An object is allowed to fall from a height R above the earth, where R is the radius of earth. Its velocity when it strikes the earth’s surface, ignoring air resistance, will be :

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As gravitational field is a conservative field and therefore mechanical energy will be conserved. Therefore,
Loss in PE = Gain in KE

$(- \frac{G M m}{2 R}) - (- \frac{G M m}{R}) = \frac{1}{2} m v^{2}$

$\Rightarrow v^{2} = \frac{G M}{R} = g R (as g = \frac{G M}{R^{2}})$

$\Rightarrow v = \sqrt{g R}$

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