GG 450 Lecture 2GRAVITY BASICS

1/17/08

GRAVITY

Gravity, the force that effects our lives more than any other we think of as a constant, but changes in gravity are significant and relatively easy to measure with modern instruments. Since gravity changes with mass and distance away from mass, it is simple to make models of gravity anomalies, but in reality, most gravity anomalies can be fit by a host of models (non-unique). Even so, gravity measurements are often important supplements to other geophysical measurements.

What IS gravity?? Mass attracts mass, but how? Where’s the “spring”?

Gravity and electro-magnetic forces generate FORCE FIELDS. All mass feels the effects of all other mass in the universe, but, since the effect decreases as 1/r2, where r is the distance between the masses, we can often ignore all but the closest and largest masses.

Many forces are POINT FORCES, acting at a point, like poking someone with a finger, force fields act uniformly over space. THIS is why gravitational force can't be felt.

You don't feel gravity pulling you down, you feel the earth pushing you up.

Geophysical studies using these fields are called the POTENTIAL FIELD METHODS, because we can find a SCALAR value at all points in space which tells us the potential effect on any body caused by all masses considered.

The effect on any body can then be determined by differentiation. Just as a body will not feel a magnetic force if it has no susceptibility, a body will feel no gravitational force if it has no mass. More about potential later.

€

F =Gm1m2

r2

Newton's Law of Gravity describes how the force of gravity behaves:

where F is the gravitational force of attraction between the two masses, G is the universal gravity constant (6.6732x10-11 nt•m2/kg2, often called “Big G”), and r is the distance between the two masses. How big is this force? What is the gravitational force between two car-sized masses about 2 meters apart?

F=6.67x10-11 103 103/(22)= 1.67x10-5 nt = 1.67 dynes

This is a VERY small force. A dyne is the force of a mosquito slamming into a wall at 1 cm/s. Yet, this is the force that holds the universe together.

(6.1)

Some points of interest:• The attraction of a SPHERE is the same as the

attraction of an equal mass located at the center of the sphere.

• The gravitational attraction of a spherical shell of mass is the same as the attraction of the same mass concentrated at the center of the shell - as long as the observation is made OUTSIDE the shell.

• The gravitational attraction of a spherical shell when the observation is made INSIDE the shell is zero. Why?

Outside a spherical shell the gravity is as though all the massof the shell were concentrated at the middle.But INSIDE the shell, the gravitycaused by the shell is zero

Inside a spherical shell, the gravitational attraction of the nearest sections are exactly compensated for by the attraction of a larger area on the opposite side of the sphere.

What is the gravitational acceleration at the center of the earth?

Newton's 2nd law of motion, F=ma, combined with the first law leads to:

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F = ma = mg =GMem

R2, so g =G

Me

R2

where g is the gravitational acceleration at the surface of a planet of radius R and mass Me. This formula says that ANY object near the surface of the planet will accelerate towards the center of the planet at the rate g, regardless of the mass of the object, so a feather will fall at the same rate as a steel ball. right?

(6.4)

If the earth is a sphere, we can calculate its mass if we know its radius and the acceleration of gravity.

r ~ 6.378x106 m, or 6.378x108 cm

g=980 cm/s2

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F = mg =GmM

r2

g =GM

r2

M = gr2 /G = 980cm /s2(6.378x108cm)2 /(6.6732x108dyne ⋅cm2 /g2)

M = 5.97x1027g

Always track UNITS, to be sure your logic is correct.

An aside:

tera- 1012

giga- 109

mega- 106

kilo- 103

centi- 10-2

milli- 10-3

micro- 10-6

nano- 10-9

We have thermometers to measure temperature, barometers to measure pressure, fathometers to measure depth, seismometers to measure motion.

We have meters, centimeters, millimeters, and so…

How do you pronounce

"kilometers"?

MEASURING GRAVITY

As we saw above, it takes a LOT of mass to generate much gravity, so we might expect that we need to be able to measure very small changes in g to be able to detect even fairly large mass anomalies. In fact, many surveys require resolutions of about 0.01 mGal to image anomalies - or roughly 10-8 or 10 billionths of g!

One way to measure gravity is with a pendulum, where the period of oscillation changes with g:

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T = 2πl

g

where l is the length of the pendulum. If l is 1 meter, T~2 s.What happens to T as g increases?

l(6.5)

How well must we be able to measure T to resolve a change in g of 1 mGal? About 1 microsecond. This is not easy, given the other problems with pendulums, like friction.

Another way is to measure the time of fall of a body through a known distance. Can be done, but takes great care, and the equipment is not easily moved.

This is an “absolute” gravity instrument that measures the time of fall of a mass. Precise to ~3 microgals.

GRAVITY METERS (GRAVIMETERS).

Nearly all gravity surveys use gravimeters to take their data. A gravimeter is identical to a long-period seismometer in most respects, the only difference being that the goal is to measure the force needed to center the mass, which is proportional to g.

This is called a Lacoste suspension.The mass deflects downwards when gravity increases. The adjusting screw changes the length of the spring until the mass is in its centered position. Such instruments are very portable but only measure gravity RELATIVE to some known value.

The length of the spring when the boom is horizontal is proportional to g.

