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The escape speed of a body on the earth’s surface is $11.2 k m s^{- 1} .$ A body is projected with thrice of this speed. The speed of the body when it escapes the gravitational pull of earth is

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$22.4 \sqrt{2} k m s^{- 1}$

$11.2 k m s^{- 1}$

$22.4 \sqrt{3} k m s^{- 1}$

$\frac{22.4}{\sqrt{2}} k m s^{- 1}$

answer is B.

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Detailed Solution

$\frac{1}{2} m u^{2} - \frac{G M m}{R} = \frac{1}{2} m v^{2} - 0$
$v^{2} - u^{2} = - \frac{2 G M}{R}, v^{2} = u^{2} - {v_{e}}^{2}$
$v^{2} = (3 {v_{e}}^{2}) - {v_{e}}^{2}$
$v = 11.2 \sqrt{8} = 22.4 \sqrt{2} K m s^{- 1}$

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