Problem 11.21

11.21.

We find the lunar escape speed from conservation of energy.
$\frac{1}{2} mv_{esc}^{2} = \frac{GM_{m}m}{R_{m}}$
or $v_{esc}=\sqrt{\frac{2GM_{m}}{R_{m}}}$
so the launch speed, which is twice the escape velocity, is given by
$v_{launch}=2\sqrt{\frac{2GM_{m}}{R_{m}}}$
or $v_{launch}^{2}=4\frac{2GM_{m}}{R_{m}}$
Now, we consider conservation of energy again. This time, the energy the mass has at the surface of the moon is the same as the energy that it has when it reaches the surface of the earth.
K_i + U_mi + U_ei = K_f + U_mf + U_ef
where the terms U_m and U_e reflect the gravitational potential energy at the moon and the earth respectively. So, substituting in, we find
$\frac{1}{2} m v_{launch}^{2} - \frac{GM_{m}m}{R_{m}} - \frac{GM_{e}m}{d_{em}} ~=~ \frac{1}{2}mv_{f}^{2} -\frac{GM_{m}m}{d_{em}} - \frac{GM_{e}m}{R_{e}}$
where d_em is the distance between the earth and the moon.
Substituting in our value for v_launch, we find
$v_{f}^{2} ~=~ 2G\left(\frac{3M_{m}}{R_{m}} + \frac{M_{m}}{d_{em}} + \frac{M_{e}}{R_{e}} - \frac{M_{e}}{d_{em}}\right)$
Plugging in the numbers, we find
v_f = 11.8 km/s