Answer
We show by using the fact that ${\bf{r}}{\rm{''}}$ is proportional to ${\bf{r}}$ leads to the conclusion that the rate at which the radial vector sweeps out area is constant. Hence, proves Kepler's Second Law.
Work Step by Step
We have the vector ${\bf{J}} = {\bf{r}} \times {\bf{r}}'$. So, ${\bf{J}}$ is orthogonal to both ${\bf{r}}\left( t \right)$ and ${\bf{r}}'\left( t \right)$.
By Eq. (1), ${\bf{r}}{\rm{''}}$ is proportional to ${\bf{r}}$. Using this fact, it is shown on page 748 that $\frac{{d{\bf{J}}}}{{dt}} = {\bf{0}}$. Thus, ${\bf{J}}$ is constant.
Using Theorem 2, we get $\frac{{dA}}{{dt}} = \frac{1}{2}||{\bf{J}}||$, a constant. Hence, this proves Kepler's Second Law.
