A) \[\frac{mgR}{2}\]
B) \[mgR\]
C) \[2mgR\]
D) \[\frac{1}{4}mgR\]
Correct Answer: A
Solution :
The potential energy on surface of earth \[U=\frac{-G{{M}_{e}}m}{R}\] The potential energy at height R from surface of earth \[U=\frac{-G{{M}_{e}}m}{2R}\] \[\therefore \] \[\Delta U=U'-U\] \[=\frac{G{{M}_{e}}m}{2R}=\frac{g{{R}^{2}}m}{2R}-\frac{mgR}{2}\]
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