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Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11} g$ , where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $2 / 3$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11 k m s^{- 1}$ , the escape speed on the surface of the planet in ${kms}^{- 1}$ will be.

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answer is 3.

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Detailed Solution

$V_{e} = \sqrt{\frac{2 GM}{R}} & g = \frac{GM}{R^{2}} \Rightarrow V_{e} α \frac{g}{\sqrt{ρ}}$

$\frac{V_{e}}{V_{ρ}} = \frac{g_{e}}{g_{ρ}} \cdot \sqrt{\frac{ρ_{p}}{ρ_{e}}} \Rightarrow \frac{11}{V_{p}} = \frac{11}{\sqrt{6}} \cdot \sqrt{\frac{2}{3}}$

$\Rightarrow V_{ρ} = 3 {kms}^{- 1}$

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