Physics - Gravitation Question with Solution | TestHub

Area covered by line joining planet and sun in time dt is$\text{ds}=\frac{1}{2}{x}^2\text{d}\mathrm{\theta }\text{;}$Areal velocity$=\frac{\text{ds}}{\text{dt}}=\frac{1}{2}{x}^2\text{d}\mathrm{\theta }/\text{dt}=\frac{1}{2}{x}^2\mathrm{\omega }\text{,}$where$x$= distance between planet and sun ω = angular speed of planet about the sun. From kepler's second low, areal velocity of planet is constant. At farthest position



$\text{A}=\frac{\text{ds}}{\text{dt}}=\frac{1}{2}{\left(\text{2R - r}\right)}^2\mathrm{\omega }=\frac{1}{2}\left(\text{2R - r}\right){\text{V}}_{\text{B}}$

$\mathrm{A}=\frac{1}{2}\left(\text{2R - r}\right){\text{V}}_{\text{B}}$

or${\text{V}}_{\text{B}}=\frac{\text{2A}}{\text{2R - r}}$

(least speed). (Using values)

${\text{V}}_{\text{B}}=\text{40 km/s}$

Thus,$\text{K = 4}$