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If mass of earth is M, radius is R, and gravitational constant is G, then work done to take 1kg mass slowly from earth surface to infinity will be (Neglect rotation of Earth)

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$\frac{G M}{2 R}$

$\sqrt{\frac{G M}{2 R}}$

$\sqrt{\frac{2 G M}{R}}$

$\frac{G M}{R}$

answer is B.

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Detailed Solution

Potential energy of the 1kg mass which is placed at the earth surface $= \frac{G M}{R}$
its potential energy at infinite=0
Work done = change in potential energy $= \frac{G M}{R}$

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