Physics - Gravitation Question with Solution | TestHub

The escape velocity from the surface of a planet is given by $v_e=\sqrt{\frac{2GM}{R}}$.

Expressing the mass $M$ in terms of density $\rho$ and radius $R$, we have $M=\frac{4}{3}\pi R^3\rho$.

Substituting this into the escape velocity formula gives:

$v_e=\sqrt{\frac{2G}{R}\left(\frac{4}{3}\pi R^3\rho \right)}=R\sqrt{\frac{8\pi G\rho }{3}}$

For a constant density $\rho$, the escape velocity is directly proportional to the radius of the planet, $v_e\propto R$.

Given that the new planet has twice the radius of Earth ($R^{\prime }=2R$) and the same mean density, its escape velocity will be twice that of Earth.

$v_e^{\prime }=2v_e=2\times 11\text{ km/s}=22\text{ km/s}$

Answer: $22\mathrm{km}/\mathrm{s}$