Question

5.8 acceleration on the moon: a) repeat the estimate of the acceleration due to gravity in example 5.7, but use the mass $m_{\text{s}} = 7.34\times10^{22}\text{ kg}$ and radius $r_{\text{s}} = 1.74\times10^{3}\text{ km}$ of the moon (see ?? for relative values) to find its surface acceleration constant $g_{\text{s}}$. b) find the ratio $g_{\text{s}}/g$, where $g_{\text{e}}\approx9.8\text{ m/s}^2$ is the earths gravitational acceleration. since your weight is the product of the acceleration due to gravity and your mass, this is the fraction that you would weigh on the moon relative to your weight on the earth.

Explanation:

Step1: Recall the formula for gravitational - acceleration

The formula for the acceleration due to gravity at the surface of a planet is $g=\frac{GM}{R^{2}}$, where $G = 6.67\times10^{- 11}\ m^{3}/(kg\cdot s^{2})$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.

Step2: Calculate the acceleration due to gravity on the Moon ($g_{M}$)

First, convert the radius of the Moon from km to m: $R_{M}=1.74\times10^{3}\ km = 1.74\times10^{6}\ m$.
Substitute $G = 6.67\times10^{-11}\ m^{3}/(kg\cdot s^{2})$, $M_{M}=7.34\times10^{22}\ kg$, and $R_{M}=1.74\times10^{6}\ m$ into the formula $g_{M}=\frac{GM_{M}}{R_{M}^{2}}$.
\[

$$\begin{align*} g_{M}&=\frac{6.67\times 10^{-11}\times7.34\times 10^{22}}{(1.74\times 10^{6})^{2}}\\ &=\frac{6.67\times7.34\times10^{-11 + 22}}{1.74^{2}\times10^{12}}\\ &=\frac{49.0578\times10^{11}}{3.0276\times10^{12}}\\ &=\frac{49.0578}{3.0276}\times10^{11 - 12}\\ &\approx1.62\ m/s^{2} \end{align*}$$

\]

Step3: Calculate the ratio $\frac{g_{M}}{g_{E}}$

Given $g_{E}\approx9.8\ m/s^{2}$, then $\frac{g_{M}}{g_{E}}=\frac{1.62}{9.8}\approx0.165$.

Answer:

a) The acceleration due to gravity on the Moon $g_{M}\approx1.62\ m/s^{2}$.
b) The ratio $\frac{g_{M}}{g_{E}}\approx0.165$.