dp dt 
=  dφ dt  dp dφ  =  l μr²  dp dφ  = (γ/r²)e_{r}  so dp/dφ = (μγ/l)e_{r} 
(b)
After you look at this is Cartesian coordinates (e.g. (e_{r})_{x} = cosφ, (e_{r})_{y} = sinφ)
you realize that e_{r} = – de_{φ}/dφ, hence the integral of the above equation is
(c)
Since e_{φ} 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 p_{0} and radius μγ/l.
(d)
Following instructions we square:
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 p_{0}·e_{φ} = p_{0} cosf and, solving for 1/r we find
1 r 
=  p_{0}² 2μγ 
+  μγ 2l² 
–  E γ 
+  p_{0} l 
cosf 
2. (Problem 8.3 **) The motion is obviously confined to a vertical line, so we can treat it as a onedimensional problem with coordinates y_{1} and y_{2} measured vertically up from the table. The Lagrangian is L = ^{1}/_{2}M Y^{·2}  M g Y + ^{1}/_{2}my^{·2} ^{1}/_{2}k (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

with solutions Y = Y_{o}+Y^{·}_{o}t ^{1}/_{2}g t^{2} (whereY_{o} = Lm_{1}/M and Y^{·}_{o} = v_{o}m_{1}/M) and y = L + A sinwt (where w = Ö{k/m} and A = v_{o}/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

Note that for small t, y_{2} » 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 v_{0} is large enough, the second mass will eventually leave the table; before that happens it is simpler to write the equation for y_{1}, namely m_{1}d²y_{1}/dt² =  mg  k(y_{1}  L) with solution
3. (Problem 8.18 **) We are given the satellite's height h_{min}= 250 km and speed v_{max}= 8500 m/s at perigee. The distance from the earth's center is then r_{min} = R_{e} + h_{min} = 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 v_{max}r_{min} and the parameter c of Eq.(8.48) is c = l^{2}/gm = (v_{max}r_{min})^{2}/GM_{e}. (Recall that g = GM_{e}m.) Putting in the given numbers, we get c = 7960 km. The rest is easy: From Eq.(8.50), r_{min} =c/(1+e), so e = (c  r_{min})/r_{min} = 0.197. Similarly, from Eq.(8.50) r_{max} = c/(1e) = 9910 km, so h_{max} = r_{max} R_{e} = 3510 km.
4. (Problem 8.29) ** When the mass of the sun is suddenly halved, the earth's PE is immediately halved, U = ^{1}/_{2}U_{o}. On the other hand, the KE is unchanged, T = T_{o}. Therefore, the total energy becomes E = T + U = T_{o}+ ^{1}/_{2}U_{o} = 0, because T_{o} = ^{1}/_{2}U_{o} 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.