Chapter 5, Problem 36P

To determine

The net force on the Moonassuming Earth and Sun areright angles to each other.

Answer to Problem 36P

Solution:The net force on the Moon is $4.77\times {10}^{20}\mathrm{N}$ at $65.5°$

Explanation of Solution

Given:

Mass of Moon = $m_M=7.35\times {10}^{22}\text{kg}$

Mass of Earth = $m_E=5.98\times {10}^{24}\text{kg}$

Mass of Sun = ${\text{m}}_S=1.99\times {10}^{30}\text{kg}$

Formula Used:

Gravitational force between two masses m and M which are r distance apart can be obtained by:

  $F=\frac{GmM}{r^2}$

Where, G is the gravitational constant.

Calculation:

Substitute the values and solve:

  $\begin{array}{l}{F}_S=G\frac{m_1{m}_2}{r^2}\\ F_S=6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{kg}}^{-2}\frac{\left(1.99\times {10}^{30}\text{kg}\right)\times \left(7.35\times {10}^{22}\text{kg}\right)}{{\left(1.50\times {10}^{11}\text{m}\right)}^2}\\ F_S=4.34\times {10}^{20}\text{N}\end{array}$

  $\begin{array}{l}{F}_S=G\frac{m_1{m}_2}{r^2}\\ F_S=6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{kg}}^{-2}\frac{\left(1.99\times {10}^{30}\text{kg}\right)\times \left(7.35\times {10}^{22}\text{kg}\right)}{{\left(1.50\times {10}^{11}\text{m}\right)}^2}\\ F_S=4.34\times {10}^{20}\text{N}\end{array}$

Force ofSun on Moon is $4.34\times {10}^{20}\text{N}$

  $\begin{array}{l}{F}_E=G\frac{m_1{m}_2}{r^2}\\ F_E=6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{kg}}^{-2}\frac{\left(5.98\times {10}^{24}\text{kg}\right)\times \left(7.35\times {10}^{22}\text{kg}\right)}{{\left(3.84\times {10}^8\text{m}\right)}^2}\\ F_E=1.98\times {10}^{20}\text{N}\end{array}$

Force ofEarth on Moon is $1.98\times {10}^{20}\text{N}$

Net Force on Moon

  $\begin{array}{l}{F}_n={\left({\left(4.34\times {10}^{20}\text{N}\right)}^2+{\left(1.98\times {10}^{20}\text{N}\right)}^2\right)}^{\frac{1}{2}}\\ F_n=4.77\times {10}^{20}\mathrm{N}\end{array}$

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{4.34\times {10}^{20}\text{N}}{1.98\times {10}^{20}\text{N}}\right)\\ \theta =65.5°\end{array}$

Conclusion:

The net force on the Moon is $4.77\times {10}^{20}\mathrm{N}$ at $65.5°$.