
3p
1
2) fBM ¼ 51
Therefore the geographical coordinates of the magnetic boreal pole are 51 N, 0. Notice
that this is different from the geomagnetic pole.
(c) The declination is given by
tanD ¼Y
X¼
B0
2sin l
B0
ffiffiffi
3p
2cosf
1
2sinf cos l
B0
4cosf
¼ sin l
ffiffiffi
3p
1
2
cosf sinf cos l
Substituting the geographical coordinates of the point (45 N, 45 E) we obtain
D ¼ 62
(d) We call feq the angle between the geographical and magnetic equators at longitude
l ¼ 45º (Fig. 92b). Then the angle to calculate will be feq þ 45.
Tocalculatefeqwe take into account that at themagnetic equator thevertical component isZ¼ 0:
Z ¼ Br ¼@F
@r¼
2CM
r3
ffiffiffi
3p
2sinfeq þ
1
2cosfeq cos l
þ N sinfeq ¼ 0
tanfeq ¼4 cos l
4ffiffiffi
3p
þ 1
GNP GMNP
fB
f
P
l
feq
Fig. 92b
163 External magnetic field
Substituting l ¼ 45º in the equation we obtain
feq ¼ 20
Then the angle we are asked for is 45º – 20º ¼ 25º.
93. The internal magnetic field is formed by two orthogonal dipoles of equal moment
M, one of them in the direction of the axis of rotation and the other contained in the 0
meridian on the equator. There is also an external field of constant intensity N ¼ B0/4
(B0 is the geomagnetic constant of the internal field) and lines of force parallel to the
axis of rotation.
(a) Calculate the potential of the total field for points on the surface.
(b) At which latitude is Z maximum on the 0 meridian?
(a) The potential of the total field is the sum of the two potentials of the internal
dipoles and the potential of the external field:
F ¼ F1 þ F2 þ Fe ¼
M cos y1
r2
M cos y2
r2
þ Fe
¼Mðcos y1 þ cos y2Þ
r2
þ Fe
In this equation r is the distance from the dipole’s centre (the Earth’s centre), y1 and y2 are
the co-latitudes relative to each dipole, and M ¼ Cm.
Dipole 1 is on the direction of the axis of rotation and so the geomagnetic co-latitude of
the point with respect to this dipole is
y1 ¼ 90 f
cos y1 ¼ sinf
Dipole 2 is in the equatorial plane (fB2 ¼ 0) and contained in the 0 meridian so the
geomagnetic co-latitude y2 is given by (71.3)
cos y2 ¼ sinfB2 sinfþ cosfB2 cosf cosðl lB2Þ
fB2 ¼ 0
lB2 ¼ 0
cos y2 ¼ cosf cos l
The equation for the potential of the external field is the same as that of Problem 92:
Fe ¼ þNr sinf
Therefore the potential of the total field is given by
F ¼Mðsinfþ cosf cos lÞ
r2þ Nr sinf
164 Geomagnetism
(b) We obtain the component Z by taking the vertical component of the gradient of the
potential (B ¼ ∇F)
Z ¼ Br ¼@F
@r¼
2Mðsinfþ cosf cos lÞ
r3þ N sinf
To calculate the maximum of Z on the 0 meridian we substitute l ¼ 0 and apply the
condition that the first derivate with respect to the latitude is null:
@Z
@f¼
2M
r3ðcosf sinfÞ þ N cosf ¼ 0
At the Earth’s surface r ¼ a and we know that
B0 ¼M
a3
Substituting this constant and solving the equation we determine the latitude at which the
Z component is maximum:
2B0ðcosf sinfÞ þB0
4cosf ¼ 0 ) f ¼ 48
94. The internal field has its northern geomagnetic pole at the coordinates 60 N, 0 E,
and B0 ¼ 30 000 nT. At a point with coordinates 30 N, 45 W, one observes an
increase of 7.7 in the value of the declination from 00:00 h to 09:00 h. There is
known to be an external field parallel to the Earth’s axis of rotation in the direction
from N to S which is null at 00:00 h and maximum at 12:00 h local times. Calculate:
(a) The components of the internal and external fields.
(b) The difference in the inclination at 00:00 h and 09:00 h.
(c) The maximum value of the declination during the day.
(a) To calculate the geomagnetic main field intensity components we obtain first the
geomagnetic latitude (Equation 71.3):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 47:7
The declination and inclination are given by (71.2):
sinD ¼ cosfB sinðl lBÞ
cosf ) D ¼ 31:7
tan I ¼ 2 tanf ) I ¼ 65:5
So the geomagnetic main field components are
Z ¼ 2B0 sinf ¼ 44 378 nT
H ¼ B0 cosf ¼ 20 190 nT
X ¼ H cosD ¼ 17 178 nT
Y ¼ H sinD ¼ 10 609 nT
165 External magnetic field
The external field is parallel to the axis of rotation so its components are in the vertical and
NS directions. This field is null at 00:00 h and maximum at 12:00 h local time (period T ¼
24 h). Its components are given by
Ze ¼N
2ð1 cosotÞ sinf
X e ¼ N
2ð1 cosotÞ cosf
Y e ¼ 0
He ¼ X e
o ¼2p
T¼
2p
24
(b) To calculate the difference in the inclination we obtain first the value of N, bearing
in mind the time variation of the declination. The observed declination as a
function of time is given by
tanD ¼Y
X¼
Y þ Y e
X þ X e¼
Y
X N
2ð1 cosotÞ cosf
For t ¼ 0 h:
tanD1 ¼Y
X ¼ tanD ) D1 ¼ D ¼ 31:7
Since we know the change in declination between 0 h and 9 h, we find the declination at
9 h, D2:
D2 D1 ¼ 7:7 ) D2 ¼ 39:4
We know that at t ¼ 9 h
tanD2 ¼Y
X N
2ð1 cosotÞ cosf
¼ 0:82
Solving for N we obtain
N ¼2
ð1 cosotÞ cosfX
Y
tanD2
¼ 5766 nT
The magnetic inclination is given by
tan I ¼Z
H¼
Z þ Ze
H þ H e¼
Z þN
2ð1 cosotÞ sinf
H N
2ð1 cosotÞ cosf
At t ¼ 0 h:
I1 ¼ I ¼ 65:5
At t ¼ 9 h:
I2 ¼ 71:2
166 Geomagnetism
The difference in the inclination is therefore
I2 I1 ¼ 5:7
(c) The declination is given by
tanD ¼Y
X N
2ð1 cosotÞ cosf
ð94:1Þ
The maximum value is at 12 h because at that time the external field has the maximum
value. Substituting ot ¼ p and the values obtained for X , Y , N, and f:
Dmax ¼ 41:0
95. At a point on the Earth with coordinates 45 N, 45 E, measurements are made of
the magnetic field components at 00:00 h and 12:00 h in nT with a 2 nT precision:
0 h X ¼ 20 732 Y ¼ 2500 Z ¼ 57 768
12 h X ¼ 24 267 Y ¼ 2500 Z ¼ 54 232
It is known that the modulus of the magnetic field intensity has a harmonic diurnal
variation, and that the geomagnetic pole is on the zero meridian. Calculate:
(a) The moment and coordinates of the main field dipole.
(b) Expressions for the potential and components of the external field.
(Earth’s radius a ¼ 6400 km, m0 ¼ 4p107
kg m s2
A2).
(a) To calculate themoment and coordinates of themainfield dipolewe need to obtain the
geomagnetic main field intensity components. The observed values are equal to the
sum of the geomagnetic main field (X , Y , Z ) and the external field (X e, Ye, Z e):
X ¼ X þ X e
Y ¼ Y þ Y e
Z ¼ Z þ Ze
The geomagnetic main field is constant but the external field changes with time. So if we
denote by (X0, Y0, Z0) and (X12, Y12, Z12) the observed values at 0 h and at 12 h respectively
then the differences are due to the variations of the external field:
X12 X0 ¼ 3535 nT ¼ X e12 X e
0
Y12 Y0 ¼ 0 nT ¼ Y e12 Y e
0
Z12 Z0 ¼ 3536 nT ¼ Ze12 Ze
0
We notice that the Y component doesn’t vary, which implies that the component Y e is zero,
and so the external field is parallel to the axis of rotation. We also notice that the NS
component increases in the time interval between 0 h and 12 h, while the vertical
component diminishes, which implies that the polarity of the external field is inverted
with respect to that of the main field.
The modulus of the magnetic field intensity has a harmonic diurnal variation and
increases with time. Therefore the components of the external field are (Fig. 95a)
167 External magnetic field
X e ¼ Nð1 cosotÞ cosf
Ze ¼ Nð1 cosotÞ sinf
o ¼2p
T¼
2p
24h
We notice that at time t ¼ 0 the external field is null and so
X e12 ¼ 3535 nT
Y e12 ¼ 0 nT
Ze12 ¼ 3536 nT
Therefore we calculate the main field components by
X ¼ X12 X e12 ¼ 20 732 nT
Y ¼ Y12 Y e12 ¼ 2500 nT
Z ¼ Z12 Ze12 ¼ 57 768 nT
H¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðX Þ2 þ ðY Þ2q
¼ 20 882 nT
If we consider the centred magnetic dipole model, the vertical and horizontal field
components are given by the equations
Z ¼ 2B0 sinf
H ¼ B0 cosf
Fe
ZeX
e
P
f
GNP
Fig. 95a
168 Geomagnetism
From these equations we obtain the geomagnetic latitude (f ) and the geomagnetic
constant B0:
tanf ¼Z
2H) f ¼ 54:1
B0 ¼Z
2 sinf ¼ 35 657 nT
From B0 we calculate the magnetic moment:
B0 ¼m04p
m
a3) m ¼ B010
7a3 ¼ 9:3 1022 A m2
The longitude of the Geomagnetic North Pole is lB ¼ 0 and we calculate the latitude fB
from the spherical triangle (Fig. 95b), but obtaining first the declination from the X and
Y components:
tanD ¼Y
X ) D ¼ 6:9
Applying the cosine rule:
cosð90 fBÞ ¼ cosð90 fÞ cosð90 fÞ þ sinð90 fÞ sinð90 fÞ cosD
sinfB ¼ sinf sinfþ cosf cosf cosD
fB ¼ 79:9
(b) We obtain the radial and transverse components of the external field from the
vertical and NS components:
Ber¼ Ze ¼ Nð1 cosotÞ sinf ¼
@Fe
@r
Bef ¼ X e ¼ Nð1 cosotÞ cosf ¼
1
r
@Fe
@f
q = 90º – f∗
GMNP
90º – fB
GNP
l – lB
180º – l∗
D∗
P
90º – f
Fig. 95b
169 External magnetic field
Therefore the potential of the external field is given by
Fe ¼ Nrð1 cosotÞ sinf
where
N ¼Ze12
2 sinf¼ 2500 nT
96. The Earth’s main magnetic field is that of a centred dipole of moment M (M ¼
Cm) in the direction of the axis of rotation, and the external field is produced by
electric currents of intensity J circulating in a clockwise sense in a ring in the plane of
the ecliptic at a distance of 10 Earth radii around the Earth.
(a) Express the potential of the total field and the components Br and Bø on the
Earth’s surface for l ¼ 0.
(b) If the inclination of the ecliptic is 30, and the external field strength is N ¼ M /
4R3, what is the latitude of the northern magnetic pole?
(a) The potential of the total field is the sum of the potentials of the dipole and of the
external field:
FT ¼ Fþ F
e ¼M cos y
r2þ F
e
where r is the distance from the dipole’s centre (Earth’s centre) and y is the geomagnetic
co-latitude. The dipole is in the direction of the axis of rotation and so the geomagnetic co-
latitude is equal to the geographic co-latitude:
y ¼ 90 f
cos y ¼ sinf
To calculate the potential of the external field we know that it is produced by electric
currents of intensity J circulating in a clockwise sense at a distance of 10 Earth radii around
the Earth. These electric currents produce, at remote points, a magnetic dipolar field whose
modulus is m0J/2R, J being the current intensity and R the radius of the circular currents;
the dipole is oriented in the direction of the ecliptic’s axes with the negative pole in the
southern hemisphere, because the currents are clockwise. So the potential of the external
field is given by
Fe ¼
Cme cos y2
r2
In this equation me ¼ Jp100a2, a is the Earth’s radius, and y2 is the angle between the axes
of the circular currents and the direction of the point from the negative pole (Fig. 96)
Therefore
y2 ¼ 180 ðyþ eÞ
cos y2 ¼ cosðyþ eÞ
where e is the angle between the axes of the circular currents and the axes of rotation of the
Earth.
170 Geomagnetism
The potential of the external field is
Fe ¼
Cme cosðyþ eÞ
r2
and the total potential is given by
FT ¼
M cos y
r2þCme cosðyþ eÞ
r2
We calculate the components of the main field intensity by taking the gradient of the
potential:
Br ¼ @F
@r¼
2M cos y
r3
By ¼ 1
r
@F
@y¼
M sin y
r3
The magnitude of the external field is given by
Be ¼m04
J
10a¼ N
The radial and tangential components are
Ber¼ N cosðyþ eÞ
Bey ¼ N sinðyþ eÞ
GNP
qP
q2
e
e
–
+
Fig. 96
171 External magnetic field
The total field is the sum of the two contributions:
Br ¼ 2M cos y
r3þ N cosðyþ eÞ
By ¼ M sin y
r3
þ N sinðyþ eÞ
(b) We know that the Earth’s dipole is in the direction of the axis of rotation and y is
the geographical co-latitude (Fig. 96). At the magnetic North Pole the total field is
vertical and the tangential component is By ¼ 0:
By ¼ M sin y
r3þ N sinðyþ eÞ ¼ 0
The values of N and e are known:
N ¼M
4r3)
M
r3¼ 4N
e ¼ 30
From these values we calculate y, the geographical co-latitude of the magnetic North Pole,
By ¼ 4N sin yþ N sinðyþ 30Þ ¼ 0
4 sin yþ
ffiffiffi
3p
2sin yþ
1
2cos y ¼ 0 ) y ¼ 9
97. Two spherical planets of radius a and separated by a centre-to-centre distance of 4a
orbit around each other and spin in the equatorial plane. Each has a magnetic field
produced by a centred dipole in the direction of the axis of rotation, with the positive
pole in the northern hemisphere andB0¼ 10 000 nT. Determine the componentsX, Y, Z,
D, and I of the totalmagnetic field at theNorth Pole of one of the planets (precision 1 nT).
The total magnetic field in each planet is the sum of its main field and the external field
created by the other planet. To determine the main field of either of the planets we need the
geomagnetic latitude, which is positive toward the negative pole, in this case, the South
Pole. Therefore the geomagnetic latitude and the components of the main field at the North
Pole are
f ¼ 90
Z ¼ 2B0 sinf ¼ 20 000 nT
H ¼ B0 cosf ¼ 0
X ¼ Y ¼ 0
The external field at one of the planets is created by the main field of the other planet and
corresponds to that of a magnetic dipole. Its components are (Fig. 97)
Ber¼
2Cm cos y
r3
Bey ¼
Cm sin y
r3
172 Geomagnetism
At the North Pole r ¼ffiffiffiffiffi
17p
a ¼ 4:12a and we calculate the geomagnetic co-latitude y by
(Fig. 97)
tanðy 90Þ ¼a
4a¼
1
4
y ¼ 14 þ 90 ¼ 104
Substituting these values in the equations for the radial and tangential components with
respect to the planet producing the external field:
Ber¼
2Cm cos y
r3
¼Cm cos y
35a3¼
B0 cos y
35¼ 69 nT
Bey ¼
Cm sin y
r3
¼Cm sin y
70a3¼
B0 sin y
70¼ 139 nT
From this value we calculate the vertical and horizontal components (Fig. 97):
Ze ¼ Bercosð180 yÞ þ Be
y cosðy 90Þ ¼ Bercos yþ Be
y sin y ¼ 118 nT
H e ¼ Bersinð180 yÞ þ Be
y sinðy 90Þ ¼ Bersin y Be
y cos y ¼ 33 nT
X e ¼ H e ¼ 33 nT
Y e ¼ 0
The components of the total field are finally
Z ¼ Z þ Ze ¼ 19882 nT
H ¼ H þ H e ¼ 33 nT
X ¼ X þ X e ¼ 33 nT
Y ¼ Y þ Y e ¼ 0
tanD ¼Y
X) D ¼ 0
tan I ¼Z
H) I ¼ 89:9
4aθ
+
–
aa
r
q – 90°
Ber
Beθ
q – 90°
Fig. 97
173 External magnetic field
Main (internal), external, and anomalous magnetic fields
98. At a point with geographical coordinates 40 N, 45 E, measurements are made of
the magnetic field components, obtaining the values (in nT):
At 06:00 h: X ¼ 19 204; Y ¼ 0; Z ¼ 38 195
At 12:00 h: X ¼ 11 544; Y ¼ 0; Z ¼ 44 623
Buried at a depth of 20 m below this point is a dipole of magnetic moment Cm ¼ 0.01
T m3, oriented in the NS plane at an angle of 45 with the vertical towards the south,
and the positive pole upwards. Given that the external field at 12:00 h is twice that at
06:00 h, determine:
(a) The geomagnetic constant Bo and the coordinates of the northern geomagnetic pole.
(b) The magnitude and direction of the external field. How does the magnitude of the
external field vary with time?
(a) The observed magnetic field is composed of three parts: the geomagnetic main
(internal) field, the anomalous field (magnetic anomaly) created by the buried
dipole, and the external field.
We determine first the magnetic anomaly produced by the buried dipole, applying Equa-
tions (82.2) and (82.3) in Problem 82, substituting a ¼ 225º and x ¼ 0. The horizontal
component is in the NS direction (DX), because the buried dipole is in the NS-vertical
plane (Fig. 98a):
Z ¼2Cm cos a
d3¼ 1768 nT
X ¼Cm sin a
d3¼ 884 nT
Y ¼ 0
P
45°
α
d
N
Z
+
–
Fig. 98a
174 Geomagnetism
The components of the total magnetic field at 06:00 h are given by
X1 ¼ X þX þ X e ¼ 19 204 nT
Y1 ¼ Y þY þ Y e ¼ 0
Z1 ¼ Z þZ þ Ze ¼ 38 195 nT
At 12:00 h given that the external field is twice that at 06:00 h:
X2 ¼ X þX þ 2X e ¼ 11 544 nT
Y2 ¼ Y þY þ 2Y e ¼ 0
Z2 ¼ Z þZ þ 2Ze ¼ 44 623 nT
If we subtract both sets of equations and obtain
X2 X1 ¼ X e ¼ 7660 nT
Y2 Y1 ¼ Y e ¼ 0
Z2 Z1 ¼ Ze ¼ 6428 nT
Now we can calculate the elements of the main field
X ¼ X1 X X e ¼ 25 980 nT
Y ¼ Y1 Y Y e ¼ 0
Z ¼ Z1 Z Ze ¼ 33 535 nT
H ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðX Þ2 þ ðY Þ2q
¼ X ¼ 25 980 nT
tanD ¼Y
H) D ¼ 0
We calculate the geomagnetic latitude of the point f and the geomagnetic constant B0
from the vertical and horizontal geomagnetic main field components by
Z ¼ 2B0 sinf
H ¼ B0 cosf
tanf ¼Z
2H) f ¼ 32:8
B0 ¼Z
2 sinf ¼ 30 953 nT
We obtain the coordinates of the Geomagnetic North Pole by (Fig. 98b)
D ¼ 0 ) lB ¼ 180þ l ¼ 225 E ¼ 135 W
90 fB ¼ f f ) fB ¼ 82:8
(b) The magnitude of the external field at 06:00 h is
B6e ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2e þ Y 2
e þ Z2e
q
¼ 10 000 nT
175 Main (internal), external, and anomalous magnetic fields
At 12:00 h the magnitude is
B12e ¼ 2B6
e ¼ 20 000 nT
The direction of the external field is in the NS-vertical plane because the EW component is
null, forming with the horizontal an angle Ie (Fig. 98c). This direction is the same at 06:00
h and at 12:00 h. We calculate the angle Ie by
tan Ie ¼Ze
Xe
Ie ¼ 40
GNPGMNP
P
f90° – fB
f∗
Fig. 98b
Z
Ze
Ie
Xe
Be
N
Fig. 98c
176 Geomagnetism
Because at
06 :00 h ðt ¼ p=2Þ ! N ¼ 10 000
12 :00 hðt ¼ pÞ ! N ¼ 20 000
the variation of the magnitude of the external field with time is given by
N ¼ 10 000ð1 cos tÞ
99. Buried at a depth of 100 m at a point with geographical coordinates 45 N, 45 W is
a dipole anomaly of Cm¼ 0.1 T m3, inclined 45 from the horizontal northwards in the
vertical plane with the negative pole downwards. Measurements gave the following
results (in nT):
09:00 h X ¼ 27 759; Y ¼ 0; Z ¼ 30 141
12:00 h X ¼ 28 052; Y ¼ 0; Z ¼ 30 141
Find:
(a) The coordinates of the magnetic dipole’s North Pole.
(b) The value of B0.
(c) An expression for the variation Sq knowing that it is zero at 00:00 h and
maximum at 12:00 h.
(a) As in Problem 98, the observed field is the result of three parts: the main (internal)
field, the buried dipole field, and the external field. To calculate the coordinates of
the magnetic dipole’s North Pole we need to obtain the components of the
geomagnetic main field from the components of the total field. With this aim we
begin by calculating the magnetic anomaly created by the buried dipole, applying
Equations (82.2) and (82.3), and substituting a ¼ 225º and x ¼ 0. The horizontal
component is in the NS direction (DX) given that the dipole is on the NS-vertical
plane (Fig. 99a).
Z ¼2Cm cos a
d3¼ 141 nT
X ¼Cm sin a
d3¼ 71 nT
Y ¼ 0
The total observed field at 09:00 h is
X1 ¼ X þX þ X e1
Y1 ¼ Y þY þ Y e1 ¼ 0 ) Y ¼ Y e
1
Z1 ¼ Z þZ þ Ze1
The total field at 12:00 h is
X2 ¼ X þX þ X e2
Y2 ¼ Y þY þ Y e2
Z2 ¼ Z þZ þ Ze2
177 Main (internal), external, and anomalous magnetic fields
Subtracting both sets of equations we obtain
X2 X1 ¼ X e2 X e
1 ¼ 293 nT
Y2 Y1 ¼ Y e2 Y e
1 ¼ 0 ) Y e2 ¼ Y e
1 ¼ Y
Z2 Z1 ¼ Ze2 Ze
1 ¼ 0 ) Ze2 ¼ Ze
1
ð99:1Þ
We assume that the time variation of the observations is due to the diurnal Sq variation
which is zero at 00:00 h and maximum at 12:00 h. Therefore the only possible values
for the components Ye and Ze are zero because these components have the same values at
09:00 h and at 12:00 h:
Y e2 ¼ Y e
1 ¼ Y ¼ 0
Ze2 ¼ Ze
1 ¼ 0
Then, the intensity of the external field is given by
X e ¼ Nð1 cosotÞ
o ¼2p
24
The NS components of this field at 09:00 h (X1) and at 12:00 h (X2) are
X e1 ¼ N 1 cos
3p
4
¼ N 1þ
ffiffiffi
2p
2
X e2 ¼ Nð1 cos pÞ ¼ 2N
Subtracting the two values and using Equation (99.1) we obtain
X e2 X e
1 ¼ 1
ffiffiffi
2p
2
N ¼ 1
ffiffiffi
2p
2
X e2
2¼ 293 nT ) X e
2 ¼ 2001 nT
X e1 ¼ 1708 nT
45° a
Z
∆X
∆Z
d
–
+
NP
Fig. 99a
178 Geomagnetism
The components of the geomagnetic main field intensity are given by
X ¼ X1 X X e1 ¼ 26 122 nT
Y ¼ Y1 Y Y e1 ¼ 0
Z ¼ Z1 Z Ze ¼ 30 000 nT
H ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðX Þ2 þ ðY Þ2q
¼ X ¼ 26 122 nT
tanD ¼Y
H) D ¼ 0
The geomagnetic latitude of the point f is determined from the vertical and horizontal
geomagnetic main field components:
Z ¼ 2B0 sinf
H ¼ B0 cosf
tanf ¼Z
2H) f ¼ 29:9
We obtain the coordinates of the Geomagnetic North Pole by (Fig. 99b)
D ¼ 0 ) lB ¼ 180þ l ¼ 135 E
90 fB ¼ f f ) fB ¼ 74:9
GMNPGNP
f
f∗
90° – fB
P
Fig. 99b
179 Main (internal), external, and anomalous magnetic fields
(b) The geomagnetic constant B0 is given by
H ¼ B0 cosf
B0 ¼H
cosf ¼ 30 133 nT
(c) We have obtained above that
Fe ¼ X e ¼ Nð1 cosotÞ
To calculate N we take into account that the Sq variation is maximum at 12:00 h:
X e2 ¼ 2N ) N ¼ 1000 nT
100. At a point with geographical coordinates 60 N, 45 E, a magnetic
observation results in the following values: FT ¼ 48 277 nT, DT ¼ 2.9, IT ¼ 63.7.
It is known that at 20 m below this point there is a horizontal magnetic dipole of
momentCm¼ 0.01 Tm3, with the positive pole oriented in the direction N 60 E. There
is also an external field parallel to the axis of rotation directed southwards of magni-
tude 1000 nT.
Determine the main (internal) field components, the constant B0, and the geocentric
geographical coordinates of the northern and southern geomagnetic poles (precision
1 nT).
We first calculate the components of the total field intensity from FT, DT, and IT(Fig. 100a) by
HT ¼ FT cos IT ¼ 21 390 nT
ZT ¼ FT sin IT ¼ 43 280 nT
XT ¼ HT cosDT ¼ 21 363 nT
YT ¼ HT sinDT ¼ 1082 nT
Y
Z
ZT
XYT
FT
IT
HT
DT
XT
Fig. 100a
180 Geomagnetism
The observed field is the result of three parts: the main field (X , Y , Z ), the buried dipole
field (DX, DY, DZ), and the external field (Xe, Ye, Ze):
XT ¼ X þX þ X e
YT ¼ Y þY þ Y e
ZT ¼ Z þZ þ Ze
We determine the magnetic anomaly created by the buried dipole from Equations (82.2)
and (82.3) substituting a¼ 90º and x¼ 0. If we call X60 the direction N 60º E the horizontal
component of this anomaly is DX60 (Fig. 100b):
Z ¼2Cm cos a
d3¼ 0 nT
X60 ¼Cm sin a
d3¼ 1250 nT
The NS and EW components will be given by (Fig. 100c)
X ¼ X60 cos 60 ¼ 625 nT
Y ¼ X60 sin 60 ¼ 1083 nT
The external field is parallel to the Earth’s axis of rotation in the southwards direction so its
components are in the vertical and NS direction and are given by (Fig. 100d)
Ze ¼ 1000 sinf ¼ 866 nT
X e ¼ 1000 cosf ¼ 500 nT
We calculate the main field components from these values:
X ¼ XT X X e ¼ 21 238 nT
Y ¼ YT Y Y e ¼ 1082 nT
Z ¼ ZT Z Ze ¼ 42 414 nT
H ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðX Þ2 þ ðY Þ2q
¼ 21 266 nT
tanD ¼Y
X ) D ¼ 2:9
PX60
d
Z
a
– +
Fig. 100b
181 Main (internal), external, and anomalous magnetic fields
The geomagnetic latitude of the point f and the geomagnetic constant B0 are found from
the vertical and horizontal geomagnetic main field components by
Z ¼ 2B0 sinf
H ¼ B0 cosf
tanf ¼Z
2H) f ¼ 44:9
B0 ¼Z
2 sinf ¼ 30 953 nT ¼ 30 044 nT
N
E
60°
∆Y
∆X60
∆X
Fig. 100c
GNP
Xe
Fe
Ze
P
f
Fig. 100d
182 Geomagnetism
To calculate the geographical coordinates of the Geomagnetic North Pole we use the
corresponding spherical triangle (Fig. 100e). We obtain the latitude fB by applying the
cosine law for the angle 90º – fB:
cosð90 fBÞ ¼ cosð90 fÞ cosð90 fÞ þ sinð90 fÞ sinð90 fÞ cosD
sinfB ¼ sinf sinfþ cosf cosf cosD
fB ¼ 75:0
We obtain the longitude lB by applying the cosine rule for the angle 90º – f :
cosð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ þ sinð90 fBÞ sinð90 fÞ cosðl lBÞ
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
cosðl lBÞ ¼sinf sinfB sinf
cosfB cosf
l lB ¼ 180:0
lB ¼ 135:0 W
Therefore the coordinates of the Geomagnetic North Pole are
fB ¼ 75 N lB ¼ 135 W
The coordinates of the Geomagnetic South Pole (the antipodal point) are:
fA ¼ fB ¼ 75 S lA ¼ 180 þ lB ¼ 45 E
101. At a point with geographical coordinates 30 N, 30 E, the observed geomagnetic
field components are (in nT): X ¼ 15 364, Y ¼ 7660, Z ¼ 48 980. The northern
geomagnetic pole is at 60 N, 0 E, andB0¼ 30 000 nT. There is also a constant external
magnetic field normal to the equatorial plane, with a southwards direction, of 1000 nT
intensity. Buried 10 m below the observation point is a magnetic dipole. Calculate:
(a) The magnetic anomalies DX, DY, DZ, DH, DF.
90º – fB
GNP
l – lB
180º – l∗
q = 90º – f∗
GMNP
D∗
90º – f
P
Fig. 100e
183 Main (internal), external, and anomalous magnetic fields
(b) The orientation and magnetic moment (Cm, in nT m3) of the buried dipole.
(a) The observed values are the sum of the geomagnetic main field, the magnetic field
due to the buried dipole, and the external field:
X ¼ X þX þ X e
Y ¼ Y þY þ Y e
Z ¼ Z þZ þ Ze
To obtain the magnetic anomalies from these equations we first calculate the geomagnetic
latitude and the vertical and horizontal components of the geomagnetic main field by (71.3):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 54
Z ¼ 2B0 sinf ¼ 48 479 nT
H ¼ B0 cosf ¼ 17 676 nT
The declination and inclination are given by
sinD ¼ cosfB sinðl lBÞ
cosf ) D ¼ 25
tan I ¼ 2 tanf ) I ¼ 70
The NS and EW components are
X ¼ H cosD ¼ 16 007 nT
Y ¼ H sinD ¼ 7498 nT
The external field is parallel to the axis of rotation directed southwards so its components
are in the vertical and NS direction and are given by (Fig. 101a)
Ze ¼ 1000 sinf ¼ 500 nT
X e ¼ 1000 cosf ¼ 866 nT
Y e ¼ 0
From these values we calculate the magnetic anomalies DX, DY, DZ, DH, DF:
X ¼ X X X e ¼ 223 nT
Y ¼ Y Y Y e ¼ 162 nT
Z ¼ Z Z Ze ¼ 1 nT
H ¼ X cosD þY sinD ¼ 271 nT
F ¼ H cos I þZ sin I ¼ 94 nT
(b) We call b
the angle between the geographical north and the buried dipole direc-
tions. Then using Equations (82.2) and (82.3) we obtain
Z ¼2Cm cos a
d3ð101:1Þ
184 Geomagnetism
X ¼Cm sin a
d3cos b ð101:2Þ
Y ¼Cm sin a
d3sinb ð101:3Þ
To solve this system of three equations in three unknowns (Cm, a, b) we divide Equation
(101.3) by (101.2):
tan b ¼Y
X) b ¼ 36 þ 180 ¼ 144
This value of the angle bimplies that the dipole is oriented in the N 144º E direction.
To calculate the angle a between the buried dipole and the vertical we divide Equation
(101.2) by (101.1):
X
Z¼
1
2tan a cos b
tan a ¼ 2
cos b
X
Z) a 90
Therefore the dipole is practically horizontal (Fig. 101b) and this explains the small value
of the vertical component DZ.
Finally we calculate the magnetic moment of the buried dipole from Equation (101.2):
Cm ¼ d3
sin a cos bX ¼ 2:8 105 nTm3
GNP
P
Xe
Ze
Fef
Fig. 101a
185 Main (internal), external, and anomalous magnetic fields
102. At a point with geographical coordinates 30 N, 30 E, magnetic measurements
give the following values: F ¼ 52 355 nT, I ¼ 70.5, and D ¼ -26.0. The terrestrial
dipole has a north pole at 60 N, 0 E and B0 ¼ 30 000 nT. There is also a constant
external field of 1000 nTwith lines of force contained in the plane of the 30 Emeridian
at an angle of 60 with the equatorial plane and directed southwards.
(a) Calculate the anomalies DX, DY, DZ.
(b) If the anomalies are produced by a dipole buried at 10 m below the point, what is
its orientation and its magnetic moment, Cm? Neglect values less than 10 nT.
