A) \[\sqrt{\frac{GM}{2R}}\]
B) \[\frac{GM}{R}\]
C) \[\sqrt{\frac{2GM}{R}}\]
D) \[\frac{GM}{2R}\]
Correct Answer: B
Solution :
Potential energy of the 1 kg mass which is placed at the earth surface = \[-\frac{GM}{R}\] its potential energy at infinite = 0 Work done = change in potential energy = \[\frac{GM}{R}\]You need to login to perform this action.
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