A) \[2mgR\]
B) \[\frac{1}{2}mgR\]
C) \[\frac{1}{4}mgR\]
D) \[mgR\]
Correct Answer: B
Solution :
Gravitational potential energy of body on earths surface \[U=-\frac{G{{M}_{e}}m}{R}\] At a height h from earths surface, its value is \[{{U}_{h}}=-\frac{G{{M}_{e}}m}{(R+h)}=-\frac{G{{M}_{e}}m}{2R}\] \[(\because h=R)\] where \[{{M}_{e}}=\]mass of earth, \[m=\]mass of body, \[R=\]radius of earth. \[\therefore \]Gain in potential energy \[={{U}_{h}}-U\] \[=-\frac{G{{M}_{e}}m}{2R}-\left( -\frac{G{{M}_{e}}m}{R} \right)\] \[=-\frac{G{{M}_{e}}m}{2R}+\frac{G{{M}_{e}}m}{R}=\frac{G{{M}_{e}}m}{2R}=\frac{g{{R}^{2}}m}{2R}\] \[\left( \because g=\frac{G{{M}_{e}}}{{{R}^{2}}} \right)\]
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