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:

  • 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] $



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