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An object is released from a height twice the radius of the earth on the surface of earth. Find the speed with which it will collide with ground by neglecting effect of air. (Take, R as radius of earth and M as mass of earth)

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$2 \sqrt{\frac{G M}{3 R}}$

$3 \sqrt{\frac{G M}{2 R}}$

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

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

answer is A.

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Detailed Solution

The initial potential energy (U_i) of object, $U_{i} = - \frac{G M m}{3 R}$

Final potential energy, $U_{f} = - \frac{G M m}{R}$

By law of conservation of energy, $∆ K E = - ∆ P E$

$\Rightarrow \frac{1}{2} m v^{2} = - (U_{f} - U_{i}) = U_{i} - U_{f}$

$\Rightarrow \frac{1}{2} m v^{2} = - \frac{G M m}{3 R} + \frac{G M m}{R}$

$\Rightarrow \frac{1}{2} v^{2} = \frac{2 G M}{3 R}$

$\Rightarrow v = \sqrt{\frac{4 G M}{3 R}} = 2 \sqrt{\frac{G M}{3 R}}$

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