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Q12 P

Expert-verified
Found in: Page 439

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth. The radius of the Moon is about $0.250{R}_{E}$ $\left({R}_{E}=\mathbf{Earths}\text{}\mathbf{radius}\right)=6.37×{10}^{6}\text{m.}$. Find the ratio of their average densities ${\rho }_{Moon/{\rho }_{Earth}}$.

Thus, the required ratio of their average densities $\frac{{\rho }_{Moon}}{{\rho }_{Earth}}=0.6667$

See the step by step solution

## Step 1: Concept

An object at a distance h above the Earth’s surface experiences a gravitational force of magnitude mg, where g is the free-fall acceleration at that elevation:

$g=\frac{G{M}_{E}}{{r}^{2}}=\frac{G{M}_{E}}{{\left(R{}_{E}+h\right)}^{2}}$

And force is given by:

$F=\frac{G{M}_{E}{m}_{a}}{{\left(R{}_{E}+r\right)}^{2}}$

${M}_{E}$ is the mass of the Earth and ${R}_{E}$is its radius. Therefore, the weight of an object decreases as the object moves away from the Earth’s surface.

## Step 2: Find the ratio of their average densities

Let free fall acceleration of moon is ${g}_{Moon}$ .

Let free fall acceleration of earth is ${g}_{Earth}$ .

Radius of moon ${R}_{Moon}$

it is Given that in question:

${g}_{Moon}=\frac{{g}_{Earth}}{6}$

According to Concept, we have

$\frac{G{m}_{Moon}}{{{R}_{Moon}}^{2}}=\frac{G{m}_{Earth}}{6×{{R}_{Earth}}^{2}}\phantom{\rule{0ex}{0ex}}\frac{\rho {}_{Moon}×\frac{4}{3}\pi {{R}^{3}}_{Moon}}{{{R}_{Moon}}^{2}}=\frac{1}{6}×\frac{\rho {}_{Earth}×\frac{4}{3}\pi {{R}^{3}}_{E}}{{{R}_{E}}^{2}}.......\left(M=\rho V;V=\frac{4}{3}\pi {R}^{3}\right)\phantom{\rule{0ex}{0ex}}\rho {}_{Moon}×{R}_{Moon}=\frac{\rho {}_{Earth}×{R}_{E}}{6}\phantom{\rule{0ex}{0ex}}\frac{{\rho }_{Moon}}{{\rho }_{Earth}}=\frac{{R}_{E}}{6{R}_{Moon}}\phantom{\rule{0ex}{0ex}}\frac{{\rho }_{Moon}}{{\rho }_{Earth}}=\frac{{R}_{E}}{6×0.250{R}_{E}}.........\left({R}_{Moon}=0.250{R}_{E}\right)\phantom{\rule{0ex}{0ex}}\frac{{\rho }_{Moon}}{{\rho }_{Earth}}=\frac{1}{1.5}\phantom{\rule{0ex}{0ex}}\frac{{\rho }_{Moon}}{{\rho }_{Earth}}=0.6667$

Thus the required ratio of their average densities $\frac{{\rho }_{Moon}}{{\rho }_{Earth}}=0.6667$