Answer
(a) We evaluate $\frac{d}{{dt}}\frac{1}{2}m{v^2}$ and $\frac{d}{{dt}}\frac{{GMm}}{{||{\bf{r}}||}}$ directly,
and obtain
$\frac{d}{{dt}}\frac{1}{2}m{v^2} = {\bf{v}}\cdot\left( {m{\bf{a}}} \right)$
$\frac{d}{{dt}}\frac{{GMm}}{{||{\bf{r}}||}} = {\bf{v}}\cdot\left( { - \frac{{GMm}}{{||{\bf{r}}|{|^3}}}{\bf{r}}} \right)$
(b) Taking the derivative of $E$ and using results in part (a) we obtain $\frac{{dE}}{{dt}} = 0$
Work Step by Step
(a)
1. Evaluate $\frac{d}{{dt}}\frac{1}{2}m{v^2}$
Since ${v^2} = {\bf{v}}\cdot{\bf{v}}$, so
$\frac{d}{{dt}}\frac{1}{2}m{v^2} = \frac{1}{2}m\frac{d}{{dt}}{\bf{v}}\cdot{\bf{v}} = \frac{1}{2}m\left( {{\bf{v}}\cdot\frac{{d{\bf{v}}}}{{dt}} + \frac{{d{\bf{v}}}}{{dt}}\cdot{\bf{v}}} \right)$
$\frac{d}{{dt}}\frac{1}{2}m{v^2} = m{\bf{v}}\cdot\frac{{d{\bf{v}}}}{{dt}}$
But ${\bf{a}} = \frac{{d{\bf{v}}}}{{dt}}$, so
$\frac{d}{{dt}}\frac{1}{2}m{v^2} = {\bf{v}}\cdot\left( {m{\bf{a}}} \right)$
2. Evaluate $\frac{d}{{dt}}\frac{{GMm}}{{||{\bf{r}}||}}$
$\frac{d}{{dt}}\frac{{GMm}}{{||{\bf{r}}||}} = GMm\frac{d}{{dt}}{\left( {{\bf{r}}\cdot{\bf{r}}} \right)^{ - 1/2}}$
$ = GMm\left( { - \frac{1}{2}} \right){\left( {{\bf{r}}\cdot{\bf{r}}} \right)^{ - 3/2}}\left( {{\bf{r}}\cdot\frac{{d{\bf{r}}}}{{dt}} + \frac{{d{\bf{r}}}}{{dt}}\cdot{\bf{r}}} \right)$
$ = - \frac{{GMm}}{{||{\bf{r}}|{|^3}}}{\bf{r}}\cdot\frac{{d{\bf{r}}}}{{dt}}$
But ${\bf{v}} = \frac{{d{\bf{r}}}}{{dt}}$, so
$\frac{d}{{dt}}\frac{{GMm}}{{||{\bf{r}}||}} = {\bf{v}}\cdot\left( { - \frac{{GMm}}{{||{\bf{r}}|{|^3}}}{\bf{r}}} \right)$
(b) Recall from Eq. (1):
${\bf{a}}\left( t \right) = - \frac{{GM}}{{||{\bf{r}}\left( t \right)|{|^2}}}{{\bf{e}}_r}$
So, we can write
$\frac{d}{{dt}}\frac{1}{2}m{v^2} = m{\bf{v}}\cdot\frac{{d{\bf{v}}}}{{dt}} = - m{\bf{v}}\cdot\frac{{GM}}{{||{\bf{r}}\left( t \right)|{|^2}}}{{\bf{e}}_r}$
Since ${{\bf{e}}_r} = \frac{{{\bf{r}}\left( t \right)}}{{||{\bf{r}}\left( t \right)||}}$, so
$\frac{d}{{dt}}\frac{1}{2}m{v^2} = {\bf{v}}\cdot\left( { - \frac{{GMm}}{{||{\bf{r}}|{|^3}}}{\bf{r}}} \right)$
We have from Eq. (8):
(8) ${\ \ \ \ }$ $E = \frac{1}{2}m{v^2} - \frac{{GMm}}{{||{\bf{r}}||}}$
Taking the derivative of $E$ gives
$\frac{{dE}}{{dt}} = \frac{d}{{dt}}\frac{1}{2}m{v^2} - \frac{d}{{dt}}\frac{{GMm}}{{||{\bf{r}}||}}$
Substituting our previous results, we obtain
$\frac{{dE}}{{dt}} = {\bf{v}}\cdot\left( { - \frac{{GMm}}{{||{\bf{r}}|{|^3}}}{\bf{r}}} \right) - {\bf{v}}\cdot\left( { - \frac{{GMm}}{{||{\bf{r}}|{|^3}}}{\bf{r}}} \right) = 0$
Hence, energy is conserved, that is, $\frac{{dE}}{{dt}} = 0$.
