A) \[\frac{mgR}{4}\]
B) \[\frac{mgR}{2}\]
C) \[mgR\]
D) \[2mgR\]
Correct Answer: B
Solution :
The potential energy on the surface of earth would be equal to \[mgR.\] So, the change in potential energy would be equal to \[\Delta PE=\frac{mgh}{1+\frac{h}{R}}=\frac{mgR}{1+\frac{R}{R}}=\frac{mgR}{2}\]You need to login to perform this action.
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