Question

A planet orbiting a distant star has a radius 3.24 × 106 𝑚. The escape speed for an object launched from this planet’s surface is 7.65 × 103 𝑚/𝑠. What is the acceleration due to gravity at the surface of the planet?

Answer 1

Let's first find the mass of the planet:

$$v=\sqrt{\dfrac{2GM}{R}},$$ $$v^2=\dfrac{2GM}{R},$$ $$M=\dfrac{v^2R}{2G},$$ $$M=\dfrac{(7.65\times10^3\ \dfrac{m}{s})^2\times3.24\times10^6\ m}{2\times6.67\times10^{-11}\ \dfrac{N\times m^2}{kg^2}}=1.42\times10^{24}\ kg.$$

Finally, we can find the acceleration due to gravity at the surface of the planet:

$$g=\dfrac{GM}{R^2},$$ $$g=\dfrac{6.67\times10^{-11}\ \dfrac{N\times m^2}{kg^2}\times1.42\times10^{24}\ kg}{(3.24\times10^6\ m)^2}=9.02\ \dfrac{m}{s^2}.$$