Q. A satellite is revolving around the planet. The gravitational force between them varies with $R^{- 5/2},$ where R is the radius of the satellite. The square of the time period $T$ will be directly proportional to
A) $R^{3}$
B) $R^{7/2}$
C) $R^{3/2}$
D) $R^{5/7}$
Explanation
(b) Gravitational force provides the required centripetal force $ie,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {m\omega}^{2}R = \frac{GMm}{R^{\frac{5}{2}}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{m4\pi^{2}}{T^{2}} = \frac{GMm}{R^{\frac{7}{2}}}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto R^{7/2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.