(a) We first calculate the components of the total field from F, D, and I by
H ¼ F cos I ¼ 17 476 nT
Z ¼ F sin I ¼ 49 352 nT
X ¼ H cosD ¼ 15 707 nT
Y ¼ H sinD ¼ 7661 nT
The magnetic anomalies are found by subtracting from the observed values the main and
external field contributions:
X ¼ X X X e
Y ¼ Y Y Y e
Z ¼ Z Z Ze
First we determine the geomagnetic latitude and declination using (71.3) and (71.2):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 53:9
sinD ¼ cosfB sinðl lBÞ
cosf ) D ¼ 25:1
The vertical and horizontal components are given by
Z ¼ 2B0 sinf ¼ 48 479 nT
H ¼ B0 cosf ¼ 17 676 nT
X ¼ H cosD ¼ 16 007 nT
Y ¼ H sinD ¼ 7498 nT
– +
a
P
d
N 144° E
Fig. 101b
186 Geomagnetism
The external field is on the plane containing the vertical and NS directions (Figs 102a
and 102b):
Ze ¼ 1000 cosð60 fÞ ¼ 866 nT
X e ¼ 1000 sinð60 fÞ ¼ 500 nT
Y e ¼ 0
GNP
P60°
Xe
Ze
f Fe
Fig. 102a
Ze
Fe
Xe
f
60°
N
E
Fig. 102b
187 Main (internal), external, and anomalous magnetic fields
Finally, the magnetic anomalies are given by
X ¼ X X X e ¼ 200 nT
Y ¼ Y Y Y e ¼ 163 nT
Z ¼ Z Z Ze ¼ 7 nT
(b) We call b the angle between the positive pole of the buried dipole and the geographical
north. Then applying Equations (82.2) and (82.3) we obtain
Z ¼2Cm cos a
d3ð102:1Þ
X ¼Cm sin a
d3cos b ð102:2Þ
Y ¼Cm sin a
d3sin b ð102:3Þ
We divide Equation (102.3) by Equation (102.2) and obtain
tan b ¼Y
X) b ¼ 39:2 þ 180 ¼ 140:8
Therefore the dipole is oriented in the N 140.8º E direction.
To calculate the inclination of the dipole from the vertical we divide Equation (102.2) by
Equation (102.1):
X
Z¼
1
2tan a cos b
tan a ¼ 2
cos b
X
Z) a ¼ 89:2
This value implies that the dipole is nearly horizontal.
Finally we calculate Cm from the total anomalous field DB
B ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þY 2 þZ2p
¼ 258 nT
B ¼Cm
d3) Cm ¼ 258 000 nTm3
103. The Earth’s magnetic field is formed by a centred dipole with a geomagnetic pole
at 60 N, 0 E, and B0 ¼ 30 000 nT, and an external field of 10 000 nT parallel to the
equatorial plane and to the zero meridian.
(a) Calculate the components X, Y, Z of the total field at a point P with geographical
coordinates 60 N, 30 W.
(b) If at 30 m in the direction of the compass needle from P there is a vertical dipole of
moment Cm ¼ 4000 nT m3 buried 40 m deep, what would be the anomaly DZ
produced at P?
188 Geomagnetism
(a) The components of the total field are the sum of the geomagnetic main field and
the external field:
X ¼ X þ X e
Y ¼ Y þ Y e
Z ¼ Z þ Ze
Let us first calculate the geomagnetic latitude, declination, and inclination using (71.3)
and (71.2):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 75:1
sinD ¼ cosfB sinðl lBÞ
cosf ) D ¼ 76:5
tan I ¼ 2 tanf ) I ¼ 82:4
The vertical and horizontal components are given by
Z ¼ 2B0 sinf ¼ 57 983 nT
H ¼ B0 cosf ¼ 7714 nT
X ¼ H cosD ¼ 1801 nT
Y ¼ H sinD ¼ 7501 nT
The external field is parallel to the equatorial plane and to the zero meridian. If its
magnitude is N ¼ 10 000 nT, its components, from Fig. 103a (plane through the point
parallel to the equator) and Fig. 103b (plane through the geographical meridian of the
point), are given by
Ze ¼ Ber¼ N cos l cos’ ¼ 4330 nT
X e ¼ Bef ¼ N cos l sin’ ¼ 7500 nT
Y e ¼ Bel ¼ N sin l ¼ 5000 nT
Therefore the components of the total field are
Z ¼ 62 313 nT
X ¼ 9301 nT
Y ¼ 12 501 nT
(b) The buried vertical dipole is in the direction of the compass, that is, in the direction
of the magnetic north (Fig. 103c). We calculate the anomaly DZ produced at
P using Equations (82.2) and (82.3) substituting a ¼ 0º and x ¼ 30:
Z ¼Cm½ðx2 2d2Þ cos aþ 3dx sin a
½x2 þ d25=2
Z ¼Cmðx2 2d2Þ
½x2 þ d25=2¼ 0:029 nT
189 Main (internal), external, and anomalous magnetic fields
Be
l
P
Ncosλ
Plane parallel
to the equator
Beλ= Nsinl
Fig. 103a
GNP
P
f
Ncosλ
Bef
Ber
Fig. 103b
190 Geomagnetism
104. The Geomagnetic North Pole is at 60 N, 150 W, with B0 ¼ 30 000 nT, and there
is an external magnetic field of intensity 3000 nT parallel to the axis of rotation
pointing away from the North Pole. Buried 10 m below a point with coordinates
30 N, 30 E there is a horizontal dipole with Cm ¼ 40 000 nT m3 and the negative
pole pointing in the direction N 45 E.
(a) What are the components X, Y, Z of the total field?
(b) Calculate the total field anomaly DF.
(c) What is the angle between the direction of the compass and geographic north?
(a) The components of the total field are the sum of the geomagnetic main field, the
magnetic field due to the buried dipole, and the external field:
X ¼ X þX þ X e
Y ¼ Y þY þ Y e
Z ¼ Z þZ þ Ze
We determine first the geomagnetic latitude, declination, and inclination using (71.3)
and (71.2):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 0
sinD ¼ cosfB sinðl lBÞ
cosf ) D ¼ 0
tan I ¼ 2 tanf ) I ¼ 0
P x
d
z
–
+
NM
Fig. 103c
191 Main (internal), external, and anomalous magnetic fields
The vertical and horizontal components are given by
Z ¼ 2B0 sinf ¼ 0 nT
H ¼ B0 cosf ¼ 30 000 nT
X ¼ H cosD ¼ 30 000 nT
Y ¼ H sinD ¼ 0 nT
We calculate the magnetic anomaly created by the buried dipole from Equations (82.2) and
(82.3), substituting a ¼ 270º and x ¼ 0. The vertical component DZ and the horizontal
component DX45 of the anomaly in the direction N 45 E are (Fig. 104)
Z ¼2Cm cos a
d3¼ 0 nT
X45 ¼Cm sin a
d3¼ 40 nT
The NS and EW components are given by
X ¼ 40 sin 45 ¼ 28:3 nT
Y ¼ 40 cos 45 ¼ 28:3 nT
The external field is parallel to the axis of rotation directed southwards, so its components
are in the vertical and NS direction and are given by
Nr ¼ Ze ¼ N cos 60 ¼ 1500 nT
N# ¼ X e ¼ N sin 60 ¼ 2598 nT
The components of the observed total magnetic field are
X ¼ 27 430 nT
Y ¼ 28 nT
Z ¼ 1500 nT
P
d
a
–+
z
X45
Fig. 104
192 Geomagnetism
(b) The total field anomaly DF is given by
F ¼ X ¼ 28 nT
(c) The direction of the compass is affected by the three fields. The angle D between
the direction of the compass and the geographic north is obtained from the
horizontal components of the total observed field
tanD ¼Y
X¼
28
27430) D ¼ 0:06
105. The geomagnetic field is that of a dipole in the direction of the axis of rotation
and B0 ¼ 30 000 nT. There is also a constant external field of 2500 nT normal to the
equatorial plane in the direction of the South Pole.
(a) Calculate the value of the inclination observed at a point P with coordinates 45 N,
45 E given that, at 10 m below it, there is a vertical dipole with the negative pole
upwards and moment Cm ¼ 40 000 nT m3.
(b) For a point 20 m north of P, calculate the observed inclination and declination,
and the total field anomaly DF.
(a) The components of the total observed field are the sum of the geomagnetic main
field, the magnetic field due to the buried dipole, and the external field:
X ¼ X þX þ X e
Y ¼ Y þY þ Y e
Z ¼ Z þZ þ Ze
The magnetic dipole is oriented in the direction of the axis of rotation and therefore
f ¼ f ¼ 45
D ¼ 0
Then the components of the geomagnetic main field are
Z ¼ 2B0 sinf ¼ 42 426 nT
H ¼ B0 cosf ¼ 21 213 nT
X ¼ H cosD ¼ 21 213 nT
Y ¼ H sinD ¼ 0
The magnetic anomaly created by the dipole is obtained from Equations (82.2) and (82.3)
substituting a ¼ 0º and x ¼ 0 (Fig. 105a):
Z ¼2Cm cos a
d3¼ 80 nT
X ¼Cm sin a
d3¼ 0
Y ¼ 0
193 Main (internal), external, and anomalous magnetic fields
The external field is parallel to the axis of rotation directed southwards so its components
are in the vertical and NS direction (Fig. 105b) and are given by
Ze ¼ 2500 sinf ¼ 1768 nT
X e ¼ 2500 cosf ¼ 1768 nT
Y e ¼ 0
P
d
z
X
–
+
Fig. 105a
Ze
GNP
f Fe
Xe
P
Fig. 105b
194 Geomagnetism
Therefore the components of the observed field are
X ¼ X þX þ X e ¼ 19 445 nT
Y ¼ Y þY þ Y e ¼ 0
H ¼ X
Z ¼ Z þZ þ Ze ¼ 44 274 nT
From these values we calculate the observed inclination
tan I ¼Z
H) I ¼ 66:3
(b) For a point Q located 20 m to the north of P we can assume that the main and
external fields have the same value as at point P and only the magnetic anomaly
created by the buried dipole is different. We calculate this anomaly from Equations
(82.2) and (82.3) substituting a ¼ 0º and x ¼ 20:
Z ¼Cm½ðx2 2d2Þ cos aþ 3dx sin a
½x2 þ d25=2¼
Cmðx2 2d2Þ
½x2 þ d25=2¼ 1 nT
X ¼Cm½ð2x2 d2Þ sin a 3dx cos a
½x2 þ d25=2¼
Cm3dx
½x2 þ d25=2¼ 4 nT
Y ¼ 0
Therefore, the components of the observed field at that point are
X ¼ X þX þ X e ¼ 19 441 nT
Y ¼ Y þY þ Y e ¼ 0
H ¼ X
Z ¼ Z þZ þ Ze ¼ 44 193 nT
From these values we calculate the observed inclination and declination by the expressions
tan I ¼Z
H) I ¼ 66:2
tanD ¼Y
X) D ¼ 0
The total field anomaly DF is given by
F ¼ X cos I þZ sin I ¼ 4 cos 66 1 sin 66 ¼ 3 nT
106. Consider a point with coordinates 30 N, 30 E under which is buried at a depth
of 100 m a horizontal dipole of moment m0 m/4p ¼ 1 T m3, with the positive pole in
the direction N 60 E. The terrestrial field is formed by a centred dipole in the
direction of the axis of rotation and a constant external field of 10 000 nT from the
Sun, B0 ¼ 30 000 nT.
(a) Calculate F, D, and I at that point on December 21 at 12 noon.
195 Main (internal), external, and anomalous magnetic fields
(b) How do D and I vary throughout the year?
(a) The components of the total field intensity are the sum of the geomagnetic main
field, the magnetic field due to the buried dipole, and the external field:
X ¼ X þXþ X e
Y ¼ Y þY þ Y e
Z ¼ Z þZ þ Ze
We first calculate the geomagnetic latitude and the declination and inclination. The
magnetic dipole is oriented in the direction of the axis of rotation and therefore
f ¼ f ¼ 30
D ¼ 0
Then the components of the geomagnetic main field are
Z ¼ 2B0 sinf ¼ 30 000 nT
H ¼ B0 cosf ¼ 25 981 nT
X ¼ H cosD ¼ 25 981 nT
Y ¼ H sinD ¼ 0
We calculate the magnetic anomaly produced by the buried dipole from Equations (82.2)
and (82.3) substituting a ¼ 90º and x ¼ 0. We call X60 the direction N 60º E and DX60 the
horizontal component of the anomaly (Fig. 106a):
Z ¼2Cm cos a
d3¼ 0 nT
X60 ¼Cm sin a
d3¼ 1000 nT
P
a
– +
d
z
X60
Fig. 106a
196 Geomagnetism
The NS and EW components are given by (Fig. 106b)
X ¼ X60 cos 60 ¼ 500 nT
Y ¼ X60 sin 60 ¼ 866 nT
The external field has a diurnal period (o ¼ 2p / 24 h) and we know that the Sun is at the
meridian point at 12:00 h (solar time). This field changes through the year because the Sun
moves on the ecliptic plane (apparent motion) which is tilted with respect to the equatorial
plane by an angle e¼ 23º. Therefore the solar declination (d), the angle from the Sun to the
celestial equator, changes through the year. On December 21 (winter solstice) this angle is
d¼e¼23º (Fig. 106c). If we call N the magnitude of the external field (N¼ 10 000nT)
its components on December 21 at 12:00 are (Fig. 106d)
Ze ¼ Br ¼ N cosðfþ eÞ ¼ 6018 nT
X e ¼ Bf ¼ N sinðfþ eÞ ¼ 7986 nT
Y e ¼ 0
60°
N
E∆Y
∆X
∆X60
Fig. 106b
GNP
P
e
f
f +e
N = Be
Sun
Fig. 106c
197 Main (internal), external, and anomalous magnetic fields
The components of the observed field are
X ¼ X þX þ X e ¼ 33 467 nT
Y ¼ Y þY þ Y e ¼ 866 nT
Z ¼ Z þZ þ Ze ¼ 36 018 nT
Therefore the total field F and the declination D are
F ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þ Y 2 þ Z2p
¼ 49 166 nT
tanD ¼Y
X) D ¼ 1:5
sin I ¼Z
F) I ¼ 47:1
(b) Any other day at 12:00 h
X ¼ X þX þ X e ¼ 25 481 nTþ N sinðf dÞ
Y ¼ Y þY þ Y e ¼ 866 nT
Z ¼ Z þZ þ Ze ¼ 30 000 nTþ N cosðf dÞ
107. The internal field of the Earth corresponds to a centred dipole with the negative
pole in the northern hemisphere at coordinates 80 N, 130 W and B0 ¼ 30 000 nT.
There is a uniform external field from the Sun of 1000 nT. Buried at 500 m depth
under a point P with geocentric coordinates 40 N, 50 E there is a positive magnetic
pole of strength CP ¼ 0.5 T m². Calculate:
(a) The total field components X, Y, and Z, and the magnetic and geomagnetic
declination at P on March 21 at 12:00 h.
GNP
N
Z = – Br
X = Bf
r
f
P
f + e
Fig. 106d
198 Geomagnetism
(b) The same parameters for a point 200 m north of P, assuming that neither the
internal nor the external fields change (precision 1 nT).
(a) The components of the intensity of the total field are the sum of the geomagnetic
main field, the external field, and the magnetic field due to the buried pole.
To calculate the main field we determine first the geomagnetic latitude and the declination
by (71.3) and (71.2):
sinf ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
f ¼ 30
sinD ¼ cosfB sinðl lBÞ
cosf ) D ¼ 0
From these values we obtain the vertical and horizontal components of the geomagnetic
main field:
Z ¼ 2B0 sinf ¼ 30 000 nT
H ¼ B0 cosf ¼ 25 981 nT
X ¼ H cosD ¼ 25 981 nT
Y ¼ H sinD ¼ 0
To calculate the external field we notice that it comes from the Sun which on March 21
(spring equinox) is on the equatorial plane so that the external field is parallel to this plane.
In addition this field changes during the day as a function of local time t with a diurnal
period (o ¼ 2p/24 h). At t ¼ 12 h, the external field is maximum given that at this time
the Sun is at the meridian point (Fig. 107a). Calling N its magnitude (N ¼ 1000 nT), the
components of the external field are given by
GNP
Xe
P
Ze
N
f
Fig. 107a
199 Main (internal), external, and anomalous magnetic fields
Ze ¼ Br ¼ N cosf ¼ 766 nT
X e ¼ Bf ¼ N sinf ¼ 643 nT
Y e ¼ Bel ¼ 0
The magnetic field anomaly created by the buried pole is derived from its potential DF,
given by
F ¼CP
r
Applying the gradient, we obtain
B ¼ F
but the only component is the vertical:
Z ¼ Br ¼@ðFÞ
@r¼
CP
r2
Substituting r ¼ d ¼ 500 m in this equation we obtain
Z ¼ 2000 nT
Therefore the components of the total field and the observed declination are given by
X ¼ X þ X e þX ¼ 26 624 nT
Y ¼ Y þ Y e þY ¼ 0
Z ¼ Z þ Ze þZ ¼ 28 766 nT
tanD ¼Y
X) D ¼ 0
(b) The radial component of the magnetic field anomaly created by the buried pole for
a point 200 m north of P (x ¼ 200 m) is given by
Br ¼ @ðFÞ
@r¼
CP
r2
From Fig. 107b the vertical and NS components are
Z ¼ Br cos a ¼CP
r2
d
r¼
CPd
ðx2 þ d2Þ3=2
X ¼ Br sin a ¼CP
r2
x
r¼
CPx
ðx2 þ d2Þ3=2
Y ¼ 0
Substituting the values given (d ¼ 500 m, x ¼ 200 m, CP ¼ 0.5 Tm2), we obtain
Z ¼ 1601 nT
X ¼ 640 nT
Y ¼ 0
200 Geomagnetism
The components of the observed total field and the declination are the sum of the three
contributions:
X ¼ X þ X e þX ¼ 27 264 nT
Y ¼ Y þ Y e þY ¼ 0
Z ¼ Z þ Ze þZ ¼ 29 165 nT
tanD ¼Y
X) D ¼ 0
Paleomagnetism
108. At a point with geographical coordinates 60 N, 60 Wa 1 cm3 sample was taken
of a rock with remanent magnetism, age 10 000 years, specific susceptibility
0.01 cm3. The magnetization components of the rock were:
X ¼ 40, Y ¼ 30, Z ¼ 50 nT (N, E, nadir).
The current field is B0 ¼ 30 000 nT and the geomagnetic pole coincides with the
geographical pole. Calculate:
(a) The coordinates of the virtual geomagnetic pole which corresponds to the
sample.
(b) The magnetic moment of the terrestrial dipole 10 000 years ago.
(c) The secular variation of F, D, and I in nT and minutes per year assuming that the
variation since that time has been constant.
Px
X
α
d
z
r a
+
Fig. 107b
201 Paleomagnetism
(a) First we determine the declination D and the geomagnetic co-latitude y, corres-
ponding to the virtual pole, from the magnetization components of the rock X, Y,
and Z:
tanD ¼Y
X) D ¼ 36:9
H ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þ Y 2p
¼ 50 nT
tan I ¼Z
H) I ¼ 45
tan I ¼ 2 cot y ) y ¼ 63:4;fvirtual ¼ 26:6
Since at present the geomagnetic pole coincides with the geographical pole, the geograph-
ical latitude of the point coincides with the present geomagnetic latitude:
f ¼ fpresent ¼ 60N
To determine the coordinates of the virtual Geomagnetic North Pole (VP), corresponding
to the magnetization of the rock, we solve the spherical triangle of Fig. 108 for ’B and lBusing the obtained values of y and D. The latitude fB applying the cosine rule is given by
cosð90 fBÞ ¼ cos y cosð90 fÞ þ sin y sinð90 fÞ cosD
sinfB ¼ cos y sinfþ sin y cosf cosD
fB ¼ 48:2
To obtain the longitude lB we again apply the cosine rule:
cos y ¼ sinð90 fÞ ¼ cosð90 fBÞ cosð90 fÞ
þ sinð90 fBÞ sinð90 fÞ cosðl lBÞ
cos y ¼ sinfB sinfþ cosfB cosf cosðl lBÞ
90º – fB
GNP
VP
l – lB
90º – f∗
180º – l∗
D∗
P
90º – f
Fig. 108
202 Geomagnetism
cosðl lBÞ ¼cos y sinfB sinf
cosfB cosf
l lB ¼ 126:4
To choose between the positive and negative solution we bear in mind that the declination
is negative and so the point is to the east of the virtual magnetic North Pole:
D < 0 ) l lB > 0
lB ¼ 173:6 E
(b) To obtain the magnetic moment we first calculate the constant B0. The susceptibility
w relates the magnetization and the magnetic field. If we call F the magnitude of the
paleomagnetic field andF0 the remanent magnetization, the relation between them is
F 0 ¼ wF ð108:1Þ
w ¼ 0:01
We calculate F0 from its components
F 0 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þ Y 2 þ Z2p
¼ 71 nT
The field F of the virtual pole is given by
F ¼ B0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ 3 cos2 yp
Substituting in Equation (108.1) we obtain
F 0 ¼ wB0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ 3 cos2 yp
B0 ¼F 0
wffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ 3 cos2 yp ¼ 5610 nT
From this value we calculate the magnetic moment of the virtual pole taking a ¼ 6370 km
for the Earth’s radius
B0 ¼Cm
a3) m ¼ 1:45 1022 Am2
(c) The magnetic field, the declination, and inclination 10 000 years ago were
F ¼F 0
w¼ 7100 nT
D ¼ 36:9
I ¼ 45
At present the values of these parameters are
Fp ¼ Ba0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ 3 sin2 fpresent
q
¼ 54 083 nT
Dp ¼ 0
tan Ip ¼ 2 tanf ) Ip ¼ 73:9
203 Paleomagnetism
The secular variation of F, D, and I, for this period of time, is
Fp F
10 000¼ 4:7 nT=yr
Dp D
10 000¼ 0:220=yr
Ip I
10 000¼ 0:170=yr
109. The following table gives the demagnetization data for a sample that was
subjected to stepwise thermal demagnetization of its natural remanent magnetization
(NRM).
Calculate the direction of each stable component identified by the demagnetization
curve.
First, construct a vector component diagram of the demagnetization data. Decompose
each observation into its north (X), east (Y), and vertical (Z) components:
H ¼ J cos I
X ¼ H cosD
Y ¼ H sinD
Z ¼ J sin I
In Fig. 109 plotting X versus Y gives the projection of the demagnetization vector
onto the horizontal plane, while plotting X versus Z gives the projection onto the vertical
plane.
A stable component of NRM is represented by collinear points on the vector component
diagrams, so that two stable components can be identified in the range 20–300 C and in
the range 400–700 C.
The declination of a stable component is determined by measuring or by calculating the
angle between the north axis and the trajectory of the stable component in the horizontal
plane.
Demagnetization
temperature (°C) Declination (D, °E) Inclination (I, °) NRM Intensity (J, mA/m1)
20 32 33 0.056
100 36 22 0.056
200 38 12 0.057
300 39 4 0.058
400 41 5 0.058
500 41 5 0.050
600 41 5 0.016
650 41 5 0.009
700 300 55 0.000
204 Geomagnetism
For the 20–300 C component:
a ¼ tan1 X300 X20
Y300 Y20
¼ tan1 0:0051
0:0115
¼ 23:9
D ¼ 180þ 23:9 ¼ 203:9
For the 400–700 C component:
D ¼ b ¼ tan1 Y400
X400
¼ tan1 0:0379
0:0436
¼ 41:0
Note: this value can be obtained directly from the declination of the observations between
400 and 650 C.
The apparent inclination, Iap, of a stable component is determined by measuring or by
calculating the angle between the north axis and the trajectory of the stable component in
the vertical plane. Iap is related to the true inclination, I, by:
tan I ¼ tan Iapj cosDj
For the 20–300 C component:
Iap ¼ g ¼ tan1 Z20 Z300
X300 X20
¼ tan1 0:0265
0:0051
¼ 79:1
I ¼ tan1ðtanð79:1Þj cosð203:9ÞjÞ ¼ 78:1
For the 400–700 C component:
Iap ¼ d ¼ tan1 Z400
X400
¼ tan1 0:0051
0:0436
¼ 6:7
I ¼ tan1ðtanð6:7Þj cosð41ÞjÞ ¼ 5:0
Note: this value can be obtained directly from the inclination of the observations between
400 and 650 C.
Therefore the stable component isolated in the range 20–300 C has D ¼ 203.9 and
I ¼ 78.1 and the stable component isolated in the range 400–700 C has D ¼ 41.0 and
I ¼ 5.0.
0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.0350.00
0.01
0.02
0.03
0.04
0.05
0.06
X X
YZ
a
b d
γ
–0.01 0.00 0.01 0.02 0.03
0.01
0.00
0.02
0.03
0.04
0.05
0.06
Fig. 109
205 Paleomagnetism
110. A palaeomagnetic study of a late Jurassic limestone outcrop near Alhama de
Granada (37 N, 4 W) in southern Spain yielded a well-defined primary remanent
magnetization whose directions are given in the table below. Calculate the mean
direction of the primary remanence of the seven samples. Compare this direction
with that defined by the reference late Jurassic palaeomagnetic pole for the stable
Iberian tectonic plate (252 E, 58 N). How much vertical axis rotation has the studied
outcrop suffered with respect to stable Iberia?
Use unit vector addition to calculate the mean direction of the primary remanence.
Calculate the direction cosines of each direction, the resultant total field vector, F, and
then the mean direction using:
X ¼ cos I cosD
Y ¼ cos I sinD
Z ¼ sin I
F ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X
N
i¼1
Xi
!2
þX
N
i¼1
Yi
!2
þX
N
i¼1
Zi
!2v
u
u
t ¼ 6:98542
Xmean ¼
P
N
i¼1
Xi
F¼ 0:6446;
Ymean ¼
P
N
i¼1
Yi
F¼ 0:37186;
Zmean ¼
P
N
i¼1
Zi
F¼ 0:66799
Dmean ¼ tan1 Ymean
Xmean
¼ 30:0
Imean ¼ sin1ðZmeanÞ ¼ 41:9
The mean direction of the primary remanence has a declination of 30.0 and an inclination
of 41.9.
Declination (D, °E) Inclination (I, °)
30 43
28 39
34 44
25 45
32 38
35 44
26 40
206 Geomagnetism
Next, calculate the expected field direction at the site using the reference palaeomagnetic
pole. The first step is to determine y ¼ 90 f, (Equation 71.3), from the pole (fp, lp) to
the site (fs, ls) using spherical triangles:
sinf ¼ sin’p sin’s þ cos’p cos’s cosðls lpÞ ) f ¼ 24:1
The expected inclination can then be calculated using:
tan Iexp ¼ 2 tanf ) Iexp ¼ 41:8
The expected declination can be calculated by (71.2):
sinDexp ¼ cos’p sinðls lpÞ
cosf ) Dexp ¼ 34:5 W ¼ 325:5
rotation about the vertical axis should give rise to a difference between the observed and
expected declinations, defined as positive for an observed declination clockwise from the
expected declination.
Therefore the outcrop has suffered 64.5 of clockwise rotation with respect to stable
Iberia.
207 Paleomagnetism
4 Seismology
Elasticity
111. Determine the principal stresses and principal axes of the stress tensor:
2 1 1
1 0 1
1 1 2
0
@
1
A
Find the invariants I1, I2, I3, the deviator tensor, its eigenvalues, and the invariants
J2 and J3.
To calculate the principal stresses (s1, s2, s3) and principal axes (n1i, n2i, n3i), we
calculate the eigenvalues and eigenvectors of the matrix. They are found through the
equation
ðtij sdijÞni ¼ 0 ð111:1Þ
The eigenvalues are the roots of the cubic equation for s resulting from putting the
determinant of the matrix in (111.1) equal to zero:
2 s 1 1
1 s 1
1 1 2 s
¼ 0
) 2 sð Þ sð Þ 2 sð Þ 1 1þ s 2 sð Þ 2 sð Þ ¼ 0
s3 4s2 þ sþ 6 ¼ 0
The three roots of the equation are the principal stresses
s1 ¼ 1
s2 ¼ 2
s3 ¼ 3
From these values we obtain s0 ¼13s1 þ s2 þ s3ð Þ ¼ 4
3.
The invariants of the matrix are the coefficients of the characteristic equation
s3 I1s2 þ I2s I3 ¼ 0
208
which, in terms of the roots of the equation, are
I1 ¼ 4 ¼ s1 þ s2 þ s3
I2 ¼ 1 ¼ s1s2 þ s1s3 þ s2s3
I3 ¼ 6 ¼ s1s2s3
The principal axes of stress are the eigenvectors ni associated with the three eigenvalues.
For s1 ¼ 1
3 1 1
1 1 1
1 1 3
0
@
1
A
n1n2n3
0
@
1
A ¼ 0 ) n11; n12; n
13
¼ 1;2; 1ð Þ
For s2 ¼ 2
0 1 1
1 2 1
1 1 0
0
@
1
A
n1n2n3
0
@
1
A ¼ 0 ) n21; n22; n
23
¼ 1; 1; 1ð Þ
For s3 ¼ 3
1 1 1
1 3 1
1 1 1
0
@
1
A
n1n2n3
0
@
1
A ¼ 0 ) n31; n32; n
33
¼ 1; 0; 1ð Þ
The deviatoric stress tensor is defined as
t0ij ¼ tij s0dij
where in our problem, s0 ¼ 4/3.
The three components of the principal diagonal of the deviatoric tensor are
t011 ¼2
3
t022 ¼ 4
3
t033 ¼2
3
To calculate its eigenvalues we proceed as we did before:
2
3 s 1 1
1 4
3 s 1
1 12
3 s
0
B
B
B
B
B
@
1
C
C
C
C
C
A
¼ 0 ) s3 13
3sþ
38
27¼ 0
Comparing with the characteristic equation
s3 J1s2 þ J2s J3 ¼ 0
209 Elasticity
the invariants are
J2 ¼ 13
3
J3 ¼ 38
27
Solving the cubic equation we obtain s1 ¼ 2.22, s2 ¼ 0.33, s3 ¼ 1.89.
112. Given the stress tensor
1
42
5
4
1
2
32
25
4
1
42
1
2
32
1
2
32
21
2
32 3
2
0
B
B
B
B
B
B
B
B
B
@
1
C
C
C
C
C
C
C
C
C
A
calculate:
(a) The principal stresses.
(b) The angles formed by the greatest of these stresses with the axes 1, 2, 3.
(a) As in the previous problem to find the principal stresses we calculate the eigen-
values of the stress matrix
1
4 s
5
4
1
2
3=2
5
4
1
4 s
1
2
3=2
1
2
3=2
1
2
3=23
2 s
0
B
B
B
B
B
B
B
B
B
@
1
C
C
C
C
C
C
C
C
C
A
¼ 0 ) s3 2s2 sþ 2 ¼ 0
Solving the cubic equation, its three roots are
s1 ¼ 2
s2 ¼ 1
s3 ¼ 1
The largest is s1. The associated eigenvector corresponds to the axis of greatest stress
whose direction cosines are (n1, n2, n3). They are found by solving the equation
7
45
4
1
2
3=2
5
47
4
1
2
3=2
1
2
3=2
1
2
3=21
2
0
B
B
B
B
B
B
B
B
@
1
C
C
C
C
C
C
C
C
A
n1n2n3
0
@
1
A ¼ 0
210 Seismology
Solving this equation with the condition that n21 þ n22 þ n23 ¼ 1, we obtain,
n1 ¼ n2 ¼1
2
n3 ¼1ffiffiffi
2p
(b) From these values we obtain #, the angle with the vertical axis (x3) and ’, the angle
which forms its projection on the horizontal plane with x1:
n1 ¼ sin# cos’ ¼1
2
n2 ¼ sin# sin’ ¼ 1
2
n3 ¼ cos# ¼1ffiffiffi
2p
9
>
>
>
>
>
=
>
>
>
>
>
;
) ’ ¼ 315; # ¼ 45
113. The stress tensor tij in a continuous medium is
3x1x2 5x22 0
5x22 0 2x30 2x3 0
0
@
1
A
Determine the stress vector Tin acting at the point (2, 1, √3) through the plane
tangential to the cylindrical surface x22 1 x23 at that point.
First we calculate the value of the stress tensor at the given point:
tijð2; 1;ffiffiffi
3p
Þ ¼6 5 0
5 0 2ffiffiffi
3p
0 2ffiffiffi
3p
0
0
@
1
A
A unit vector normal to the surface f ¼ x22 þ x23 4 ¼ 0 at the given point is
ni ¼grad f
grad fj j¼
@f
@x1@f
@x1
;
@f
@x2@f
@x2
;
@f
@x3@f
@x3
0
B
B
@
1
C
C
A
¼ 0;1
2;
ffiffiffi
3p
2
Then, the stress vector acting at the point through that surface is given by
T ni ¼ tijnj ¼
6 5 0
5 0 2ffiffiffi
3p
0 2ffiffiffi
3p
0
0
@
1
A
01
2ffiffiffi
3p
2
0
B
B
B
B
@
1
C
C
C
C
A
¼5
2; 3;
ffiffiffi
3p
Wave propagation. Potentials and displacements
114. The amplitudes of P- and S-waves of frequency 4/2p Hz are
uP ¼4
3ffiffiffi
2p ;
4
3ffiffiffi
2p ;
4ffiffiffi
3p
uS ¼ ffiffiffi
3p
;ffiffiffi
3p
;ffiffiffi
2p
211 Wave propagation. Potentials and displacements
and their speeds of propagation are 6 km s1 and 4 km s1, respectively. Find the
scalar and vector potentials. Displacements are always given in µm.
