Answer
We show that the planet's speed is $v = \omega R$. Then, obtaining $vT = 2\pi R$.
Using Kepler's Third Law from Exercise 8 we prove that
$v = \sqrt {\frac{k}{R}} $, where $k = GM$.
Work Step by Step
Since the orbit is circular of radius $R$, we can parametrize it by ${\bf{r}}\left( t \right) = \left( {R\cos \omega t,R\sin \omega t} \right)$. So,
$||{\bf{r}}\left( t \right)|| = R$
${\bf{r}}'\left( t \right) = \left( { - \omega R\sin \omega t,\omega R\cos \omega t} \right)$
The planet's speed is
$v = ||{\bf{r}}'\left( t \right)||$
$ = \sqrt {\left( { - \omega R\sin \omega t,\omega R\cos \omega t} \right)\cdot\left( { - \omega R\sin \omega t,\omega R\cos \omega t} \right)} $
$v = \omega R$
Since $\omega = \frac{{2\pi }}{T}$, so $v = \frac{{2\pi R}}{T}$. Hence, $vT = 2\pi R$.
From Exercise 8, we obtain Kepler's Third Law for circular orbit:
${T^2} = \left( {\frac{{4{\pi ^2}}}{k}} \right){R^3}$
Substituting $T = \frac{{2\pi R}}{v}$ in the equation above gives
$\frac{{4{\pi ^2}{R^2}}}{{{v^2}}} = \left( {\frac{{4{\pi ^2}}}{k}} \right){R^3}$
$\frac{1}{{{v^2}}} = \frac{R}{k}$
Hence, $v = \sqrt {\frac{k}{R}} $, where $k = GM$.
