## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

The free-fall acceleration on the surface of the moon is $1.58~m/s^2$
Let $g_m$ be the free-fall acceleration on the surface of the moon; $g_m = \frac{G~M_m}{r^2}$ To find an expression for $g_m$, we can use the equation of the period $T$ for an object's orbit. $T^2 = \frac{4\pi^2~r^3}{G~M_m}$ $\frac{G~M_m}{r^2} = \frac{4\pi^2~r}{T^2}$ $g_m = \frac{4\pi^2~r}{T^2}$ $g_m = \frac{(4\pi^2)(1.74\times 10^6~m)}{[(110~min)(60~s/min)]^2}$ $g_m = 1.58~m/s^2$ The free-fall acceleration on the surface of the moon is $1.58~m/s^2$.