Surface gravity on the Moon

Problem

The Moon has $M_L = 7{,}35\cdot 10^{22}$ kg and radius $R_L = 1{,}74\cdot 10^6$ m. Calculate $g$ at the surface and compare it with the Earth’s value.

Solution

The surface acceleration due to gravity is obtained from the law of gravitation, setting $r = R_L$: $g_L = \frac{G M_L}{R_L^2}$ Substituting: $g_L = \frac{6{,}67\cdot 10^{-11}\cdot 7{,}35\cdot 10^{22}}{(1{,}74\cdot 10^6)^2}$ Numerator: $6{,}67\cdot 10^{-11}\cdot 7{,}35\cdot 10^{22} = 4{,}90\cdot 10^{12}$. Denominator: $(1{,}74\cdot 10^6)^2 = 3{,}03\cdot 10^{12}$ m$^2$. Hence: $g_L = \frac{4{,}90\cdot 10^{12}}{3{,}03\cdot 10^{12}} \approx 1{,}62~\text{m/s}^2$ Comparing with the Earth’s value $g_T \approx 9{,}81$ m/s$^2$: $\frac{g_L}{g_T} = \frac{1{,}62}{9{,}81} \approx 0{,}165 \approx \frac{1}{6}$ On the Moon one weighs about a sixth of what one weighs on Earth. $\ev{g_L \approx 1{,}62~\text{m/s}^2 \approx \tfrac{1}{6}\, g_T}$

Links

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