The adjustment screw changes the length of the spring until the capacitor output is centered. The adjustment dial is then read and converted to a gravity reading in mgals.

boom

What's a zero-length spring, and why is it important?

A zero-length spring in a Lacoste suspension can be adjusted so that the force need to change the location of the mass does not depend on the length of the spring. This means that there is no "restoring force" and the displacement of the mass relative to its zero displacement position depends only on spring length. It will also have an infinite oscillation period around its zero position.

This system is very sensitive to changes in g.

Changing from dial reading to mGals.

Your data will have a dial reading like “2045.74”. On the calibration sheet, look for a value of about 2050. You will likely find a number like “0.725” mGals/division. Multiply your dial reading by this number to get RELATIVE mGals.

We can correct our readings to true gravitational acceleration by taking a measurement at a place where the absolute gravity is known. We have such a location outside of the HIG building.

After changing the reading you get at HIG to mGals, subtract it from the absolute gravity at HIG (given on the sheet for the HIG gravity station). This will give you the constant to add to all your readings to correct them to real values of observed gravity.

Gravimeters are extremely sensitive and can drift as the spring(s) age and temperature changes. Thus they are housed in a vacuum, and some (like ours) are kept at a constant temperature. Some meters work only in a limited range of elevation and latitudes, and must be pre-set at the factory for the expected range. Geodetic meters (like ours) have a range which allows them to be used all over the world, sacrificing some precision.

Gravity measurements were a huge research field here in the 1960’s when gravity readings were very important for calculating the orbits of satellites. Now it's the other way around - satellite orbits are known so precisely that orbit perturbations are used to obtain changes in gravity around the world.

Instrument drift is caused by instrument problems. When an instrument measures such small changes, problems are not unusual.

How do you remove drift?

A) Is the difference between the two base station readings different by a significant amount? If not - no drift.

B) Calculate the difference between the two base station readings.

C) Calculate the amount of time between the two base station readings.

D) Calculate a drift rate (dial reading per minute)

E) Add or subtract the appropriate drift correction from each gravity reading.

GRAVITY ADJUSTMENTS

Above I said that gravitational force is caused only by changes in mass and location relative to mass, but if this is true, why do we need "adjustments"?

What factors might affect the reading we get on a gravimeter?

• change in mass distribution- in the solar system: TIDES (what causes tides?)- lateral changes in density:

+ in the rocks: (THIS IS USUALLY WHAT WE ARE AFTER !)

+ topography:

• change in meter location:- changes in distances from each mass:

+ elevation:

as we get farther from the mass, gravity decreases.

- changes in ‘aspect’ (vertical component)

Gravity is measured in the VERTICAL direction. Masses that are off to the side have no effect.

You place a gravity meter (red) at each of the points above, what effect does the anomalous mass have on gravity?

up

Because the force of gravity is proportional to the square of the distance between two bodies, and since the gravimeter is only sensitive to the vertical component of gravity, it can be shown that 2-dimensional bodies like each element in the cross section above will have the SAME EFFECT on the gravity measured at the central point (A).

For example the effect of equal density bodies of the two brown regions will contribute equally to the gravity measured at A.

A

LATITUDE EFFECTS

Where we are on earth changes gravity considerably in three ways:

1) Shape of the earth: The earth is actually an “oblate spheroid”, , not a sphere. So you are closer to the center (higher gravity) at the poles than at the equator.

2) Because of the bulge at the equator, there is more mass between you and the center of the earth (increasing gravity) at the equator than at the poles.

3) Earth rotation: The rotation lowers gravity at the equator.

NET EFFECT OF ALL THREE: g~ 5,200 mGals less at the equator than at the poles.

changes in acceleration:speed of ship or plane: Eötvös effectWhy should speed change gravity?!

Think about it: what if you got in a very fast vehicle that actually put you in orbit around the earth at the surface - what would your gravimeter read?

What way does the Eötvös effect act?

What radius would the earth have to have for the gravity at the equator to be zero, assuming that it’s mass and rotation rate don’t change?

Your ship is moving in the directions shown. What will the Eötvös effect be in each case?

waves on ocean: "gravity" is far from constant as the ship moves up and down. How can gravimeters possibly work at sea??

Is a gravity survey worth the effort?

How small a hole in the ground can be detected?

To answer this, let’s determine the gravity change expected from a hollow sphere buried directly below our gravity meter.

A sphere with a density contrast of 3000 kg/m3 (3 gm/cm3) needs to be ~30m in radius at a depth of 50 m before it generates a 1 mGal anomaly! (next page)

Lesson: don’t use gravity to find buried pipes!

Gravity IS useful, however, as we shall see. In the islands we see gravity anomalies as high as 300 mGals, so there is likely a story to be told…

%% GG450 Gravity from a buried void% g=(4/3)*pi*R^3*rho*G/z^2% g= vertical acceleration caused by density difference rho, % sphere radius R, centered z below the gravity meterz=50; % metersR=[1:5:45]; % meters, plot from 1 to 45 meter radius stepping 5 mrho=3000; % kg/m^3% 1 Gal=1 cm/sec^2; 1 m/s^2= 100Gals=100000 mGals g=((4/3)*pi*R.^3*rho*6.67e-11/z^2)*100000;plot(R,g)xlabel('sphere radius, m');ylabel('mGal');title('buried sphere at 50 m depth, rho=3000');