Physics 326 Homework #8 due in course homework box by FRIDAY, 1 pmAll solutions must clearly show the steps and/or reasoning you used to arrive at your result. You will lose points for poorly written solutions or incorrect reasoning. Answers given without explanation will not be graded: NoWork = No Points. However you may always use any relation on the 1DMath, 3DMath or exam formula sheets or derived in lecture / discussion. Write your NAME and DISCUSSION SECTION on your solutions.Here is our growing collection of useful formulae concerning the two-body central force problem. Remember from our study of particle collections in 325: unscripted CAPITAL letters denote TOTALs for the system if they are additive quantities (e.g. M = m1 + m2 ) or CM properties if they are not (e.g.

R = CM position).

Coordinates & Reduced Mass :

r1 =R + m2M

r ,

r2 =R m1M

r , = m1m2M

Centrifugal force & PE :

Fcf =

L2r3 r , Ucf =

L22r2 , effective U*=U +Ucf

Angular EOM : = L

r2 Radial EOM : r = F(r) + Fcf (r) , Energy E = 12 r

2 +U (r)+Ucf (r)

Path Equation : u() 1 / r() u + u = F(1 /u)L2u2 and u = rL

Ellipse / Hyperbola with (r,) centered on a focal point : 1r =ab2 (1+ ecos) ,

e = ca =a2 b2a

Kepler Orbits for F = r2 : r() =r0

1+ ecos with r0 =L2

=b2a ,

E = 2a = (e2 1)2r0

, 2 = 42

a3

If you are doing this homework before Mondays lecture, we havent yet derived that last equation for the period (Keplers 3rd Law) but do use it: we will derive it first thing on Monday!

Problem 1 : Satellite I Taylor 8.19The height of a satellite at perigee is 300km above the earths surface and it is 3000km at apogee. EARTH DATA: For this problem and the ones following, you may need the radius of the earth; it is R =6.4106m. If you need the mass M of the earth it will almost certainly be in the combination GM which is equal to gR2 ; use that and the familiar value g = 9.8 m/s2.

(a) Find the orbits eccentricity e.(b) If we take the orbit to define the xy plane and the major axis in the x direction with the earth at the origin, what is the satellites height above the earths surface when it crosses the y axis?

Problem 2 : Satellite II Taylor 8.18An earth satellite is observed at perigee to be 250 km above the earths surface and travelling at about 8500m/s.(a) Find the eccentricity of its orbit. (b) Find its height above the earths surface at apogee.

Problem 3 : SputnikIn Yuri Gagarins first manned space flight in 1961, the perigee and apogee were 181km and 327km above the earths surface. (a) Find the period of his orbit.(b) Find his maximum speed in the orbit.

Problem 4 : The Earth Becomes a Star Taylor 8.7(a) Using simple mechanics, find the period of a mass m1 in a circular orbit of radius r around a fixed mass m2. (b) Using the separation into CM and relative motions, find the corresponding period for the case that m2 is not fixed (i.e. it can move) and the masses circle each other a constant distance r apart. Discuss the limit of this result if m2 / m1 (which is often true, but we should not get too used to assuming it!)

(c) What would be the orbital period if the earth were replaced by a star of mass equal to the solar mass, in a circular orbit, with the distance between the sun and star equal to the present earth-sun distance? (The mass of the sun is more than 300,000 times that of the earth.) Express your answer in years.

Problem 5 : The Virial Theorem adapted from Taylor 8.17The virial theorem is a rather well-known theorem in physics. The most common way it is stated is that a particle held in a bound state by an inverse-square central force field (e.g. gravity, the electric force) has kineticenergy equal to half the magnitude of its potential energy. Since the particle is bound, the force is necessarily attractive (F~1/r2), so the potential energy U ~ 1/r is negative. The total energy is thus E = T +U = U / 2 = T its negative, as it must be for a bound state. This rule of thumb that the kinetic energy fills up half of the negative potential-energy well works wonderfully for a large number of common bound states (planets in orbit, electrons in atoms); it is a really useful fact to know! Lets examine the virial theorem in greater detail.(a) A mass m moves in a circular orbit (centered on the origin) in the field of a stationary, attractive central force with potential energy U = krn . Prove the virial theorem that T = nU / 2 . (The most common appearance of the theorem, described in the preceding paragraph, is for the most common case: n = 1.) (b) We can be even more general than that! Lets derive a form of the theorem that applies to any periodic orbit of a particle, not just a circle. First, find the time derivative of the quantity G =

r p and, by integrating from

time 0 to t, show that

G(t) G(0)t = 2 T +

F r , where

F is the net force on the particle and f denotes

the average over time of any quantity f. (c) Explain why, if the particles orbit is periodic and if we make t sufficiently large, we can make the left-hand side of this equation as small as we please. That is, the left side approaches zero as t . FYI: Notice that the left-hand side can also be made zero if we choose t = n = any integer multiple of the orbital period. (d) Use this result to prove that if

F comes from the potential energy U = krn , then T = n U / 2 , if now

f denotes the time average over a very long time. FYI: Following the FYI from part (c), you can see that the theorem also applies if f denotes instead the time average over a cycle of the periodic orbit. That is the more common interpretation of T = n U / 2 . Taylors point is that if you average a periodic quantity over a very long time, you get the cycle average anyway without having to stop your integral at t = exactly n.