A) \[\frac{5GmM}{6R}\]
B) \[\frac{2GmM}{3R}\]
C) \[\frac{GmM}{2R}\]
D) \[\frac{GmM}{3R}\]
Correct Answer: A
Solution :
The kinetic energy at altitude\[2R=\frac{GMm}{6R}\] The gravitational potential energy at altitude \[2R=-\frac{GMm}{3R}\] \[\therefore \]Total energy\[=k\varepsilon +PE=-\frac{GMm}{6R}\] Potential energy at the surface is \[-\frac{GMm}{R}\] \[\therefore \]Req. kinetic energy \[=\frac{GMm}{R}-\frac{GMm}{6R}=\frac{5GMm}{6R}\]
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