Gravity on the Moon and an astronaut's weight

Problem

Calculate the acceleration due to gravity at the surface of the Moon ($M = 7{,}35\cdot 10^{22}\ \text{kg}$, $R = 1{,}74\cdot 10^6\ \text{m}$). How much does an $80\ \text{kg}$ astronaut weigh on the Moon? Given: $G = 6{,}67\cdot 10^{-11}\ \text{N}\,\text{m}^2/\text{kg}^2$.

Solution

Surface gravity. From the equality $mg = GMm/R^2$ we get $g = GM/R^2$: $g_M = \frac{G M}{R^2} = \frac{6{,}67\cdot 10^{-11}\cdot 7{,}35\cdot 10^{22}}{(1{,}74\cdot 10^6)^2} = \frac{4{,}90\cdot 10^{12}}{3{,}03\cdot 10^{12}}$

$\ev{g_M \approx 1{,}6\ \text{m/s}^2}$

This is about a sixth of Earth’s gravity ($9{,}8\ \text{m/s}^2$), consistent with the famous “bounces” of the Apollo astronauts.

The astronaut’s weight on the Moon. $P = m\,g_M = 80\cdot 1{,}6 \approx 1{,}3\cdot 10^{2}\ \text{N}$

$\ev{P \approx 130\ \text{N}}$

The mass stays $80\ \text{kg}$ (invariant), but the weight drops from about $784\ \text{N}$ on Earth to only $\sim 130\ \text{N}$ on the Moon.

Links

Topics: Gravitation Concepts: Law of universal gravitation · Weight force Skills: Symbolic set-up