Question

1. acceleration due to gravity on the moons surface is known to be about one - sixth the acceleration due to gravity on earth. given that the radius of the moon is roughly one - fourth that of earth, find the mass of the moon in terms of the mass of earth.

Explanation:

Step1: Recall gravitational - acceleration formula

The gravitational acceleration $g$ at the surface of a planet is given by $g=\frac{GM}{R^{2}}$, where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet. Let $g_E$ and $M_E$ be the gravitational acceleration and mass of Earth respectively, and $R_E$ be the radius of Earth. Let $g_M$ and $M_M$ be the gravitational acceleration and mass of the Moon respectively, and $R_M$ be the radius of the Moon. We know that $g_M=\frac{1}{6}g_E$ and $R_M = \frac{1}{4}R_E$.

Step2: Express $g_E$ and $g_M$

$g_E=\frac{GM_E}{R_E^{2}}$ and $g_M=\frac{GM_M}{R_M^{2}}$.

Step3: Substitute the given relationships

Since $g_M=\frac{1}{6}g_E$, we have $\frac{GM_M}{R_M^{2}}=\frac{1}{6}\times\frac{GM_E}{R_E^{2}}$. Substitute $R_M=\frac{1}{4}R_E$ into the equation:
\[

$$\begin{align*} \frac{GM_M}{(\frac{1}{4}R_E)^{2}}&=\frac{1}{6}\times\frac{GM_E}{R_E^{2}}\\ \frac{GM_M}{\frac{1}{16}R_E^{2}}&=\frac{1}{6}\times\frac{GM_E}{R_E^{2}}\\ GM_M\times16&=\frac{1}{6}GM_E\\ M_M&=\frac{1}{96}M_E \end{align*}$$

\]

Answer:

The mass of the Moon is $\frac{1}{96}$ of the mass of Earth.