Chapter 5, Problem 5SP

(a)

To determine

The average force exerted on the Earth by the sun using Newton’s law of gravitation.

Answer to Problem 5SP

The average force exerted on the Earth by the sun is $3.53\times {10}^{22}\text{ N}$.

Explanation of Solution

Given Info: The mass of sun is $1.99\times {10}^{30}\text{ kg}$ and the mass of Earth is $5.98\times {10}^{24}\text{ kg}$. The average distance between the Earth and the sun is $1.50\times {10}^{11}\text{m}$.

Write the mathematical expression for Newton’s law of universal gravitation.

$F=G\frac{m_1{m}_2}{r^2}$ (1)

Here,

$F$ is the gravitational force between the two masses $m_1$ and $m_2$

$G$ is the universal gravitational constant

$r$ is the distance between the two masses $m_1$ and $m_2$

The value of $G$ is $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$.

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$,$1.99\times {10}^{30}\text{ kg}$ for $m_1$,$5.98\times {10}^{24}\text{ kg}$ for $m_2$ and $1.50\times {10}^{11}\text{m}$ for $r$ in the above equation to find $F$.

$\begin{array}{c}F=\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(1.99\times {10}^{30}\text{ kg}\right)\left(5.98\times {10}^{24}\text{ kg}\right)}{{\left(1.50\times {10}^{11}\text{m}\right)}^2}\\ =3.53\times {10}^{22}\text{ N}\end{array}$

Conclusion:

Thus, the average force exerted on the Earth by the sun is $3.53\times {10}^{22}\text{ N}$.

(b)

To determine

The average force exerted on the Earth by the moon.

Answer to Problem 5SP

The average force exerted on the Earth by the moon is $2.01\times {10}^{20}\text{ N}$.

Explanation of Solution

Given Info: The mass of Earth is $5.98\times {10}^{24}\text{ kg}$ and the mass of moon is $7.36\times {10}^{22}\text{ kg}$. The average distance between the Earth and the moon is $3.82\times {10}^8\text{m}$.

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$,$7.36\times {10}^{22}\text{ kg}$ for $m_1$,$5.98\times {10}^{24}\text{ kg}$ for $m_2$ and $3.82\times {10}^8\text{m}$ for $r$ in the above equation to find $F$.

$\begin{array}{c}F=\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(7.36\times {10}^{22}\text{ kg}\right)\left(5.98\times {10}^{24}\text{ kg}\right)}{{\left(3.82\times {10}^8\text{m}\right)}^2}\\ =2.01\times {10}^{20}\text{ N}\end{array}$

Conclusion:

Thus, the average force exerted on the Earth by the moon is $2.01\times {10}^{20}\text{ N}$.

(c)

To determine

The ratio of the force exerted on the Earth by the sun to that exerted by the moon and whether the moon will have much of an impact on the Earth’s orbit about the sun.

Answer to Problem 5SP

The ratio of the force exerted on the Earth by the sun to that exerted by the moon is $175:1$ and the moon will not have much of an impact on the orbit of Earth around sun.

Explanation of Solution

Find the ratio of the force exerted on the Earth by the sun to that exerted by the moon.

$\begin{array}{c}\frac{F_{\text{es}}}{F_{\text{em}}}=\frac{3.53\times {10}^{22}\text{ N}}{2.01\times {10}^{20}\text{ N}}\\ =175.62\\ =175\end{array}$

The value of the force exerted on the Earth by the sun is much greater than the force exerted on Earth by the moon. Therefore the moon will not have much of an impact on the orbit of Earth around sun.

Conclusion:

Thus, the ratio of the force exerted on the Earth by the sun to that exerted by the moon is $175:1$ and the moon will not have much of an impact on the orbit of Earth around sun.

(d)

To determine

The average force exerted on the moon by the sun and whether the sun will have much impact on the orbit of the moon about the Earth.

Answer to Problem 5SP

The average force exerted on the moon by the sun is $4.34\times {10}^{20}\text{ N}$ and the sun will have some impact on the orbit of the moon about the Earth.

Explanation of Solution

Given Info: The mass of sun is $1.99\times {10}^{30}\text{ kg}$ and the mass of moon is $7.36\times {10}^{22}\text{ kg}$. The average distance between the moon and the sun is $1.50\times {10}^{11}\text{m}$.

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$,$1.99\times {10}^{30}\text{ kg}$ for $m_1$,$7.36\times {10}^{22}\text{ kg}$ for $m_2$ and $1.50\times {10}^{11}\text{m}$ for $r$ in the above equation to find $F$.

$\begin{array}{c}F=\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(1.99\times {10}^{30}\text{ kg}\right)\left(7.36\times {10}^{22}\text{ kg}\right)}{{\left(1.50\times {10}^{11}\text{m}\right)}^2}\\ =4.34\times {10}^{20}\text{ N}\end{array}$

The Earth and sun exert strong forces on moon. This is because, even though the Earth is much closer to the moon than the sun is, the sun has much larger mass than the Earth. The force of the sun on moon keeps it to remain in the annual orbit about the sun as it moves along with the sun.

Conclusion:

Thus, the average force exerted on the moon by the sun is $4.34\times {10}^{20}\text{ N}$ and the sun will have some impact on the orbit of the moon about the Earth. Even though the earth is much closer to the moon than the sun the heavier mass of the sun makes more influence on the moon’s orbit.