Answer
We take the derivatives of ${\bf{r}}\left( t \right)$ and show it satisfies the differential equation, Eq. (1):
(1) ${\ \ \ }$ ${\bf{r}}{\rm{''}}\left( t \right) = - \frac{k}{{||{\bf{r}}\left( t \right)|{|^2}}}{{\bf{e}}_r}$
provided that ${\omega ^2} = k{R^{ - 3}}$.
Then, we deduce Kepler's Third Law:
${T^2} = \left( {\frac{{4{\pi ^2}}}{k}} \right){R^3}$.
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {R\cos \omega t,R\sin \omega t} \right)$. So,
$||{\bf{r}}\left( t \right)|| = R$
${\bf{r}}'\left( t \right) = \left( { - \omega R\sin \omega t,\omega R\cos \omega t} \right)$
${\bf{r}}{\rm{''}}\left( t \right) = \left( { - {\omega ^2}R\cos \omega t, - {\omega ^2}R\sin \omega t} \right)$
It follows that
${\bf{r}}{\rm{''}}\left( t \right) = - {\omega ^2}\left( {R\cos \omega t,R\sin \omega t} \right) = - {\omega ^2}{\bf{r}}\left( t \right)$
Write ${\omega ^2} = k{R^{ - 3}}$. So,
${\bf{r}}{\rm{''}}\left( t \right) = - \frac{k}{{{R^3}}}{\bf{r}}\left( t \right)$
Since $||{\bf{r}}\left( t \right)|| = R$ and the unit radial vector ${{\bf{e}}_r} = \frac{{{\bf{r}}\left( t \right)}}{{||{\bf{r}}\left( t \right)||}}$, so
${\bf{r}}{\rm{''}}\left( t \right) = - \frac{k}{{||{\bf{r}}\left( t \right)|{|^3}}}{\bf{r}}\left( t \right) = - \frac{k}{{||{\bf{r}}\left( t \right)|{|^2}}}{{\bf{e}}_r}$
Thus, ${\bf{r}}\left( t \right)$ satisfies the differential equation, Eq. (1).
Since $T = \frac{{2\pi }}{\omega }$, so
${T^2} = \frac{{4{\pi ^2}}}{{{\omega ^2}}} = 4{\pi ^2}\frac{{{R^3}}}{k}$
Thus, Kepler's Third Law becomes ${T^2} = \left( {\frac{{4{\pi ^2}}}{k}} \right){R^3}$.
