Chapter 7, Problem 46A

To determine

The ratio of the gravitational forces exerted by Earth and the Sun on the Moon.

Answer to Problem 46A

  $4.44\times {10}^{-7}$

Explanation of Solution

Given:

The distance between Moon’s center and Earth’s center is $r_{ME}=3.9\times {10}^8\text{m}$ .

The distance between Moon’s center and Sun’s center is $r_{MS}=1.5\times {10}^8\text{m}$ .

The mass of the Moon is $m_M=7.3\times {10}^{22}\text{kg}$ .

Formula used:

Gravitational force:

  $F=\frac{G{m}_1{m}_2}{r^2}$

Where, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between the masses.

Calculation:

The value of gravitational constant is:

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

The mass of the Sun is $m_S=1.99\times {10}^{30}\text{kg}$ .

The mass of the Earth is $m_E=5.98\times {10}^{24}\text{kg}$ .

Now, the gravitational force between the Moon and the Earth is,

  $F_{ME}=\frac{G{m}_M{m}_E}{r_{ME}{}^2}$

And, the gravitational force between the Moon and the Sun is,

  $F_{MS}=\frac{G{m}_M{m}_S}{r_{MS}{}^2}$

Then, the ratio of the two forces will be,

  $\begin{array}{c}\frac{F_{ME}}{F_{MS}}=\frac{\left(\frac{G{m}_M{m}_E}{r_{ME}{}^2}\right)}{\left(\frac{G{m}_M{m}_S}{r_{MS}{}^2}\right)}\\ \frac{F_{ME}}{F_{MS}}=\frac{m_E{r}_{MS}{}^2}{m_s{r}_{ME}{}^2}\end{array}$

Using the given and the known values in above,

  $\begin{array}{c}\frac{F_{ME}}{F_{MS}}=\frac{m_E{r}_{MS}{}^2}{m_s{r}_{ME}{}^2}\\ =\frac{\left(5.98\times {10}^{24}\right){\left(1.5\times {10}^8\right)}^2}{\left(1.99\times {10}^{30}\right){\left(3.9\times {10}^8\right)}^2}\\ =4.44\times {10}^{-7}\end{array}$

Conclusion:

Thus, the ratio of the gravitational forces exerted by Earth and the Sun on the Moon is $4.44\times {10}^{-7}$ .