Physics - Gravitation Question with Solution | TestHub

The escape velocity is given by $v=\sqrt{\frac{2GM}{R}}$

Substituting mass $M=\frac{4}{3}\pi R^3\rho$, we get:

$v=\sqrt{\frac{2G}{R}\times \frac{4}{3}\pi R^3\rho }=R\sqrt{\frac{8}{3}\pi G\rho }$

This implies $v\propto R\sqrt{\rho }$

For the earth, $v_e\propto R_e\sqrt{{\rho }_e}$

For the planet, $v_p\propto R_p\sqrt{{\rho }_p}$

Given $R_p=4R_e$ and ${\rho }_p=4{\rho }_e$

$\frac{v_e}{v_p}=\frac{R_e}{R_p}\sqrt{\frac{{\rho }_e}{{\rho }_p}}=\frac{1}{4}\sqrt{\frac{1}{4}}=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$

The ratio of escape velocity at the earth to the escape velocity at the planet is $1:8$.

Answer: $1:8$