Q. In a satellite, if the time of revolution is $T,$ then KE is proportional to
A) $\frac{1}{T}$
B) $\frac{1}{T^{2}}$
C) $\frac{1}{T^{3}}$
D) $T^{- 2/3}$
Explanation
(d) $Velocity\ of\ satellite\ v = \sqrt{\frac{GM}{r}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ KE \propto v^{2} \propto \frac{1}{r}$
$and\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto r^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ KE \propto T^{- 2/3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.