A) \[\frac{1}{4}mgR\]
B) mgR
C) 2mgR
D) \[\frac{1}{2}\]mgR
Correct Answer: D
Solution :
Gravitational potential energy of an object of mass m at the surface of earth is \[{{E}_{1}}=-\frac{GMm}{R}\] ?. (i) Gravitational potential energy of an object of mass m at height h is \[{{E}_{2}}=-\frac{GMm}{R+h}\] Given, \[h=R\] \[\therefore \] \[{{E}_{2}}=-\frac{GMm}{R+R}=-\frac{GMm}{2R}\] ?. (ii) \[\therefore \]Gain in potential energy is\[{{E}_{2}}-{{E}_{1}},\] \[=-\frac{GMm}{2R}+\frac{GMm}{R}=\frac{1}{2}\frac{GMm}{R}\] Also,\[g=\frac{GM}{{{R}^{2}}}\]acceleration due to gravity. \[\therefore \]Gain in potential energy\[=\frac{1}{2}mgR\]
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