An object of mass 100 kg starts to fall from a height equal to radius of earth (R). Its speed when it has lost half the initial height is [ $M$ is mass of earth]

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$\sqrt{\frac{50 G}{3 R}}$

$\sqrt{\frac{100 G}{3 R}}$

$\sqrt{\frac{G M}{6 R}}$

$\sqrt{\frac{G M}{3 R}}$

answer is D.

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Detailed Solution

By Law of Conservation of Mechanical Energy
$\begin{array}{l} U_{i} + K_{i} = U_{f} + K_{f} \\ - \frac{G M m}{R + h_{i}} + 0 = - \frac{G M m}{R + h_{f}} + \frac{1}{2} m v^{2} \\ - \frac{G M}{R + R} + 0 = - \frac{G M}{R + 0.5 R} + \frac{1}{2} v^{2} \\ \Rightarrow - \frac{G M}{2 R} + \frac{2 G M}{3 R} = \frac{1}{2} v^{2} \\ \Rightarrow \frac{G M}{6 R} = \frac{1}{2} v^{2} \\ \Rightarrow v = \sqrt{\frac{G M}{3 R}} \end{array}$