Question

The radius of a planet is twice that of the earth but its average density is the same, If the escape speed at the planet and at the earth are $v_p$ and $v_e$, respectively, then prove that $v_p=2v_e$.

Answer 1

For earth the escape velocity is given by,

$V_e=\sqrt{\frac{2GM}{R}}$

$V_e=\sqrt{\frac{2G\frac{4}{3}\pi R^3\rho }{R}}$

($\because M=V\rho =\frac{4}{3}\pi R^3\rho$)

$V_e=R\sqrt{\frac{8}{3}\pi \rho G}$

$\Rightarrow V_e\propto R\sqrt{\rho }$

Therefore,

$\frac{V_p}{V_e}=\frac{R_p{\sqrt{\rho }}_p}{R_e{\sqrt{\rho }}_e}$

$\frac{V_p}{V_e}=\frac{2R_e{\sqrt{\rho }}_e}{R_e{\sqrt{\rho }}_e}$

$\frac{V_p}{V_e}=\frac{2}{1}$

$2V_e=V_p$.