Gravitation Question 131
Question: A body of mass m kg. starts falling from a point2R 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] [MP PMT 2002]
Options:
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}] $
Show Answer
Answer:
Correct Answer: B
Solution:
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}] $