The displacements of P-waves can be deduced from a scalar potential function ’ such that
uPi ¼ ðr’Þi. The general form of the potential function for P-waves for harmonic motion is
’ ¼ A exp ikaðnixi atÞ ð114:1Þ
where A is the amplitude, ni the direction cosines of the ray or propagation direction, a the
velocity of propagation, and ka the wavenumber. If uj is given in µm and ka in km1, then A
is given in 103 m2. Taking the derivatives in (114.1) we obtain for the components of the
displacement
uPj ¼@’
@xj¼ Aikanj exp ika nkxk atð Þ
Their amplitude is
uPj ¼ Akanj ð114:2Þ
and the wavenumber is
ka ¼o
a¼
2p4
2p6
¼2
3km1
By substitution in (114.2), we obtain for the two horizontal components
uP1 ¼ Akan1 ¼4
3ffiffiffi
2p mm
uP2 ¼ Akan2 ¼4
3ffiffiffi
2p mm
Dividing these two expressions and writing the direction cosines of the ray in terms of the
incident angle i and azimuth az,
n1 ¼ sin i cos az
v2 ¼ sin i sin az
n3 ¼ cos i
ð114:3Þ
we have
uP1uP2
¼ 1 ¼n1
n2¼
sin i cos az
sin i sin az) a ¼ 45
Using the uP3 and uP1 components,
uP3uP1
¼Akan3
Akan1¼
cos i
sin i cos az¼
ffiffiffi
3p
) i ¼ 30
From the values of the direction cosines and the amplitude of the displacement we calculate
the amplitude of the potential A:
A ¼uP1kan1
) A ¼ 4 103 m2
212 Seismology
Finally, the expression for the scalar potential of P-waves is
’ ¼ 4 exp2
3i
ffiffiffi
2p
4x1 þ
ffiffiffi
2p
4x2 þ
ffiffiffi
3p
2x3 6t
Displacements of the S-wave are obtained from a vector potential ci of null divergence,
whose general form is
ci ¼ Bi exp ikb njxj bt
where b is the velocity of propagation and kb the wavenumber. The displacements are
given by
uS ¼ r c ð114:4Þ
The wavenumber is kb ¼ o/b ¼ 1 km1. According to (114.4) the relation between the
components of the displacement and of the amplitude of the potential is
uS1 ¼ B3
ffiffiffi
2p
4 B2
ffiffiffi
3p
2¼
ffiffiffi
3p
mm
uS2 ¼ B1
ffiffiffi
3p
2 B3
ffiffiffi
2p
4¼
ffiffiffi
3p
mm
uS3 ¼ B2
ffiffiffi
2p
4 B1
ffiffiffi
2p
4¼
ffiffiffi
2p
mm
The potential must have null divergence,
r c ¼
ffiffiffi
2p
4B1 þ
ffiffiffi
2p
4B2 þ
ffiffiffi
3p
2B3 ¼ 0
From these equations we obtain, in units of 103 m2,
B1 ¼ 2
B2 ¼ 2
B3 ¼ 0
The S-wave vector potential is
cj ¼ ð2; 2; 0Þ exp i
ffiffiffi
2p
4x1 þ
ffiffiffi
2p
4x2 þ
ffiffiffi
3p
2x3 4t
Note: These units will be used for all problems but not explicitly given.
115. The components of the S-wave with respect to the axes (x1, x2, x3) are (6, 3.25, 3)
where x2 is the vertical axis, the azimuth is 60, and the angle of incidence is 30.
Determine the amplitude and direction cosines of the SV and SH components.
From the azimuth and incident angles we calculate the direction cosines (x2 is the
vertical axis)
213 Wave propagation. Potentials and displacements
n1 ¼ sin i cos az ¼1
4
n2 ¼ cos i ¼
ffiffiffi
3p
2
n3 ¼ sin i sin az ¼
ffiffiffi
3p
4
Since the SV and SH components are on a plane normal to the direction of the ray r
(Fig. 115a) unit vectors in the direction of SV (a1, a2, a3) and of SH (b1, 0, b3) must satisfy
the equations
SV r ¼ 0 )a1
4þa2
ffiffiffi
3p
2þa3
ffiffiffi
3p
4¼ 0
SH r ¼ 0 )b1
4þb3
ffiffiffi
3p
4¼ 0
SH SV ¼ 0 ) a1b1 þ a3b3 ¼ 0
ð115:1Þ
The projections on the horizontal plane R of the ray r and SH are perpendicular
(Fig. 115b). Then SH forms an angle of 180 – 30 with the x1 axis. The direction
cosines of SH are
b1 ¼
ffiffiffi
3p
2
b3 ¼1
2
SV
SVH
SH
RX3
X2
X1
90°
60°
30°
r
Fig. 115a
214 Seismology
SV forms an angle of 60 with the vertical axis x2 (Fig. 115a). Then a2 ¼ sin i ¼ 12
From a2 using Equations (115.1) and a21 þ a22 þ a23 ¼ 1, we calculate a1 and a2:
a1 ¼
ffiffiffi
3p
4
a3 ¼3
4
116.Given the potentialci ¼
ffiffiffi
7p
ffiffiffi
5p ;
5ffiffiffi
7p
ffiffiffi
5p ;6
exp 4i 1ffiffiffi
5p x1 þ
1ffiffiffi
3p x2 þ
ffiffiffi
7p
ffiffiffiffiffi
15p x3 4t
,
calculate the polarization angle.
First we calculate the amplitudes of the components of the displacement of the S-wave
from the vector potential
uiS ¼ r ci )
uS1 ¼ c3;2 c2;3 ¼ 413ffiffiffi
3p ¼ 30:02
uS2 ¼ c1;3 c3;1 ¼ 47
5ffiffiffi
3p þ
6ffiffiffi
5p
¼ 13:97
uS3 ¼ c2;1 c1;2 ¼ 4ffiffiffi
7p
ffiffiffi
7p
ffiffiffiffiffi
15p
¼ 7:85
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
The modulus is
uS ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u21 þ u22 þ u23
q
¼ 34:03
From the vertical component uS3 we calculate the SV component (Fig. 116) knowing that
n3 ¼ cos i ¼
ffiffiffi
7p
ffiffiffiffiffi
15p ) sin i ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
17
15¼
rffiffiffi
8p
ffiffiffiffiffi
15p
Then, uS3 ¼ uSVcos 90 ið Þ ) uSV ¼ 10:75.
SH
R
SVH
X3
X130°
60°
30°
Fig. 115b
215 Wave propagation. Potentials and displacements
To find the SH component we use the relation
uS ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðuSVÞ2 þ ðuSHÞ2q
) uSH ¼ 32:29
From the SV and SH components we obtain the polarization angle e:
tan e ¼uSH
uSV) e ¼ 71:6
117. Given the amplitudes of the P- and S-waves (ka ¼ 1):
uP1 ¼ 4 uS1 ¼ 8
uP2 ¼ 4 uS2 ¼ 2ffiffiffi
2p
uP3 ¼ 8 uS3 ¼ 4þffiffiffi
2p
determine the angle of incidence i, azimuth az, polarization angle « of the S-wave,
and apparent polarization angle g.
Given that the displacements of the P-wave are on the incident plane, in the direction of the
ray r, the angle of incidence i can be obtained from the modulus and the vertical component:
uP ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
42 þ 42 þ 82p
¼ 4ffiffiffi
6p
cos i ¼uP3uP
¼8
4ffiffiffi
6p ) i ¼ 35:3
The azimuth, the angle between the horizontal projection of the ray and the north (x1), is
obtained from the horizontal components, uP1 and uP2 :
tan az ¼uP2uP1
) az ¼ 45
SV
R
X3
r
i
90° – i
Fig. 116
216 Seismology
We calculate the SV component from the vertical component uS3 , as in the previous
problem:
uSV ¼uS3
cos 90 i0ð Þ¼
uS3sin i0
¼ 9:37
The SH component is found from the horizontal components (Fig. 117)
uSH ¼ uS2 cos az uS1 sin az ¼ 3:66
From SV and SH we find the polarization angle e (Fig. 117):
tan e ¼uSH
uSV¼
3:66
9:37) e ¼ 21:3
To calculate the apparent polarization angle g, the angle between the horizontal component
of S, (uSH), and the radial direction R (Fig. 117), we use the relation
tan e ¼ cos i0 tan g ) g ¼ 25:5
118. Given the values az ¼ 60, « ¼ 30, g ¼ 45, and ub
¼ 5, calculate the
amplitudes of the components 1, 2, and 3 of the S-wave.
From the modulus of the displacement of S-waves and the polarization angle, we calculate
the SV and SH components (Fig. 118a):
uSV ¼ uS cos e ¼5ffiffiffi
3p
2
uSH ¼ uS sin e ¼5
2
SH
R
X1
X2
az
az
γ
uS2
uS1
uSH
90° – az
Fig. 117
217 Wave propagation. Potentials and displacements
From the angles e and g we obtain the incidence angle i:
tan e ¼ tan g cos i ) cos i ¼1ffiffiffi
3p ) sin i ¼
ffiffiffiffiffiffiffiffiffiffiffi
11
3
r
¼
ffiffiffi
2p
ffiffiffi
3p
The vertical component u3 of the S-wave is obtained from the value of SV (Fig. 118b):
uS3 ¼ uSV cos 90 ið Þ ¼5ffiffiffi
3p
2
ffiffiffi
2p
ffiffiffi
3p ¼ 3:53
To calculate the horizontal components we have to take into account the horizontal
component of SV (Fig. 118b):
uSVH ¼ uSV cos i ¼5
2
SH
SV
e
Fig. 118a
SV
R
X3
r
i
i
uS3
uSVH
Fig. 118b
218 Seismology
From SH and the horizontal component of SV we find the horizontal components
of S (Fig. 118c):
uS1 ¼ uSH sin az þ uSVH cos az
¼ 5
41þ
ffiffiffi
3p
¼ 3:42
uS2 ¼ uSH cos az uSVH sin az ¼5
41
ffiffiffi
3p
¼ 0:92
119. For a scalar potential w and a vector potential ci, it is known that A ¼ 3, Bi ¼
(2, 2, 0), ka ¼ 2/3, T ¼ p/2, and Poisson’s ratio is s ¼ 1/4. Calculate the amplitudes
on the free surface (x3 ¼ 0) of the components u1, u2, and u3 of the P- and SV-waves.
The direction cosines are1
2ffiffiffi
2p ;
1
2ffiffiffi
2p ;
ffiffiffi
3p
2
.
The general expressions for the scalar and vector potentials are
’ ¼ A exp ika n1x1 þ n2x2 þ n3x3 atð Þ
ci ¼ Bi exp ikb n1x1 þ n2x2 þ n3x3 btð Þ
Since ka ¼ o/a then a ¼ o/ka ¼ 6 km s1.
If Poisson’s ratio is 0.25 then
s ¼1
4¼
l
2 lþ mð Þ) l ¼ m
Substituting this condition in the equation for the P-wave velocity a, we find for the
velocity b of S-waves is given by
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
s
¼
ffiffiffiffiffiffi
3m
r
s
¼ bffiffiffi
3p
) b ¼affiffiffi
3p ¼ 2
ffiffiffi
3p
km s1
Then the wavenumber of S-waves is
kb ¼o
b¼
2ffiffiffi
3p
RX2
X1
az
az
az
u SH
SH
uSH
uSVH
Fig. 118c
219 Wave propagation. Potentials and displacements
Then we can write the complete expressions for the potentials:
’ ¼ 3 exp i2
3
1
2ffiffiffi
2p x1 þ
1
2ffiffiffi
2p x2 þ
ffiffiffi
3p
2x3 6t
ci ¼ 2; 2; 0ð Þ exp i2ffiffiffi
3p
1
2ffiffiffi
2p x1 þ
1
2ffiffiffi
2p x2 þ
ffiffiffi
3p
2x3 2
ffiffiffi
3p
t
ð119:1Þ
To determine the displacements of the P- and S-waves we use the relations
uP ¼ r’
uS ¼ r c
obtaining
uP1 ¼1ffiffiffi
2p uS1 ¼ 2
uP2 ¼1ffiffiffi
2p uS2 ¼ 2
uP3 ¼ffiffiffi
3p
uS3 ¼2ffiffiffi
2p
ffiffiffi
3p
From the components of the displacement of the S-wave we can obtain the SV component.
The values of the angles of incidence and azimuth are found from the direction cosines,
n1 ¼ sin i cos az ¼1
2ffiffiffi
2p
n2 ¼ sin i sin az ¼1
2ffiffiffi
2p ) i ¼ 30; az ¼ 45
n3 ¼ cos i ¼
ffiffiffi
3p
2
and from the angle i we obtain the SV component from uS3:
uS3 ¼ uSV cos 90 ið Þ ) uSV ¼4ffiffiffi
2p
ffiffiffi
3p ¼ 3:27
From the horizontal components of the S-waves and the azimuth we calculate the SH
component:
uSH ¼ uS2 cos az uS1 sin az ¼ 0
120. In an elastic medium of density r¼ 3 g cm3
and Poisson ratio 1/3 there
propagate P- and S-waves of frequency 1 Hz in the direction1
3;1ffiffiffi
2p ;
ffiffiffi
7p
3ffiffiffi
2p
. Given
that the pressure of the P-wave is 5000 dyn cm2, the magnitude of its displacement,
10 mm, is twice that of the S-wave, and the angle g ¼ 45, find all the parameters
involved in the expression of the potentials ’ and ci. (It is not necessary to solve the
equations to obtain the coefficients Bi)
220 Seismology
Given that Poisson’s ratio is 1/3 the relation between the elastic coefficients l and m is,
s ¼l
2 lþ mð Þ¼
1
3) l ¼ 2m
and between the velocities of P (a) and S (b)-waves is
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
s
¼
ffiffiffiffiffiffi
4m
r
s
) a ¼ 2b
The bulk modulus K is
K ¼ lþ2
3m ¼
8
3m
Expressing m in terms of b and a, we get for K:
b ¼
ffiffiffi
m
r
r
) m ¼ rb2 ) K ¼8
3rb2 ¼
8
3ra2
4¼
2
3ra2
Taking into account that the bulk modulus K is defined as the applied pressure divided by
the change in volume per unit volume y,
P ¼ Ky ¼2
3ra2y
y ¼ r2’ ¼ Ak2a
y ¼3P
2ra2¼ Ak2a ) a2 ¼
3P
2rAk2a
ð120:1Þ
The expression for the scalar potential is
’ ¼ A exp ika n1x1 þ n2x2 þ n3x3 atð Þ ð120:2Þ
where the wavenumber for P-waves is
ka ¼o
a¼
2pf
a
uP
¼ r’j j ¼ kaA ) A ¼uPj j
ka¼
uPj ja
2pf
By substitution in (120.1) we obtain the values of a and b:
a2 ¼3P
2rAk2a¼
3Pa
2r uPj j2pf) a ¼
3P
2r uPj j2pf
where
P ¼ 5000 dyn cm2
r ¼ 3 g cm3
f ¼ 1 Hz
uP ¼ 10 mm
221 Wave propagation. Potentials and displacements
We obtain a ¼ 3.98 km s1 and b ¼ 1.99 km s1.
Since we know A, a, and ka we can write the complete expression for the scalar
potential
’ ¼ 10 exp i1:581
3x1 þ
1ffiffiffi
2p x2 þ
ffiffiffi
7p
3ffiffiffi
2p x3 3:98t
The vector potential of the S-waves is given by
ci ¼ Bi exp ikb n1x1 þ n2x2 þ n3x3 btð Þ ð120:3Þ
We calculate kb:
kb ¼o
b¼ 3:16 km1
The displacements are given by
uS ¼ r c
and
uS1 ¼@c3
@x2@c2
@x3¼ kb B3n2 B2n3ð Þ
uS2 ¼@c1
@x3@c3
@x1¼ kb B1n3 B3n1ð Þ
uS3 ¼@c2
@x1@c1
@x2¼ kb B2n1 B1n2ð Þ
ð120:4Þ
The incidence angle i is found from n3 and, using tane¼ cosi tang, we find the polarization
angle e:
n3 ¼ cos i ¼
ffiffiffi
7p
3ffiffiffi
2p ) i ¼ 31:95 ) e ¼ 31:95
The azimuth is
az ¼ tan1 n2
n1¼ 64:76
Since the amplitude of the S-wave displacement is 5 mm, knowing the value of e we can
find the values of the SV and SH components:
uSV ¼ uS cos e ¼ 4:24 mm
uSH ¼ uS sin e ¼ 2:65 mm
From uSV we calculate its vertical and horizontal components:
uS3 ¼ uSV cos 90 ið Þ ¼ 2:25 mm
uSVH ¼ uSV cos i ¼ 3:61 mm
222 Seismology
Using uHSV and uSH we find the two horizontal components:
uS1 ¼ uSH sin az þ uSVH cos az
¼ 3:94 mm
uS2 ¼ uSH cos az uSVH sin az ¼ 2:14 mm
Using the values found for the displacements and Equations (120.4) and∇ ·c ¼ 0 (n1 B1þ
n2 B2 þ n3 B3 ¼ 0) we find the values of B1, B2, B3. Substituting all the values in (120.3) we
obtain for the vector potential
ci ¼ ð1:17; 2:5;3:44Þ exp 3:16i1
3x1 þ
1ffiffiffi
2p x2 þ
ffiffiffi
7p
3ffiffiffi
2p x3 1:99t
121. At the origin in an infinite medium in which s, Poisson’s ratio, is 0.25, and the
density is 3 g cm3, there is an emitter of elastic plane waves of frequency 0.5 cps.
Calculate:
(a) The equation of the P- and S-waves in exponential form and with arbitrary
amplitudes for the wave arriving at the point A(500, 300, 141) km.
(b) The arrival time.
(a) First we calculate the distance to point A and the direction cosines of the direction
of the ray (r) (Fig. 121):
r ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5002 þ 3002 þ 1412p
¼ 599:90 600 km
n1 ¼500
600¼
5
6
n2 ¼300
600¼
1
2
n3 ¼141
600¼
ffiffiffi
2p
6
X2
X3
X1
141
300
500
r
Fig. 121
223 Wave propagation. Potentials and displacements
The S-wave velocity is
b ¼
ffiffiffi
m
r
r
¼ffiffiffiffiffi
10p
km s1
Given that Poisson’s ratio is 0.25,
s ¼ 0:25 ) l ¼ m ) a ¼ bffiffiffi
3p
¼ffiffiffiffiffi
30p
km s1
The wavenumbers of the P- and S-waves are
ka ¼2pf
a¼
2p 0:5
5:5¼
pffiffiffiffiffi
30p km1
kb ¼2pf
b¼
2p 0:5ffiffiffiffiffi
10p ¼
pffiffiffiffiffi
10p km1
The general expressions for the scalar and vector potentials are
’ ¼ A exp ika n1x1 þ n2x2 þ n3x3 atð Þ
ci ¼ Bi exp ikb n1x1 þ n2x2 þ n3x3 btð Þ
Leaving the amplitudes A and Bi in arbitrary form and substituting the obtained values we
have for the potentials,
’ ¼ A exp ipffiffiffiffiffi
30p
5
6x1 þ
1
2x2 þ
ffiffiffi
2p
6x3
ffiffiffiffiffi
30p
t
ci ¼ Bi exp ipffiffiffiffiffi
10p
5
6x1 þ
1
2x2 þ
ffiffiffi
2p
6x3
ffiffiffiffiffi
10p
t
(b) The travel times for P- and S-waves from the origin to the given point are
ta ¼
r
a¼
600ffiffiffiffiffi
30p ¼ 109:5 s
tb ¼
r
b¼
600ffiffiffiffiffi
10p ¼ 189:7 s
Reflection and refraction
122. A P-wave represented by the potential
w ¼ 4 exp 0:25ix1ffiffiffi
6p þ
x2ffiffiffi
3p þ
x3ffiffiffi
2p 4t
is incident on the surface x3 ¼ 0 separating two liquids of densities 3 g cm3
and
4 g cm3. If the speed of propagation in the second medium is 2 km s1, write the
expressions for the potentials of the reflected and transmitted waves.
224 Seismology
The potentials of the reflected and transmitted waves are given by
’refl ¼ A exp ik tan e x3 þ x1 ctð Þ
’trans ¼ A0 exp ik tan e0 x3 þ x1 ctð Þð122:1Þ
where e ¼ 90 – i is the emergence angle and i the incidence angle, and k ¼ ka cos e is the
wavenumber corresponding to the apparent horizontal velocity, c ¼ a /cos e. These
expressions are written for rays contained on the incidence plane (x1, x3). Then, we have
to rotate the given potential to refer it to the incidence plane. First, from the direction
cosines we calculate the incidence angle i and the azimuth az:
n1 ¼ sin i cos az ¼1ffiffiffi
6p
n2 ¼ sin i sin az ¼1ffiffiffi
3p
n3 ¼ cos i ¼1ffiffiffi
2p ) i ¼ 45 ¼ e
cos az ¼1ffiffiffi
3p ) sin az ¼
ffiffiffi
2p
ffiffiffi
3p
Using the rotation matrix we obtain the direction cosines on the plane of incidence (x1, x3):
n01n02n03
0
@
1
A ¼cos az sin az 0
sin az cos az 0
0 0 1
0
@
1
A
n1n2n3
0
@
1
A)
1ffiffiffi
2p
01ffiffiffi
2p
0
B
B
B
@
1
C
C
C
A
The values of c and k are
c ¼a
cos e¼
a0
cos e0¼
8ffiffiffi
2p ¼ 4
ffiffiffi
2p
km s1
k ¼ ka cos e ¼ ka0 cos e0 ¼
ffiffiffi
2p
8km1
Then the potential of the incident wave is now given by
’inc ¼ 4 exp i1
4ffiffiffi
2p x3 þ x1 4
ffiffiffi
2p
t
The angle i0 of the transmitted or refracted ray is found from Snell’s law:
sin i
a¼
sin i0
a0) sin i0 ¼
ffiffiffi
2p
4¼ cos e0
) cos i0 ¼ sin e0 ¼
ffiffiffiffiffi
14p
4) tan e0 ¼
ffiffiffi
7p
Using the expressions for the reflection and refraction coefficients, V and W, we can
calculate the amplitude of the reflected and refracted potentials:
V ¼A
A0
¼r0 tan e r tan e0
r0 tan eþ r tan e0¼
4 3ffiffiffi
7p
4þ 3ffiffiffi
7p ) A ¼
16 12ffiffiffi
7p
4þ 3ffiffiffi
7p ¼ 1:07
W ¼A0
A0
¼2rtge0
r0 tan eþ r tan e0¼
6
4þ 3ffiffiffi
7p ) A0 ¼
24
4þ 3ffiffiffi
7p ¼ 2:01
225 Reflection and refraction
By substitution in (122.1) we obtain
’refl ¼ 1:07 exp
ffiffiffi
2p
8i x3 þ x1
8ffiffiffi
2p t
’trans ¼ 2:01 exp
ffiffiffi
2p
8i
ffiffiffi
7p
x3 þ x1 8ffiffiffi
2p t
123. A P-wave of amplitude 5ffiffiffi
2p
; 5ffiffiffi
6p
; 10ffiffiffi
2p
and frequency v ¼ 12 rad s1 in a
semi-infinite medium of speed of propagation a ¼ 6 km s1 and Poisson’s ratio 0.25 is
incident on the free surface. Calculate:
(a) The potential of the incident P-wave.
(b) The potential of the reflected S-wave.
(c) The components u1, u2, u3 of the reflected S-wave.
(a) The displacements of the P-wave can be deduced from its scalar potential:
’ ¼ A exp ika n1x1 þ n2x2 þ n3x3 atð Þ
uP ¼ r’ð123:1Þ
where A is the amplitude, ka is the wavenumber (P), ni are the direction cosines, and a is the
P-wave velocity. The wavenumber is found from the given angular frequency and velocity:
ka ¼o
a¼
12
6¼ 2 km1
Since we know the amplitudes of the components of the displacements we can find the
incidence angle i and the azimuth az:
uP1 ¼@’
@x1¼ Akan1 ¼ A2 sin i cos az ¼ 5
ffiffiffi
2p
uP2 ¼@’
@x2¼ Akan2 ¼ A2 sin i sin az ¼ 5
ffiffiffi
6p
uP3 ¼@’
@x3¼ Akan3 ¼ A2 cos i ¼ 10
ffiffiffi
2p
Dividing the two first equations we obtainffiffiffi
3p
¼ tan az ) az ¼ 60, and dividing the
last two,
5ffiffiffi
2p
10ffiffiffi
2p ¼ tan i cos az ¼
1
2tan i ) i ¼ 45
The amplitude A is given by
uP
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
uP1 2
þ uP2 2
þ uP3 2
q
¼ Aka ) A ¼uPj j
ka¼ 102 m2
Finally the potential is given by
’inc ¼ 102exp i2
ffiffiffi
2p
4x1 þ
ffiffiffi
6p
4x2 þ
ffiffiffi
2p
2x3 6t
m2
226 Seismology
If we express the potential referred to the plane (x1, x3) as the incidence plane, as we did in
Problem 122, then
’ ¼ A exp ik x1 þ tan e x3 ctð Þ
where
e ¼ 90 i ¼ 45; k ¼ ka cos e ¼ 21ffiffiffi
2p ¼
ffiffiffi
2p
c ¼a
cos e¼
6
1ffiffiffi
2p
¼ 6ffiffiffi
2p
km s1
and the potential is
’inc ¼ 102 exp iffiffiffi
2p
x1 þ x3 6ffiffiffi
2p
t
m2 ð123:2Þ
(b) Since Poisson’s ratio is 1/4, l ¼ m, and
b ¼affiffiffi
3p ¼ 2
ffiffiffi
3p
km s1
The angle f of the reflected S-wave is obtained by Snell’s law:
cos e
a¼
cos f
b) cos f ¼
b
acos e ¼
ffiffiffi
2p
2ffiffiffi
3p ¼
1ffiffiffi
6p ) sin f ¼
ffiffiffi
5
6
r
From the values of e and f we calculate the P-to-S reflection coefficient VPS, using equation
VPS ¼4a 1þ 3a2ð Þ
4abþ 1þ 3a2ð Þ2
where we substitute
a ¼ tan e ¼ 1 and b ¼ tan f ¼ffiffiffi
5p
so
VPS ¼4 1þ 3ð Þ
4ffiffiffi
5p
þ 1þ 3ð Þ2¼ 0:64
From this coefficient we calculate the proportion of the incident P-wave which is reflected
as an S-wave (only with SV component; the negative sign indicates the opposite sense of
the reflected ray):
B ¼ AVPS ¼ 10 0:64 ¼ 6:4 103 m2
When the ray is contained in the (x1, x3) plane we use a scalar potential for the S-wave
which in this case is given by
c ¼ B exp ik x1 tan f x3 ctð Þ
¼ 6:4 103 exp iffiffiffi
2p
x1 ffiffiffi
5p
x3 6ffiffiffi
2p
t
m2
227 Reflection and refraction
(c) To calculate the amplitudes of the total displacements in terms of the two scalar
potentials, we remember that for this orientation of the axes the displacements are
given by
u1 ¼@’
@x1
@c
@x3¼ uP1 þ uSV1
u3 ¼@’
@x3þ
@c
@x1¼ uP3 þ uSV3
The displacements of the SV reflected wave in this case are
uSV1 ¼ @c
@x3¼ 6:4
ffiffiffi
5p
uSV3 ¼@c
@x1¼ 6:4
ffiffiffi
2p
If we want to determine the components 1 and 2, referred to the original system of axes, we
project uSV1 ¼ uSVR using the azimuth 60:
uSV1 ¼ uSVR cos az ¼ 3:2ffiffiffi
5p
uSV2 ¼ uSV2 sin az ¼ 3:2ffiffiffiffiffi
15p
124. An S-wave of vector potential
c i ¼ 10ffiffiffi
3p
; 2; 4
exp 5ix1
4þ
ffiffiffi
3p
4x2 þ
ffiffiffi
3p
2x3 4t
is incident on the free surface x3 ¼ 0. Find the SV and SH components of the reflected
S-wave referred to the same coordinate system as the incident wave, and the coeffi-
cient of reflection. Poisson’s ratio is 3/8.
According to the value of Poisson’s ratio the relation between l and m is
s ¼l
2 lþ mð Þ¼
3
8) l ¼ 3m
and the relation between the velocities of the P-waves and S-waves is
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r¼
s ffiffiffiffiffiffi
5m
r
s
¼ffiffiffi
5p
b
The incidence angle i and the azimuth az are obtained from the direction cosines:
n3 ¼ cos i ¼
ffiffiffi
3p
2) i ¼ 30
n1 ¼ sin i cos az ¼1
4¼ sin 30 cos az ) az ¼ 60
Using Snell’s law we find the value of the critical angle
sin ic
b¼
1
a) sin ic ¼
b
a¼
1ffiffiffi
5p ) ic ¼ 26:5
228 Seismology
Since i > ic, there is no reflected P-wave. The components of the incident S-wave are
obtained from the potential
uSi ¼ r c )uS1 ¼ 0
uS2 ¼ 80
uS3 ¼ 40
8
<
:
The modulus of the displacement is uS
¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
80ð Þ2þ 40ð Þ2q
¼ 40ffiffiffi
5p
:
The SV component is given by (Fig. 124a)
uSV ¼u3
cos 90 i0ð Þ¼ 80
and the SH component is
uSH ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
uSð Þ2 uSVð Þ2q
¼ 40
The minus sign corresponds to that of uS2The amplitude of the reflected SH wave is equal to that of the incident SH wave. We find
the 1 and 2 components using the azimuth 60 (Fig. 124b):
uSHi ¼ 40
ffiffiffi
3p
2;1
2; 0
exp 5i1
4x1 þ
ffiffiffi
3p
4x2
ffiffiffi
3p
2x3 4t
For total reflection the amplitude of the reflected SV is equal to that of the incident one, but
with a phase shift d. The components are given by (Figs 124a and 124b)
uSV1 ¼ uSV cos i cos az
uSV2 ¼ uSV cos i sin az
uSV3 ¼ uSV sin i
X3
i
r
60°
R
uSV
Fig. 124a
229 Reflection and refraction
The components of the reflected SV wave are
uSVi ¼ 80
ffiffiffi
3p
4;
3
4;1
2
exp 5i1
4x1 þ
ffiffiffi
3p
4x2
ffiffiffi
3p
2x3 4t
þ id
To determine the phase shift d we have to determine the reflection coefficients for a free
surface. Textbooks usually give those for Poisson’s ratio s = 1/4, but since in this problem
s ¼ 3/8 we have to calculate them. On a free surface the boundary conditions are that
stresses are null, which in terms of the scalar potentials ’ and c, are given by
t31 ¼ 0 ¼ m u3;1 þ u1;3
¼ 2’;31 c;11 þ c;33
t33 ¼ 0 ¼ l u1;1 þ u3;3
þ 2mu3;3 ¼ 3’;11 5’;33 2c;13
ð124:1Þ
where, using (x1, x3) as the plane of incidence, the scalar potential ’ of the reflected
P-waves is
’ ¼ A exp ik ax3 þ x1 ctð Þ
and the scalar potential of the incident and reflected S-wave is
c ¼ B0 exp ik bx3 þ x1 ctð Þ þ B exp ik bx3 þ x1 ctð Þ
Substituting in (124.1) we obtain for the coefficient of the reflected S-waves,
VSS ¼B
B0
¼i4ab 1 b2ð Þ 3þ 5a2ð Þ
i4abþ 1 b2ð Þ 3þ 5a2ð Þ
The phase shift is given by
d ¼ tan1 4ab
1 b2ð Þ 3 5a2ð Þ
60°
30°R
X2
X1
uHSV
uSH
Fig. 124b
230 Seismology
By substitution of the values of the problem,
b ¼ tan f ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b2 1
s
¼1ffiffiffi
3p
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
a2
r
¼
ffiffiffiffiffi
22
30
r
and finally we obtain d ¼ 77.33.
125. An S-wave represented by the potential
c ¼ 10 exp i31
2x1 þ
ffiffiffi
3p
2x3 4
ffiffiffi
2p
t
is incident from an elastic medium with l¼ 0 onto a liquid with velocity a0 ¼ 4 km s1
(the two media have the same density). Derive the equations relating the amplitudes of
the potentials of the incident, reflected, and transmitted waves.
Given that thewave is incident froman elasticmedium onto a liquidmedium, there are reflected
S- and P-waves in the elastic medium and transmitted P-waves in the liquid (Fig. 125).
