Q. A body of mass $m\ kg$. starts falling from a point $2R$ above the Earth’s surface. Its kinetic energy when it has fallen to a point $'R'$ above the Earth’s surface [$R$-Radius of Earth, $M$-Mass of Earth, $G$-Gravitational Constant]
A) $\frac{1}{2}\frac{GMm}{R}$
B) $\frac{1}{6}\frac{GMm}{R}$
C) $\frac{2}{3}\frac{GMm}{R}$
D) $\frac{1}{3}\frac{GMm}{R}$
Explanation
(b) Potential energy $U = \frac{- GMm}{r} = - \frac{GMm}{R + h}$
$U_{initial} = - \frac{GMm}{3R}$ and $U_{final} = - \frac{GMm}{2R}$
Loss in $PE =$ gain in $KE = \frac{GMm}{2R} - \frac{GMm}{3R} = \frac{GMm}{6R}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.