Chapter 3, Problem 3AYK

To determine

The expression for $g$ on Earth and the acceleration due to gravity on the surface of the Moon is one-sixth that of acceleration due to gravity on Earth.

Answer to Problem 3AYK

The expression for $g$ on Earth is $g_E=\frac{G{M}_E}{R_E^2}$. The acceleration due to gravity on the surface of the Moon is $0.16$ or one-sixth that of acceleration due to gravity on Earth.

Explanation of Solution

The expression for force of gravity on Earth according to Universal law of Gravitation is,

$\begin{array}{c}m{g}_E=\frac{G{M}_{E}m}{R_E^2}\\ g_E=\frac{G{M}_E}{R_E^2}\end{array}$

Here,

$M_E$ is mass of Earth

$R_E$ is the radius of Earth

$G$ is universal gravitational constant

$m$ is mass of object

$g_E$ is acceleration due to gravity on Earth

The expression for force of gravity or weight on Earth is,

$F=mg$

Here,

$m$ is mass of object

$g_E$ is acceleration due to gravity on earth

The expression for acceleration due to gravity can be obtained by equating the two forces as both are gravitational forces acting on the object on Earth.

The expression for acceleration due to gravity or $g_E$ on Earth by equating the two forces is,

$\begin{array}{c}m{g}_E=\frac{G{M}_{E}m}{R_E^2}\\ g_E=\frac{G{M}_E}{R_E^2}\end{array}$

Let $g_E$ represent the acceleration due to gravity or $g$ on Earth. So the expression for $g_E$ is,

$g_E=\frac{G{M}_E}{R_E^2}$

The value of $g_E$ on Earth and moon is obtained as the following.

Substitute $5.972\times {10}^{24}\text{kg}$ for $M_E$, $6371\text{km}$ for $R_E$ and $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$.

$\begin{array}{c}{g}_E=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(5.972\times {10}^{24}\text{kg}\right)}{{\left(6371\text{km}\left(\frac{1000\text{m}}{1\text{ km}}\right)\right)}^2}\\ =9.81{\text{m/s}}^{\text{2}}\end{array}$

The expression for acceleration due to gravity on moon is,

$g_M=\frac{G{M}_M}{R_M^2}$

Here,

$M_M$ is mass of moon

$R_M$ is the radius of moon

$G$ is universal gravitational constant

Substitute $7.34\times {10}^{22}\text{kg}$ for $M_M$, $1737\text{km}$ for $R_M$ and $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$.

$\begin{array}{c}{g}_M=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(7.34\times {10}^{22}\text{kg}\right)}{{\left(1737\text{km}\left(\frac{1000\text{m}}{1\text{ km}}\right)\right)}^2}\\ =1.625{\text{m/s}}^{\text{2}}\end{array}$

The expression for the ratio of acceleration due to gravity on Moon and Earth is,

$\frac{g_M}{g_E}$

Substitute $1.625{\text{m/s}}^{\text{2}}$ for $g_M$ and $9.81{\text{m/s}}^{\text{2}}$ for $g_E$.

$\begin{array}{c}\frac{g_M}{g_E}=\frac{1.625{\text{m/s}}^{\text{2}}}{9.81{\text{m/s}}^{\text{2}}}\\ \frac{g_M}{g_E}=0.1656\\ g_M=0.16g_E\end{array}$

Therefore, acceleration due to gravity on the surface of the Moon is $0.16$ or one-sixth that of acceleration due to gravity on Earth.

Conclusion:

Therefore, the expression for $g_E$ on Earth is $g_E=\frac{G{M}_E}{R_E^2}$. The acceleration due to gravity on the surface of the Moon is $0.16$ or one-sixth that of acceleration due to gravity on Earth.