## University Physics with Modern Physics (14th Edition)

The acceleration due to gravity at the surface of this planet is $9.03~m/s^2$.
Let $M$ be the mass of the planet. We can use the expression for the escape speed $v$ to find an expression for the mass. $v = \sqrt{\frac{2GM}{R}}$ $M = \frac{v^2~R}{2~G}$ We can find the acceleration due to gravity $g'$ at the surface of this planet. $g' = \frac{G~M}{R^2}$ $g' = \frac{G~(\frac{v^2~R}{2~G})}{R^2}$ $g' = \frac{v^2}{2~R}$ $g' = \frac{(7.65\times 10^3~m/s)^2}{(2)(3.24\times 10^6~m)}$ $g' = 9.03~m/s^2$ The acceleration due to gravity at the surface of this planet is $9.03~m/s^2$.