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The escape velocity of a body on the earth’s surface is $V_{e}$ . A body is thrown up with a speed of ${kv}_{e}$ , where k>1, Assuming that the sun and planets do not influence the motion of the body, then the velocity of the body at the infinite distance in

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$v_{e} \sqrt{k^{2} - 1}$

$v_{e} / \sqrt{k^{2} - 1}$

$v_{e} (k^{2} - 1)$

$\frac{v_{e}}{k - 1}$

answer is B.

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Detailed Solution

Since $V > V_{e}$ , after crossing the earth gravity field, gravitational P.E is zero.

Apply law of conservation of energy at A& B

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

$\Rightarrow - \frac{GMm}{R} + \frac{1}{2} {mK}^{2} \frac{2 GM}{R} = \frac{1}{2} {mV}^{' 2}$

$\Rightarrow - \frac{GM}{R} + \frac{K^{2} GM}{R} = \frac{V^{' 2}}{2}$

$\Rightarrow V^{' 2} = K^{2} \frac{2 GM}{R} - \frac{2 GM}{R}$

$∴ V^{'} = V_{e} \sqrt{K^{2} - 1}$

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