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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, where v is its escape velocity from the surface of the earth. The value of f is :

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1/3

$1 / \sqrt{2}$

1/2

$\sqrt{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} V = V_{e} = \sqrt{\frac{2 GM}{R}} \to on the surface \\ fV = \sqrt{\frac{2 GM}{R + R}} = \frac{V}{\sqrt{2}} \to at platform \\ f = \frac{1}{\sqrt{2}} \end{array}$

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