Gravitation - Escape Velocity

Determine the escape velocity on a planet whose mass is $8$ times that of Earth and whose radius is $2$ times that of Earth. (Take escape velocity of Earth $v_{e(earth)} = 11.2 \text{ km/s}$)

The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$. For the new planet: $M' = 8M$ and $R' = 2R$. Substituting these into the formula: $v_e' = \sqrt{\frac{2G(8M)}{2R}} = \sqrt{4 \times \frac{2GM}{R}} = 2 \times \sqrt{\frac{2GM}{R}}$. Thus, $v_e' = 2 \times v_{e(earth)} = 2 \times 11.2 \text{ km/s} = 22.4 \text{ km/s}$.

Since escape velocity is proportional to $\sqrt{M/R}$, increasing mass by a factor of $8$ and radius by a factor of $2$ results in the escape velocity increasing by a factor of $\sqrt{8/2} = \sqrt{4} = 2$.

Calculate the escape velocity from the surface of the Moon. Given: $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$, Mass of Moon $M = 7.35 \times 10^{22} \text{ kg}$, Radius of Moon $R = 1.74 \times 10^6 \text{ m}$.

Using $v_e = \sqrt{\frac{2GM}{R}}$: $v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^6}} = \sqrt{\frac{9.805 \times 10^{12}}{1.74 \times 10^6}} = \sqrt{5.635 \times 10^6} \approx 2373.8 \text{ m/s} \approx 2.37 \text{ km/s}$.

This calculation uses the fundamental constants of the Moon to determine the speed required to overcome its gravitational field. Because this speed is low, light gases cannot be retained by the Moon's gravity.