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
maddy

1 Answer

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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Asked: 3 years ago

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