A) \[4\,mgR\]
B) \[mgR\text{/}4\]
C) \[mg\,R\text{/2}\]
D) \[2mgR\]
E) \[mgR\]
Correct Answer: C
Solution :
Gravitational potential energy is \[U=-\int_{R}^{2R}{\frac{GMm}{{{r}^{2}}}}dr=-\left[ \frac{GMm}{r} \right]_{R}^{2R}\] \[U=-\left[ \frac{GMm}{r} \right]_{R}^{2R}=-\left[ \frac{GMm}{2R}-\frac{GMm}{R} \right]\] \[U=+\frac{GMm}{2R}\] Also, \[GM=g{{R}^{2}}\] \[\therefore \] \[U=\frac{gmR}{2}\]
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