If l ¼ 0, the P-wave velocity in the elastic medium is
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
s
¼
ffiffiffiffiffiffi
2m
r
s
¼ bffiffiffi
2p
¼ 4ffiffiffi
2p ffiffiffi
2p
¼ 8 km s1
Assuming (x1, x3) is the incidence plane, we use the scalar S-wave potential which, for the
incident and reflected waves in the solid medium, is given by
c ¼B0 exp ikb x1 cos f þ x3 sin f btð Þ
þ B exp ikb x1 cos f x3 sin f btð Þ
M
M ff
X3
X1
P
P
e
e
SS
a= 4
a = 8
b= 0
b = 4√2
Fig. 125
231 Reflection and refraction
From the potential of the incident wave given in the problem,
cos f ¼1
2) f ¼ 60; B0 ¼ 10; kb ¼ 3
The potential can also be written in the form (Problem 122)
c ¼ B0 exp ik x1 þ tan f x3 ctð Þ þ B exp ik x1 tan f x3 ctð Þ
where
k ¼ kb cos f ¼3
2km1
c ¼b
cos f¼ 8
ffiffiffi
2p
km s1
and the potential is given by
c ¼ 10 exp i3
2x1 þ x3
ffiffiffi
3p
8ffiffiffi
2p
t
þ B exp i3
2x1 x3
ffiffiffi
3p
8ffiffiffi
2p
t
Applying Snell’s law we determine the angle e of the reflected P-wave in the solid medium
and the angle e0 of the transmitted P-wave onto the liquid (Fig. 125):
cos f
b¼
cos e
a)
cos 60
4ffiffiffi
2p ¼
cos e
8) e ¼ 45
cos e0
a0¼
cos f
b) cos e0 ¼
1
2ffiffiffi
2p ) e0 ¼ 69
sin e0 ¼
ffiffiffiffiffiffiffiffiffiffiffi
11
8
r
¼
ffiffiffi
7p
2ffiffiffi
2p ) tan e0 ¼
ffiffiffi
7p
The potential of the reflected P-wave in the solid medium (M) is
’ ¼ A exp ik x1 x3 tan e ctð Þ ¼ A exp i3
2x1 x3 8
ffiffiffi
2p
t
and that of the transmitted P-wave in the liquid (M0) is
’0 ¼ A0 exp ik x1 þ x3 tan e0 ctð Þ ¼ A0 exp i
3
2x1 þ x3
ffiffiffi
7p
8ffiffiffi
2p
t
The relation between the amplitude of the potential of the incident S-wave (B0 = 10) and
those of the reflected and refracted P-waves A, A0 and the reflected S-wave B can be
obtained from the conditions at the boundary between the two media (x3 ¼ 0), that is,
continuity of the normal component of the displacements (u3) and of the stress (t33) and
null tangential stress (t31):
u3 ¼ u03 ) ’;3 þ c;1 ¼ ’0;3
t31 ¼ 0 ) 2’;13 c;33 þ c;11 ¼ 0
t33 ¼ t033 ) l0 ’0;33 þ ’0
;11
¼ 2m ’;33 þ c;13
232 Seismology
By substitution of the potentials we obtain the equations (in units of 103 m2)
Aþ 10þ B ¼ A0ffiffiffi
7p
Aþ Bþ 10 ¼ 0
2A0 ¼ Aþ 10ffiffiffi
3p
ffiffiffi
3p
B
Solving the system of equations we obtain
A0 ¼40
ffiffiffi
3p
4þffiffiffi
7p
þffiffiffiffiffi
21p ¼ 6:2
A ¼20
ffiffiffiffiffi
21p
4þffiffiffi
7p
þffiffiffiffiffi
21p ¼ 8:2
B ¼10 4
ffiffiffi
7p
þffiffiffiffiffi
21p
4þffiffiffi
7p
þffiffiffiffiffi
21p ¼ 1:8
126. An S-wave incident on the free surface of a semi-infinite medium with s ¼ 0.25 is
given by (in units of 103 m2)
c i ¼ 10ffiffiffi
3p
; 2; 4
exp 5i1
4x1 þ
ffiffiffi
3p
4x2 þ
ffiffiffi
3p
2x3 4t
Calculate:
(a) The amplitude of the components of the reflected P-wave.
(b) The components of the reflected S-wave.
(a) From the direction cosines we calculate the incidence i and emergence f angles and
azimuth az of the incident S-wave:
n1 ¼1
4¼ sin i cos az
n2 ¼
ffiffiffi
3p
4¼ sin i sin az
n3 ¼
ffiffiffi
3p
2¼ cos i ) i ¼ 30 ) f ¼ 60
1
4¼
1
2cos az ) az ¼ 60
Since Poisson’s ratio is 0.25 then l ¼ m ) a ¼ffiffiffi
3p
b ¼ 4ffiffiffi
3p
, and
sin ic
a¼
1
b) sin ic ¼
1ffiffiffi
3p ) ic ¼ 35
Since i < ic we have a reflected P-wave. The reflection coefficient at a free surface for a
reflected P-wave from an incident S-wave is given by
VSP ¼4b 1þ 3a2ð Þ
4abþ 1þ 3a2ð Þ2ð126:1Þ
233 Reflection and refraction
where a ¼ tan e and b ¼ tan f, and f is the emergence angles of the incident S-wave and e is
that of the reflected P-wave (Fig. 126). The relation between f and e according to Snell’s law is
cos f
b¼
cos e
a) cos e ¼
a
bcos f ¼
ffiffiffi
3p 1
2) e ¼ 30
Substituting in Equation (126.1):
VSP ¼4ffiffiffi
3p
1þ 1ð Þ
4þ 1þ 1ð Þ2¼
ffiffiffi
3p
We can write the potential of the incident S-wave referred to the incidence plane (x1, x3) by
means of the rotation matrix
cos az sin az 0
sin az cos az 0
0 0 1
0
@
1
A
and substituting
1
2
ffiffiffi
3p
20
ffiffiffi
3p
2
1
20
0 0 1
0
B
B
B
@
1
C
C
C
A
10ffiffiffi
3p
2
4
0
@
1
A ¼B1
B2
B3
0
@
1
A)B1 ¼ 4
ffiffiffi
3p
B2 ¼ 16 ¼ B0
B3 ¼ 4
8
<
:
the potential is
c0i ¼ B1;B2;B3ð Þ exp ikbðcos fx1 þ sin fx3 4tÞ
¼ 4ffiffiffiffi
3;p
16; 4
exp 5i1
2x1 þ
ffiffiffi
3p
2x3 4t
ð126:2Þ
The scalar potential of the SV component is
cSV ¼ B0 exp ikbðcos fx1 þ sin fx3 4tÞ
¼ 16 exp 5i1
2x1 þ
ffiffiffi
3p
2x3 4t
S S
P
f f e
X3
X1
Fig. 126
234 Seismology
The amplitude of the potential of the reflected P-wave is
A ¼ B0VSP ¼ 16ffiffiffi
3p
From
kb ¼o
b¼ 5 ¼
o
4we have o ¼ 20 s1
and
ka ¼o
a¼
20
4ffiffiffi
3p km1
The scalar potential of the reflected P-wave referred to the original system of axes is
given by
’ ¼ A exp ika njxj at
where the direction cosines are now
n1 ¼ sin i cos az ¼ cos e cos az ¼
ffiffiffi
3p
2
1
2¼
ffiffiffi
3p
4
n2 ¼ sin i sin az ¼ cos e sin az ¼
ffiffiffi
3p
2
ffiffiffi
3p
2¼
3
4
n3 ¼ cos i ¼ sin e ¼1
2
Substituting these values we obtain
’ ¼ 16ffiffiffi
3p
exp i5ffiffiffi
3p
ffiffiffi
3p
4x1 þ
3
4x2
1
2x3 4
ffiffiffi
3p
t
The components of the displacement in mm are
uP ¼ r’ )
uP1 ¼ 20ffiffiffi
3p
uP2 ¼ 60
uP3 ¼ 40
8
>
>
<
>
>
:
(b) For the reflected S-wave we have to separate the SV and SH components. The SV
component can be deduced from the scalar potential
cr
SV ¼ B exp ikbðcos f x1 sin f x3 4tÞ ¼ B exp 5i1
2x1
ffiffiffi
3p
2x3 4t
We obtain B by means of the reflection Vss:
VSS ¼4ab 1þ 3a2ð Þ
2
4abþ 1þ 3a2ð Þ2¼ 0 ¼
B
B0
) B ¼ 0
The reflected S-wave doesn’t have an SV component.
235 Reflection and refraction
For the reflected SH component we use the displacement of the u2 instead of the
potential. The displacements of the SH component of the incident wave are obtained from
(126.2):
uiSH ¼ c01;3 c0
3;1 ¼ 40
Referred to the reference of the plane of incidence, the displacement is given by
uiSH ¼ 40 exp 5i1
2x1 þ
ffiffiffi
3p
2x3 4t
The amplitude of the reflected SH wave is equal to that of the incident SH wave. Referred
to the incidence plane system of reference,
urSH ¼ 40 exp 5i1
2x1
ffiffiffi
3p
2x3 4t
The displacement of the reflected S-wave referred to the original system of axes is
uri ¼ ðBr
1;Br
2;Br
3Þ exp i51
4x1 þ
ffiffiffi
3p
4x2
ffiffiffi
3p
2x3 4t
Since the SV component is zero, Br
3 ¼ 0;Br
1 and Br
2 are found using the equations
urj j ¼ urSH ) ðBr
1Þ2 þ ðBr
2Þ2 ¼ 1600
uri ni ¼ 0 )1
4Br
1 þ
ffiffiffi
3p
4Br
2 ¼ 0
resulting in
Br
1 ¼ 20ffiffiffi
3p
Br
2 ¼ 20
127. A wave represented by the potential
w ¼ 4 exp 0:25ix1ffiffiffi
6p þ
x2ffiffiffi
3p þ
x3ffiffiffi
2p 4t
is incident on the surface x3 ¼ 0 of separation between two liquids. If the speed of
propagation in the second medium is 2 km s1, the pressure exerted by the incident
wave on the surface of separation is 5 109 Pa, and the transmitted energy is four
times greater than the reflected energy, calculate:
(a) The energy transmitted to the second medium.
(b) The potentials of the transmitted and reflected waves referred to the same
coordinate system as the incident potential.
(a) The intensity or energy per unit surface area of the wavefront of an incident
P-wave is given in units of J m2 by
Iinc ¼ A20o
2k2aar ð127:1Þ
236 Seismology
From the given potential we have
A0 ¼ 4 103 m2
ka ¼ 0:25 km1
a ¼ 4 km s1
ka ¼o
a) o ¼ 1 s1
We need to know the value of the density r. Since the medium is liquid l ¼ K (bulk
modulus) and then l ¼ P/y, where P is the pressure and y the cubic dilatation (change of
volume per unit volume). For liquids the shear modulus m is zero and from the velocity of
P-waves we obtain,
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
s
¼
ffiffiffi
l
r
s
) l ¼ a2r
The cubic dilatation is obtained from the potential ’:
y ¼ r2’ ¼ k2aA ¼1
164 ¼
1
4
Then, we obtain
l ¼ a2r ¼P
y¼
5 109
1
4 16 106
Pa
m2 s2) r ¼
5
4g cm3
By substitution in (127.1) the incident energy is
Iinc ¼ 16 11
16 4
5
4¼ 5 Jm2
The energy transmitted into the second medium is
Itras ¼ A02o2k 02a a0r0 ¼ W 2A2
0k02a a
0r0 ð127:2Þ
and the energy reflected is
Irefl ¼ Ao2k 02a a0r0 ¼ V 2A2
0k02a a
0r0 ð127:3Þ
where W and V are the transmission and reflection coefficients, respectively:
W ¼A0
A0
¼2r tan e
r0 tan eþ r tan e0
V ¼A
A0
¼r0 tan e r tan e0
r0 tan eþ r tan e0
ð127:4Þ
The emergence angle e of the incident wave is
n3 ¼1ffiffiffi
2p ¼ cos i ¼ sin e ) i ¼ e ¼ 45
237 Reflection and refraction
and from Snell’s law the emergence angle of the transmitted wave e0 is
cos e
a¼
cos e0
a0) cos e0 ¼
a0
acos e ¼
2
4
1ffiffiffi
2p ¼
1
2ffiffiffi
2p ) sin e0 ¼
ffiffiffiffiffiffiffiffiffiffiffi
11
8
r
¼
ffiffiffi
7p
2ffiffiffi
2p
Given that the transmitted energy is four times the reflected energy,
Itras
Iref¼ 4 ¼
W 2A20r
0o4
a0
V 2A20ro
4
a
)V 2
W 2¼
2r0a
5ra0
If we substitute in (127.4)
V ¼r0
5
4
ffiffiffi
7p
r0 þ5
4
ffiffiffi
7p
W ¼25
4
r0 þ5
4
ffiffiffi
7p
We have three equations for r’, V, and W. The solution for positive values of the variables
is
r0 ¼ 7:7 )W ¼ 0:23V ¼ 0:40
(b) The potential of the reflected P-wave is
’ref ¼ VA0 exp ika n1x1 þ n2x2 n3x3 atð Þ
¼ 1:6 exp i1
4
1ffiffiffi
6p x1 þ
1ffiffiffi
3p x2
1ffiffiffi
2p x3 4t
To determine the potential of the transmitted wave we have to calculate the direction
cosines of the transmitted ray. The azimuth is the same as that of the incident wave which
can be deduced from the direction cosines and the value of i:
n1 ¼ sin i cos az ¼1ffiffiffi
6p ) cos az ¼
1ffiffiffi
3p
n2 ¼ sin i sin az ¼1ffiffiffi
3p ) sin az ¼
ffiffiffi
2p
ffiffiffi
3p
The direction cosines of the transmitted ray in the medium M0 are
n01 ¼ sin i0 cos az ¼1
2ffiffiffi
2p
1ffiffiffi
3p ¼
1
2ffiffiffi
6p
n02 ¼ sin i0 sin az ¼1
2ffiffiffi
3p
n03 ¼ cos i0 ¼
ffiffiffi
7p
2ffiffiffi
2p
238 Seismology
and the potential of the transmitted wave is
’tras ¼ 0:92 exp i1
2
1
2ffiffiffi
6p x1 þ
1
2ffiffiffi
3p x2 þ
ffiffiffi
7p
2ffiffiffi
2p x3 2t
128. Two liquid media are separated at x3 ¼ 0, the first of volumetric coefficient
K ¼1
2 109 Pa and density 1 g cm3. The amplitudes of the components of an
incident wave of frequency 3 Hz are ui ¼ 18p 1; 1;ffiffiffi
6p
mm and those of the wave
transmitted to medium 2 are63
ffiffiffi
2p
p
7
ffiffiffiffi
2;p ffiffiffi
2p
;ffiffiffi
3p
mm. Given that the amplitude of
the transmitted potential is twice that of the reflected potential, find expressions for
the incident, reflected, and transmitted potentials.
In liquids only P-waves are propagated and their displacements can be deduced from the
scalar potential
’ ¼ A0 exp ika n1x1 þ n2x2 þ n3x3 atð Þ ) uP ¼ r’
Then in our case the components of the displacement in mm are
uP1 ¼@’
@x1¼ A0kan1 ¼ A0ka sin i cos az ¼ 18p
uP2 ¼@’
@x2¼ A0kan2 ¼ A0ka sin i sin az ¼ 18p
uP3 ¼@’
@x3¼ A0kan3 ¼ A0ka cos i ¼ 18p
ffiffiffi
6p
and we find az ¼ 45; i ¼ 30; e ¼ 60, A0 ¼ 6 103 m2.
The emergence angle of the refracted wave, e0, can be found from its displacements,
uPtras ¼63p
ffiffiffi
2p
ffiffiffi
7p
ffiffiffi
2p
ffiffiffi
7p ;
ffiffiffi
2p
ffiffiffi
7p ;
ffiffiffi
3p
ffiffiffi
7p
) n03 ¼ sin e0 ¼
ffiffiffi
3p
ffiffiffi
7p ) cos e0 ¼
2ffiffiffi
7p
The P-wave velocity in the medium of the incident wave is
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
s
¼
ffiffiffiffi
K
r
s
¼1ffiffiffi
2p km s1
Using Snell’s law we find the velocity of the medium of the refracted wave,
cos e
cos e0¼
a
a0) a0 ¼
2ffiffiffi
2p
ffiffiffi
7p
From the values of the velocities in the two media we calculate their densities:
a2
a02¼
7
16¼
Kr0
K 0r¼
1
2r0
K 0
a0 ¼ffiffiffiffi
K 0
r0
q
¼2ffiffiffi
2p
ffiffiffi
7p ) K 0 ¼
8
7r0
9
>
>
>
>
>
=
>
>
>
>
>
;
) r0 ¼3
2r ¼
3
2g cm3
239 Reflection and refraction
The reflection Vand transmissionW coefficients are found using their expressions and from
them we get the relation between the amplitude A0 of the incident wave potential and those
of the reflected A and refracted A0 waves, and substituting the value for A0 ¼ 6, we obtain
A0 ¼ 6 ) A ¼ 3 103 m2
From these values we can write the potentials of the incident, reflected, and transmitted
waves:
’inc ¼ 6 exp i6pffiffiffi
2p 1
2ffiffiffi
2p x1 þ
1
2ffiffiffi
2p x2 þ
ffiffiffi
3p
2x3
1ffiffiffi
2p t
’ref ¼ 3 exp i6pffiffiffi
2p 1
2ffiffiffi
2p x1 þ
1
2ffiffiffi
2p x2
ffiffiffi
3p
2x3
1ffiffiffi
2p t
’tras ¼ 6 exp i3p
ffiffiffi
7p
ffiffiffi
2p
ffiffiffi
2p
ffiffiffi
7p x1 þ
ffiffiffi
2p
ffiffiffi
7p x2 þ
ffiffiffi
3p
ffiffiffi
7p x3 2
ffiffiffi
2p
ffiffiffi
7p t
129. Two liquids in contact have speeds of propagation of 4 and 6 km s1. The density
of the first is 2 g cm3 and is less than that of the second. For waves of normal
incidence, the reflected and transmitted energies are equal. A wave of v ¼ 1 s1 and
with a potential of amplitude A0 ¼ 2103 cm2 is incident from the first onto the
second at an angle of 30. Calculate:
(a) The transmitted and reflected energies.
(b) An expression for the transmitted potential.
(a) For normal incidence, the reflection and transmission coefficients in terms of the
refractive index m ¼ a/a0 and the density contrast m ¼ r0/r are given by
Vn ¼m n
mþ n
Wn ¼2
mþ n
If the reflected energy is equal to the transmitted energy, then
V 2n ¼ mnW 2
n )ðm nÞ2
ðmþ nÞ2¼
4mn
ðmþ nÞ2ð129:1Þ
Substituting n ¼ a/a0 ¼ 2/3, from (129.1) we obtain the value of m:
m2 4mþ4
9¼ 0 )
m ¼ 3:9 4
m ¼ 0:1
Trying both values we obtain for r0
m ¼r0
r¼ 0:1 ) r0 ¼ 0:2 g cm3
m ¼ 4 ) r0 ¼ 8 g cm3
240 Seismology
But the problem states that r ¼ 4 < r0, so r0 ¼ 8 g cm3.
For a wave with incidence angle 30 (e ¼ 60) the emergence angle of the transmitted
wave e0 is, according to Snell’s law (Fig. 129), given by
cos e
a¼
cos e0
a0) cos e0 ¼
3
4) sin e0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
19
16
r
¼
ffiffiffi
7p
4
The partition of energy between the reflected and refracted waves is given by
mnsin e0
sin eW 2 þ V 2 ¼ 1
The reflection V and transmission W coefficients are
V ¼m sin e n sin e0
m sin eþ n sin e0¼
4
ffiffiffi
3p
22
3
ffiffiffi
7p
4
4
ffiffiffi
3p
2þ2
3
ffiffiffi
7p
4
¼ 0:8
W ¼2 sin e
m sin eþ n sin e0¼
ffiffiffi
3p
4
ffiffiffi
3p
2þ2
3
ffiffiffi
7p
4
¼ 0:46
The incident, reflected, and transmitted energies per unit time and surface area are (Problem127)
Einc ¼ro4
aA20 sin e ¼ 17:3 erg cm2 s
Eref ¼ro4
aA2 sin e ¼
ro4
aA20V
2 sin e ¼ 11:1 erg cm2 s
Etrans ¼r0o4
a0A02 sin e0 ¼
r0o4
a0A20W
2 sin e0 ¼ 7:5 erg cm2 s
M
M
e
e
i = 30°
r = 2 g cm–3
a = 4 km s–1
a = 6 km s–1
Fig. 129
241 Reflection and refraction
(b) The potential of the transmitted wave is
’tras ¼ A0 exp ik x3 tan e0 þ x1 ctð Þ
where
A0 ¼ A0W ¼ 2 103 0:46 ¼ 920 cm2
k ¼ k 0a0 cos e0 ¼
o
a0cos e0 ¼
1
8km1
c ¼a0
cos e0¼ 8 km s1
’tras ¼ 920 exp i1
8
ffiffiffi
7p
3x3 þ x1 8t
130. An SV wave is incident on the free surface of an elastic medium of Poisson ratio
0.25. If the potential of the wave is (in units of 103m2)
ci ¼5ffiffiffi
2p ;
5ffiffiffi
2p ; 0
exp i1
2ffiffiffi
2p x1 þ
1
2ffiffiffi
2p x2 þ
ffiffiffi
3p
2x3 4t
find the components of the amplitude of the reflected P-wave referred to this set of axes.
From the direction cosines we find the incidence angle i, the emergence angle f, and the
azimuth az of the incident SV wave (Fig. 130):
n3 ¼ cos i ¼ sin f ¼
ffiffiffi
3p
2) i ¼ 30and f ¼ 60
n1 ¼ sin i cos az ¼1
2cos az ¼
1
2ffiffiffi
2p ) az ¼ 45
Bearing in mind that Poisson’s ratio is 0.25, from Snell’s law we find the emergence angle
e of the reflected P-wave:
s ¼ 0:25 ) l ¼ m ) a ¼ffiffiffi
3p
b ¼ 4ffiffiffi
3p
km s1
cos f
b¼
cos e
a) cos e ¼
ffiffiffi
3p
2) e ¼ 30
X3
SV S
ff
e
iP
Fig. 130
242 Seismology
The reflection coefficient for the reflected P-wave gives us the relation between the
amplitude of the potential, B0, of the incident SV wave and A, that of the reflected P wave:
VSP ¼A
B0
¼4 tan f 1þ 3 tan2 eð Þ
4 tan e tan f þ 1þ 3 tan2 eð Þ2¼
ffiffiffi
3p
To find B0 we write the potential of the incident SV wave referred to the (x1, x3) plane of
incidence using the rotation matrix
cos az sin az 0
sin az cos az 0
0 0 1
0
B
@
1
C
A
5ffiffiffi
2p
5ffiffiffi
2p
0
0
B
B
B
B
@
1
C
C
C
C
A
¼
0
5
0
0
B
@
1
C
A
B0 ¼ 5 103m2 ) A ¼ 5ffiffiffi
3p
103m2
Referred to this system of axes the potential of the reflected P-wave is given by
’ ¼ A exp ik x3 tan eþ x1 a tð Þ
k ¼kb
cos f¼
112
¼ 2 km1
’ ¼ 5ffiffiffi
3p
exp i2 x31ffiffiffi
3p þ x1 4
ffiffiffi
3p
t
The amplitudes of the displacements of the reflected P-wave referred to this set of axes are
u1 ¼@’
@x1¼ 10
ffiffiffi
3p
mm
u3 ¼@’
@x3¼ 10 mm
Referred to the original set of axes the horizontal components are
u10 ¼ u1 cos az ¼10
ffiffiffi
3p
ffiffiffi
2p mm
u20 ¼ u1 sin az ¼10
ffiffiffi
3p
ffiffiffi
2p mm
Ray theory. Constant and variable velocity
131. Assume that the Earth’s crust consists of a single layer of thickness H and a
constant speed of propagation of seismic waves of v1 on top of a mantle of velocity of
propagation 20% greater than the crust. Given that a focus on the surface produces a
reflected wave that takes 17.2 s to reach a distance of 99 km, and that this is the
243 Ray theory. Constant and variable velocity
critical distance, calculate the values of H, v1, and v2. Plot the travel-time curve (t, x)
for this specific case with numerical values.
The critical distance xc is the distance at which a ray that is reflected with the critical angle
at the top of the mantle arrives at the surface and is given by the equation (Fig. 131a)
xc ¼ 2H tan ic ¼ 99 km ð131:1Þ
where H is the thickness of the crust. Since we know the relation between the velocities in
the crust and the mantle, we can calculate the critical angle
v2 ¼ 1:2v1 )sin ic
v1¼
1
v2) sin ic ¼
1
1:2) ic ¼ 56:44
If we substitute in Equation (131.1) we obtain the thickness of the crust, H:
99 ¼ 2H tan 56:44 ) H ¼ 32:8 km
The travel time of the critically reflected ray is
t ¼ 2H
v1 cos ic¼ 2
H
v1 cos 56:44¼ 17:2
) v1 ¼ 232:84
17:2 cos 56:44¼ 6:9 km s1 ) v2 ¼ 8:3 km s1
To draw the travel-time curve for different distances of the direct, reflected, and critically
refracted waves we use the equations
t1 ¼x
v1
t2 ¼2
v1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2
4þ H2
r
t3 ¼x
v2þ2H
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 v21
p
v1v2
v2
v1
xc
icic
SF
H
Fig. 131a
244 Seismology
We obtain the following values
The travel-time curves are drawn in Fig. 131b.
x(km) t1 (s) t2 (s) t3 (s)
0 0 9.5 –
30 4.3 10.4 –
60 8.7 12.9 –
90 13.0 16.1 –
99 14.3 17.2 17.2
120 17.4 19.8 19.7
150 21.7 23.7 23.4
0
0
10
20
30
t (S
)
40
50 xc 150
x (km)
200
2
3
1
250 300
ti
Fig. 131b
245 Ray theory. Constant and variable velocity
132. In a seismogram recorded at a regional distance, the S-P time lag is 5.5 s, and the
focus is at a depth x/2, where x is the epicentral distance. The model Earth has a single
layer of Poisson ratio 0.25 and constant S-wave velocityffiffiffi
3p
km s1. Calculate:
(a) The depth of the focus.
(b) The epicentral distance.
(a) For a direct wave from point F to point P (Fig. 132) the difference of the arrival
times of the P- and S-waves (the S-P interval) is
tS-P ¼ 5:5 ¼
FP
bFP
a
The distance FP can be expressed in terms of x as (Fig. 132)
FP ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ h2p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þx
2
2r
¼ x
ffiffiffi
5p
2
The S-P interval is given by
5:5 ¼ x
ffiffiffi
5p
2
a b
ab
Since Poisson’s ratio is 0.25 and knowing the S-wave velocity we obtain
s ¼ 0:25 ) a ¼ bffiffiffi
3p
) 5:5 ¼ x
ffiffiffi
5p
2
ffiffiffi
3p
1ffiffiffi
3p ffiffiffi
3p
x ¼ 21 km
h ¼x
2¼ 10:5 km
133. The Earth consists of a layer of thickness 20 km and seismic wave velocity
6 km s1 on top of a medium of speed of propagation 8 km s1. A seismic focus is
P
x
h
F
Fig. 132
246 Seismology
located at a depth of 10 km. Calculate the difference in travel times between the
reflected and the critical refracted waves observed on the surface at a distance of
150 km from the epicentre.
This problem is similar to Problem 131, but now the focus is at depth h ¼10 km.
The critical distance in this case is given by (Fig. 133)
xc ¼ 2H hð Þ tan ic
sin ic ¼v1
v2) ic ¼ sin1 6
8
¼ 48:6
) xc ¼ 2 20 10ð Þ tan 48:6ð Þ ¼ 34:0 km
Since the distance 150 km is greater than the critical distance there arrive critically
refracted rays. The travel times of the reflected (t2), and critically refracted (t3) rays at that
distance are
t2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 2H hð Þ2q
v1¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1502 þ 2 20 10ð Þ2q
6¼ 25:5 s
t3 ¼x
v2þ
2H hð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 v21
p
v1v2¼
150
8þ
2 20 10ð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
82 62p
8 6¼ 22:1 s
The time difference between the travel times of the two rays is
t3 t2 ¼ 22:06 24:49 ¼ 3:4 s
134. Consider a crust of thicknessH and constant speed of propagation v1 on a mantle
of constant speed of propagation v2. A seismic focus is located at depth H/2, the
critical distance is 51.09 km, the delay time is 4.96 s, and the critical angle is 48.59.
Calculate the values of H, v1 and v2, and the depth of the focus.
For a focus at depth h ¼ H/2, the travel times of the critically refracted (t3) rays and the
critical distance are given by the expressions (Fig. 133).
H
h
x S
v1
v2
ic
icic
i
F
Fig. 133
247 Ray theory. Constant and variable velocity
t3 ¼x
v2þ
2H hð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 v21
p
v1v2
xc ¼ 2H hð Þ tan ic ) 51:09 ¼ 2H H
2
tan 48:59ð Þ
) H ¼ 30 km; h ¼ 15km
Knowing the depth and thickness of the crust, using Snell’s law, the value of the critical
angle, and the delay time ti, we find the velocities v1 and v2:
sin ic
v1¼
1
v2) v1 ¼ 0:75v2
ti ¼2H hð Þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 v21
p
v1v2) 4:96 ¼
2 30 15ð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 0:75v2ð Þ2q
0:75v22
) v2 ¼ 8 km s1
v1 ¼ 6 km s1
135. In a seismogram, the S-P time difference is equal to 5.31 s, and corresponds to a
regional earthquake that occurred at a depth h ¼ 2H, where H is the thickness
of the crust. Given that the crust is formed by a layer of constant P-wave velocity
of 3 km s1, that below it there is a semi-infinite mantle of double that speed of
propagation, and that Poisson’s ratio is 0.25, determine:
(a) An expression for the travel-time of the P- and S-waves.
(b) The epicentral distance for an emerging P-wave with a take-off angle of 30 at the
focus.
(a) The travel time corresponding to the ray given in Fig. 135 is given by
t ¼FA
2vþAS
v
F
F
S
ih
i0i0H
2H
2v
xS
v
A
Fig. 135
248 Seismology
where v ¼ a for P-waves and v ¼ b for S-waves. Using Snell’s law we find the relation
between the incidence angle at the focus ih and at the station i0:
sin ih
2v¼
sin i0
v) sin i0 ¼
1
2sin ih
From Fig. 135 we obtain
cos ih ¼H
FA) FA ¼
H
cos ih
cos i0 ¼H
AS) AS ¼
H
cos i0
From these equations we deduce the expression for the travel time:
t ¼H
2v cos ihþ
H
v cos i0ð135:1Þ
For the epicentral distance we obtain
x ¼ F0Aþ AS0 ¼ H tan ih þ H tan i0 ð135:2Þ
Using Equations (135.1) and (135.2) and putting v ¼ a we obtain the travel time and
epicentral distance for P-waves and putting v ¼ b for S-waves.
(b) For a P-wave with take-off angle at the focus (ih) of 30, we first find the value of
the angle at the station i0,
sin ih
2a¼
sin i0
a) i0 ¼ 14:47
Substituting in (135.1) we obtain
tP ¼
H
2a
ffiffiffi
3p
2
þH
a0:97¼ 1:61
H
a
The travel time of the S-wave with take-off angle jh ¼30 can also be calculated using
(135.1). Since Poisson’s ratio is 0.25, the velocity of the S-wave is
s ¼ 0:25 ) a ¼ffiffiffi
3p
b ) b ¼affiffiffi
3p ¼
ffiffiffi
3p
km s1
The incidence angle at the station, j0, using Snell’s law, is given by
sin jh
2b¼
sin j0
b) j0 ¼ 14:47
and the travel time is
tS ¼
H
2b cos jhþ
H
b cos j0¼ 2:79
H
a
Since we know the S-P time interval we can obtain the value of h:
tS-P ¼ 5:31 ¼ 2:79
H
3 1:61
H
3) h ¼ 13:61 km
249 Ray theory. Constant and variable velocity
The epicentral distance is found using (135.2):
x ¼ 13:61 tan 30 þ 13:61 tan 14:47 ¼ 11:41 km
136. Consider a crust composed of two layers of thickness 12 and 18 km, and constant
P-wave speeds of propagation of 7 and 6 km s1, respectively, on top of a semi-infinite
mantle of constant speed of propagation 8 km s1. There is a seismic focus at a depth
of 6 km below the surface. For a station located at 100 km epicentral distance,
calculate the travel time of the direct, reflected, and critical refracted waves (neglect-
ing waves with more than a single reflection or critical refraction).
