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The escape velocity from the earth is about $11 k m s^{- 1}$ . The escape velocity from a planet having twice the radius and the same mean density as the earth is

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$22 k m s^{- 1}$

$15.5 k m s^{- 1}$

$11 k m s^{- 1}$

$5.5 k m s^{- 1}$

answer is A.

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Detailed Solution

On the earth, $V_{e} = 11 k m s^{- 1}$

$V_{e} = \sqrt{\frac{2 G M}{R}}$

$V_{e} \propto R \sqrt{ρ}$

$R_{p l a n t} = 2 R_{e a r t h}$

$V_{e_{p l a n e t}} = 2 V_{e_{e a r t h}}$

$= 2 \times 11 = 22 k m s^{- 1}$

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