Question

The Sun and Earth each exert a gravitational force on theMoon. What is the ratio${\mathit{F}}_{\mathbf{s}\mathbf{u}\mathbf{n}}\mathbf{/}{\mathit{F}}_{\mathbf{E}\mathbf{a}\mathbf{r}\mathbf{t}\mathbf{h}}$of these two forces? (The average Sun–Moon distance is equal to the Sun–Earth distance.)

Short answer

The value of the ratio $\frac{F_{sun}}{F_{earth}}\mathrm{is}2.16\times {10}^4$

Step-by-step solution

The given data

Mass of Sun,$m_s=1.99\times {10}^{30}kg$

Mass of Earth,$m_s=5.98\times {10}^{24}kg$

Distance between Earth and Moon,$r_{ms}=3.82\times {10}^{10}m$

Distance between Sun and Moon,$r_{em}=1.5\times {10}^{11}$

Understanding the concept of Newton’s law of gravitation

According to Newton’s law of gravitation, the force between two objects is directly proportional to their masses and inversely proportional to the square of distance between them.

Formula:

Gravitational force between two particles, ${}^{F=\frac{G{m}_1{m}_2}{r^2}}$ (i)

Calculating the value of the ratio,FsunFearth

Using equation (i), Force between Sun & Moon can be given as:

$F_{sm}=\frac{G\times m_s\times m_m}{r_{ms}^2}$

Using equation (i), Force between Earth & Moon can be given as:

$F_{em}=\frac{G\times m_m\times m_e}{r_{em}^2}$

Ratio of two forces becomes:

$$\begin{gathered} \frac{F_{sm}}{F_{em}}=\frac{G\times m_s\times m_m}{r_{ms}^2}\times \frac{r_{em}^2}{G\times m_m\times m_e} \\ =\frac{m_s}{m_e}\times \frac{r_{em}^2}{r_{sm}^2} \\ =\frac{1.99\times {10}^{30}}{5.98\times {10}^{24}}\times \frac{\left(3.82\times {10}^{10}\right)}{\left(1.5\times {10}^{11}\right)} \\ =2.16\times {10}^4 \end{gathered}$$

The value of the ratio is$2.16\times {10}^4$