Question

During a solar eclipse, the Moon, the Earth, and the Sun all lie on the same line, with the Moon 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

(a)Force exerted by sun on the moon$F_{sm}\simeq 4.33\times {10}^{20}\text{N}$

(b) Force exerted by earth on the moon$F_{em}\simeq 1.97\times {10}^{20}\text{N}$

(c) Force exerted by sun on the earth$F_{se}\simeq 3.56\times {10}^{22}\text{N}$

(d) Force exerted by sun on the moon is greater as compare to Force exerted by earth on the moon and hence because of this reason the sun does not capture the moon away from the earth.

Step-by-step solution

Step 1: Newton’s law

Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses $M_1$and $M_2$separated by a distance r has the magnitude:

$F_g=G\frac{m_1.m_2}{r^2}$

Where,$G=6.674\times {10}^{-11}\text{N}·{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ is the universal gravitational constant. This equation enables us to calculate the force of attraction between masses under many circumstances.

(a): Determine the force exerted by the Sun on the Moon.

We know that:

Mass of sun$M_s=2\times {10}^{30}\text{kg}$

Mass of earth$M_e=6\times {10}^{24}\text{kg}$

Mass of moon$M_m=7.3\times {10}^{22}\text{kg}$

Distance between sun and earth is$d_{se}=1.5\times {10}^{11}\text{m}$

Distance between earth and moon$d_{em}=3.85\times {10}^8\text{m}$

Force exerted by sun on the moon

$$\begin{gathered} F_{sm}=G\frac{m_s.m_m}{{\left(d_{se}-d_{em}\right)}^2} \\ {F}_{sm}=6.67\times {10}^{-11}\frac{2\times {10}^{30}\times 7.3\times {10}^{22}}{{\left(1.5\times {10}^{11}-3.85\times {10}^8\right)}^2} \\ {F}_{sm}=\frac{6.67\times 14.6\times {10}^{41}}{{\left(1.5-0.00385\right)}^2{.10}^{22}} \\ {F}_{sm}=\frac{97.382\times {10}^{41-22}}{\left(2.25\right)} \\ {F}_{sm}\simeq 4.33\times {10}^{20}\text{N} \end{gathered}$$

Hence, the Force exerted by sun on the moon$F_{sm}\simeq 4.33\times {10}^{20}\text{N}$

(b): Determine the force exerted by the Earth on the Moon.

Force exerted by earth on the moon

$$\begin{gathered} F_{em}=G\frac{m_e.m_m}{{\left(d_{em}\right)}^2} \\ {F}_{em}=6.67\times {10}^{-11}\frac{6\times {10}^{24}\times 7.3\times {10}^{22}}{{\left(3.85\times {10}^8\right)}^2} \\ {F}_{em}=\frac{6.67\times 43.8\times {10}^{35}}{{\left(3.85\right)}^2{.10}^{16}} \\ {F}_{em}=\frac{292.146\times {10}^{35-16}}{14.8225} \\ {F}_{em}\simeq 1.97\times {10}^{20}\text{N} \end{gathered}$$

Hence, the force exerted by the Earth on the Moon$F_{em}\simeq 1.97\times {10}^{20}\text{N}$

(c): Determine the force exerted by the Sun on the Earth

Force exerted by sun on the earth

$$\begin{gathered} F_{se}=G\frac{m_s.m_e}{{\left(d_{se}\right)}^2} \\ {F}_{se}=6.67\times {10}^{-11}\frac{2\times {10}^{30}\times 6\times {10}^{24}}{{\left(1.5\times {10}^{11}\right)}^2} \\ {F}_{se}=\frac{6.67\times 12\times {10}^{54-11}}{{\left(3.85\right)}^2{.10}^{22}} \\ {F}_{se}=\frac{80.04\times {10}^{43-22}}{2.25} \\ {F}_{se}\simeq 3.56\times {10}^{22}\text{N} \end{gathered}$$

Hence, the Force exerted by sun on the earth$F_{se}\simeq 3.56\times {10}^{22}\text{N}$

(d): Compare the answers to parts (a) and (b)

From part (a) Force exerted by sun on the moons $F_{sm}\simeq 4.33\times {10}^{20}\text{N}$and from part (b) Force exerted by earth on the moon $F_{em}\simeq 1.97\times {10}^{20}\text{N}$, So it is clearly seen that Force exerted by sun on the moon is greater as compare to Force exerted by earth on the moon and hence because of this reason the sun does not capture the moon away from the earth.