A planet revolves about the sun in elliptical orbit of semi-major axis $2\times {10}^{12}m.$  The areal velocity of the planet when it is nearest to the sun is  $4.4\times {10}^{16}{m}^2/s.$  The least distance between planet and the sun is $1.8\times {10}^{12}m.$ The minimum speed of the planet in km/s is 10k. Determine the value of k

Area covered by line joining planet and sun in time $dtisds=\frac{1}{2}{X}^{2}d\theta ;$

 Areal  velocity
$\frac{dS}{dt}=\frac{1}{2}{X}^2\frac{d\theta }{dt}=\frac{1}{2}{X}^2\omega$

where X = distance between planet and sun and  $\omega =$ angular speed of planet about sun. 

From Kepler’s second law areal velocity of planet is constant. At farthest position
$\begin{array}{l}\mathrm{A}=\frac{\mathrm{dS}}{\mathrm{dt}}=\frac{1}{2}{(2\mathrm{R}-\mathrm{r})}^2=\frac{1}{2}(2\mathrm{R}-\mathrm{r})\left[(2\mathrm{R}-\mathrm{r})\mathrm{\omega }\right]\\ =\frac{1}{2}(2\mathrm{R}-\mathrm{r}){\mathrm{v}}_{\mathrm{B}}\mathrm{or}{\mathrm{v}}_{\mathrm{B}}=\frac{2\mathrm{A}}{2\mathrm{R}-\mathrm{r}}\left(\mathrm{least}\mathrm{speed}\right),\end{array}$
(using values)  ${\mathrm{v}}_{\mathrm{B}}=40\mathrm{km}/\mathrm{s}\mathrm{thus},\mathrm{k}=4$