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 the formula for gravitational acceleration

The formula for gravitational acceleration at the surface of a planet is $g = \frac{GM}{R^{2}}$, where $g$ is the gravitational - acceleration, $G$ 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. So, $g_E=\frac{GM_E}{R_E^{2}}$ and $g_M=\frac{GM_M}{R_M^{2}}$.

Step2: Substitute the given relationships

We know that $g_M=\frac{1}{6}g_E$ and $R_M = \frac{1}{4}R_E$. Substitute these into the gravitational - acceleration formulas:
$\frac{1}{6}g_E=\frac{GM_M}{(\frac{1}{4}R_E)^{2}}$. Since $g_E=\frac{GM_E}{R_E^{2}}$, we can substitute $g_E$:
$\frac{1}{6}\times\frac{GM_E}{R_E^{2}}=\frac{GM_M}{\frac{1}{16}R_E^{2}}$.

Step3: Solve for $M_M$ in terms of $M_E$

First, cross - multiply the equation $\frac{1}{6}\times\frac{GM_E}{R_E^{2}}=\frac{GM_M}{\frac{1}{16}R_E^{2}}$:
$\frac{1}{6}GM_E\times\frac{1}{16}R_E^{2}=GM_M\times R_E^{2}$.
Cancel out $G$ and $R_E^{2}$ on both sides of the equation:
$\frac{1}{6}\times\frac{1}{16}M_E = M_M$.
So, $M_M=\frac{1}{96}M_E$.

Answer:

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