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Q.

A body starts from rest from a point distance $R_{0}$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be ( $R \to$ radius of the earth) ( $R_{0} > R)$

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a

$\sqrt{2 GM (\frac{1}{R} - \frac{1}{R_{0}})}$

b

$GM (\frac{1}{R} - \frac{1}{R_{0}})$

c

$2 G M \sqrt{(\frac{1}{R} - \frac{1}{R_{0}})}$

d

$2 GM (\frac{1}{R} - \frac{1}{R_{0}})$

answer is B.

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Detailed Solution

Apply law of conservation of Energy

$O - \frac{GMm}{R_{0}} = \frac{- GMm}{R} + \frac{1}{2} {mV}^{2}$

$\Rightarrow V = \sqrt{2 GM [\frac{1}{R} - \frac{1}{R_{0}}]}$

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A body starts from rest from a point distance R0 from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (R→radius of the earth) (R0>R)