Hint 7

The orbit has a semimajor axis of 9000 km and an eccentricity of 0.167. The energy, which is a constant of motion, is given by

$$E={1\over 2}mv^2 - {GMm\over r}=-{GMm\over 2a}$$

(which you can rederive by considering a circular orbit whose radius is equal to $a$.) From this you should get:

(a) 81 GJ,

(b) $7.9\ \hbox{\rm km s}^{-1}$ and $5.6\ \hbox{\rm km s}^{-1}$,

(c) $1.18\times 10^{14}\ \hbox{\rm kg m$^2$\space s}^{-1}$, which is most easily obtained from one of your answers to part (b).