Question

During a solar eclipse, the moon, the earth, and the sun all lie on the same line, with the moon lying between the earth and the sun. (a) What force is exerted by the sun on the moon? (b) What force is exerted by the Earth on the moon? (c) What force is exerted by the sun on the Earth? (d) Compare the answers to parts (a) and (b) .Why doesn’t the sun capture the moon away from the Earth?

Short answer

$$\begin{gathered} \left(a\right)F_{SM}=4.39\times 10{ }^{20} N towards the sun \\ \left(b\right)F_{EM}=1.99\times 10{ }^{20} N towards the Earth \\ \left(c\right)F_{SE}=3.55\times 10{ }^{22} N towards the sun \end{gathered}$$

(d) The force exerted by the Sun on the Moon is greater than the force exerted by the Earth on the moon. Therefore, the Moon is orbiting the Sun more than it orbits the Earth. But, it only appears to be orbiting the Earth as the Earth is also orbiting the Sun and the Moon is orbiting the center of mass of the Sun-Earth system.

Step-by-step solution

Given data

$$\begin{gathered} {\mathrm{M}}_{\mathrm{s}}\left(\mathrm{massoftheSun}\right)=1.991\times {10}^{30}\mathrm{kg} \\ {\mathrm{M}}_{\mathrm{E}}\left(\mathrm{massoftheEarth}\right)=5.98\times {10}^{24}\mathrm{kg} \\ {\mathrm{M}}_{\mathrm{M}}\left(\mathrm{massoftheMoon}\right)=7.36\times {10}^{22}\mathrm{kg} \end{gathered}$$

$$\begin{gathered} {\mathrm{r}}_{\mathrm{SE}}\left(\mathrm{distancebetween} \mathrm{Sun}-\mathrm{Earth} \mathrm{system}\right)=1.496\times {10}^{11} \mathrm{m} \\ {\mathrm{r}}_{\mathrm{EM}}\left(\mathrm{distancebetween} \mathrm{Earth}-\mathrm{Moon} \mathrm{system}\right)= 3.84\times {10}^8\mathrm{m} \\ {\mathrm{r}}_{\mathrm{SM}}\left(\mathrm{distancebetween} \mathrm{Sun}-\mathrm{Moon} \mathrm{system}\right)=1.496\times {10}^{11} \mathrm{m}-3.84\times {10}^8\mathrm{m} \end{gathered}$$

Definition and concept of newton’s law of gravitation

According to Newton's Law of Universal Gravitation, every particle in the universe is drawn to every other particle with a force that is directly proportional to the product of their masses and inversely proportional to their distance from one another.

During a solar eclipse, the moon, the earth, and the sun all lie on the same line, with the moon lying between the earth and the sun.

The two particles have masses $m_1 and m_2$are separated by a distance r, the force exerted by particle 1 on particle 2 is:

$F_g=G\frac{m_{m}m{m}_2}{r^2}$

G is a constant, called the universal gravitational constant, and has a value of

$6.674\times {10}^{-11}N{m}^2/K{g}^2$

(a) Determining the force exerted by the sun on the moon

This is a simple substitution problem. Apply equation (i) to this part

$F_{SM}=G\frac{m_S{m}_M}{{r^2}_{SM}}$

Substitute numerical values:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{SM}}=\mathrm{m}\frac{\left({\displaystyle 6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{Kg}}^2}\right){\displaystyle \left(1.991\times {10}^{30}\mathrm{Kg}\right)}{\displaystyle \left(7.36\times {10}^{22}\mathrm{Kg}\right)}}{{\displaystyle {\left({\displaystyle 1.496\times {10}^{11}\mathrm{m}-3.84\times {10}^8\mathrm{m}}\right)}^2}} \\ =\left(4.39\times {10}^{20} \mathrm{N} \right) \mathrm{towards} \mathrm{the} \mathrm{sun} \end{gathered}$$

(b) Determining the force exerted by the Earth on the moon

$F_{EM}=G\frac{m_E{m}_M}{{r^2}_{EM}}$

Substitute numerical values:

$$\begin{gathered} \frac{\left({\displaystyle 6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{Kg}}^2}\right){\displaystyle \left(5.98\times {10}^{24}kg\right)}{\displaystyle \left(7.36\times {10}^{22}\mathrm{Kg}\right)}}{{\displaystyle {\left(3.84\times {10}^8\mathrm{m}\right)}^2}} \\ =1.99\times {10}^{20} \text{N\hspace{0.05em}\hspace{0.05em}towards\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}the\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}Earth} \end{gathered}$$

(c) Determining the force exerted by the sun on the Earth

$F_{SE}=G\frac{m_S{m}_M}{{r^2}_{SE}}$

Substitute numerical values:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{SM}}=\mathrm{m}\frac{\left({\displaystyle 6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{Kg}}^2}\right){\displaystyle \left(1.991\times {10}^{30}kg\right)}{\displaystyle \left(5.98\times {10}^{24}kg\right)}}{{\displaystyle {\left(1.496\times {10}^{11}\mathrm{m}\right)}^2}} \\ =3.55\times {10}^{22} \text{N\hspace{0.05em}\hspace{0.05em}towards\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}the\hspace{0.05em}\hspace{0.05em}\hspace{0.05em}Earth} \end{gathered}$$

(d) Comparing the answer of part (a) and (b)

The force exerted by the Sun on the moon is greater than the force exerted by the Earth on the moon. Therefore, the Moon orbits the Sun more than it orbits the Earth.

But, it only appears to be orbiting the Earth as the Earth is also orbiting the Sun and the Moon is orbiting the center of mass of the Sun-Earth system. So that Sun does not capture the moon away from the earth