The travel times of the direct ray t1 and the ray reflected on the bottom of the first layer
t2 are given by (Fig.136)
t1 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
h2 þ x2p
v1¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
62 þ 1002p
7¼ 14:3 s
t2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2H1 hð Þ2þx2q
v1¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 12 6ð Þ2þ1002q
7¼ 14:5 s
As the velocity of the second layer is less than that of the first layer there is no critical
refraction at that boundary. There is critical refraction at the boundary between the second
layer and the mantle where the velocity is greater. Using Snell’s law, we can calculate the
H2
H1
h
E Sx
F
ic
i1i1
ic
2 (
H1–
h)
F ∗
n2
n3
n1
Fig. 136
250 Seismology
critical angle ic and from this value the incidence angle at the focus i1 for the critically
refracted ray:
sin i1
v1¼
sin ic
v2¼
1
v3) sin ic ¼
6
8) ic ¼ 48:6
sin i1 ¼7
8) i1 ¼ 61:0
The travel time t3 of the critically refracted ray at the bottom of the second layer is given by
t3 ¼FA
v1þAB
v2þBC
v3þCD
v2þDS
v1ð136:1Þ
If the epicentral distance x is 100 km, the different segments of (136.1) are
cos i1 ¼H1 h
FA¼
H1
DS) FA ¼ 12:4 km; DS ¼ 24:8 km
cos ic ¼H2
AB) AB ¼ CD ¼ 27:2 km
BC ¼ x 2H2 tan ic H1 hð Þ tan i1 H1 tan i1 ¼ 26:8 km
Finally, by substitution in (136.1) we obtain,
t3 ¼ 17:7 s
137. Consider a two-layered structure of thickness H and speed of propagation v and
3v on top of a half-space medium of speed of propagation 2v. At a depth 3H below the
surface there is a seismic focus. Write the expressions (as functions of H, v, and ih) for
the travel times of waves that reach the surface without being reflected. Give the
range of values of ih.
In this problem the focus is located at the half-space medium at depth h¼3H under its
boundary. Applying Snell’s law we can find the relation between the velocities, the
incidence angles at the focus and at the bottom of each layer, and the critical angle at the
boundary between the second layer and the half-space (Fig. 137):
sin ih
2v¼
sin i2
3v¼
sin i1
vsin ic
2v¼
1
3v) ic ¼ 41:8
ð137:1Þ
The rays which leave the focus and arrive at the surface at a distance x are only those with
angles less than the critical angle (Fig. 137). The travel time for these rays is
t ¼FA
2vþAB
3vþBS
v¼
H
2v cos ihþ
H
3v cos i2þ
H
v cos i1ð137:2Þ
According to Equation (137.1) we have the relation between the incidence angles:
sin i2 ¼3
2sin ih ) cos i2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 sin2 ih
q
¼1
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 9 sin2 ih
q
sin i1 ¼1
2sin ih ) cos i1 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 sin2 ih
q
¼1
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 sin2 ih
q
251 Ray theory. Constant and variable velocity
Substituting in (137.2) we write the travel time as function of the take-off angle ih:
t ¼H
v
1
2 cos ihþ
2
3ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 9 sin2 ihp þ
2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 sin2 ihp
!
We find a similar expression for the epicentral distance x:
x ¼ H tan ih þ H tan i2 þ H tan i1
x ¼ H tan ih þsin ihffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 sin2 ihp þ
3 sin ihffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 9 sin2 ihp
!
The range of values of the take-off angle for rays which arrive at the surface is
0< ih <42.
138. A semi-infinite medium consists of two media of velocities v and 3v separated by a
vertical surface. In the first medium there is a focus of seismic waves at a depth a
below the free surface and at the same distance a from the surface separating the two
media. Write the expressions for the direct, reflected, and transmitted waves arriving
at the free surface, and plot the travel time curve (t, x) in units of a/v and a (neglecting
waves with more than a single reflection).
In this situation we have the following rays arriving at the surface: direct in the first
medium, reflected at the boundary, and critically refracted and refracted to the
second medium. We consider two cases for rays arriving at distances 0 < x < a and
distances x > a.
E
B
A
x S
v
3v
F
H
H
H
2v
i2
i1
i0
ic
ih
Fig. 137
252 Seismology
(a) For 0 < x <a, the travel time t1 of the direct wave is (Fig. 138a)
t1 ¼FS
v¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ a2p
v
The travel time t2 of the reflected ray is
t2 ¼FPþ PS
v¼
APþ PS
v¼
AS
v¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2a xð Þ2þa2q
v
The reflected rays exist also for negative distances, but we will not consider them.
The travel time t3 of the critically refracted ray (Fig. 138b) is
t3 ¼FA
vþAB
3vþAS
vð138:1Þ
E
AF
S
P
x
i
ia
a a
i
v 3v
Fig. 138a
E
B
A
3v
F
a
a
v
S
ic
ic
ic
x
Fig. 138b
253 Ray theory. Constant and variable velocity
Using Snell’s law the critical angle is given by
sin ic
v¼
1
3v) ic ¼ 19:47
and
cos ic ¼a
FA¼
a x
AS
The distance AB is given by
AB ¼ a a tan ic a xð Þ tan ic ¼ a tan ic x 2að Þ
By substitution in (138.1) we obtain
t3 ¼ 2:22a
v0:94
vx
The critical distance (distance of the ray reflected with the critical angle) is given by
(Fig. 138c)
xc ¼ a SB ð138:2Þ
From Fig. 138c we obtain
tan ic ¼BP
SB¼
a a tan ic
SB) SB ¼ 1:83a
and substituting in (138.2)
xc ¼ 0:83a
Critically refracted rays exists for distances 0.83a < x < a. Here we consider only those
for 0 < x < a.
ES
F
xc
ic
a
a
v 3v
ic
ic
ic
Fig. 138c
254 Seismology
(b) For distances x > a, we have the rays refracted at the boundary between the two
media when the incidence angle is less than the critical angle (Fig. 138d), that is,
i <19.47:
a
a
e
E
F
A
SX
∗
i
v 3v
Fig. 138d
0.00.0
0.0
1.0
t (a
/v)
1.5
3
1
2
2.0
0.2 0.4 0.6
x (a)
0.8 1.0
Fig. 138e
255 Ray theory. Constant and variable velocity
t4 ¼FA
vþAS
3v
FA ¼a
cos i
AS ¼x a
cos e
Using Snell`law
sin i
v¼
sin e
3v) sin e ¼ 3 sin i
The travel-time is given by
t4 ¼a
v cos iþ
x a
3vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 9 sin2 ip
The travel-time curves for direct (1), reflected (2), critically refracted (3) and transmitted
(4) waves are given in Fig. 138e.
139. Given the structure in the diagram, calculate the arrival times of the direct and
(non-reflected) transmitted waves for x 0, where x ¼ 0 is a point on the free surface
in the vertical above the focus.
At x ¼ 0 )ih ¼ 0, the travel-time of the vertical ray is (Fig. 139a)
t ¼a
2vþa
v¼
3a
2v
For rays arriving at x > 0 and leaving the focus with take-off angles 0 < ih < 45, the
travel-times are given by
t ¼FA
2vþAS
v¼
a
2v cos ihþ
a
v cos r
cos ih ¼a
FA
cos r ¼a
AS
ð139:1Þ
2a
Focus
a
a2v
ν
Fig. 139
256 Seismology
Applying Snell’s law,
sin ih
2v¼
sin r
v) sin r ¼
1
2sin ih ð139:2Þ
Substituting in (139.1):
t ¼a
v
1
2 cos ihþ
2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 sin2 ihp
!
ð139:3Þ
The relation between the epicentral distance x and the incidence angle ih is
x ¼ F0Aþ A0S
F0A ¼ a tan ih
A0S ¼ a tan r
x ¼ a tan iþ tan rð Þ ¼ a tan iþsin ihffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 sin2 ip
h
!
From Fig. 139a we deduce:
tan ih ¼F0A
a
tan r ¼x F0A
a¼
x
a tan ih
Using Equation (139.2),
sin r
cos r¼
sin ih2
cos r¼
x
a tan ih )
1
cos r¼
2x
asin ih
2
cos ih
A
r
2a
v
a
a
F
F
X
ih
2v
A SE
Fig. 139a
257 Ray theory. Constant and variable velocity
By substitution in (139.3),
t ¼2x
v sin ih
3a
2v cos ih
For ih ¼ 45 the epicentral distance is
x ¼ a 1þ1ffiffiffi
7p
This is the limit of the epicentral distance at which these rays arrive. The corresponding
time limit is
t ¼a
v
1ffiffiffi
2p þ
4ffiffiffiffiffi
14p
For angles ih > 45 (Fig. 139b), the travel time and epicentral distance, as a function of the
take-off angle ih, are
t ¼FA
2vþAS
v¼
a
2v sin ihþ
x a
v sin r¼
a
2v sin ihþ
x a
vsin ih
2
x ¼ aþ A0S
tan r ¼A0S
2a F0A
F0A ¼a
tan ih
x ¼ aþ 2aa
tan ih
sin ihffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 sin2 ihp
E A
x
S
n
A2n
F
a
a
F
ih
ih
r
2a
Fig. 139b
258 Seismology
Using
cos ih
2v¼
cos r
v) cos r ¼
1
2cos ih
sin r ¼1
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 cos2 ihp
tan r ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 cos2 ihp
cos ih
we obtain for x and t,
x ¼ aþ 2aa
tan ih
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 cos2 ihp
cos ih
t ¼a
2v sin ihþ
2ðx aÞ
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4 cos2 ihp
For i ¼ 90, as expected the ray doesn’t arrive at the free surface.
140. For the structure in Fig. 140a, write the equations of the travel times of the
direct, reflected, and transmitted waves (neglecting waves with more than a single
reflection) as a function of the epicentral distance. Determine the times of intersection,
and the minimum and maximum distances in each case in terms of a/v and v. Plot the
travel-time curves.
The travel-time of the direct wave for distance 0 < x < 1 is (Fig. 140a)
t ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4x2 þ a2p
2v
For the ray reflected on the horizontal surface at depth a the travel time is
t ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ3a
2
2s
v¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4x2 þ 9a2p
2v
ν
2ν
a
a
xS
a /2
F∗
Fig. 140a
259 Ray theory. Constant and variable velocity
The range of distances for this ray is
xmin ¼ 0
xmax )aa
2
¼xmax
3a
2
) xmax ¼ 3a
For the reflected ray on the surface at depth 2a the travel time is (Fig. 140b)
t ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ7a
2
2s
v¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4x2 þ 49a2p
2v
The range of distances is
3a
3
2a¼
xmin
7
2a
) xmin ¼ 7a ) tmin ¼7ffiffiffi
5p
2
a
v
xmax ¼ 1
The critically refracted ray on the surface at depth a, using the general expression, is given by
t ¼x
v2þ
2H hð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 v21
p
v1v2¼
x
2vþa
v
3ffiffiffi
3p
4
The minimum distance for this ray corresponds to the critical distance:
sin ic
v¼
1
2v) ic ¼ 30
xc ¼a
2tan ic þ a tan ic ¼
ffiffiffi
3p
2a
and the maximum distance is
xmax ¼ aþaffiffiffi
3p ¼
a 3þffiffiffi
3p
3
2v
v
a
a
S
Fa/2
x
Fig. 140b
260 Seismology
On the surface at depth 2a there is no critically refracted ray, since the minimum take-off
angle ih at that surface is
tan ih ¼aa
2
¼ 2 ) ih ¼ 63:4
greater than the critical angle 30.
The travel-time curves are drawn after rewriting the equations in units of a/v and a, and
are represented in Fig. 140c
1. Direct ray:tv
a
2
x
a
2
¼ 14; 0
x
a< 1 (a hyperbola)
2. Reflected ray on the surface at depth a:tv
a
2
x
a
2
¼9
4; 0
x
a 3 (a hyperbola)
3. Reflected ray on the surface at depth 2a:tv
a
2
x
a
2
¼49
4; 7
x
a 1 (a hyperbola)
4. Critically refracted ray on surface at depth a:tv
a¼
1
2
x
aþ3ffiffiffi
3p
4; 0:87
x
a 1:58 (due
to the short range of distances this is not noticeable in the figure)
0
1
3
2
42
4
6
t (a
/v)
8
10
2 4 6
X (a)
8 10 12
Fig. 140c
261 Ray theory. Constant and variable velocity
141. For the structure in the diagram, assume a seismic focus at the surface, and
calculate the travel time of the direct, reflected, and critical refracted waves for
epicentral distances between 0 and a. Calculate the critical distance, and the expres-
sion for the transmitted wave.
Since the focus is at the free surface, the travel time of the direct ray is simply given by
(Fig. 141a)
t ¼x
v0 < x < a
a
a3ν
45°
ν
F
*
Fig. 141
a
a
a√
45°
45°
2
P
ν3ν
S
x
F
F
Fig. 141a
262 Seismology
The travel time of the reflected wave is (Fig. 141a)
t ¼FP
vþPS
v¼
F0P
vþPS
v¼
1
v
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ 2a2 2axp
The critical angle is given by
sin ic
v¼
1
3v) sin ic ¼
1
3) cos ic ¼
2ffiffiffi
2p
3) ic ¼ 19:47
The travel-time of the critically refracted ray is (Fig. 141b)
t ¼FA
vþAB
3vþBS
v
FA ¼
affiffiffi
2p
cos ic¼
3
4a
The distance BS can be found using the sine law in triangle SBD:
sin 90 icð Þ
a x¼
sin 45
BS) BS ¼ a xð Þ
3
4
and the distance AB (calling d ¼ BD) is
AB ¼ affiffiffi
2p
a cos 45 a
ffiffiffi
2p
2tan ic d ¼ a
ffiffiffi
2p
2 a
ffiffiffi
2p
2tan ic d
a a
a
45°
45°
45°
ν
A
2
2
B
DSF
x
ic
ic
√
*β
Fig. 141b
263 Ray theory. Constant and variable velocity
The distance d, using the sine law, is given by
sin 45þ icð Þ
d¼
sin 90 icð Þ
a x) d ¼ a xð Þ
sin 45þ icð Þ
sin 90 icð Þ
AB ¼ a
ffiffiffi
2p
21 tan icð Þ a xð Þ
sin 45þ icð Þ
sin 90 icð Þ¼ 0:96x 0:50a
The travel time is given by
t ¼ 0:43x
vþ 1:33
a
v
Finally we determine the critical distance xc from the triangle AS0D (Fig. 141c):
bþ 90 icð Þ þ 45 ¼ 180 ) b ¼ 64:47
sin 90 icð Þ
a xc¼
sin baffiffiffi
2p
affiffiffi
2p tan ic
) xc ¼ 0:52a
142. A medium consists of a flat crust of thicknessH and constant speed of propagation
v1 on a semi-infinite mantle of constant speed of propagation v2. For a focus at the
surface, at a distance x the direct wave arrives at a time t1 ¼ x/a, the critical distance is
xc ¼2affiffiffi
3p , and the direct and critical refracted waves intersect at the distance x ¼ 2a
ffiffiffi
3p
.
(a) Calculate the crust’s thickness, its speed of propagation, the mantle’s speed of
propagation, and the critical angle.
a
a
45°
45°
45°
ν
a2
2
ic
ic
β
A
DF
xc
s
Fig. 141c
264 Seismology
(b) Assume now that this is a layer which dips downwards at 45 with the parameters
of the model being those determined in the previous part. Calculate the travel
times of the reflected and critical refracted waves at x ¼ a, 3a, and 5a.
(a) We determine the velocity of the crust from the travel time of the direct ray:
t1 ¼x
v1¼
x
a) v1 ¼ a
The critical distance is given by
xc ¼ 2H tan ic ¼ 2Hv1ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 v21
p ¼ 2Haffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 a2p ¼ 2
affiffiffi
3p ð142:1Þ
Equating the travel times of the direct and critically refracted ray for the value of the
distancex ¼ 2affiffiffi
3p
we obtain
t1 ¼ t3 )x
v1¼
x
v2þ2H
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 v21
p
v1v2ð142:2Þ
From Equations (142.1) and (142.2) we obtain the values of H and v2:
H ¼ a
v2 ¼ 2a
The critical angle may be estimated from
x0c ¼ 2H tan ic )2affiffiffi
3p ¼ 2a tan ic ) ic ¼ 30
(b) We now consider the case of a dipping layer with dip angle y ¼ 45. The critical
distance is now given by the equation (Fig. 142)
S
xc
xc sinθ
H
F
ic
θ
θ
*
Fig. 142
265 Ray theory. Constant and variable velocity
xc cos y ¼ H tan ic þ H þ xc sin yð Þ tan ic ð142:3Þ
so
xc ¼2H tan ic
cos y sin y tan ic
The critical distance along the horizontal free surface xc is found by substituting H, y, and icin (142.3)
xc ¼ 3:86a
The travel times of the reflected and critically refracted rays for a dipping layer are given by
the equations
t2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4H2 þ x2 þ 4Hx sin yp
v1
t3 ¼x cos y
v2þx sin yþ 2H
v1v2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v22 v21
q
By substitution of the values of H, v1, v2, and y, we obtain
t2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4a2 þ x2 þ 2ffiffiffi
2p
axp
a
t3 ¼xffiffiffi
2p
4aþx
ffiffiffi
2p
2þ 2a
2a
ffiffiffi
3p
¼xffiffiffi
2p
4a1þ
ffiffiffi
3p
þffiffiffi
3p
For the required values of x, we obtain the following values of the travel time in units
of a/v:
Since the critical distance is 3.86a, it only exists for x ¼5a.
143. In a flat medium, the velocity increases linearly with depth according to the
expression v ¼ v0 þ kz, where v0 is the velocity at the surface, k is a constant, and z is
the depth. For a focus at depth h, calculate an expression for the take-off angle at the
focus ih in terms of the epicentral distance x, and v0, h, and k.
At the maximum depth of penetration of the ray r the incidence angle of the ray with the
vertical is equal to 90 (Fig. 143). Using Snell’s law we can relate the angles at the focus ih,
at the bottom of the ray 90, and i0, the incidence angle at the station on the surface:
sin i0
v0¼
1
v0 þ kr¼
sin ih
v0 þ kh) sin ih ¼
v0 þ kh
v0 þ krð143:1Þ
x t2 (a/v) t3 (a/v)
a 2.80 –
3a 4.64 –
5a 6.57 6.56
266 Seismology
The problem is solved if we can express r as a function of x, v0, h, and k. The epicentral
distance x is the sum of a and b (Fig. 143):
b ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R2 v0
k
2r
But we know that for a distribution of velocities which increases linearly with depth the
rays are circular and their radius R is
R ¼v0
kþ r ) b ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v0
kþ r
2
v0
k
2r
and
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v0
kþ r
2
v0
kþ h
2r
Therefore,
b ¼ x a )
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v0
kþ r
2
v0
k
2r
¼ x
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v0
kþ r
2
v0
kþ h
2r
Solving for r we obtain
r ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 h2 2v0
kh
2x
0
@
1
A
2
þv0
kþ h
2
v
u
u
u
t v0
k
By substitution of this expression for r in (143.1) we obtain the required expression
for ih.
F
E
R
ba
x
S
hr
V0
i0
ih
ν0/ k
∗
Fig. 143
267 Ray theory. Constant and variable velocity
144. Consider a semi-infinite medium in which the velocity increases linearly with depth
according to the expression v ¼ 4 þ 0.1 z. There is a seismic focus at a depth of 10 km.
Calculate the epicentral distance reached by a wave leaving the focus at an angle of 30.
The velocity at the focus is found directly by putting in the equation for the distribution of
velocity, z ¼ h:
vh ¼ 4þ 0:1 10 ¼ 5 km s1
According to Snell’s law we find the velocity at the point of greatest depth penetration
(i ¼ 90) of the ray (Fig. 144):
sin ih
vh¼
1
vm) vm ¼
5
sin 30¼ 10 km s1
From this value we find the depth to that point:
vm ¼ 10 ¼ 4þ 0:1 r ) r ¼ 60 km
Knowing that the rays are circular of radius R,
R ¼ r þv0
k¼ 60þ 40 ¼ 100 km
As in the previous problem, the epicentral distance (from point E to S) is (Fig. 144):
x ¼ aþ b
where
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R2 v0
kþ h
2r
¼ 86:60 km
b ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R2 v0
k
2r
¼ 91:65 km
so
x ¼ 178:25 km
x
ν0/k
a b
R
r
ih
F
E
h
V0
S
i0
Fig. 144
268 Seismology
145. A flat medium consists of a layer of thickness H and constant speed of propaga-
tion v on top of a medium of variable speed of propagation v ¼ v0 þ k(z H) where z
is the depth and k is a constant. If there is a focus at the surface:
(a) Write expressions for the epicentral distance x and the travel time t as functions of
the angle of incidence i0 at the surface.
(b) If H ¼ 10 km, k ¼ 0.1 s1, and v0 ¼ 6 km s1, calculate the angle of incidence of a
wave that reaches an epicentral distance of 140 km.
(a) As we saw in Problem 143, for a distribution with a linear increase of velocity with
depth, now in themediumunder the layer, v¼ v0þ k(zH), the rays are circular with
radius R ¼v0
kþ r where r is the maximum depth of penetration of the ray (Fig. 145).
The travel-time of the ray that crosses the layer and penetrates the medium is given by
t ¼ 2FP
v0þ2
ksinh1 kx
0
2v0ð145:1Þ
In the layer of constant velocity the path is a straight line and in the medium it is circular.
The epicentral distance x (from F to S) is (Fig. 145)
x ¼ x0 þ 2H tan i0
The length of the straight ray in the layer is
FP ¼H
cos i0
Substituting in (145.1):
t ¼2H
v0 cos i0þ2
ksinh1 kx0
2v0
ð145:2Þ
Since the layer has constant velocity the angle i0 is the same at the focus as at the bottom of
the layer at the boundary with the medium. According to Snell’s law
x
x
i0
i0
r
Q
S
P
F
H
i0ν0
ν0/k
Fig. 145
269 Ray theory. Constant and variable velocity
sin i0
v0¼
1
v0 þ kr) vm ¼ v0 þ kr ¼
v0
sin i0
where r is the maximum depth reached by the ray in the medium and vm the velocity at that
depth. According to Fig. 145,
x0
2
2
þv0
k
2
¼v0
kþ r
2
¼v2mk2
¼v20
k2 sin2 i0) x0 ¼ 2
v0
kcot i0 ð145:3Þ
The epicentral distance x is given by
x ¼ x0 þ 2H tan i0 ¼2v0
kcot i0 þ 2H tan i0 ð145:4Þ
Substituting in (145.2) the expression for x 0 in terms of i0 (143.3) we obtain
t ¼2H
v0 cos i0þ2
ksinh1 cot i0ð Þ
(b) By substituting the given values in (145.4), we obtain
140 ¼2 6
0:01cot i0 þ 2 10 tan i0 )
i0 ¼ 45
i0 ¼ 80:5
146. Beneath a layer of thickness H of velocity distribution v ¼ v0 þ kz there is a semi-
infinite medium of speed of propagation v1 ¼ 2(v0 þ kH).
(a) Determine expressions (as functions of the above parameters) for the critical
distance, the time of intersection of the reflected wave, and the maximum distance
of the direct wave.
(b) For H ¼ 10 km, v0 ¼ 1 km s1, and k ¼ 0.1 s1, calculate these parameters and
plot the travel-time curves.
(a) In a layer of thickness H with variable velocity the epicentral distance x for a
reflected ray is given by
x ¼ 2
ðH
0
tan i dz
Using the ray parameter p ¼ sin i/v we can write
sin i ¼ vp ) tan i ¼pvffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 p2v2p
The epicentral distance for a ray reaching the bottom the layer is given by
x ¼ 2
ðH
0
vpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 p2v2p dz ¼
2
k
ðH
0
kpðv0 þ kzÞffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 p2ðv0 þ kzÞ2q dz ð146:1Þ
For a ray incident at the bottom of the layer at the critical angle, we have
p ¼sin ic
v0 þ kH¼
1
2ðv0 þ kHÞ
270 Seismology
Substituting this expression in (146.1) and evaluating the integral, making the change of
variable u ¼ v0 þ kz, we obtain the critical distance
xc ¼ 4 v0 þ kHð Þ
k
ffiffiffi
3p
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1v20
4 v0 þ kHð Þ2
s" #
ð146:2Þ
The intercept time for x ¼ 0, corresponding to the time of the reflected vertical ray (p ¼ 0),
is given by
ti ¼ 2
ðH
0
dz
v0 þ kz¼
2
klnv0 þ kH
v0ð146:3Þ
The maximum distance xmax corresponds to the last ray propagated inside the layer
without penetrating into the medium and has a circular path of radius R ¼ H þv0
k(Fig. 146a):
xmax
2
2
þv0
k
2
¼ H þv0
k
2
) xmax ¼ 2H
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ2v0
kH
r
ð146:4Þ
(b) For the particular case with the values, H ¼10 km, v0 ¼ 1 km s1, and k ¼ 0.1 s1,
the velocity at the bottom of the layer H is
vH ¼ v0 þ kH ¼ 1þ 10 0:1 ¼ 2 km s1
The velocity of the medium is
v1 ¼ 2 v0 þ kHð Þ ¼ 2 1þ 10 0:1ð Þ ¼ 4 km s1
The critical distance, using Equation (146.2), is
xc ¼ 4 2
0:1
ffiffiffi
3p
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
11
4 2ð Þ2
s" #
¼ 20ffiffiffi
3p
2ffiffiffi
5p
¼ 8:2 km
S
xmax
xc
ic
F
H
n0/k
n1
Fig. 146a
271 Ray theory. Constant and variable velocity
The intercept time (146.3) of the reflected ray is
ti ¼2
klnv0 þ kH
v0¼
2
0:1ln1þ 0:1 10
1¼ 13:9 s
and the maximum distance for the ray in the layer (146.4) is
xmax ¼ 2 10
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ2 1
0:1 10
r
¼ 20ffiffiffi
3p
¼ 34:6 km
The travel-time curve for rays inside the layer is calculated using the expression
t ¼2
ksinh1 kx
2v0
¼ 20 sinh1ð0:05xÞ
and is represented in Fig. 146b.
30
25
20
t (s
)
15
10
5
0 10 20
x (km)
30 40
Fig. 146b
272 Seismology
147. A medium has a distribution of velocity with depth of the form v ¼ v0 eaz, with
0 < a <1. Write as functions of the epicentral distance x the expressions for the ray
parameter, travel-time, and maximum depth reached.
If r is the maximum depth reached for a ray with ray parameter p (Fig. 147), the epicentral
distance x is given by
x ¼
ð
r
0
pvdzffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 p2v2p ¼
ð
r
0
pv0eazdz
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 p2v20e2az
p
¼2
asin1ðpv0e
azÞ
r
0
¼2
a½sin1ðpv0e
arÞ sin1ðpv0Þ
and, as p ¼1
v0ear, we have
x ¼2
a
p
2 sin1ðpv0Þ
h i
From this expression we obtain
p ¼1
v0cos
ax
2
The travel-time is given by
t ¼
ðx
0
pdx ¼1
v0
ðx
0
cosax
2dx ¼
2
v0asin
ax
2
To find the maximum depth of penetration r of a ray arriving at distance x, we write
p ¼1
v0ear¼
1
v0cos
ax
2) r ¼
1
aln cos
ax
2
148. In a semi-infinite medium of speed of propagation v ¼ 6 expz
2
, the P-wave
emerges with an angle of incidence of 30. Calculate the difference in arrival times at
a given station of the P-wave and the PP-wave (the wave reflected once at the free
surface). At what angle of incidence does the PP-wave emerge?
For a velocity distribution increasing with depth of the type v ¼ v0ea z (in our case with
v0 ¼ 6, a¼ 1/2) rays follow a curved path. For a focus on the free surface the ray parameter
p and the travel times t are given by (Problem 147)
p ¼1
v0cos
ax
2
t ¼2
av0sin
ax
2
ð148:1Þ
x
r
SF
Fig. 147
273 Ray theory. Constant and variable velocity
For a ray with incidence angle at the surface i0 ¼ 30, the ray parameter of the direct
P-wave is given by
p ¼sin i0
v0¼
1
2 6¼
1
12
Substituting this value in (148.1) we obtain, for the epicentral distance x,
1
12¼
1
6cos
x
2 2
) x ¼4p
3km
The corresponding travel time is
tP ¼
2
1
26
sin1
2
4p
3
1
2
¼ 0:58 s
The travel time of the reflected PP-wave (Fig. 148) is double that of the direct P-wave
arriving at the distance x/2:
tPP ¼ 2
2
1
26
sin1
2
4p
6
1
2
¼ 0:67 s:
The difference between the two times is
tPP t
P ¼ 0:67 0:58 ¼ 0:09 s:
To calculate the incidence angle of the PP-wave, we determine first the ray parameter
p corresponding to the distance x/2:
x
2¼
4p
3 2¼
4p
6
so
p ¼1
6cos
1
2
4p
6
1
2
¼
ffiffiffi
3p
12
x
SF
Fig. 148
274 Seismology
From the value of p, using Snell’s law, we obtain i0:
p ¼sin i0
v0) sin i0 ¼
ffiffiffi
3p
126 ) i0 ¼ 60
149. A layer of thickness H has a velocity distribution v ¼ v0 exp(az) where a <1.
Beneath it there is a semi-infinite medium of speed of propagation v1 ¼ 2v0 exp(aH).
Determine in terms of v0, v1, a, and H:
(a) The distance and critical angle.
(b) The time of intersection of the reflected wave.
(c) The maximum distance of the direct wave.
(d) Calculate the values of these parameters if v ¼ 1 km s1, H ¼ 10 km, and
a ¼ 0.1 km1.
(a) The distance for a ray reaching a depth H is given by (Fig. 149)
x ¼ 2
ðH
0
tan idz ð149:1Þ
and using the definition of the ray parameter p,
p ¼sin i
v) sin i ¼ vp
cos i ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 v2p2p
tan i ¼vpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 v2p2p
Substituting in (149.1) we obtain
x ¼ 2
ðH
0
vpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 v2p2p dz ¼ 2
ðH
0
pv0eaz
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 p2v20e2az
p dz ð149:2Þ
FS
ic
xc
xmax
H
n1
Fig. 149
275 Ray theory. Constant and variable velocity
Introducing the change of variable
u ¼ v0eaz
du ¼ v0eazadz
we find
x ¼2
asin1pv0e
aH sin1pv0
For the critical angle ic at the bottom of the layer we have
p ¼sin i0
v0¼
sin ic
vH¼
1
v1) ic ¼ sin1 vH
v1¼ sin1 v0e
aH
v1
By substitution in (149.2) we find, for the critical distance xc,
xc ¼2
asin1 v0e
aH
v1 sin1 v0
v1
¼2
asin1 vH
v1
sin1 v0
v1
¼2
aic i0ð Þ
(b) The intercept timeof the reflected ray corresponding to the vertical ray (x¼ 0 andp¼ 0) is
ti ¼ 2
ðH
0
dz
v¼ 2
ðH
0
1
v0eazdz ¼
2
av01 eaH
(c) The maximum distance of a ray contained in the layer is given by (149.2)
xmax ¼ 2
ðH
0
vpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 p2v2p dz ¼
2
asin1pv0e
aH sin1pv0
ð149:3Þ
At the point of greatest depth penetration the incidence angle is 90 and, according to
Snell’s law,
p ¼sin 90
vH¼
eaH
v0¼
1
vH¼ p
By substitution in (149.3),
xmax ¼2
a
p
2 sin1eaH
h i
¼2
acos1eaH
(d) Substituting the data of the problem we obtain
ic ¼ sin1 1 e0:110
5:62¼ 30
ti ¼2
0:1 11 e0:110
¼ 12:6 s
xmax ¼2
0:1cos1e0:110 ¼ 23:9 km
xc ¼2
0:1sin1 1
2 sin1 1
5:62
¼ 6:8 km
276 Seismology
Ray theory. Spherical media
150. Assume that the Earth consists of two concentric regions of constant velocity: the
core of radius R/2 and the mantle. The speed of propagation in the core is twice that of
the mantle. Calculate:
(a) The maximum angular distance of the direct ray in the mantle.
(b) The critical angular distance of the refracted ray in the core.
(c) Plot the paths of the waves that propagate through the Earth’s interior, and the
travel-time curves of these waves in units of R/v, where v is the speed of propaga-
tion in the mantle.
(a) The travel time of the direct ray in the mantle in terms of the angular distance is
given by
t1 ¼ 2R
vsin
2
where 0 D Dmax and Dmax is the maximum distance for a ray contained in the mantle.
According to Fig. 150a the last ray which propagates in the mantle without entering the
core corresponds to angular distance Dmax which in our case is
cosmax
2¼
R
2R) max ¼ 120
(b) The critical angle for a ray incident at the core is
sin ic
v¼
1
2v) ic ¼ 30
To calculate the critical distance Dc we consider the relation (Fig. 150b)
Δmax
R/2
F
FR
R
S∗
∗
n
2n
Δ/2
Fig. 150a
277 Ray theory. Spherical media
yþ bþ a ¼ 180
sin a
R
2
¼sin b
R
ic þ b ¼ 180 ) b ¼ 150 ) a ¼ 14:5 ) y ¼ 15:5
c ¼ 2y ¼ 31
(c) The travel time of a ray reflected at the mantle–core boundary is (Fig. 150c)
t2 ¼FP
vþPS
v¼ 2
FP
vð150:1Þ
SF
R
P
ic
n
2n
ΔcR/2
β
α
q
Fig. 150b
F
Δ/22ν
nP
S
R
R/2
Fig. 150c
278 Seismology
According to the cosine law,
FP2¼
R
2
2
þ R2 2R
2R cos
2
Substituting in (150.1):
t2 ¼R
v
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5 4 cos
2
r
The travel times for the minimum and maximum angular distances are
min ¼ 0 ) t ¼R
v
max ¼ 120 ) t ¼ffiffiffi
3p R
v
The travel time for a ray which enters the core, that is, with i1 < ic, can be determined
according to Fig. 150d using Snell’s law:
sin i1
v¼
sin i2
2v
sin i2 ¼ 2 sin i1
ð150:2Þ
Adding the times of the paths through the mantle and the core:
t3 ¼FP
vþPQ
2vþQS
v¼
2FP
vþPQ
2v
FP2¼ R2 þ
R2
4 2R
R
2cos
a
2
PQ ¼ 2R
2sin
a
2
Because a ¼ 180 – 2i2, we obtain
t3 ¼2R
v
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5
4 sin
2þ i2
s
þR
2vcos i2 ð150:3Þ
for values of the incidence angle 0 < i < ic, corresponding to distances Dc < D < 180.
