The escape speed from earth ' s surface is 11km/s . A certain planet has a radius twice that of earth but its mean density is same as that of the earth. Find the value of the escape speed from the planet.

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answer is 22.

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Detailed Solution

$v_{e} = \sqrt{\frac{2 GM}{R}} = \sqrt{\frac{2 G}{R} \times \frac{4}{3} {πR}^{3} ρ} = \sqrt{\frac{8 {πGρR}^{2}}{3}}$

$Escape speed from planet' s surface is$

$v_{p} = \sqrt{\frac{8 πGρ {(2 R)}^{2}}{3}} = 2 \sqrt{\frac{8 {πGρR}^{2}}{3}} = 2 v_{e}$

$\Rightarrow v_{p} = 2 \times 11 = 22 km / s$

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