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If $v_{e}$ and $v'_{e}$ be the escape velocity of earth and a planet respectively, radius of planet is twice that of earth and density is same for both. Then the value of $\frac{v'_{e}}{v_{e}}$ is____

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Detailed Solution

$v_{e} = \sqrt{\frac{2 G M}{R},} v'_{e} = \sqrt{\frac{2 G M'}{R'}} M' = \frac{4}{3} π {(2 R)}^{3} ρ a n d M = \frac{4}{3} π R^{3} ρ \frac{M'}{M} = \frac{8}{1} a n d \frac{R'}{R} = \frac{2}{1} ∴ \frac{v'_{e}}{v_{e}} = \sqrt{\frac{M'}{M} \times \frac{R}{R'}} = \sqrt{\frac{8}{1} \times \frac{1}{2}} = 2 \Rightarrow v'_{e} = 2 v_{e} \Rightarrow \frac{v'_{e}}{v_{e}} = 2$

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