The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth.

# The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth.

Asked by 3 years ago
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2. The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth.  If the radius of the Moon is about 0.29 Re (radius of the Earth), find the ratio $\frac{\rho_{Moon}}{\rho_{Earth} }$ of their densities.

The free

Answered by 3 years ago
389.5k points

Given that

$R_m = 0.29 R_e$

$g_m = \frac{1}{6} g_e$

and since Density is mass per unit volume

$\rho = \frac{m}{V}\,\, \to \,\, V=\frac{m}{\rho}$

where

$\frac{m}{\rho} = \frac{4}{3} \pi R^3\,\, \to \,\, \frac{m}{R^2} = \frac{4}{3} \pi \rho R$

using Newton's Law of Universal Gravitation

$F_g = G \frac{M m}{R^2}\,\, \to \,\, Mg = G \frac{M m}{R^2}\,\, \to \,\, g=G\frac{m}{R^2}$ and from the previous step $g=G(\frac{4}{3} \pi \rho R) \propto R \rho$

hence

$\frac{g_{m}}{g_{e}} = \frac{R_{m} \rho_{Moon}}{R_{e} \rho_{Earth}}\,\,\to\,\,\frac{\rho_{Moon}}{\rho_{Earth}}=\frac{R_e g_m}{R_m g_e}=\frac{(R_e) (\frac{1}{6} g_e)}{(0.29 R_e) (g_e)}=0.5747$

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