The escape speed from earth’s surface is $11 {\mathrm{kms}}^{–1}$. A certain planet has a radius twice that of earth but its mean density is the same as that of the earth. Find the value of the escape speed from the planet.

Let M,R,
be the mass, radius and density of earth and M,R, $\rho$ be the corresponding values
for the planet. Then escape speed form earth’s surface is

${\mathrm{V}}_{\mathrm{e}}=\sqrt{\frac{2\mathrm{GM}}{\mathrm{R}}}=\sqrt{\frac{2\mathrm{G}}{\mathrm{R}}\times \frac{4}{3}{\mathrm{\pi R}}^3\mathrm{\rho }}=\sqrt{\frac{8{\mathrm{\pi G\rho R}}^2}{3}}$

Escope speed from planet’s surface is

${\mathrm{V}}_{\mathrm{e}}^1=\sqrt{\frac{8\mathrm{\pi G\pi }(2\mathrm{R}{)}^2}{3}}=2\sqrt{\frac{8\mathrm{\pi }{\mathrm{G\rho R}}^2}{3}}=2{\mathrm{V}}_{\mathrm{e}}$

$=2\times 11=22 Km/s$