### Assignment 7 Solutions

(a)
 dpdt = dφdt dpdφ = l μr² dpdφ = -(γ/r²)er so   dp/dφ = -(μγ/l)er

(b)
After you look at this is Cartesian coordinates (e.g. (er)x = cosφ, (er)y = sinφ) you realize that er = – deφ/dφ, hence the integral of the above equation is

p = p0 + (μγ/l)eφ
where p0 is a constant vector.

(c)
Since eφ is a unit vetor, its endpoint lies along a unit circle. So the equation for p says that the endpoint of p lies on a circle with center p0 and radius μγ/l.

(d)
Following instructions we square:

p² = p0² + 2(μγ/l)p0·eφ + (μγ/l)² = 2μE + 2μγ/r

where for the last step we have used conservation of energy, E = p²/2μ – γ/r. Define a new angle f = φ + p/2, then p0·eφ = p0 cosf and, solving for 1/r we find
 1r = p0²2μγ + μγ2l² – Eγ + p0l cosf
Comparing this to the equation in the book shows p0 = γμε/l and (using 8.58) the constant terms combine to 1/c. (The direction of p0 along the minor axis of the elliptical orbit. It is a kind of average momentum, but averaged over angles rather than time. The time average of p is of course zero, otherwise the orbit would "drift" in the direction of the time average.)

2. (Problem 8.3 **)   The motion is obviously confined to a vertical line, so we can treat it as a one-dimensional problem with coordinates y1 and y2 measured vertically up from the table. The Lagrangian is L = 1/2M Y·2 - M g Y + 1/2my·2 -1/2k (y - L)2, where M and m are the total and reduced masses, and Y and y are the CM and relative positions. The two Lagrange equations are

 ×× Y = -g      and     m ××y = -k(y - L),

with solutions Y = Yo+Y·ot -1/2g t2 (whereYo = Lm1/M and Y·o = vom1/M) and y = L + A sinwt (where w = Ö{k/m} and A = vo/w). That is, the CM moves up then down like a body in free fall, while the relative position oscillates in SHM. Using Eq.(8.9), we can find the individual positions

 y1 = L + m1 M vot - 1/2g t2 + m2 vo Mw sinwt      and     y2 = m1M vot - 1/2gt2 - m1 vo M w sinwt.

Note that for small t, y2 » -½gt², so the second particle is initially in free fall, and if the table were there it would remain on the table -- with the table present the solution is not valid even initially. If v0 is large enough, the second mass will eventually leave the table; before that happens it is simpler to write the equation for y1, namely   m1d²y1/dt² = - mg - k(y1 - L) with solution

y1 = L + gω²(1–cosωt) + (v0/ω)sinωt
The second mass will leave the table when k(y1 - L) > m2g.

3. (Problem 8.18 **)   We are given the satellite's height hmin= 250 km and speed vmax= 8500 m/s at perigee. The distance from the earth's center is then rmin = Re + hmin = 6650 km. For any known satellite, we can certainly ignore the difference between the mass m and the reduced mass m » m. Thus the angular momentum is l = m vmaxrmin and the parameter c of Eq.(8.48) is c = l2/gm = (vmaxrmin)2/GMe. (Recall that g = GMem.) Putting in the given numbers, we get c = 7960 km. The rest is easy: From Eq.(8.50), rmin =c/(1+e), so e = (c - rmin)/rmin = 0.197. Similarly, from Eq.(8.50) rmax = c/(1-e) = 9910 km, so hmax = rmax- Re = 3510 km.

4. (Problem 8.29) **  When the mass of the sun is suddenly halved, the earth's PE is immediately halved, U = 1/2Uo. On the other hand, the KE is unchanged, T = To. Therefore, the total energy becomes E = T + U = To+ 1/2Uo = 0, because To = -1/2Uo in a circular orbit (virial theorem). Since the final orbit has E = 0, it is a parabola, and the earth would eventually escape from the sun.

5. Take the masses equal (= M) on circular orbits. One equilibrium point is (obviously) at the center of mass. To find another, let the distance to the common center of mass be R, then the angular velocity ω of the masses is given by ω² = GM²/4R³. The small body (mass m) at r from the CM of the large ones must revolve with the same angular velocity. Assume the three bodies are on a line, then putting the necessary centrifugal force, mrω², equal to the force from the two large bodies, GMm[(r–R)-2 + (r+R)-2] yields a fifth degree equation for r that can be solved numerically for r ~ 2.4 R.