Answer
Using the results from Exercise 13 and Exercise 19; and the conservation of total energy, we prove that
$E = - \frac{{GMm}}{{2a}}$
Work Step by Step
The total energy of a planet of mass $m$ orbiting a sun is given by Eq. (8):
(8) ${\ \ \ \ }$ $E = \frac{1}{2}m{v^2} - \frac{{GMm}}{{||{\bf{r}}||}}$
It is proved in Exercise 17, the total energy is conserved, that is, $\frac{{dE}}{{dt}} = 0$. Therefore, we can choose a point in the orbit to compute this energy. For convenience, we choose the perihelion. Thus, the total energy is
$E = \frac{1}{2}mv_{per}^2 - \frac{{GMm}}{{{r_{per}}}}$
From Exercise 13 and Exercise 19, we obtain ${r_{per}} = a\left( {1 - e} \right)$ and ${v_{per}} = \sqrt {\left( {\frac{{GM}}{a}} \right)\frac{{1 + e}}{{1 - e}}} $. Thus,
$E = \frac{1}{2}m\left( {\frac{{GM}}{a}} \right)\frac{{1 + e}}{{1 - e}} - \frac{{GMm}}{{a\left( {1 - e} \right)}}$
$ = \frac{{GMm}}{a}\left( {\frac{{1 + e}}{{2 - 2e}} - \frac{1}{{1 - e}}} \right)$
$ = \frac{{GMm}}{a}\frac{{1 + e - 2}}{{2 - 2e}}$
$ = - \frac{{GMm}}{a}\frac{{1 - e}}{{2 - 2e}}$
Hence, the total mechanical energy is $E = - \frac{{GMm}}{{2a}}$.
