A) \[\frac{1}{2}\frac{GMm}{R}\]
B) \[\frac{1}{6}\frac{GMm}{R}\]
C) \[\frac{2}{3}\frac{GMm}{R}\]
D) \[\frac{1}{3}\frac{GMm}{R}\]
Correct Answer: B
Solution :
Potential energy \[U=\frac{-GMm}{r}=-\frac{GMm}{R+h}\] \[{{U}_{initial}}=-\frac{GMm}{3R}\] and \[{{U}_{final}}=-\frac{-GMm}{2R}\] Loss in PE = gain in \[KE=\frac{GMm}{2R}-\frac{GMm}{3R}=\frac{GMm}{6R}\]
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