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Q.

The ratio of escape velocity at earth $(v_{e})$ to the escape velocity at a planet $(v_{p})$ whose radius and mean density are twice as that of earth is:

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a

$1 : 2$

b

$1 : 4$

c

$1 : \sqrt 2$

d

$1 : 2 \sqrt 2$

answer is D.

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Detailed Solution

As escape velocity $v = \sqrt{\frac{2 G M}{R}}$

$= \sqrt{\frac{2 G}{R} \cdot \frac{4 π R^{3}}{3} ρ} = R \sqrt{\frac{8 π G}{3} ρ}$

$∴ \frac{v_{e}}{v_{p}} = \frac{R_{e}}{R_{p}} \times \sqrt{\frac{ρ_{e}}{ρ_{p}}} = \frac{1}{2} \times \sqrt{\frac{1}{2}} = \frac{1}{2 \sqrt{2}}$

$(∵ R_{p} = 2 R_{e} and ρ_{p} = 2 ρ_{e})$

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