A)
B)
C)
D)
Correct Answer: C
Solution :
\[{{V}_{in}}=\frac{-GM}{2R}\left[ 3-{{\left( \frac{r}{R} \right)}^{2}} \right],\] \[{{V}_{surface}}=\frac{-GM}{R},\] \[{{V}_{Out}}=\frac{-GM}{r}\]
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