Gravitation Question 294
Question: Assuming the radius of the earth as R, the change in gravitational potential energy of a body of mass m, when it is taken from the earth’s surface to a height 3R above its surface, is
Options:
A) $ [3mgR] $
B) $ [\frac{3}{4}mgR] $
C) $ [\text{1 }mgR] $
D) $ [\frac{3}{2}mgR] $
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Answer:
Correct Answer: B
Solution:
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Gravitational potential energy (GPE) on the surface of earth,
$ [E_{1}=-\frac{GMm}{R}] $ GPE at $ [3R,E_{2}=-\frac{GMm}{\left( R+3R \right)}=-\frac{GMm}{4R}] $
$ [\therefore ] $ Change in GPE $ [=E_{2}-E_{1}=-\frac{GMm}{4R}+\frac{GMm}{R}=\frac{3GMm}{4R}] $
$ [=\frac{3gR^{2}m}{4R}\left( \because g=\frac{GM}{R^{2}} \right)=\frac{3}{4}mg,,R] $
BETA
