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 11kms^-1 , the escape speed on the surface of the planet in kms^-1 will be

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Detailed Solution

$g = \frac{G M}{R^{2}} = \frac{G (\frac{4}{3} π R^{3}) ρ}{R^{2}}$
$g \infty ρ R; R \infty \frac{g}{ρ}$
Now escape velocity, $v_{e} = \sqrt{2 gR}$
$\begin{array}{l} v_{e} \infty \sqrt{gR}; v_{e} \infty \sqrt{g \times \frac{g}{ρ}} \infty \sqrt{\frac{g^{2}}{ρ}} \\ {(v_{e})}_{planet} = (11 {kms}^{- 1}) \sqrt{\frac{6}{121} \times \frac{3}{2}} = 3 {kms}^{- 1} \end{array}$

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