A) \[\frac{1}{4}\]mgR
B) \[\frac{1}{2}\]mgR
C) mgR
D) 2 mgR
Correct Answer: C
Solution :
For minimum KE, the velocity at infinite should be zero. Apply energy conservation \[\frac{m{{v}^{2}}}{2}-\frac{GMm}{R}=0\] \[\Rightarrow \] \[\frac{m{{v}^{2}}}{2}-\frac{GMm}{R}\] \[\Rightarrow \] \[KE=\frac{GMm}{R}\] \[=\frac{GMm}{{{R}^{2}}}\times R=mgR\]You need to login to perform this action.
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