Surface gravity on the Moon

Problem

On the surface of the Moon, $M_L = 7{,}3\cdot 10^{22}\ \text{kg}$ and $R_L = 1{,}7\cdot 10^6\ \text{m}$. Calculate the surface gravitational acceleration using $g = GM/R^2$ with $G = 6{,}67\cdot 10^{-11}\ \text{N}\,\text{m}^2/\text{kg}^2$.

Solution

Formula. The gravitational acceleration at the surface of a spherical body: $g_L = \frac{GM_L}{R_L^2}$

Numerical substitution. $g_L = \frac{6{,}67\cdot 10^{-11}\cdot 7{,}3\cdot 10^{22}}{(1{,}7\cdot 10^{6})^2} = \frac{4{,}87\cdot 10^{12}}{2{,}89\cdot 10^{12}}$

$\ev{g_L \approx 1{,}7\ \text{m/s}^2}$

This agrees with the real value $g_L \approx 1{,}6\ \text{m/s}^2$, about a sixth of the Earth’s value: this is why the Apollo astronauts bounced around.

Links

Topics: Gravitation Concepts: Law of universal gravitation · Weight force Skills: Symbolic setup