Question

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 $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms ${}^{-1}$, the escape speed on the surface of the planet in kms ${}^{-1}$ will be

Answer 1

let the gravitational acceleration on the surface be $g{}^{\prime }$ and desity be $\rho {}^{\prime }$

$\frac{g^{\prime }}{g}=\frac{\sqrt{6}}{11};\frac{{\rho }^{\prime }}{\rho }=\frac{2}{3}$

Hence, $\frac{R^{\prime }}{R}=\frac{3\sqrt{6}}{22}$

$\frac{v_{esc}^{\prime }}{v_{esc}}\propto \sqrt{\frac{R^{\prime 2}{\rho }^{\prime }}{R^2\rho }}=\frac{3}{11}$

$v_{esc}^{\prime }=3km/s.$