The relation between the incidence angle i1 and angular distance D is given by
i2i1
i1
i2
nF
R/2 R
S
P
Q
α
Δ
2n
Fig. 150d
279 Ray theory. Spherical media
sin i1
R¼
sin 90 þ i1 i2
2
R=2
sin i1 ¼ 2 cos i1 i2
2
) ¼ 2 i1 i2 cos1 sin i1
2
ð150:4Þ
Using Equations (150.2), (150.3), and (150.4) we can calculate the travel times of the direct
ray in the mantle, the reflected ray, and the transmitted ray through the core. Some values
for the transmitted rays in the core are given in the table.
The travel-time curves for direct rays (1), reflected rays (2), and rays refracted in the core
(3) are shown in Fig. 150e.
Δº
00.0
0.5
t (R
/v)
1.0
1.5
2.0
2
3
1
50 100 150
Fig. 150e
i1 () i2 (
) D () t3 (R/v)
0 0 180.0 1.50
10 20.3 169.9 1.53
20 43.2 153.4 1.60
30 90 89.0 1.46
280 Seismology
151. Assume that the Earth consists of two concentric regions of constant velocity: the
core of radius R/2 and the mantle. The speed of propagation in the core is half that
of the mantle. Plot the travel-time curves of the waves that propagate in the interior of
the Earth in units of R/v where v is the speed of propagation in the mantle.
This problem is similar to Problem 150, but now the velocity of the core is less than that of
the mantle. In the mantle we have direct and reflected rays. As in Problem 150 the
maximum angular distance for the direct wave is 120. The travel times for the direct
(t1) and reflected (t2) rays are
t1 ¼ 2R
vsin
2ð151:1Þ
t2 ¼R
v
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5 4 cos
2
r
ð151:2Þ
Since the velocity of the core is less than that of the mantle there is no critical angle. All
rays incident at the core are refracted into it. According to Snell’s law the refracted angle i2is less than the incident angle i1 (Fig. 151a):
sin i1
v¼
sin i2v
2
sin i2 ¼1
2sin i1
i2
i1
i2
i1
F ∗
n n/2
P
S
R/2
R
QΔ
a
Fig. 151a
281 Ray theory. Spherical media
The travel-time for a ray crossing the mantle and the core is (Fig. 151a)
t3 ¼FP
vþPQv
2
þQS
v¼
2FP
vþ2PQ
v
where
FP2¼ R2 þ
R2
4 2R
R
2cos a ¼ R2 þ
R2
4 R2 cos 90þ i2
2
FP2¼
5R2
4þ R2 sin i2
2
PQ ¼ 2R
2cos i2 ¼ Rcos i2
so
t3 ¼2R
v
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5
4þ sin i2
2
s
þ2R
vcos i2 ð151:3Þ
Δ (º)
t (R
/n
)
2
1
0
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
25 50 75 100 125 150 175
3
Fig. 151b
282 Seismology
The relation between the incidence angle at the mantle–core boundary, i1, and the angular
distance, D, of a ray which crosses the core is
sinð180 i1Þ
R¼
sin i1 i2 þ
2 90
R=2
sin i1 ¼ 2 cos i1 i2 þ
2
) ¼ 2 cos1 sin i1
2
i1 þ sin1 1
2sin i1
ð151:4Þ
The range of distance for this ray is 120 < D <180.
From Equations (151.1), (151.2), and (151.3) we can calculate the values for the travel-
times of the direct, reflected, and refracted rays. Some values for t3 are given in the table.
The travel-time curves for rays that are direct (1), reflected (2), and refracted in the core (3)
are shown in Fig. 151b.
152. Consider a spherical Earth of radius R formed by two hemispherical media of
constant velocities of propagation v and 2v. For a focus on the surface of the
hemisphere of velocity v at the point of intersection of the diameter perpendicular
to the plane that separates the two media, calculate the travel times and travel-time
curves of the direct, reflected, and critical refracted waves at the surface of separation
of the two media, in units of R/v. Calculate the expression for the travel time of waves
that propagate through the medium of speed of propagation 2v.
The travel time for angular distances D 90 are given by (Fig. 152a)
t1 ¼FS
v¼ 2
R
vsin
2ð152:1Þ
The travel time of the ray reflected at the plane boundary between the two hemispheres is
(Fig. 152a)
t2 ¼FP
vþPS
v¼
F0S
v
According to Fig. 152a (triangle OSF0) the relation between the angles a and D is
ð180 Þ þ 2a ¼ 180 ) a ¼
2
sin a ¼SS
0
F0S) F0S ¼
SS0
sin a
SS0 ¼ R sin ) F0S ¼R sin
sin a¼ 2R cos
2
i1 () i2 (
) D () t3 (R/v)
0 0 180.0 1.50
10 20.3 149.4 1.48
20 43.2 114.0 1.40
30 90 31.0 1.07
283 Ray theory. Spherical media
Then for D < 90
t2 ¼2R
vcos
2ð152:2Þ
The critical angle, according to Snell’s law, is given by
sin ic
v¼
1
2v) ic ¼ 30
The travel-time of the critically refracted ray is (Fig. 152b)
t3 ¼FA
vþAB
2vþBS
vð152:3Þ
According to Fig. 152b
FA ¼R
cos ic
BS ¼SS0
cos ic¼
R cos
cos ic
AB ¼ OS0 OA BS
0
OS0¼ R sin
OA ¼ R tan ic
BS0¼ BS sin ic
F
S
PO
R
2ν
ν
α
α
F
S
∆
Fig. 152a
284 Seismology
Substituting ic ¼30:
FA ¼2Rffiffiffi
3p ; BS ¼
2R cosffiffiffi
3p ; AB ¼ R sin
Rffiffiffi
3p
R cosffiffiffi
3p
and substituting in (152.3) we obtain, for Dc D 90,
t3 ¼R
2vsinþ
ffiffiffi
3p
1þ cosð Þh i
ð152:4Þ
The critical distance can be calculated from ic¼ 30 using Fig. 152a (triangleOSP) anda¼D/2
90 c þ aþ ic þ 90 ¼ 180 ) c ¼ ic þ a ) c ¼ 60
The travel time of the rays that cross the boundary and penetrate in to the medium of
velocity 2v is given by (Fig. 152c)
t4 ¼FP
vþPS
2vð152:5Þ
According to Snell’s law,
sin i
v¼
sin i0
2v) sin i0 ¼ 2 sin i
and we have that
FP ¼R
cos i
PS ¼SP0
sin i0
SP0¼ R sin 180 ð Þ R tan i
ic
ic ic
ic
F
S
S
F
P
2n
n
O
R
A B
Δ
∗
Fig. 152b
285 Ray theory. Spherical media
F
F
O
2n
n
S
P
R
P
i
i
i
Δ
Fig. 152c
Δ (°)0
0
0.5
1.0
1.5
2.0
1
2
3
20 40 60 80
t (R
/n
)
Fig. 152d
286 Seismology
Substituting in Equation (152.5) we obtain, for D > 90,
t4 ¼3R
4v cos iþR sin
4v sin ið152:6Þ
Travel-time curves for t1, t2, and t3 are shown in Fig. 152d.
153. Assume an Earth formed by a mantle of thickness R and a core of radius R/2,
with velocities v and 2v. There is a seismic focus at depth R/4 below the surface.
Calculate the travel-time curves of the direct and reflected waves.
The travel time of the direct ray is (Fig. 153a)
t1 ¼FP
v
The distance FP can be expressed in terms of R and D using the cosine law in triangle FOP:
FP ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
3R
4
2
þ R2 2R3R
4cos
s
Then, we obtain
t1 ¼R
v
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
25
163
2cos
r
ð153:1Þ
The maximum distance for the direct ray is
max ¼ 1 þ2
cos1 ¼
R
23R
4
) 1 ¼ 48:2
cos2 ¼
R
2R) 2 ¼ 60:0
max ¼ 48:2þ 60:0 ¼ 108:2
Δ1
Δ2
P
O
S
F
RR/2R /4∗
Δ
ΔMAX2n n
Fig. 153a
287 Ray theory. Spherical media
The travel time for the reflected ray (Fig. 153b) is
t2 ¼FP
vþPS
v
The distances FP and PS are expressed in terms of R, D1, and D2 using the cosine law in
triangles FOP and SOP (Fig. 153b):
FP ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
3
4R
2
þR
2
2
2R
2
3R
4cos1
s
PS ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R2 þR
2
2
2R
2cos2
s
Then, we obtain
t2 ¼R
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
13
163
4cos1
r
vþR
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5
4 cos2
r
vð153:2Þ
Now we need to express D1 and D2 in terms of the take-off angle i at the focus (F). Using
Snell’s law for a spherical medium, we relate i and i0, the incidence angle at the station (S):
3R
4sin i
v¼
R
2sin r
v¼
R sin i0
v) sin i0 ¼
3
4sin i ð153:3Þ
According to Fig. 153b for triangle FOP we have
iþ aþ1 ¼ 180
sin a
3R
4
¼sin i
R
2
and we obtain
2 sin 1 þ ið Þ ¼ 3 sin i ð153:4Þ
S
F∗
R /4 R/2 O R
2n n
Δ2
Δ1
P
b
a
r
i
r
i
Fig. 153b
288 Seismology
and for triangle POS
b ¼ 180 2 i0
sin i0
R
2
¼sin b
R
and
2 sin i0 ¼ sin 2 þ i0ð Þ ð153:5Þ
Equations (153.3), (153.4), and (153.5) allow us to calculate D1 and D2 from the take-off
angle i at the focus. The travel-times are given in the following table.
For angular distance D greater than 108.2 there are no reflected rays. The travel-time
curves corresponding to the direct (1) and reflected (2) rays are shown in Fig. 153c.
0
0.4
0.6
0.8
0.8
0.8
1.4
20
1
2
40 60
Δ ( )
80 100
t (R
/n
)
Fig. 153c
i () i0 () D1 () D2 (
) D () t2 (R/v)
0 0 0 0 0 0.75
10 7.5 5.1 7.6 12.7 0.76
30 22.0 18.6 26.6 45.2 0.92
40 28.8 34.6 45.8 80.4 1.19
41.8 30.0 47.1 58.9 106.0 1.41
289 Ray theory. Spherical media
154. Consider a spherical Earth of radius R ¼ 3000 km and constant P-wave speed of
propagation of 4 km s1. Within it there is a core of radius R/2 and constant velocity
v1. At a station at epicentral distance D from an earthquake with focus at the surface,
the observed time interval is tS-P ¼ 547.0 s. Given that Poisson's ratio is 1/6, and that
the arrival of the P-wave is at 12 h 23 m 20.4 s, calculate:
(a) The epicentral distance.
(b) The time of the earthquake.
(a) For a spherical Earth of constant velocity the travel time of the direct ray is given by
t ¼ 2FO0
v¼ 2
R
vsin
2ð154:1Þ
Taking into account the presence of the core themaximumdistance for thedirect ray is (Fig. 154)
cosmax
2¼
R
2R) max ¼ 120
Since Poisson’s ratio is 1/6 we have
s ¼1
6¼
l
2 lþ mð Þ) m ¼ 2l
and we can calculate the relation between the velocities of the P-wave (a) and the S-wave (b),
a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
r
b ¼
ffiffiffi
m
r
r
9
>
>
=
>
>
;
) a ¼
ffiffiffi
5
2
r
b ) b ¼ 2:53 km s1
Using (154.1) and assuming the same path for P- and S-waves, from the time interval S-P
we obtain the distance:
tS-P ¼ 2R sin
2
1
b1
a
) ¼ 77:7
Δmax
F
O
Δ
n1 n
ROR/2∗
S
Fig. 154
290 Seismology
Notice that D < Dmax.
(b) To calculate the time of origin we subtract from the arrival time of the P-wave the
value of the travel time for that distance:
tP ¼
2R
asin
2¼ 2
3000
4sin
77:7
2¼ 940:9 s
The time of origin is then given by
t0 ¼ 12 h 23 m 20:4 s 940:9 s ¼ 12 h 07m 39:5 s
155. Consider the Earth formed by a sphere of radius R ¼ 4000 km, Poisson's ratio
1/8, and constant S-wave speed of propagation 3 km s1. Within it, there is a liquid
core of radius R/2. There occurs an earthquake with a focus in the interior of the
Earth. At a station of epicentral distance D the observed time interval is tS-P ¼ 600 s.
This focus may be at a depth of either R/10 or 2R/5. Calculate the correct depth of the
focus, and the epicentral distance.
First we determine the maximum distance for direct rays which don’t penetrate into the
core, which according to Fig. 155 corresponds to Dmax ¼ D1 þ D2:
cos1 ¼
R
2R h
cos2 ¼
R
2R) 2 ¼ 60
If the depth of the focus is R/10 then
h ¼R
10) cos1 ¼
5
9) 1 ¼ 56 ) max ¼ 116
and if it is 2R/5, then
h ¼2
5R ) cos1 ¼
5
6) 1 ¼ 33:56 ) max ¼ 94
P
F
h R/2 R
S
O
Δ2
Δ1Δ
ν2 ν1
Fig. 155
291 Ray theory. Spherical media
For a point on the surface at distance D, the S-P time interval implies, assuming the same
path for P- and S-waves,
tS-P ¼
FP
bFP
a¼
FP
aba bð Þ ð155:1Þ
From the value of Poisson’s ratio the relation between the P and S velocities is given by
s ¼1
8¼
l
2 lþ mð Þ) l ¼
m
3) a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
s
¼
ffiffiffiffiffiffi
7m
3r
s
¼
ffiffiffi
7
3
r
b
Substituting in (155.1) the S-P interval equal to 600 s we obtain the length of the ray:
FP ¼ 600ab
a bð Þ¼ 5212 km
Using the cosine law for triangle FOP
FP2¼ R hð Þ2 þ R2 2R R hð Þ cos
cos ¼FP
2 ðR hÞ2 R2
2RðR hÞ
If h ¼2
5R then D ¼ 106, but this result is not possible because the maximum distance of
the direct ray for that depth is 94. If h ¼R
10then D ¼ 86, this result is possible because
this distance is less than the maximum distance. The depth is, then, 400 km.
156. Consider the Earth of radius R and constant velocity v with a core of radius 6R/10
and constant speed of propagation 2v. An earthquake occurs with focus at 8R/10 from
the centre of the Earth. A wave emerges from that focus with a take-off angle of 15.
(a) Will it pass through the core?
(b) What epicentral distance will it reach?
(c) What will be the travel time of the wave (in units of R/v)?
(a) First we calculate the maximum epicentral distance for a ray which doesn’t
penetrate the core. According to Fig. 156a the maximum distance is
max ¼ 1 þ2
cos1 ¼
6
10R
8
10R
) 1 ¼ 41:4
cos2 ¼
6
10R
R) 2 ¼ 53:1
max ¼ 41:4þ 53:1 ¼ 94:5
From this value we calculate the take-off angle ih for this ray:
1 þ ih ¼ 90 ) ih ¼ 48:6
292 Seismology
For take-off angles less than 48.6 the rays travel through the core.
Since the velocity in the core is greater than in the mantle, to find out which rays
penetrate into the core, we also need to know the critical angle. Rays with incidence angle
at the core–mantle boundary with i > ic are totally reflected and don’t penetrate into the
core. According to Snell’s law the critical angle is given by
sin ic
v¼
1
2v) ic ¼ 30:0
We calculate, using Snell’s law, the angle of incidence i corresponding to the take-off angle
of 15 (Fig. 156b):
8
10R sin ih
v¼
6
10R sin i
v) i ¼ 20:2
Since the incidence angle i (20.2) is less than the critical angle (30) and less than the
angle corresponding to the maximum distance (48.6), the ray with take-off angle of 15
penetrates into the core.
P
O
2nF
6R/10 R
8R/10
ih
S
Δ2
Δ1 nΔ
Fig. 156a
A
B
R
S
2n
6R/10
8R/10
O
F
iih
i2
i2
i
i0
Δ1Δ2
Δ3
n
Fig. 156b
293 Ray theory. Spherical media
(b) Applying Snell’s lawwe find the angle of the transmitted ray in the core i2 (Fig. 156b):
8
10R sin ih
v¼
6
10R sin i2
2v) i2 ¼ 43:7
By consideration of triangles FOA and AOB, we determine D1 and D2 (Fig. 156b):
ih þ1 þ 180 i ¼ 180 ) 1 ¼ 5:2
i2 þ2 þ i2 ¼ 180 ) 2 ¼ 92:6
and using Snell’s law we determine i0 the incidence angle at the surface and D3:
6
10R sin i2
2v¼
6
10R sin i
v¼
R sin io
v) io ¼ 11:9
3 ¼ i io ¼ 8:3
The epicentral distance of the ray is
¼ 1 þ2 þ3 ¼ 106
(c) The travel time is
t ¼FA
vþAB
2vþBS
v
where
FA ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
8R
10
2
þ6R
10
2
28
10R
6
10R cos1
s
¼ 0:21R
AB ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
6
10R
2
þ6
10R
2
26
10
6
10R2 cos2
s
¼ 0:87R
BS ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
6
10R
2
þ R2 26
10RR cos3
s
¼ 0:42R
so
t ¼ 1:07R
v
157. Assume a spherical Earth of radius R ¼ 6000 km and constant S-wave speed of
propagation 4.17 km s1. Poisson's ratio is 1/4. At a station at epicentral distance 60
an earthquake is recorded with a time interval tS-P ¼ 554 s. Calculate the depth of the
earthquake.
Given that Poisson’s ratio is 0.25, the P-wave velocity is
s ¼1
4¼
l
2 lþ mð Þ) l ¼ m ) a ¼
ffiffiffi
3p
b ¼ 7:22 km s1
294 Seismology
From the time interval S-P we can calculate the length of the ray FS (Fig. 157):
tS-P ¼
FS
bFS
a¼ FS
a b
ab
FS ¼tS-Pab
a b¼ 5469 km
ð157:1Þ
The distance along the ray in terms of the angular epicentral distance D, using the cosine
law, is
FS ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R2 þ R hð Þ2 2R R hð Þ cos
q
Substituting FS from (157.1):
5370 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R2 þ R hð Þ2 2R R hð Þ cos
q
ð157:2Þ
We substitute in (157.2) the values R ¼ 6000 km and D ¼ 60 and solve for h, finding two
possible solutions:
h1 ¼ 4706 km
h2 ¼ 1294 km
158. Assume a spherical Earth of radius R and P-wave velocity which can be
expressed by the equation v(r) ¼ a br2. The speed of propagation at the surface
of the Earth is v0 and at the centre of the Earth it is 2v0. What angular distance D does
a wave reach which penetrates to a depth equal to half the Earth’s radius?
If the velocity distribution inside the Earth is v(r) ¼ a br2, the ray paths are circular with
radius given by (Fig. 158)
F
Δ
R – h
S
R
Fig. 157
295 Ray theory. Spherical media
r ¼r
pdv
dr
ð158:1Þ
From the conditions of the problem
r ¼ R ) v ¼ v0 ¼ a bR2
r ¼ 0 ) v ¼ 2v0 ¼ a
and
b ¼v0
R2
a ¼ 2v0
)
) v ¼ v0 2r2
R2
The radius of curvature of the ray which penetrates to r ¼ R/2 is that corresponding to the
ray parameter
p0 ¼r0
v0¼
R
2v0ð158:2Þ
The velocity at depth R/2 is
v0 ¼ v0 2
R
2R
0
B
@
1
C
A
20
B
B
@
1
C
C
A
¼ v07
4
Substituting the ray parameter in (158.2):
p0 ¼R
v07
2
¼2R
7v0
7/4 R
7/4 R
R/2 O R
SF
Δ/2Δ/2
Fig. 158
296 Seismology
The derivative of the velocity is
dv
dr¼
d
dra br2
¼ 2br ¼ 2v0
R2r
Substituting in (158.1) we obtain, for the radius of curvature (Fig. 158),
r ¼r
p 2brð Þ¼
7R
4
The epicentral (angular) distance corresponding to this ray is found by applying the cosine
law to the triangle POS:
7
4R
2
¼9
4R
2
þ R2 29R
4R cos
2) ¼ 96:4
159. The Earth consists of a mantle of radius R and speed of propagation v ¼ a=ffiffi
rp
,
and a core of radius R/2 and speed of propagation 4v0, where v0 is the speed of
propagation at the Earth’s surface. Calculate:
(a) The maximum epicentral distance corresponding to a wave that travels only
through the mantle.
(b) The critical angle of the wave reflected at the core, and the angle at which it leaves
the surface.
(c) The epicentral distance Dc corresponding to the critical angle.
(d) Plot the travel-time curve, specifying Dc and Dm.
(a) Thevalue ofa in the velocity distribution is found from the value of velocity at the surface:
r ¼ R ! v ¼ v0 ¼ aR12 ) a ¼ v0R
12
The velocity distribution is
v ¼ v0R
r
a
¼ v0R
r
12
For this general type of distribution of velocity with depth a < 1, the angular distance for a
surface focus (Fig. 159a) is given by
¼2
1þ acos1 p
0
ð159:1Þ
where
¼r
v) 0 ¼
R
v0
The maximum distance for a ray which travels only through the mantle, that is that reaches
depth R/2, can be calculated from the velocity at that depth, vm:
vm ¼ v0R
R
2
0
B
@
1
C
A
12
¼ v0ffiffiffi
2p
) p ¼
R
2vm
¼
R
2
v0ffiffiffi
2p ¼
1
2ffiffiffi
2p
R
v0¼ p
297 Ray theory. Spherical media
By substitution in (159.1) we obtain, for the maximum distance,
m ¼2
1þ1
2
cos1
1
2ffiffiffi
2p
R
v0R
v0
0
B
B
@
1
C
C
A
¼ 92:4
(b) The critical angle of a reflected ray at the mantle-core boundary applying Snell’s law is
sin ic
v0ffiffiffi
2p ¼
1
4v0) ic ¼ 20:7
The take-off angle at the surface i0 for this ray is found by again applying Snell’s law:
R sin i0
v0¼
R
2sin 20:7ð Þ
v0ffiffiffi
2p ) i0 ¼ 7:2
(c) To find the critical distance we use the expression
c ¼ 2
ð
r0
rp
p
r
drffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 p2p ð159:2Þ
where
rp ¼R
2
r0 ¼ R
p ¼r0 sin i0
v0¼
R
8v0
¼r
v¼
r
v0Rr
12
¼r3=2
v0ffiffiffi
Rp
Substituting the values of the problem in (159.2):
c ¼1
4
ðR
R=2
dr
r
R3=2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r3
R3
64
r
i0ic
Δc
R/2 RO
Δ/2Δ/2
SF ∗
Fig. 159a
298 Seismology
This integral is of the type
ð
dx
xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
xn anp ¼
2
nffiffiffiffiffi
anp cos1
ffiffiffiffiffi
an
xn
r
so we can write the solution
c ¼8
6cos1 1
8
cos1 1ffiffiffi
8p
¼ 18
(d) The travel time of the rays in the mantle is given by
t ¼201þ a
sin 1þ að Þ
2
¼R
v0
4
3sin
3
4for 0 92:4
By substitution of values of D we find the travel time curve given in Fig. 159b.
160. A spherical medium of radius R has a constant speed of propagation v0 from the
surface down to R/2, and from R/2 to the centre a core of variable speed of propaga-
tion v ¼ v0R
r
1=2
.
(a) What value should i0 have for the waves to penetrate into the core?
(b) Calculate the epicentral distance reached by a wave leaving a focus at the surface
at angle i0.
(a) The velocity at the top of the core (r ¼ R/2) is
1.2
1.0
0.8
0.6
t (R
/n)
0.4
0.2
0
0 20 40
Δ ()
Δ mΔ c
60 80
Fig. 159b
299 Ray theory. Spherical media
v1 ¼ v0R
R
2
0
B
@
1
C
A
12
¼ v0ffiffiffi
2p
Applying Snell’s law we find the critical angle ic for incident rays at the core (Fig. 160):
R sin ic
v0¼
R
2ffiffiffi
2p
v0) ic ¼ 45
The take-off angle i0, for a focus at the surface corresponding to the critical angle, using
Snell’s law, is
R sin i0
v0¼
R
2sin ic
v0) i0 ¼ 20:7
The rays that penetrate into the core must leave the focus with take-off angles less than
20.7.
(b) For a ray with take-off angle i0 which penetrates the core the epicentral distance is the
sum of that corresponding to the part that has travelled through the mantle, D1, plus the
part that has travelled through the core, D2:
¼ 21 þ2
Since in the core the velocity varies with depth with the law given in the problem, the
epicentral distance is given by
2 ¼2
1þ acos1 p
1
where a is the exponent of the velocity distribution
v ¼ v0ra ) a ¼
1
2
i0
Δ1 Δ2
Δ1
n0
∗F
R/2 R
P
S
l
O
Fig. 160
300 Seismology
and p is the ray parameter, which can be obtained using Snell’s law:
p ¼R sin i0
v0¼
R
2sin i1ffiffiffi
2p
v0
and
¼r
v) 1 ¼
R
2ffiffiffi
2p
v0
Then, we find
2 ¼2
1þ 12
cos1
R sin i0
v0R
2ffiffiffi
2p
v0
0
B
B
@
1
C
C
A
¼4
3cos1 2
ffiffiffi
2p
sin i0
The distance D1 can be determined using the sine and cosine laws for the triangle FOP
(Fig. 160):
R
2sin i0
¼l
sin1
) 1 ¼ sin1 2l sin i0
R
l2 þ R2 2Rl cos i0 ¼R2
4) l ¼ R cos i0
1
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4R2cos2i0 3R2p
and we find the expression in terms of i0:
¼ 2sin1 2 cos i0 1
2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4cos2i0 3p
sin i0
þ4
3cos1 2
ffiffiffi
2p
sin i0
161. Consider a spherical Earth of radius 6000 km and surface velocity of 6 km s1, with
a velocity distribution of the type v rð Þ ¼ a=ffiffi
rp
. At the distance reached by a wave
emerging at a take-off angle of 45 from a focus on the surface, calculate the interval
between the arrival times of the direct P- and reflectedPP-waves (the PP-wave is one that
is reflected at the surface at themidpoint between the focus and the point of observation).
The velocity distribution is of the type v ¼ v0R
r
a
where
r ¼ R ) v ¼ v0 ¼affiffiffi
Rp ) a ¼ v0
ffiffiffi
Rp
v ¼ v0R
r
12
and the ray parameter p for a ray with take-off angle of 45 is
p ¼R sin i0
v0¼
60001ffiffiffi
2p
6¼ 707 s
301 Ray theory. Spherical media
For this type of velocity distribution the relation between the ray parameter and the
epicentral distance is
p ¼ 0 cos1þ a
2
ð161:1Þ
In this problem the value of 0 is
¼r
v) 0 ¼
R
v0¼ 1000 s
Substituting the values in (161.1) we obtain the distance for the ray with take-off angle
of 45:
707 ¼ 1000 cos3
4
) ¼ 60
The PP-wave which arrives at D¼ 60 travels twice the distance which a P-wave does for a
distance of 30. For this type of velocity distribution the travel time for a distance D is
given by
t ¼201þ a
sin1þ að Þ
2
In our case for the P- and PP-waves at distance 60 we substitute the values of the problem
and find
tPð60Þ ¼ 943 s
tPPð60Þ ¼ 2tPð30
Þ ¼ 1021 s
The time interval between the PP- and P-waves 60 is
tPP tP ¼ 1021 943 ¼ 78s
162. In an elastic spherical medium of radius r0, the velocity increases with depth
according to v ¼ arb. If v0 ¼ 6 km s1, r0 ¼ 6000 km, and, at a point at distance
D ¼ 90, the slope of the travel-time curve is 500 s, determine:
(a) The value of b.
(b) The value of rp and of vp of the wave reaching an epicentral distance of 90.
(a) For this type of velocity distribution the travel time in terms of the epicentral
distance is given by
t ¼201þ b
sin 1þ bð Þ
2
ð162:1Þ
As we know the velocity at the surface,
¼r
v) 0 ¼
r0
v0¼ 1000 s
302 Seismology
The ray parameter p is known, because it is equal to the slope of the travel-time curve
which for D ¼ 90 is given as 500 s. Using the relation between p and D for this type of
velocity distribution,
p ¼dt
d¼ 0 cos 1þ bð Þ
2
500 ¼ 1000 cos 1þ bð Þp
4
h i
) cos 1þ bð Þp
4
h i
¼1
2) 1þ bð Þ
p
4¼
p
3) b ¼
1
3
(b) At the point of greatest penetration rp for D ¼ 90, we have the relation
p ¼rp sin 90
vp¼
rp
vp) rp ¼ pvp
and also
vp ¼ v0r0
rp
13
From these two equations we obtain rp and vp:
rp ¼ v0pr130r
13
p ) rp ¼ v340r
140p
34 ¼ 3564 km
vp ¼rp
p¼
3564
500¼ 7:1 km s1
163. Consider a spherical Earth of radius R, the northern hemisphere with a constant
speed of propagation v0, and the southern hemisphere with a speed of propagation of
v ¼ v0R
r
12
.
(a) Calculate the travel time of seismic waves for a focus on the equator and stations
on the same meridian.
(b) In which hemisphere does the wave at a distance of 60 arrive first?
(a) In the northern hemisphere the velocity is constant and the rays have straight paths
and their travel time is (Fig. 163)
tN ¼
2R
v0sin
20 < < 90 ð163:1Þ
In the southern hemisphere the velocity increases with depth and the rays have curved
paths. Their travel time is given by
tS ¼
201þ b
sin 1þ bð Þ
2ð163:2Þ
where
0 ¼R
v0and v ¼ v0
R
r
b
) b ¼1
2
Substituting in (163.2) we obtain for the travel time in the southern hemisphere
303 Ray theory. Spherical media
tS ¼
4
3
R
v0sin
3
40 < < 90 ð163:3Þ
(b) The travel times for waves in the northern and southern hemisphere are given by
Equations (163.1) and (163.3). By substitution of D ¼ 60 we obtain
tN ¼
R
v0
tS ¼
R
v0
2ffiffiffi
2p
6¼ 0:47
R
v0
The waves arrive first in the southern hemisphere.
164. Consider a spherical medium of radius R consisting of two concentric regions
(mantle and core), the core of radius R/2. The speeds of propagation are v ¼ ar1=2
for the mantle and v ¼ aR1=6r1=3 for the core. The surface velocity is v0. For a wave
leaving a focus with angle of incidence 14.5, calculate the angular distance D at
which it reaches the surface.
We calculate a by applying the boundary conditions
r ¼ R ) v ¼ v0 ) v0 ¼ aR12 ) a ¼ v0R
12
Mantle : v ¼ v0R
r
12
Core : v ¼ v0R
r
13
We determine the ray parameter corresponding to the ray with take-off angle i0 ¼ 14.5,
using Snell’s law (Fig. 164a):
p ¼r sin i
v¼
R sin 14:5
v0¼
R
4v0
S2
R
n0
S1
Δ
F ∗
Fig. 163
304 Seismology
At the bottom of the mantle at depth R/2 the velocity is
v1 ¼ v0R
R
2
0
B
@
1
C
A
12
¼ v0212
The incident angle i of this ray on the mantle–core boundary, applying Snell’s law, is given by
R
2sin i
v0212
¼R sin i0
v0) i ¼ 45
On the top of the core the velocity is
v2 ¼ v0R
R
2
0
B
@
1
C
A
13
¼ v0213
which is less than at the bottom of the mantle and there is no critical angle. Applying
Snell’s law again we obtain the angle i2 of the refracted ray in the core:
sin i
v1¼
sin i2
v2) i2 ¼ 39
The take-off angle of the last ray which travels only in the mantle is given by
R sin i0
v0¼
R
2
v0212
) i0 ¼ 20:7
In our case the angle 14.5 is less and the ray penetrates into the core.
The epicentral distance is the sum of the distances corresponding to the paths in the
mantle and in the core:
¼ 21 þ2
Δ1 Δ2Δ1
i0i2
i0
R/2 O R
F∗
S
i
Fig. 164a
305 Ray theory. Spherical media
The distance corresponding to the path in the core is given by
2 ¼2
1þ bcos1 p
0
where
p ¼R
4v0
0 ¼
R
2
v0213
and where p is the ray parameter and ¼ r/v. Substituting the values we obtain
2 ¼ 76:4
To calculate D1 we suppose that there is no core and a ray with take-off angle i0 ¼ 14.5
would arrive at distance D3 which is related with D1 by (Fig. 164b)
21 ¼ 3 4
The distances D3 and D4 can be determined using the equation
¼2
1þ bcos1 p
0
where for D3
0 Rð Þ ¼R
v0
R
2
¼
R
2
v0R
R
2
0
B
@
1
C
A
12
¼R
v02
32
Δ1 Δ4
Δ3
Δ1
F∗
O R
S
Fig. 164b
306 Seismology
and we obtain
3 ¼2
1þ 12
cos1
R
4v0R
v
0
B
B
@
1
C
C
A
¼ 100:6
and by similar substitutions for D4
4 ¼2
1þ1
2
cos1
R
4v0
R
v023
2
0
B
B
B
B
@
1
C
C
C
C
A
¼ 60:0
Then, 2D1 ¼ 100.6 – 60 ¼ 40.6 and the epicentral distance is
¼ 40:6þ 76:4 ¼ 117:0
Surface waves
165. A Rayleigh wave in a semi-infinite medium has a 20 s period. If the P-wave
velocity is 6 km s1 and Poisson’s ratio is 0.25, calculate the depth at which u1 ¼ 0,
and at which depth the particle movement becomes prograde.
