Question

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$0.250R_E$$(R_E=\mathbf{Earths}\mathbf{radius})=6.37\times {10}^6\text{m.}$. Find the ratio of their average densities${\rho }_{Moon/{\rho }_{Earth}}$.

Short answer

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

Step-by-step solution

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

Letfreefallaccelerationofmoonis$g_{Moon}$ .

Letfreefallaccelerationofearthis$g_{Earth}$ .

Radius of moon$R_{Moon}$

it is Given that in question:

$g_{Moon}=\frac{g_{Earth}}{6}$

According to Concept, we have

$$\begin{gathered} \frac{G{m}_{Moon}}{{R_{Moon}}^2}=\frac{G{m}_{Earth}}{6\times {R_{Earth}}^2} \\ \frac{\rho {}_{Moon}\times \frac{4}{3}\pi {R^3}_{Moon}}{{R_{Moon}}^2}=\frac{1}{6}\times \frac{\rho {}_{Earth}\times \frac{4}{3}\pi {R^3}_E}{{R_E}^2}.......\left(M=\rho V;V=\frac{4}{3}\pi R^3\right) \\ \rho {}_{Moon}\times R_{Moon}=\frac{\rho {}_{Earth}\times R_E}{6} \\ \frac{{\rho }_{Moon}}{{\rho }_{Earth}}=\frac{R_E}{6R_{Moon}} \\ \frac{{\rho }_{Moon}}{{\rho }_{Earth}}=\frac{R_E}{6\times 0.250R_E}.........\left(R_{Moon}=0.250R_E\right) \\ \frac{{\rho }_{Moon}}{{\rho }_{Earth}}=\frac{1}{1.5} \\ \frac{{\rho }_{Moon}}{{\rho }_{Earth}}=0.6667 \end{gathered}$$

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