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Q.

Given that mass of earth is M and its radius is R. A body is dropped from a height equal to the radius of earth above the surface of earth. When it reaches the ground, its speed will be:

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a

$\sqrt{\frac{GM}{R}}$

b

$\sqrt{\frac{GM}{2 R}}$

c

$\sqrt{\frac{2 Gm}{R}}$

d

$\frac{2 Gm}{R}$

answer is A.

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Detailed Solution

By using conservation of mechanical energy
$- \frac{GMm}{2 R} + 0 = - \frac{GMm}{R} + \frac{1}{2} {mv}^{2} \Rightarrow v = \sqrt{\frac{GM}{R}}$

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