Since Poisson’s ratio is 0.25, we find the relation between P- and S-waves:
s ¼1
4¼
l
2 lþ mð Þ) l ¼ m ) a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
s
¼
ffiffiffiffiffiffi
3m
r
s
¼ffiffiffi
3p
b
a ¼ 6 ) b ¼6ffiffiffi
3p ¼ 3:4 km s1
For a half-space the velocity of Rayleigh waves is
cR ¼ 0:919b ¼ 3:2 km s1
The displacement u1 is given by
u1 ¼@’
@x1
@c
@x3
where the potentials are given by
’ ¼ A exp ikrx3 þ ik x1 cRtð Þð Þ
c ¼ B exp iksx3 þ ik x1 cRtð Þð Þ
r ¼ i 1c2Ra2
1=2
¼ 0:85i
s ¼ i 1c2R
b2
1=2
¼ 0:39i
307 Surface waves
Then,
u1 ¼ 0 ) ikA exp 0:85kx3ð Þ 0:39ikB exp 0:39kx3ð Þ ¼ 0 ð165:1Þ
so
k ¼2p
l¼
2p
TcRffi 0:1 km1
We can write B in terms of A using the boundary condition of zero stress at the free surface:
t31 ¼ 0jx3¼0 ) 2rA 1 s2
B ¼ 0
so
B ¼ 1:47iA
Substituting in (165.1) we obtain the value of x3:
exp 0:85kx3ð Þ ¼ 0:39 1:47 exp 0:39kx3ð Þ
x3 ¼ 12 km
At 12 km depth u1 is null and for greater values of depth the particle motion is prograde
while for lesser values of depth it is retrograde.
166. Given a layer of thickness H and shear modulus m¼ 0 on top of a half-space or
semi-infinite medium in which l¼ 0, study (without expanding the determinant)
whether there exist surface waves that propagate in the x1-direction. Are they
dispersive waves?
In the liquid layer (m ¼ 0) the P- and S-velocities are
b0 ¼ 0 ¼
ffiffiffiffi
m0
r
s
) a0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
l0 þ 2m0
r
s
¼
ffiffiffiffi
l0
r
s
and in the solid half-space
l ¼ 0 ) a ¼
ffiffiffiffiffiffi
2m
r
s
¼ffiffiffi
2p
b
The relation between the stress and strain is
tij ¼ lydij þ 2meij
where eij ¼12ui; j þ uj;i
.
In the layer: m0 ¼ 0 )t0ii ¼ l0 e011 þ e022 þ e033ð Þt0ij ¼ 0
In the half-space: l ¼ 0 ) tij ¼ 2 meijIf there are surface waves propagating in the x1-direction, their displacements in terms of
the potentials are given by (Fig. 166)
u1 ¼ ’;1 c;3
u2 ¼ u2
u3 ¼ ’;3 þ c;1
308 Seismology
The boundary conditions at the free surface are null normal stresses:
x3 ¼ H )t033 ¼ 0
t031 ¼ 0
and at the boundary between the liquid layer and the solid half-space continuity of the
normal component of the displacement and stress and zero tangential stresses,
x3 ¼ 0 )
u3 ¼ u03
t33 ¼ t033
t32 ¼ t032 ¼ 0
t031 ¼ 0
8
>
>
>
<
>
>
>
:
In the liquid layer there is only the P-wave potential ’. Taking (x1, x3) as the incidence
plane
’0 ¼ A exp ikr0x3 þ ik x1 ctð Þð Þ þ B exp ikr0x3 þ ik x1 ctð Þð Þ
r0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
a02 1
r
where c is the velocity of wave propagation in the x1-direction
In the half-space
c ¼ C exp iksx3 þ ik x1 ctð Þð Þ
u2 ¼ E exp iksx3 þ ik x1 ctð Þð Þ
’ ¼ D exp ikrx3 þ ik x1 ctð Þð Þ
r ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
a2 1
r
s ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b2 1
s
In the layer we have only guided P-waves and r0 is real, while in the half-space for surface
waves, r and s must be imaginary. Then a > b > c > a0 must be satisfied.
x3
x1H
0
μ = 0
λ = 0
Fig. 166
309 Surface waves
From the boundary conditions we obtain the following equations:
x3 ¼ H
t033 ¼ 0 ) A 1þ r02
eikr0H þ B 1þ r
02
eikr0H ¼ 0
x3 ¼ 0
t31 ¼ 0 ) 2Dr C þ Cs2 ¼ 0
u03 ¼ u3 ) Ar0 Br0 ¼ Dr þ C
t033 ¼ t33 ) l0 1þ r02
Aþ Bð Þ ¼ 2m Dr2 þ Cs
For a solution of the system the determinant must be zero:
eikr0H eikr0H 0 0
0 0 s2 1 2r
r0 r
0 1 r
l0 1þ r02ð Þ l0 1þ r
02ð Þ 2ms 2mr2
¼ 0
Expanding the determinant and working r0, r, and s in terms of the variable c, we obtain c
(k), the velocity of waves in the x1-direction. They have the form of guided waves in the
liquid layer and surface waves in the half-space. Since the velocity c(k) is a function of the
wavenumber the waves are dispersive.
167. There is a liquid layer of density r and speed of propagation a on top of a rigid
medium (half-space). Derive the dispersion equation of waves in the layer by bound-
ary conditions and by constructive interference in terms of v. Plot the dispersion
curve for the different modes.
Given that the layer is liquid the only potential is ’:
’ ¼ A exp ikrx3 þ B exp ikrx3ð Þð Þ exp ik x1 ctð Þ ð167:1Þ
The boundary conditions at the free surface are zero normal stress and at the boundary
between the liquid layer and the rigid half-space zero normal component of displacement
(Fig. 167a):
x3 ¼ H ) t33 ¼ 0
x3 ¼ 0 ) u3 ¼ 0
where
t33 ¼ ly ¼ r2’ ¼ ro2’ ¼ 0
u3 ¼ ’;3
r ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
a2 1
r
By substitution of (167.1) we obtain
AeikrH þ BeikrH ¼ 0
A B ¼ 0 ) A ¼ B
310 Seismology
Then, 2A cos krH ¼ 0. For waves propagating in the layer, r must be real and c > a. The
solution is given by
krH ¼ nþ1
2
p; n ¼ 0; 1; 2; . . . ð167:2Þ
The solution can also be found by the method of constructive interference. The condition
of constructive interference implies that waves coinciding at a given wavefront (AB) are in
phase, that is, the distance along the ray must be an integer multiple of the wavelength,
taking into account possible phase shifts (Fig. 167b). In our case on the free surface,
x3 ¼ H, there is a phase shift of p (l/2) and we write the condition as (Fig. 167b)
A Pþ PQþ QBla
2¼ nla
or
2p
laðAPþ PQþ QBÞ p ¼ 2pn
x1
x3
H
0
liquid layer
rigid half-space
Fig. 167a
x3
x1
H
0P
A
B
O
Fig. 167b
311 Surface waves
Substituting
APþ PQþ QB ¼ 2H cos i
we obtain
2p
la2H cos i p ¼ 2p nþ 1ð Þ ð167:3Þ
According to Snell’s law,
sin i ¼a
c) cos i ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1a2
c2
r
¼a
c
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
a2 1
r
¼a
cr
anda
cka ¼ k.
Substituting in (167.3), we obtain the same solution obtained in (167.2):
kaa
cHr ¼ nþ
1
2
p
This expression can also be written as
krH ¼ nþ1
2
p)oH
c
ffiffiffiffiffiffiffiffiffiffiffiffi
c2
a21
r
¼ nþ1
2
p) c¼1
a2 nþ
1
2
2p2
H2o2
" #12
ð167:4Þ
The fundamental mode (FM) corresponds to n ¼ 0, and n 1 to the higher modes (HM).
In the FM and the higher modes, the frequency oc corresponding to the zero in the
denominator in (167.4) is called the cut-off frequency, as there are no values of c for o <
oc. For a mode of order n the cut-off frequency is given by
oc ¼
p nþ1
2
a
H
The dispersion curve is shown in Fig. 167c.
c
FM 1 HM 2 HM
ω
α
πα
2H3πα
2H5πα
2H
Fig. 167c
312 Seismology
168. Consider an elastic layer of coefficients l and m, thickness H, and density r on a
rigid semi-infinite medium. Derive the dispersion equation c(v) for P-SVand SH-type
channelled waves for the fundamental mode (FM) and the first higher mode (1HM).
Plot the dispersion curve for the SH motion.
For a SH-wave which propagates in the x1-direction its displacement is given by
u2 ¼ E exp iksx3ð Þ þ F exp iksx3ð Þð Þ exp ik x1 ctð Þ
The P- and SV-waves are given by their scalar potentials ’ and c:
’ ¼ A exp ikrx3ð Þ þ B exp ikrx3ð Þð Þ exp ik x1 ctð Þ
c ¼ C exp iksx3ð Þ þ D exp iksx3ð Þð Þ exp ik x1 ctð Þ
where r and s were defined in Problem 166.
The boundary conditions for SH-waves are null stress at the free surface and null
displacement at the boundary with the rigid medium (Fig 168a):
x3 ¼ H ) t32 ¼ 0 ¼ m@u2@x3
x3 ¼ 0 ) u2 ¼ 0
By substitution we have
EeiksH FeiksH ¼ 0
E þ F ¼ 0 ) F ¼ E
EðeiksH þ eiksH Þ ¼ 0 ) cosðksHÞ ¼ 0 ) ksh ¼ nþ1
2
p
Substituting s and putting k ¼ o/c:
Ho
c
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b2 1
s
¼ nþ1
2
p ) c ¼1
b2 nþ
1
2
2p2
H2o2
!12
ð168:1Þ
This equation give us, for the SH component, the frequency dependence of the velocity c(o).
The boundary conditions for P and SV are similarly
α, β, µ
rigid medium
0
H x1
x3
Fig. 168a
313 Surface waves
x3 ¼ H ) t31 ¼ 0; t33 ¼ 0
x3 ¼ 0 ) u1 ¼ 0; u3 ¼ 0
where
t33 ¼ lðe11 þ e33Þ þ 2me33
t31 ¼ 2me31
u1 ¼ ’;1 c;3
u3 ¼ ’;3 þ c;1
Substituting the expression for the potentials we obtain
ðlþ 2mÞðr2AeikrH þ r2BeikrH Þ þ lðAeikrH þ BeiksH Þ þ 2msðCeiksH DeiksH Þ ¼ 0
2rðAeikrH BeikrH Þ þ ð1 s2ÞðCeiksH þ DeiksH Þ ¼ 0
ðAþ BÞ sðC DÞ ¼ 0
rðA BÞ þ C þ D ¼ 0
For a solution we put the determinant of the system of equations equal to zero:
1 1 s s
r r 1 1
2reikrH 2reikrH ð1 s2ÞeiksH ð1 s2ÞeiksH
½lþ r2ðlþ2mÞeikrH ½lþ r
2ðlþ2mÞeikrH 2mseiksH 2mseiksH
¼ 0 ð168:2Þ
Expanding the determinant and putting it equal to zero, we obtain the dependence with
frequency of the velocity c(o) which gives us the dispersion curve.
For the wave with SH component the dispersion curve is given in Fig. 168b:
c ¼1
b2 nþ
1
2
2p2
H2o2
!12
c
FM 1 HM 2 HM
ω
β
πβ
2H3πβ
2H5πβ
2H
Fig. 168b
314 Seismology
For n ¼ 0 the curves correspond to the fundamental mode and for 1 n to the higher
modes. For all modes, including the fundamental mode, there is a cut-off frequency oc ¼
(nþ1)pb/2H, with n ¼ 0 for the fundamental mode and n 1, for higher-order modes.
169. For a liquid layer of thickness H with a rigid medium above and below, derive
the dispersion equation c(v) of the fundamental and higher modes. For the FM, at
what height above the layer is the motion circular?
Given that the medium is a liquid, motion is represented only by the scalar potential f:
’ ¼ A exp ikrx3 þ B exp ikrx3ð Þð Þ exp ik x1 ctð Þ ð169:1Þ
where r ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
a2 1
r
.
The boundary condition at the two boundaries between the liquid and rigid solid is that
the normal component of the displacement is null (Fig. 169):
x3 ¼ 0 ) u3 ¼ 0
x3 ¼ H ) u3 ¼ 0
Substituting u3 ¼ ’;3 we have
A B ¼ 0
AeikrH BeikrH ¼ 0ð169:2Þ
which leads to the equation
A eikrH eikrH
¼ 0 ð169:3Þ
Consider first that r is real, that is, c >a. Then, from (169.1)
2iA sin krH ¼ 0 ) krH ¼ np; n ¼ 0; 1; 2; :::
with n ¼ 0, fundamental mode (FM), and n 1 for higher modes.
For the FM, n ¼ 0 and r ¼ 0, and then
Hk
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
a2 1
r
¼ 0 ) c ¼ a
The displacements from (169.1) and (169.2) are
x3
x1
H
0
rigid medium
liquid layer
rigid medium
Fig. 169
315 Surface waves
u3 ¼@’
@x3¼ Aikr exp ikrx3 expikrx3ð Þ exp ikðx1 ctÞ
u1 ¼@’
@x1¼ Aik exp ikrx3 þ expikrx3ð Þ exp ikðx1 ctÞ
For the FM r ¼ 0, then u3 ¼ 0 and this is a P-wave, with only a u1 component, which
propagates in the x1-direction. For all HM the displacements have both components
For the first higher mode (1HM), n ¼ 1:
Ho
c
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
a2 1
r
¼ p
c2 ¼1
1
a2
p2
o2H2
If
1
a2
p2
o2H2¼ 0 then o ¼
ap
H) c ! 1
The cut-off frequency is oc > ap/H. For each higher mode there is a cut-off frequency onc
> nap/H.
If r ¼ ir_is imaginary, then c < a and this implies that 2 sinhðkr
_HÞ ¼ 0 which is
impossible (1< sinhx <1).
The particle motion inside the layer is circular when
u1 ¼ u3 ð169:4Þ
so
u1 ¼ u3 ) ð1 rÞ exp ikrx3 þ ð1þ rÞ expikrx3 ¼ 0
Taking only the amplitudes of the displacements,
ð1 rÞ cos krx3 þ ð1þ rÞ cos krx3 ¼ 0 ) cos krx3 ¼ 0
krx3 ¼ nþ1
2
p ) x3 ¼ nþ1
2
p
kr
For the FM, we have seen that u3 ¼ 0, so there is no circular motion. For the 1HM, the
height in the layer at which the motion is circular is
x3 ¼3p
2kr
The height inside the layer at which the motion is circular depends in each higher mode of
the frequency.
170. In the hypothetical case of a layer of thickness H and speed of propagation b0 on
top of a semi-infinite medium of speed of propagation b, the phase shifts at the free
surface and the contact plane are
p
4and tan1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b0 1
s
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s
2
6
6
6
6
6
4
3
7
7
7
7
7
5
:
316 Seismology
Determine:
(a) The dispersion equation using constructive interference.
(b) The cut-off frequency of the fundamental mode and first higher mode.
(c) Plot the dispersion curve of the FM and 1HM using units of c/b and H/l for
b ¼ 2b0.
(a) The distance from A to B along the ray path is (Fig. 170a)
AB ¼ 2H cos i
According to Snell’s law
sin i ¼b0
c
cos i ¼b0
c
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
and the wavenumber k associated with velocity c is
kb0 sin i ¼ kb0b0
c¼ k
As explained in Problem 167, the condition for constructive interference is that the distance
AB along the ray path be an integer multiple of the wavelength, taking into account the
phase shift at the free surface and the boundary surface between the two media:
2kB0H cos iþp
4 tan1 s0
s
¼ 2pn ð170:1Þ
where
s0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
s ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s
By substitution in (170.1) we obtain the dispersion equation
HA
B
β
β
i
Fig. 170a
317 Surface waves
tan 2kHs0 2n1
4
p
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
1c2
b2
v
u
u
u
u
u
u
t
ð170:2Þ
(b) For the fundamental mode (FM), n ¼ 0:
tan 2kH
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
þp
4
" #
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s > 0 where b0 < c < b
Given that tan B > 0 )p
4 B
p
2so that b’< c < b, then
If k ¼ 0 then
tanp
4
¼ 1 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s ) c ¼
ffiffiffi
2p
bb0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 þ b02p
so
tan 2kHs0 þp
4
¼ tan Bð Þ ¼ 1 ) B ¼p
2
If c ¼ b then
2kH
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
þp
4¼
p
2) k ¼
p
8Hs0) s0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2
b02 1
s
so
tan 2kHs0 þp
4
¼ 0
If c ¼ b0 then
s0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
¼ 0 ) tan kHs0 þp
4
¼ 0 ) k ¼ 1 ) kHs0 ¼ p
4
But kHs0 must be positive, so this last solution is not possible.
For the first higher mode (1HM), n ¼ 1:
tan 2kHs0 7
4p
¼s0
s
The tangent function is positive for the range [0, p/2]:
0 2kHs0 7p
4
p
2
For c ¼ b0
318 Seismology
tan 2kHs0 7p
4
¼ 0 ) kHs0 ¼7p
8) k ¼ 1
For c ¼ b
tan 2kHs0 7p
4
¼ 1 ) 2kHs0 7p
4¼
p
2) k ¼
9p
8Hs0
(c) Taking b’¼ b/2, the dispersion equation for the FM is
tan 2kH
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4c2
b2 1
s
þp
4
" #
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4c2
b2 1
s
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s )c
b
2
¼2
5 3 sin 8pH
l
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4c
b
2
1
s
" #
But we have seen that
2kHs0 þp
4
p
2)
H
l
1
16s0
c ¼ b ) s0 ¼ffiffiffi
3p
H
l
1
16ffiffiffi
3p
Giving values to H/l we can calculate the corresponding values of c/b by means of a
numerical method. For example, we obtain,
For the first higher mode (1HM), we arrive at the same equation, given that tan(a þ p/4)
¼ tan(a þ 7p/4), but vary the intervals of H/l and kHs0:
7p
8 kHs0
9p
8
7
16s0
H
l
9
16s0
ForH
l¼
7
16s0at one limit we have the value s0¼ 0 which corresponds to c ¼ b0 and c/b ¼
0.5.
ForH
l¼
9
16s0we have s ¼ 0 and c ¼ b, and consequently
H
l¼
9
16ffiffiffi
3p .
The dispersion curves for the fundamental mode and the first higher mode are shown in
Fig.170b.
H/l c/b
0.0 0.63
0.01 0.68
0.02 0.78
0.036 1
319 Surface waves
171. In a structure with a layer of thickness H and speed of propagation b0 on top of a
medium of speed of propagation b, the phase shifts at the free surface and the contact
plane are:
d1 ¼ p
2and d2 ¼ sin1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
8
>
>
>
>
>
<
>
>
>
>
>
:
9
>
>
>
>
>
=
>
>
>
>
>
;
Calculate:
(a) The dispersion equation of the Love wave.
(b) For the fundamental and first higher mode, and the minimum and maximum
frequencies as functions of H, b, and b0.
(c) For this mode, given b0 ¼ b/2, the maximum and minimum frequencies, and the
corresponding values of c.
(a) As in Problem 170, the condition of constructive interference with the phase shifts
given in this problem results in (Fig. 171)
4p
lH cos i
p
2 sin1 s
s0
¼ 2pn ð171:1Þ
where
c
1.0
0.5
1 HMFM
β
Hλ0.10 0.20 0.301
316 316
9
Fig. 170b
320 Seismology
s ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s
s0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
According to Snell’s law
sin i ¼b0
c
cos i ¼b0
c
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
Substituting in (171.1) we find the dispersion equation for Love waves:
2Hks0 p
2 2pn ¼ sin1 s
s0
) cos 2Hk
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s !
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s
(b) The fundamental mode (FM) corresponds to the values
0 2Hks0 p
2) 1
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s 0
The velocity at the limit of lowest frequencies, k ¼ 0, is given by
k ¼ 0 )s
s0¼
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s ¼ 1 ) c2 ¼2b02b2
b02 þ b2
k ¼p
4Hs0)
s
s0¼ 0 ) s ¼ 0 ) c ¼ b ) s0 ¼
1
b0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 b02q
H
A
B
β
β
i
Fig. 171
321 Surface waves
For the first higher mode (1HM)
3p
2 2Hks0 2p ) 0
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1c2
b2
s
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s 1
In the lowest limit
k1 ¼3p
4Hs0)
s
s0¼ 0 ) s ¼ 0 ) c ¼ b ) s0 ¼
1
b0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 b02q
The minimum value of the frequency is
k1 ¼3pb0
4Hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 b02p
In the highest frequency limit
k2 ¼p
Hs0)
s
s0¼ 1 ) s ¼ s0 ) c2 ¼
2b02b2
b2 þ b02) s0 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 b02
b2 þ b02
s
The maximum value of the frequency is given by
k2 ¼pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 þ b02p
Hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 b02p
(c) If we put b0 ¼ b=2 in the 1MS,
k1 ¼3p
b
2
4H
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 b2
4
r ¼pffiffiffi
3p
H4¼
2p
l)
H
l¼
ffiffiffi
3p
8; c ¼ b
k2 ¼p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 þb2
4
r
4H
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
b2 b2
4
r ¼pffiffiffi
5p
Hffiffiffi
3p ¼
2p
l)
H
l¼
ffiffiffi
5p
2ffiffiffi
3p ; c ¼ b
ffiffiffi
2
5
r
172. Consider a layer of thickness H and parameters m0 and r0 on top of a semi-
infinite medium of parameters m ¼ 4m0 and r ¼ r0. Ifc
b¼ a and
H
l¼ b:
(a) Write the dispersion equation of the Love wave in terms of a and b.
(b) Calculate the values of b corresponding to a ¼1
2;3
4and 1 for the fundamental
mode and first higher mode.
(c) For which values of b is the node of the amplitude of the first higher mode at a
depth of H/2?
(a) The dispersion equation for Love waves is
322 Seismology
tan kH
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
s( )
¼mffiffiffiffiffiffiffiffiffiffiffiffi
1 c2
b2
q
m0ffiffiffiffiffiffiffiffiffiffiffiffiffi
c2
b02 1
q ð172:1Þ
In this problem,
m ¼ 4m0
r ¼ r0
then,
b ¼
ffiffiffi
m
r
r
! b0 ¼
ffiffiffiffiffiffi
m
4r
r
¼b
2
We now introduce a and b:
a ¼c
b)
c
b0¼ 2a
b ¼H
l) k ¼
2p
l¼
2pb
H
Substituting in (172.1):
tan 2pbffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4a2 1p
¼ 4
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 a2p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4a2 1p
(b) For the FM
a ¼ 1; b ¼ 0 and c ¼ b
a ¼1
2; b ! 1 and c ¼ b0
a ¼3
4) tan 2pb
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
49
16 1
r
" #
¼ 4
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
19
16
r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
49
16 1
r ) b ¼ 0:17
For the 1HM,
a ¼1
2; b ! 1
a ¼ 1 ) tan 2pbffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4a2 1p
¼ 0 ) 2pbffiffiffi
3p
¼ p ) b ¼1
2ffiffiffi
3p ¼ 0:29
a ¼3
4) tan 2pb
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
43
4
2
1
s0
@
1
A ¼ 4
ffiffiffiffiffi
7
20
r
) pbffiffiffi
5p
¼ tan1 4
ffiffiffiffiffi
7
20
r
!
þ p ) b ¼ 0:61
(c) Inside the layer the amplitude of the displacements of the Love wave are given by
u02 ¼ 2A0 cos ks0H 1x3
H
h i
cos k s0H þ x1 ctð Þ
323 Surface waves
The nodes are the points where the amplitude is zero. For the 1HM the node is located at
the value of x3 which satisfies the relation
ks0H 1x3
H
¼3p
2
If we want a node located at x3 ¼ H/2 then
k ¼3p
s0H) b ¼
3
2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4a2 1p
If we substitute the values of a, ½, 1, and ¾ we obtain for b infinity, 0.86, and 1.34. The
infinite value of b corresponds to l¼ 0.
Focal parameters
173. Consider three stations with coordinates:
St1 ¼ 36.2 N, 4.8 E; St2 ¼37.0 N, 2.4 E; St3 ¼38.6 N, 4.0 E
An earthquake is recorded at the three stations with the following respective S-P
time intervals:
tSP1 ¼ 26:7 s
tSP2 ¼ 27:0 s
tSP3 ¼ 22:5 s
Given that the focus is at the surface, the P-wave velocity is constant and equal to 6
km s1, and Poisson’s ratio is 1/3, calculate the coordinates of the epicentre.
Given that Poisson’s ratio is 1/3,
s ¼1
3¼
l
2 lþ mð Þ) l ¼ 2m ) a ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
lþ 2m
r
s
¼ 2b
where a and b are the velocities of the P- and S-waves, respectively.
From the S-P time intervals, calling x the epicentral distance (Fig. 173):
tSP ¼
x
b
x
a¼
x
2b) x ¼ 2tSPb
Given that a ¼ 6 km s1 then b ¼ 3 km s1.
The epicentral distances corresponding to each station in kilometres and degrees are
x1 ¼ 26:7 6 ¼ 160 km ¼160
111:11¼ 1:44
x2 ¼ 27 6 ¼ 162 km ¼162
111:11¼ 1:46
x3 ¼ 22:5 6 ¼ 135 km ¼135
111:11¼ 1:22
324 Seismology
If xe and ye are the geographical coordinates of the epicentre,
36:2 xeð Þ2 þ 4:8 yeð Þ2 ¼ 1:442
37:0 xeð Þ2 þ 2:4 yeð Þ2 ¼ 1:462
38:6 xeð Þ2 þ 4:0 yeð Þ2 ¼ 1:222
Solving the system we find two possible solutions:
xe; yeð Þ ¼37:40 N; 3:83 Eð Þ37:63 N; 4:71 Eð Þ
174. A focus is at a depth h, in a medium with constant P- and S-wave speeds of
propagation a and b. Calculate an expression for the depth h in terms of the time
interval between the P and sP phases.
If L is the distance along the ray, the travel-time of the P-wave from the focus to the station
is given by (Fig. 174)
tP ¼
L
a
36°N
35°N
37°N
38°N
39°N
40°N
1°E
St3
St2
St1
E1
E2
3°E2°E 4°E 5°E 6°E
Fig. 173
325 Focal parameters
An sP-wave leaves the focus upward as an S-wave is reflected at the Earth’s surface and
converted into a P-wave that travels to the station. Its travel time is (Fig. 174)
tsP ¼
FS
bþSF
0
aþL
a
If we consider the length along the ray L from F and F0 to the station to be the same for both
waves, then the sP-P time interval is (Fig. 174)
tsP t
P ¼ t0 ¼
FS
bþSF
0
að174:1Þ
At the focus at depth h, the take–off angle of the direct P-wave is ih and the take-off angle
of the sP-wave is jh. We can write
FS ¼h
cos jh
cosðjh þ ihÞ ¼SF
0
FS) SF
0¼
h
cos jhðsin jh sin ih cos jh cos ihÞ
sin ih ¼a sin jh
b
By substitution in (174.1) we obtain
t0 ¼ h
cos jh
bþcos ih
a
175. The displacement vector l of an earthquake is (0, 1, 0) and the vector normal to
the plane of displacement n is (0, 1, 0). Determine:
(a) The components of the P-wave displacement at the point of azimuth 45 and angle
of incidence 30.
S
h
AF
ih
ih
jh F
A
Fig. 174
326 Seismology
(b) The kind of mechanism it represents.
(a) The elastic displacements due to a dislocation Du in the direction of li on a plane of
normal ni (Fig. 175) are given by
uPk ¼ u lnk lkdij þ m linj þ nilj
Gki;j
where Green’s function corresponding to the P-wave in the far field for an infinite medium
is given by
Gki ¼1
4pra2rgigkd t
r
a
and its derivative is
Gki;j ¼1
4pra3rgigkgj
_d t r
a
where gi are the direction cosines of the line from the focus to the observation point.
The amplitude of the displacement is then
uPk ¼u
4pra3rlnslsdij þ m linj þ ljni
gigkgj ð175:1Þ
In our problem the direction cosines of the ray of the waves arriving at the point are
g1 ¼ sin i cos az ¼
ffiffiffi
2p
4
g2 ¼ sin i sin az ¼
ffiffiffi
2p
4
g3 ¼ cos i ¼
ffiffiffi
3p
2
X3
X1
X2
l i= n i e
az
ir
P
Fig. 175
327 Focal parameters
The orientation of the source is given by li ¼ (0, 1, 0) and ni ¼ (0, 1, 0), and substituting in
(175.1) gives
uP1 ¼ A lþ 2mg22
g1
uP2 ¼ A lþ 2mg22
g2
uP3 ¼ A lþ 2mg22
g3
A ¼u
4pa3rr
Substituting the direction cosines of the ray we obtain,
uP1 ¼ A lþ1
4m
ffiffiffi
2p
4
uP2 ¼ A lþ1
4m
ffiffiffi
2p
4
uP3 ¼ A lþ1
4m
ffiffiffi
3p
2
(b) The mechanism corresponds to a fault on the (x1, x3) plane which opens in the
direction of its normal, x2, under tensional forces in that direction.
176. The focal mechanism of an earthquake can be represented by a double-couple
(DC) model. The orientation of the fault plane is azimuth 30, dip 90, and slip angle
0. Calculate:
(a) What kind of fault it is. Sketch it, indicating the direction of motion.
(b) The auxiliary plane.
(c) The azimuth of the stress axis.
(d) A wave incident at a station has azimuth 180 and angle of incidence at the focus
of 90. Calculate the amplitude of the components of the P-wave at that station.
(a) Given that the dip of the plane is 90, the fault plane is vertical, and since the slip
angle is 0, the motion is horizontal. Thus, it corresponds to a right lateral strike-
slip fault (Fig. 176).
Δu
ϕA
δA
n
l
N
Fig. 176
328 Seismology
(b) From the azimuth (’ ¼ 30), dip (d ¼ 90), and slip (l ¼ 0) we can calculate the
unit vectors li and ni which give the direction of the fracture and of the normal to
the fault plane:
n1 ¼ sin d sin’ ¼ 1
2¼ sinYn cosFn
n2 ¼ sin d cos’ ¼
ffiffiffi
3p
2¼ sinYn cosFn
n3 ¼ cos d ¼ 0 ¼ cosYn
l1 ¼ cos l cos’þ cos d sin l sin’ ¼
ffiffiffi
3p
2¼ sinYl cosFl
l2 ¼ cos l sin’ cos d sin l cos’ ¼1
2¼ sinYl cosFl
l3 ¼ sin l sin d ¼ 0 ¼ cosYl
’A ¼ Fn þ 90 ) Fn ¼ 120
dA ¼ Yn ¼ 90
lA ¼ sin1 cosYl
sinYn
¼ 0 ) Yi ¼ 90
whereY and F are the spherical coordinates for the vectors n and l (r ¼ 1, unitary vectors)
and for the auxiliary plane,
’B ¼ 30þ 90 ¼ 120
dB ¼ 90
lB ¼ sin1 cosYn
sinYl
¼ 0
(c) The T-axis is on the same plane as ni and li at 45 between them, and the direction
cosines are
T1T2T3
0
@
1
A ¼n1 n2 n3l1 l2 l3Z1 Z2 Z3
0
@
1
A
1ffiffi
2p
1ffiffi
2p
0
0
@
1
A ð176:1Þ
where Zi is the axis normal to ni and li, that is, Zi ¼ ni li which results in Zi ¼ (0, 0, 1).
Substituting in (176.1) we obtain Ti ¼
ffiffiffi
3p
1
2ffiffiffi
2p ;
ffiffiffi
3p
þ 1
2ffiffiffi
2p ; 0
. The azimuth of the T-axis is
FT ¼ tan1 T2
T1
¼ tan1
ffiffiffi
3p
þ 1ffiffiffi
3p
1
¼ 75
(d) The direction cosines of the direction from the focus to the station are
g1 ¼ sin ih cos az ¼ 1
g2 ¼ sin ih sin az ¼ 0
g3 ¼ cos ih ¼ 0
329 Focal parameters
The amplitude of the displacements for a shear fracture or double-couple (DC) source in an
infinite medium is given by
R ih; azð Þ ) uPj ¼ A nilk þ nk lið Þgigkgj
uP1 ¼ 0:84A
uP2 ¼ 0
uP3 ¼ 0
A ¼M0
4pa3rr
177. An earthquake is caused by a shear fracture. The vectors n and l (normal and
direction of travel) are, in terms of the angles F and Y,
n ¼ ð57:13; 66:44Þ
l ¼ ð305:96; 50:09Þ
Calculate the orientation of the fault plane, the auxiliary plane, and the tension (T )
and pressure (P) stress axes.
