Question

The ratio of escape velocity at the earth $\left(v_{e}\right)$ to the escape velocity at a planet $\left(v_{p}\right)$ whose radius and mean density are twice as that of the earth is $\frac{x}{y \sqrt{z}}$. Find $(x+y+z)$.

Answer 1

Escape velocity $=\sqrt{\frac{2 G M}{R}}$
$=\sqrt{\frac{2 G}{R} \cdot\left(\frac{4}{3} \cdot \pi R^{3} \rho\right)} \propto R \sqrt{\rho}$
Ratio, $\frac{v_{e}}{v_{p}}=1: 2 \sqrt{2}$