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The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fv_e, where v_eis its escape velocity from the surface of the earth. The value of f is

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$\sqrt{2}$

$\frac{1}{\sqrt{2}}$

$\frac{1}{3}$

$\frac{1}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{GMm}{2 R} = \frac{1}{2} {mf}^{2} v_{e}^{2} \Rightarrow f^{2} v_{e}^{2} = \frac{GM}{R} \\ \Rightarrow f^{2} (\frac{2 GM}{R}) = \frac{GM}{R} \Rightarrow f^{2} = \frac{1}{2} \Rightarrow f = \frac{1}{\sqrt{2}} \end{array}$

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