Chapter 3, Problem 7C

To determine

Show that the gravitational force from the Sun on the Moon is about $2.2\text{times}$ that from the Earth on the moon.

Answer to Problem 7C

The gravitational force from the Sun on the Moon is about $2.2\text{times}$ that from the Earth on the moon is proved below.

Explanation of Solution

According to Newton’s law of gravitation; gravity is a force between two bodies having masses which causes the attraction between them, and it is directly proportional to product of mass of bodies and inversely proportional to square of the distance between them.

Write the expression for gravitational force.

    $F=G\frac{Mm}{r^2}$        (I)

Here, $F$ is the gravitational force between two objects, $M$ is the mass of first object, $m$ is the mass of another object, $G$ is the gravitational constant and $r$ is the distance between them.

Compare gravitational acceleration on Earth and station.

    $\text{Ratio}=\frac{F_{SM}}{F_{EM}}$        (II)

Here, $F_{EM}$ is the gravitational force from Earth on the moon and $F_{SM}$ is the gravitational force from Sun on the moon.

Conclusion:

The gravitational force from Earth on the moon is calculated below.

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}\cdot {\text{kg}}^{\text{-2}}$ for $G$ , $5.972\times {10}^{24}\text{kg}$ for $M$ , $7.35\times {10}^{22}\text{kg}$ for $m$ and $3.85\times {10}^8\text{m}$ for $r$ in equation (I).

    $\begin{array}{c}{F}_{EM}=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}\cdot {\text{kg}}^{\text{-2}}\right)\frac{\left(5.972\times {10}^{24}\text{kg}\right)\left(7.35\times {10}^{22}\text{kg}\right)}{{\left(3.85\times {10}^8\text{m}\right)}^2}\\ =1.975\times {10}^{20}{}^{}\text{N}\end{array}$

The gravitational force from Sun on the moon is calculated below.

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}\cdot {\text{kg}}^{\text{-2}}$ for $G$ , $1.98\times {10}^{30}\text{kg}$ for $M$ , $7.35\times {10}^{22}\text{kg}$ for $m$ and $1.5\times {10}^{11}\text{m}$ for $r$ in equation (I).

    $\begin{array}{c}{F}_{SM}=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}\cdot {\text{kg}}^{\text{-2}}\right)\frac{\left(1.98\times {10}^{30}\text{kg}\right)\left(7.35\times {10}^{22}\text{kg}\right)}{{\left(1.5\times {10}^{11}\text{m}\right)}^2}\\ =4.317\times {10}^{20}{}^{}\text{N}\end{array}$

Substitute  $1.975\times {10}^{20}{}^{}\text{N}$ for $F_{EM}$ and $4.317\times {10}^{20}{}^{}\text{N}$ for $F_{SM}$ in equation (II).

    $\begin{array}{c}\frac{F_{SM}}{F_{EM}}=\frac{4.317\times {10}^{20}{}^{}\text{N}}{1.975\times {10}^{20}{}^{}\text{N}}\\ F_{SM}=2.18F_{EM}\\ =2.2F_{EM}\end{array}$

Thus, the gravitational force from the Sun on the Moon is about $2.2\text{times}$ that from the Earth on the moon is proved above.