Answer
Using the result from Exercise 20:
$E = - \frac{{GMm}}{{2a}}$
we prove that
${v^2} = GM\left( {\frac{2}{r} - \frac{1}{a}} \right)$
Work Step by Step
The total energy of a planet of mass $m$ orbiting a sun is given by Eq. (8):
(8) ${\ \ \ \ }$ $E = \frac{1}{2}m{v^2} - \frac{{GMm}}{{||{\bf{r}}||}}$
From Exercise 20, we obtain $E = - \frac{{GMm}}{{2a}}$. Since $r = ||{\bf{r}}||$, thus,
$ - \frac{{GMm}}{{2a}} = \frac{1}{2}m{v^2} - \frac{{GMm}}{r}$
$\frac{1}{2}m{v^2} = \frac{{GMm}}{r} - \frac{{GMm}}{{2a}}$
Multiplying both sides by $\frac{2}{m}$, we get
${v^2} = \frac{{2GM}}{r} - \frac{{GM}}{a}$
Hence, ${v^2} = GM\left( {\frac{2}{r} - \frac{1}{a}} \right)$ at any point on an elliptical orbit.