The orientation of the fault plane and auxiliary plane in terms of the angles’, d, and l (azimuth,
dip, and slip) are found directly from the given values (Fig. 177), using the following relations:
’A ¼ Fn þ 90 ¼ 147:13
dA ¼ Yn ¼ 66:64
lA ¼ sin1 cosYl
sinYn
¼ 135:68
’B ¼ Fl þ 90 ¼ 35:96
dB ¼ Yl ¼ 50:09
lB ¼ sin1 cosYn
sinYl
¼ 31:12
Z
X
P
N
n
Y
T
l
fn
qp
qn
Fig. 177
330 Seismology
To calculate the T and P axes, we calculate first the direction cosines of the l and n axes
from the given angles:
x1 ¼ sinY cosF
x2 ¼ sinY sinF
x3 ¼ cosY
)
n1 ¼ 0:50; l1 ¼ 0:45
n2 ¼ 0:77; l2 ¼ 0:62
n3 ¼ 0:40; l3 ¼ 0:64
Given that the T and P axes are on the same plane as n and l at 45 between them, we can
write, as in Problem 176,
T1T2T3
0
@
1
A ¼l1 n1 Z1l2 n2 Z2l3 n3 Z3
0
@
1
A
1ffiffiffi
2p
1ffiffiffi
2p
0
0
B
B
@
1
C
C
A
ð177:1Þ
where the Z-axis is normal to n and l and is found by Z ¼ n l. Its direction cosines
are (Z1, Z2, Z3) ¼ (0.72, 0.14, 0.66). By substitution of ni, li, and Zi in (177.1) we
obtain
T1 ¼ 0:67 ¼ sinYT cosFT
T2 ¼ 0:11 ¼ sinYT sinFT
T3 ¼ 0:74 ¼ cosYT
9
>
=
>
;
) TðYT ¼ 42:27;FT ¼ 9:32Þ
In the same way for the axis P
P1
P2
P3
0
@
1
A ¼l1 n1 Z1l2 n2 Z2l3 n3 Z3
0
@
1
A
1ffiffiffi
2p
1ffiffiffi
2p
0
0
B
B
@
1
C
C
A
! PðYP ¼ 80:02;FP ¼ 268:03Þ
178. The seismic moment tensor relative to the geographical axes (X1, X2, X3) (north,
east, nadir) is
Mij ¼2 1 1
1 0 1
1 1 2
0
@
1
A
Find the values of the principal stresses, and the orientation of the tension and
pressure stress axes.
First we calculate the eigenvalues ofMij. SinceMij is a symmetric tensor its eigenvalues are
real and the corresponding eigenvectors mutually orthogonal (Problem 111):
2 s 1 1
1 0 s 1
1 1 2 s
¼ 0 ) s ¼3
2
1
0
@
1
A
Ordered by magnitude, the three eigenvalues are
s1 ¼ 3; s2 ¼ 2; s3 ¼ 1
331 Focal parameters
The diagonalized matrix is
Mij ¼3 0 0
0 2 0
0 0 1
0
@
1
A
In this formMij is referred to the coordinate system formed by the eigenvectors or principal
axes. Given that the sum of the elements of the principal diagonal is not zero, the source
has net volume changes. Then, we can separate Mij into two parts: an isotropic part with
volume changes (ISO) and a deviatoric part without volume changes. The second part can
be separated into two parts: a part corresponding to a double-couple or shear fracture (DC)
and a part corresponding to a non-double-couple source usually expressed as a compen-
sated linear vector dipole (CLVD). Thus the moment tensor is separated into three parts,
namely
M ¼ M ISO þMDC þMCLVD
The isotropic part is given by
M ISO ¼ s0 ¼1
3s1 þ s2 þ s3ð Þ ¼
4
3
The deviatoric part (DCþCLVD) is given by
M 0ij ¼ Mij dijso
and in our case
M 0ij ¼
s1 0 0
0 s2 0
0 0 s3
0
@
1
A ¼
5
30 0
02
30
0 0 7
3
0
B
B
B
B
@
1
C
C
C
C
A
Now we separate this part into two parts, DC and CLVD:
M 0ij ¼ MDC þMCLVD
M 0ij ¼
1
2s1 s3ð Þ 0 0
0 0 0
0 0 1
2s1 s3ð Þ
0
B
B
B
@
1
C
C
C
A
þ
s2
20 0
0 s2 0
0 0 s2
2
0
B
B
B
@
1
C
C
C
A
M 0ij ¼
2 0 0
0 0 0
0 0 2
0
B
@
1
C
Aþ
1
30 0
02
30
0 0 1
3
0
B
B
B
B
B
@
1
C
C
C
C
C
A
332 Seismology
The orientation of the P and T axes is calculated from the double-couple part MDC:
MDCij ¼ 2
1 0 0
0 0 0
0 0 1
0
@
1
A
We can find the n and l axes:
MDCij ¼ M0ðlinj ljniÞ
~n :1ffiffiffi
2p ; 0;
1ffiffiffi
2p
) nðYn ¼ 45;Fn ¼ 0Þ
~l :1ffiffiffi
2p ; 0;
1ffiffiffi
2p
) lðYl ¼ 45;Fl ¼ 0Þ
In the same way as in Problems 176 and 177, we determine T and P from n and l, finding
first Z ¼ n l:
T1T2T3
0
@
1
A ¼l1 n1 Z1l2 n2 Z2l3 n3 Z3
0
@
1
A
1ffiffiffi
2p
1ffiffiffi
2p
0
0
B
B
B
@
1
C
C
C
A
T1 ¼ 1 ¼ sinYT cosFT
T2 ¼ 0 ¼ sinYT sinFT
T3 ¼ 0 ¼ cosYT
! TðYT ¼ 90;FT ¼ 0Þ
For the P-axis,
P1
P2
P3
0
@
1
A ¼l1 n1 Z1l2 n2 Z2l3 n3 Z3
0
@
1
A
1ffiffiffi
2p
1ffiffiffi
2p
0
0
B
B
B
@
1
C
C
C
A
PðYP ¼ 0;FP ¼ 0Þ
179. The magnitude Ms of an earthquake as calculated for surface waves of period
20 s is 6.13.
(a) Calculate the amplitude of these waves at a station 3000 km away. If the instru-
ment’s amplification is 1500, what will be the amplitude of the seismogram’s
waves and the seismic energy?
(b) IfMs ¼Mw, and the area of the fault is 12 km 8 km with m¼ 4.4104 MPa, find
the fault slip Du.
(a) The surface wave magnitude Ms is given by
Ms ¼ logA
Tþ 1:66 logþ 3:3
333 Focal parameters
where A is the ground motion amplitude, T is the period of the wave, and D is epicentral
distance in degrees. Knowing the magnitude and period of the waves we can calculate the
wave amplitude:
6:13 ¼ logA
Tþ 1:66 log
3000
111:11þ 3:3
) logA
T¼ 0:454 ) A ¼ 2:84 20 1500 ¼ 8:5 cm
We have reduced the ground motion to the amplitude of the seismogram using the
amplification of the instrument (1500).
Knowing the magnitude we can calculate the seismic energy:
logEs ¼ 11:8þ 1:5Ms ) Es ¼ 1021ergs ¼ 1014 J
(b) Mw ¼ 6.13 ¼ 2/3 log M0 – 6.1; M0 ¼ 2.191018 Nm
If M0 ¼ mDu S, with S ¼ 12 8 ¼ 9.6107 m2, then
u ¼M0
mS¼
2:19 1018 Nm
4:4 1010 Nm2 9:6 107 m2¼ 0:52m
334 Seismology
5 Heat flow and geochronology
Heat flow
180. Assume that the temperature variation within the Earth is caused by gravita-
tional forces under adiabatic conditions. Knowing that the coefficient of thermal
expansion at constant pressure is aP ¼ 2105 K1 and the specific heat at constant
pressure is cP ¼ 1.3 kJ kg1 K1, determine an expression for the gradient of the
temperature with depth. Compare it with the value observed at the surface which is
30 K km1, knowing that, at 200 km depth, T ¼ 1600 K.
Under adiabatic conditions, there is no heat flow and the variation of pressure with depth z
is a function of gravity g and density r:
dP ¼ rgdz ð180:1Þ
Using the first and second laws of thermodynamics
dU ¼ dQ PdV
dQ ¼ TdS
where Q is the heat, U is the internal energy, S is the entropy, T is the absolute temperature,
P is the pressure, and V is the volume.
If we use the specific variables (variables divided by mass) we can write
du ¼ dq pdv
dq ¼ Tds
Considering that
dS ¼@S
@T
P
dT þ@S
@P
T
dP
we can write
dq ¼ Tds ¼ T@s
@T
P
dT þ T@s
@P
T
dP
According to the definition of specific heat at constant pressure, cP and the increase in heat
dq are given by
335
cP ¼ T@S
@T
P
dq ¼ cPdT þ T@s
@P
T
dP
ð180:2Þ
The Gibbs function G is defined as
G ¼ u Tsþ pv
Taking the differential in this expression, and taking into account the second law of
thermodynamics
du ¼ Tds pdv
we obtain
dG ¼ vdp sdT
If we compare this expression to the differential of the Gibbs function
dG ¼@G
@p
T
dpþ@G
@T
p
dT
we differentiate again and using the Schwartz theorem we obtain
@v
@T
P
¼ @s
@P
T
But the coefficient of thermal expansion is defined as
ap ¼1
v
@v
@T
p
In consequence we can write Equation (180.2) as
dq ¼ cpdT Tvapdp
In our case the process is adiabatic and in consequence using this equation and Equation
(180.2) we obtain
dT
dz¼
Tapg
cpð180:3Þ
where we have taken into account that the variables are by unit mass so rv ¼ 1, and
substituting the values we obtain:
dT
dz¼
1600 2 105 10KK1 ms2
1:3 103 J kg1 K1¼ 0:25K km1
We observe that this result is two orders of magnitude lower than the observed values. This
shows that observations correspond to heat flow at the lithosphere and are not satisfied by
purely adiabatic conditions.
336 Heat flow and geochronology
181. If the Earth’s temperature gradient is 1 C/30 m, calculate the heat loss per
second due to conduction from its core. Compare this with the average power
received from the Sun.
Data:
Thermal conductivity K ¼ 4 Wm1 C1.
Earth’s radius R ¼ 6370 km.
Solar constant: 1.35 kWm2.
The heat flow is given by
_q ¼ KAdT
dr
where A is the area of the Earth’s surface
A ¼ 4pR2 ¼ 5:10 1014 m2
and
dT
dr¼
1C
30m
Substituting the values
_q ¼ 6:80 1013 J s1 ¼ 1:63 1013 cal s1
the average power received on the Earth’s surface by the radiation from the Sun is
1:35 103W
m25:10 1014 m2 ¼ 6:89 1017 J s1 ¼ 1:65 1017 cal s1
In consequence, from these values we can see that on the Earth’s surface the average solar
power is much larger than that due to the heat flow from inside the Earth.
182. At the Earth’s surface, the heat flow is 60 mWm2 and T0 ¼ 0 C. If all the heat is
generated by the crust at whose base the thermal conductivity is K¼ 4Wm1 C1, and
T is 1000 C, determine the thickness of the crust and the heat production per unit volume
If we assume that all the heat is generated at the crust and there is no heat flow from the
mantle at the crust base, then we can write
_qjz¼H ¼ 0
Using the temperature equation for a flat Earth for one-dimensional heat-flow and the
stationary case we can write
T ¼ e
2Kz2 þ
_q0Kzþ T0
e ¼_q0H
For z ¼ H:
TH ¼_q0H
2Kþ T0 ) H ¼
ðTH T0Þ2K
_q0
337 Heat flow
Substituting the values given in the problem we obtain
H ¼ 133:3 km
and the heat production by unit of volume is
e ¼_q0
H¼ 4:5 104 mWm3
183. Consider the crust to be H ¼ 30 km thick and the heat flow at the surface to be
60 mWm2.
(a) If all the heat is generated in the crust, what is the value of the heat generated per
unit volume? (Take K ¼ 3 W m1 K1)
(b) If all the heat is generated in the mantle with a distribution Aez=H mWm3, what
is the value of A? What is the temperature at 100 km depth?
(a) We solve the heat equation for a stationary one-dimensional case, assuming a flat
Earth with one-dimensional flow in the z-direction (vertical) positive downward.
In this case the solution of the heat equation is given by
T ¼ e
2Kz2 þ
_q0Kzþ T0
where e is the heat generated by unit volume and time, K is the thermal conductivity, q0 and
T0 are the heat flow and temperature at the surface of the Earth, respectively
If all the heat is generated at the crust we can write (Fig. 183)
z ¼ H ! _q0 ¼ 0
The heat generated by unit volume is
_qjz¼H ¼ 0 ¼ KdT
dz
z¼H
¼ K e2z
2Kþ
_q0K
z¼H ¼ 0
e ¼_q0H
¼60 103
30 103¼ 2 106 Wm3
(b) If all heat is generated in the mantle with distribution
e ¼ Aez=H
the heat equation is
d2T
dz2¼
e
K¼
A
Kez=H
and the solution is given by
T ¼ A
KH2ez=H þ Czþ D ð183:1Þ
where C and D are constants of integration. They may be estimated from the boundary
conditions at the surface
338 Heat flow and geochronology
z ¼ 0 ! T ¼ T0 ¼ A
KH2 þ D ! D ¼ T0 þ H2 A
K
z ¼ 0 ! _q ¼ _q0 ¼ 0 ! C ¼ A
KH
Substituting in (183.1) we obtain
T ¼ A
KH2ez=H
AH
Kzþ T0 þ H2 A
K¼
AH
KHez=H zþ H
þ T0 ð183:2Þ
If the heat has its origin in the mantle, the flow at the base of the crust is
_qjz¼H ¼ _qo ¼ KdT
dz
z¼H
) A ¼_q0
H e1 1ð Þ
The temperature at z = 100 km may be estimated from (183.2) assuming that T0 = 0:
TðzÞ ¼_q0
K e1 1ð ÞHez=H zþ H
Tðz ¼ 100Þ ¼60 103
3 e1 1ð Þ30 103e100=30 100 103 þ 30 103
¼ 2249K
184. Calculate the thickness of the continental lithosphere if its boundary coincides
with the 1350 C geothermal, knowing that the surface temperature is 15 C,
the heat flow at the surface is _q0 ¼ 46mWm2, the lithospheric mantle’s thermal
conductivity is K = 3.35 Wm1 K1, and the radiogenic heat production is
P = 0.01 103
mWm3
The geothermal equation at depth z is given by:
Tz ¼ T0 þ_q0Kðz z0Þ
P0
2Kðz z0Þ
2 ð184:1Þ
T0
TH
H
q0
•
qH
•
Fig. 183
339 Heat flow
whereK is the thermal conductivity, T0 is the temperature at the surface of the Earth (in K), _q0
is the heat flow at the surface, and P0 is the radiogenic heat production at the Earth’s surface.
At the Earth’s surface z0 = 0, and Equation (184.1) becomes:
Tz ¼ T0 þ_q0Kz
P0
2Kz2
P0
2Kz2
_q0Kzþ ðTz T0Þ ¼ 0
So
z ¼
_q0K
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
_q0K
2
4P0
2KðTz T0Þ
s
2P0
2K
¼_q0
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
_q20 2P0KðTz T0Þp
P0
Substituting the data given in the problem:
K ¼ 3.35 W m1 K1
_q0 ¼ 46 103 Wm2
P0 ¼ 0.01106 mW m3
Tz ¼ 1623 K
T0 ¼ 288 K
we obtain two solutions, but only z = 98.27 km is realistic (the second one gives a depth
larger than the Earth’s radius).
185. On the surface of an Earth of radius 6000 km, the temperature is 300 K, the heat
flow is 6.7mWm2, and the thermal conductivity is 3Wm1K1. If the heat produc-
tion per unit volume inside the Earth is homogeneously distributed, what is the
temperature at the centre of the planet?
We begin solving the problem of heat conduction inside a sphere with constant internal
heat generation per unit volume e and conductivity K. The differential equation for heat
conduction with spherical symmetry is
1
r2
d
drr2 dT
dr
¼ e
Kð185:1Þ
Integrating twice and using the boundary conditions:
Surface: r = R ! T = T0Center: r = 0 ! T finite
we obtain the solution
T ¼ T0 þe
6KR2 r
2
ð185:2Þ
The heat flow is given by
_q ¼ KdT
dr) _q0ðr ¼ RÞ ¼ 6:7mWm2
¼ Kd
drT0 þ
e
6KR2 r
2
¼e r
3
340 Heat flow and geochronology
Solving for the heat production e
e ¼ 3:35 109 Wm3
From Equation (185.1) the temperature at the Earth’s centre is
Tr¼0 ¼ 300þ3:35 109
6 3 62 1012 ¼ 7000K
186. Consider a spherical Earth of radius R = 6000 km and a core at R/2, in which
there is a uniform and stationary distribution of ε heat sources per unit volume. The
heat flow at the surface is 5 mW m2, the thermal conductivity is 3 W m1 K1, and
the temperature at the core–mantle boundary is 4000 K. Calculate the temperature at
the Earth’s surface.
We consider the problem as one of heat conduction inside a sphere with conductivity K and
constant heat generation per unit volume e inside the core (radius R/2). We begin with
Equation (185.1)
1
r2
d
drr2 dT
dr
¼ e
K
The boundary conditions at the core–mantle boundary and its centre are
r ¼ R=2 ! T ¼ TN
r ¼ 0 ! T finite
where TN is the temperature at the core–mantle boundary
Integrating twice we obtain
T rð Þ ¼ TN þe
6K
R
2
2
r2
!
The heat flow is given by
_q0 r ¼ Rð Þ ¼ KdT
dr
r¼R
¼eR
3) e ¼
_q0 3
R¼ 2:5 109 Wm3
Then, the temperature at the Earth’s surface is
T r ¼ Rð Þ ¼ 4000þ2:5 109
6 3
6000 103
2
2
60002 106
!
¼ 250K
187. Consider the Earth of radius R0 = 6000 km formed by a spherical crust
with its base at 500 km and constant thermal conductivity K. If the temperature
at the base of the crust is T1 and at the surface of the Earth is T0 = 0 C,
determine:
(a) An expression for the heat flow through the crust.
341 Heat flow
(b) An expression for the temperature distribution within the Earth.
(c) The temperature at the base of the crust if, at that depth, _q ¼ 5:5 1013 W and
K = 4 W m1 C1.
(a) We assume a spherical Earth where the temperature varies only in the radial
direction. Then we can solve the problem as one of spherical unidirectional flow.
For the stationary case, when the conductivity and heat generation are constant, the
Fourier law may be written as
_q ¼ KAdT
drð187:1Þ
where A ¼ 4pr2 is the area in the normal direction to the heat flow. Integrating this
equation:
_q
4p
ðR0
r1
dr
r2¼ K
ðT0
T1
dT
where the conditions at the Earth’s surface are, r = R0 !T = T0 and at the base of the crust,
r = r1 !T = T1.
Solving Equation (187.1), assuming that K is constant, we obtain
_q ¼ 4pK1r1 1
R0
ðT0 T1Þ ¼4pKr1R0
R0 r1
ðT0 T1Þ ð187:2Þ
(b) The temperature distribution inside of the Earth may be obtained by integration of
Equation (187.1):
_q
4p
ð
r
r1
dr
r2¼ K
ðT
T1
dT
T rð Þ ¼ T1 þR0ðr r1Þ
ðR0 r1ÞrðT0 T1Þ
(c) The radial distance to the base of the crust is r2 = 5500 km, so, using expression
(187.2), we obtain
T1 ¼ T0 þ_qðR0 r1Þ
4pKr1R0
¼ 0þ5:5 1013 500 103
4p 4 5500 103 6000 103¼ 16579 C
This result implies a constant increase of temperature from the Earth’s surface
of 1 ºC each 33.2 m similar to the observed gradient in the real Earth of 1 ºC
per 30 m
188. Assume that the heat flow inside the Earth is due to solar heating of the
Earth’s surface. Calculate the maximum penetration of this flow in the diurnal and
annual cycles. Take as typical values for the Earth K = 3 Wm1 K1, r = 5.5 g cm3,
Cv = 1 kJ kg1
K1.
342 Heat flow and geochronology
We assume the heat propagation inside the Earth coming from the solar radiation on its
surface as unidirectional flow thermal diffusion (inside the Earth) with periodic variation of
surface temperature. The diffusivity equation is
k@2T
@z2¼
@T
@tð188:1Þ
where the thermal diffusivity is k ¼K
rCv
, K is the thermal conductivity, r is the density,
and Cv is the specific heat at constant volume. We solve Equation (188.1) using the
separation of variables
T z; tð Þ ¼ Z zð Þy tð Þ
Substituting in (188.1) we obtain the solution
Z zð Þ ¼ Aeaz þ Beaz
y tð Þ ¼ Ceka2t
where a is the constant of separation of variables. Using the boundary condition of periodic
flow and the temperature T0 at the Earth’s surface,
z ¼ 0 ) T ¼ T0eiot
and as Z(z) exists only inside the Earth, B = 0. At the surface, z = 0, so
ACeka2t ¼ T0e
iot
Then
AC ¼ T0
ka2 ¼ io
But putting, i ¼ 121þ ið Þ2, we have
a ¼ 1þ ið Þ
ffiffiffiffiffiffi
o
2k
r
Then, we can write the temperature variation inside of the Earth as:
T z; tð Þ ¼ T0 exp
ffiffiffiffiffiffi
o
2k
r
zþ i
ffiffiffiffiffiffiffiffi
o
2kz
r
þ ot
This equation corresponds to a periodic wave, with angular frequency o propagating for
positive z values (to the Earth’s interior) and with the amplitude decreasing with depth. The
propagation velocity and wavelength are given by
v ¼
ffiffiffiffiffiffi
2k
o
r
l ¼ 2pv ¼ p
ffiffiffiffiffiffi
8k
o
r
343 Heat flow
The values of l corresponding to the daily and annual cycles give their maximum penetration:
Daily cycle:
o ¼2p
24 60 60¼ 7:2 105 s1
k ¼K
rCv
¼3Wm1 K1
5:5 103 K gm3 103 J Kg1 K1¼ 0:5 106 m2 s1
Then
l ¼ p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
8 0:5 106
7:2 105
r
¼ 0:74m
Annual cycle:
o ¼2p
365 24 60 60¼ 2 107 s1
l ¼ 14m
The penetration of the solar radiation as periodic heat conduction inside the Earth is very
shallow due to the poor heat conduction.
189. Consider a lithospheric plate of 100 km thickness created from asthenospheric
material originating from a ridge in the asthenosphere with constant temperature Ta and
in which no heat is generated. Given that k = 106m² s1, that the temperature at the base
of the lithosphere is 1100 C, and in the asthenosphere is 1300 C, calculate the age of the
plate, and, if the velocity of drift is 2 cm yr1, how far it has moved away from the ridge.
The heat propagation inside the plate is given by:
K
rcv
@2T
@x2þ@2T
@z2
¼ u@T
@xð189:1Þ
where T is the temperature, r is the density, cv is the specific heat at constant volume, and u
is the horizontal velocity of the plate in the x-direction (normal to the plate front). If we
assume that the horizontal conduction of heat is insignificant in comparison with the
horizontal advection and vertical conduction, we can write, using the following change
of variable t = x/u,
K
rcv
@2T
@z2¼
@T
@t
Integrating this equation and using the boundary conditions at the ridge and surface:
x ¼ 0 ! T ¼ Ta
z ¼ 0 ! T ¼ 0
we obtain for the temperature distribution
T z; tð Þ ¼ Taerfz
2ffiffiffiffiffi
ktp
344 Heat flow and geochronology
where
k ¼K
rcv
erf xð Þ ¼2ffiffiffi
pp
ðx
0
ey2dy
Substituting the data of the problem,
1100 ¼ 1300erfL
2ffiffiffiffiffi
ktp
) erfL
2ffiffiffiffiffi
ktp
¼ 0:846
Values of the error function, erf(x), may be obtained from tables. If erf(x) = 0.846, x = 1.008, then
L
2ffiffiffiffiffi
ktp
¼ 1:008 ) t ¼L2
4k 1:0082¼
1010
4 106 1:0082
¼ 2:5 1015 s ¼ 79Myr
If the displacement velocity is 2 cm yr1, the plate has moved 1580 km.
190. If the concentrations of 235U and 235Th in granite are 4 ppm and 17 ppm,
respectively, and the respective values of heat production are 5.7 104 W kg1
and 2.7 105 W kg1, respectively, calculate the heat flow at the base of a granite
column of 1 m² cross-section and 30 km height (the density of granite is 2.65 g cm3).
We estimate first the mass of the granite column:
M ¼ rV ¼ 2:65 103 1 30 103 ¼ 7:95 107 kg
If the concentration of 235U in the granite is 4 ppm, its quantity in the column is
235U :4 7:95 107
106¼ 318 kg
Then the heat flow due to the 235U is
_q ¼ 5:7 104 318 ¼ 181:26mWm2
For 235Th, the heat flow is
235Th : 7:95 107 17 106 ¼ 1351:5 kg
_q ¼ 1351:5 2:7 105 ¼ 36:49mWm2
Geochronology
191. The mass of 1 millicurie of 214Pb is 3 1014 kg. Calculate the value of the decay
constant of 214Pb.
The mass of the sample is
345 Geochronology
3 1014 kg ¼ 3 1017 g ¼N
N0
M
where N is the number of atoms in the sample, N0 is Avogadro’s number ¼ 6.02 1023,
and M is atomic number ¼ 214. Solving for N we obtain
N ¼3 1017 6:02 1023
214¼ 8:44 1038 atoms
The correspondence of a curie is
1 curie ¼dN
dt¼ lN ¼ 3:7 1010 disintegrations s1
where l is the decay constant and t is the time. Then
l ¼3:7 107
8:44 1038¼ 0:44 1031 s1
192. The isotope 40K decays by emission of b particle with a half-life of 1.83 109
years. How many b decays occur per second in one gram of pure 40K?
The average life t of a radioactive material is a function of the decay constant l:
t ¼1
l! l ¼
1
1:83 109¼ 0:55 109 yr1 ¼ 1:73 102 s1
The number of atoms N contained in 1 g of 40K may be estimated from Avogadro’s number
N0 and the atomic number M:
1 ¼N
N0
M ! N ¼1 6:02 1023
40¼ 0:15 1023 atoms
The rate of disintegration is given by
dN
dt¼ lN ¼ 0:26 1021 Bq
193. The half-life of 238U is 4468 106 yr and of 235U is 704 106 yr. The ratio235U/238U in a sample is 0.007257. Given that the ratio was 0.4 at the time of
formation, calculate the sample’s age.
From the half-life T1/2 we can obtain the decay constant l:
T1=2 ¼0:693
l
l238 ¼0:693
4468 106¼ 1:5510 1010 yr1
l235 ¼0:693
704 106¼ 9:8434 1010 yr1
The number N of disintegrating atoms at time t is given by
N ¼ N0 elt ð193:1Þ
346 Heat flow and geochronology
where N0 is the number of atoms at time t ¼ 0. Then
N235 ¼ N 2350 el235t
N238 ¼ N 2380 el238t
If we divide these equations:
N235
N238
¼N 2350
N 2380
eðl238l235Þt
Substituting the values given in the problem we obtain
4:0095 ¼ 8:2924 1010t ! t ¼ 4:8 109 yr
194. Date a meteorite which contains potassium knowing that its content of 40K is
1.19 1014 atoms g1, of 40Ar is 4.14 1017 atoms g1, and that the half-life of40K!40Ar is 1.19 109 years.
We obtain the decay constant of the 40K from its half-life T1/2:
l ¼0:693
T1=2
l ¼0:693
1:19 109¼ 0:58 109 yr1
We can solve the problem considering it as a case of radioactive ‘parent’ atom disinte-
grating to a ‘daughter’ stable atom. At time t ¼ 0 we have n0 ‘parent’ atoms at the sample,
and at time t there remain NR radioactive atoms in the sample and NE daughter atoms,
from the disintegration of the n0 parent atoms:
n0 ¼ NRþ NE
But
n0
nt¼
NRþ NE
NR¼ 1þ
NE
NR
From Equation (193.1) we obtain the age of the sample:
nt ¼ n0 elt !
n0
nt¼ elt ¼ 1þ
NR
NRð194:1Þ
so
t ¼1
lln 1þ
NE
NR
NE, the number of atoms of 40K in the sample, can be estimated from the number of atoms
contained in 1 g of potassium:
1 ¼N
6:023 102340 ! N ¼ 1:506 1022 atoms g1
347 Geochronology
so
NE ¼ 1:506 1022 1:19 1014 ¼ 1:792 1036 atoms g1
Then the age of the meteorite from (194.1) is
t ¼1
lln 1þ
1:792 1036
4:41 1017
¼ 7:4 1010 yr
195. At an archaeological site, human remains were found and assigned an age of
2000 years. One wants to confirm this with 14C dating whose half-life is 5730 yr. If the
proportion of 14C/12C in the remains is 6 1013, calculate their age. (Assume that at
the initial time the 14C/12C ratio was 1.2 1012.)
The decay constant l may be obtained from the half-life:
T1=2 ¼0:693
l! l ¼
0:693
5730¼ 1:2094 104 yr1
The activity in a sample is given by
R ¼ R0 elt
where R0 ¼ (14C/12C)t¼ 0 and R ¼ (14C/12C).
Then the age of the remains is
t ¼1
lln
R0
R
¼1
1:2094 104ln
1:2 1012
9 1013
¼ 2379 yr
196. Mass spectrometry of the different minerals in an igneous rock yielded the
following table of values for the concentrations of 87Sr originating from the radio-
active decay of 87Rb and of 87Rb, with the concentration expressed relative to the
concentrations of 86Sr of non-radioactive origin.
Express on a 87Sr/86Sr–87Rb/86Sr diagram the isochron corresponding to the forma-
tion of the rock, and calculate the age of the rock. Take l ¼ 1.42 1011 yr1.
For the decay of 87Rb
87Sr
now¼ 87Sr
0þ 87Rb
nowelt 1
ð196:1Þ
Mineral 87Sr /86Sr 87Rb /86Sr
A 0.709 0.125
B 0.715 0.418
C 0.732 1.216
D 0.755 2.000
E 0.756 2.115
F 0.762 2.247
348 Heat flow and geochronology
where [87Sr]now and [87Rb]now are the number of atoms of each isotope at time t, [87Sr]0 is the
amount of original number of atoms of the isotope 87Sr [87Sr]now, and l is the decay constant.
Equation (196.1) may be written as
87Sr86Sr
now
¼87Sr86Sr
0
þ87Rb86Sr
now
elt 1
ð196:2Þ
This equation corresponds to a line (y ¼ a þ bx) with intercept87Sr86Sr
0
and slope (elt 1),
which is called an isochron.
If we plot the values given in the problem (Fig. 196) we can obtain the equation of the
line by least-squares fitting:
y ¼ 0:025xþ 0:705
The age of the sample can be obtained from the slope b ¼ 0.025:
el t 1 ¼ b
t ¼lnð1þ bÞ
l
Substituting the values of b and l:
t ¼ 1:72 109 yr:
0.0 0.5 1.0 1.5 2.0 2.50.70
0.72
0.74
0.76
0.80
87S
r8
6S
r
0.78
87Rb86Sr
Fig. 196
349 Geochronology
197. Magma with a material proportion of 87Sr/86Sr equal to 0.709 crystallizes
producing a series of rocks with different concentrations of 87Rb with respect to the
content of 86Sr:
(a) Calculate the proportions of 87Sr/86Sr and 87Rb/86Sr that these rocks will have
after 500 Myr. Take l ¼ 1.421011 yr1.
(b) Express in a 87Sr/86Sr–87Rb/86Sr diagram the isochrons corresponding to t ¼ 0
and t ¼ 500 Myr.
(a) Using the same method as in the previous problem, we can write
87Sr86Sr
¼87Sr86Sr
0
þ87Rb86Sr
elt 1
¼ 0:709þ87Rb86Sr
e1:4210115108 1
Sample 87Rb/86Sr
A 1.195
B 2.638
C 4.892
D 5.671
0 1 2 3 4 5 6 70.70
0.72
0.76
0.74
0.78
t = 500 Ma
t = 0
87Rb86Sr
87S
r86S
r
Fig. 197
350 Heat flow and geochronology
The results for each rock are given in the following table
(b) For t ¼ 500 Myr, we carry out a least-squares fitting to obtain the isochron,
which results in
y = 0.007x þ 0.709
In Fig. 197 the isochrones corresponding to t ¼ 0 and t ¼ 500 Myr are shown.
Sample 87Sr/86Sr 87Rb/86Sr
A 0.717 1.187
B 0.728 2.619
C 0.744 4.857
D 0.749 5.631
351 Geochronology
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