A) \[3\text{ }mgR\]
B) \[\frac{3}{4}mgR\]
C) \[\text{1 }mgR\]
D) \[\frac{3}{2}mgR\]
Correct Answer: B
Solution :
[b] 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 \[={{E}_{2}}-{{E}_{1}}=-\frac{GMm}{4R}+\frac{GMm}{R}=\frac{3GMm}{4R}\] \[=\frac{3g{{R}^{2}}m}{4R}\left( \because g=\frac{GM}{{{R}^{2}}} \right)=\frac{3}{4}mg\,\,R\]
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