Q. Imagine a light planet revolving around a very massive star in a circular orbit of radius $r$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{- 3/2}$, then $T^{2}$ is proportional to
A) $R^{3}$
B) $R^{5/2}$
C) $R^{3/2}$
D) $R^{7/2}$
Explanation
(b) Gravitational force $\left( = \frac{GM\ m}{R^{3/2}} \right)$ provides the necessary
centripetal force $\left( ie,m\ R\ \omega^{2} \right)$
So, $\frac{GM\ m\ }{R^{3/2}} = mR\omega^{2} = mR\left( \frac{2\pi}{T} \right)^{2} = \frac{4\pi^{2}mR}{T^{2}}$
or $T^{2} = \ \frac{4\pi^{2}R^{5/4}}{GM}ie,T^{2} \propto R^{5/2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.