Answer
First, we show that the area swept out by the radial vector at time $t$ is
$A\left( t \right) = \frac{1}{2}||{{\bf{r}}_0} \times {{\bf{v}}_0}||t$
Since ${{\bf{r}}_0}$ and ${{\bf{v}}_0}$ are constant vectors, so $\frac{{dA\left( t \right)}}{{dt}}$ is constant.
Hence, Kepler's Second Law continues to hold.
Work Step by Step
According to Eq. (7) of Section 13.4, the area of the triangle spanned by ${{\bf{r}}_0}$ and ${\bf{r}}\left( t \right)$ is
$A\left( t \right) = \frac{1}{2}||{{\bf{r}}_0} \times {\bf{r}}\left( t \right)||$
So,
$A\left( t \right) = \frac{1}{2}||{{\bf{r}}_0} \times \left( {{{\bf{r}}_0} + t{{\bf{v}}_0}} \right)|| = \frac{1}{2}||{{\bf{r}}_0} \times {{\bf{r}}_0} + t{{\bf{r}}_0} \times {{\bf{v}}_0}||$
Since ${{\bf{r}}_0} \times {{\bf{r}}_0} = 0$, therefore the area swept out by the radial vector at time $t$ is
$A\left( t \right) = \frac{1}{2}||{{\bf{r}}_0} \times {{\bf{v}}_0}||t$
The rate of the area swept out by the radial vector is
$\frac{{dA\left( t \right)}}{{dt}} = \frac{1}{2}||{{\bf{r}}_0} \times {{\bf{v}}_0}||$
Since ${{\bf{r}}_0}$ and ${{\bf{v}}_0}$ are constant vectors, so $\frac{{dA\left( t \right)}}{{dt}}$ is constant.
Hence, Kepler's Second Law continues to hold.
