Escape velocity of a body from the surface of earth is $11.2 k m / s e c$ . If the mass of earth becomes double of its present mass and radius becomes half of its present radius, then escape velocity will become

5.6 k m / s e c

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11.2 k m / s e c

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22.4 k m / s e c

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44.8 k m / s e c

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Solution

The correct option is C $22.4 k m / s e c$
Escape velocity , $v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}} = 11.2 k m / s$
For $R^{'} = \frac{R_{e}}{2} M^{'} = 2 M_{e}$

$v_{e}^{'} = \sqrt{\frac{2 G (2 M_{e})}{\frac{R_{e}}{2}}} = 2 v_{e} = 22.4 k m / s$

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