(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 p₀ 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 p₀·e_φ = p₀ cosf and, solving for 1/r we find

1 r

=

p0² 2μγ

+

μγ 2l²

–

E γ

+

p0 l

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 = Y_o+Y^·_ot -¹/₂g t² (whereY_o = Lm₁/M and Y^·_o = v_om₁/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

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

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_maxr_min and the parameter c of Eq.(8.48) is c = l²/gm = (v_maxr_min)²/GM_e. (Recall that g = GM_em.) 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/(1-e) = 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 = ¹/₂U_o. On the other hand, the KE is unchanged, T = T_o. Therefore, the total energy becomes E = T + U = T_o+ ¹/₂U_o = 0, because T_o = -¹/₂U_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.