A) 7.9 m/s
B) 11.2 km/s
C) 15.7 km/s
D) None of these
Correct Answer: C
Solution :
\[{{v}_{e}}=\sqrt{2gR}\] \[=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2GM}{{{(GM/g)}^{1/2}}}}\] \[=\sqrt{2}.{{(GM)}^{1/4}}{{g}^{1/4}}\] \[\left[ \because g=\frac{GM}{{{R}^{2}}} \right]\] \[{{v}_{e}}\propto {{M}^{1/4}}\] \[\therefore \] \[\frac{{{v}_{p}}}{{{v}_{e}}}={{\left( \frac{4M}{M} \right)}^{1/4}}=\sqrt{2}\] \[{{v}_{p}}=\sqrt{2}.{{v}_{e}}=\sqrt{2}\times 11.2=15.7\,km/s\]
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