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 $11$ $km{s}^{-1}$, the escape speed on the surface of the planet is $km{s}^{-1}$ will be?

Answer 1

$\frac{V_{eP}}{V_{eE}}=\sqrt{\frac{M_P}{M_E}\times \frac{R_E}{R_P}}=\sqrt{\frac{{\rho }_P{R_P}^2}{{\rho }_E{R_E}^2}}⟹\left(1\right)$

$g=\frac{GM}{R^2}=G\frac{4\pi }{3}\frac{R^3\rho }{R^2}$

$\frac{R_P}{R_E}=\frac{g_P}{g_E}\frac{{\rho }_E}{{\rho }_P}Longrightarrow\left(2\right)$

$\frac{V_{eP}}{V_{eE}}=\sqrt{\frac{3}{2}\times {\left(\frac{\sqrt{6}}{11}\right)}^2}$

$V_{eP}=3\frac{km}{s}$