Chapter 4, Problem 58Q

(a)

To determine

The gravitational force exerted by the Moon on a $1\text{-kg}$ rock at a point on the Earth’s surface closest to the Moon. Given that the average distance from the Moon to the center of Earth is $384,400\text{km}$ and the diameter of Earth is $12,756\text{ km}$.

Answer to Problem 58Q

Solution:

$3.43\times {10}^{-5}\text{N}\text{.}$

Explanation of Solution

Given data:

Average distance from the Moon to the center of Earth is $384,400\text{km}$.

Diameter of Earth is $12,756\text{ km}$.

Mass of the rock is $1\text{-kg}\text{.}$

Formula used:

Write the expression for universal law of gravitation between two objects.

$F=\frac{G{m}_1{m}_2}{r^2}$

Here $G$ is the universal law of gravitation, $m_1$ is the mass of first object, $m_2$ is the mass of second object and $r$ is the separation between the two objects.

Explanation:

The the average distance from the Moon to the center of Earth is $384,400\text{km}$ and the diameter of Earth is $12,756\text{ km}$.

For a point on Earth closest to the Moon, separation between the Moon $\left(d^{\prime }\right)$ and the rock is calculated as,

$\begin{array}{c}{d}^{\prime }=\left(\text{Average distance between earth and moon}\right)-\frac{1}{2}\left(\text{ diameter of earth}\right)\\ =384,400\text{ km }-\frac{1}{2}\left(12,756\text{ km}\right)\\ =3.78\times {10}^5\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\\ =3.78\times {10}^8\text{ m}\end{array}$

Refer to the expression for universal law of gravitation between two objects.

$F=\frac{G{m}_1{m}_2}{r^2}$

Upon substituting $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}$ for $G$, $7.348\times {10}^{22}\text{kg}$ for $m_1$, $1\text{ kg}$ for $m_2$ and $d^{\prime }$ for r,

$F\text{ = }\frac{\text{(6}\text{.67}\times {\text{10}}^{-11}{\text{ Nm}}^2{\text{kg}}^{-2})(7.348\times {10}^{22}\text{ kg)(1 kg)}}{{(d^{\prime }\text{)}}^2}$

Substitute $3.78\times {10}^8\text{ m}$ for $d^{\prime }$.

$\begin{array}{c}F\text{ = }\frac{\text{(6}\text{.67}\times {\text{10}}^{-11}{\text{ Nm}}^2{\text{kg}}^{-2})(7.348\times {10}^{22}\text{ kg)(1 kg)}}{{(3.78\times {10}^8\text{ m)}}^2}\\ =3.43\times {10}^{-5}\text{ N}\end{array}$

Conclusion:

Hence, the gravitational force of attraction that the Moon would exert on a $1\text{-kg}$ rock at a point on the Earth’s surface closest to the Moon is $3.43\times {10}^{-5}\text{ N}$.

(b)

To determine

The gravitational force exerted by the Moon on a $1\text{-kg}$ rock at a point on the Earth’s surface farthest to the Moon. Given that the average distance from the Moon to the center of Earth is $384,400\text{km}$ and the diameter of Earth is $12,756\text{ km}$.

Answer to Problem 58Q

Solution:

$3.21\times {10}^{-5}\text{N}\text{.}$

Explanation of Solution

Given data:

Average distance from the Moon to the center of Earth is $384,400\text{km}$

Diameter of Earth is $12,756\text{ km}$.

Mass of the rock is $1\text{-kg}\text{.}$

Formula used:

Write the expression for universal law of gravitation between two objects

$F=\frac{G{m}_1{m}_2}{r^2}$

Here $G$ is the universal law of gravitation, $m_1$ is the mass of first object, $m_2$ is the mass of second object and $r$ is the distance of separation between the two objects.

Explanation:

The the average distance from the Moon to the center of Earth is $384,400\text{km}$ and the diameter of Earth is $12,756\text{ km}$.

For a point on Earth farthest to the Moon, separation between the Moon $\left(d^{\prime }\right)$ and the rock is calculated as,

$\begin{array}{c}d=\left(\text{Average distance between earth and moon}\right)+\frac{1}{2}\left(\text{ diameter of earth}\right)\\ =384,400\text{ km + }\frac{1}{2}\left(12,756\text{ km}\right)\\ =3.91\times {10}^5\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\\ =3.91\times {10}^8\text{ m}\end{array}$

Refer to the expression for universal law of gravitation between two objects.

$F=\frac{G{m}_1{m}_2}{r^2}$

Upon substituting $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}$ for $G$, $7.348\times {10}^{22}\text{kg}$ for $m_1$, $1\text{ kg}$ for $m_2$ and $d$ for r,

$F^{\prime }\text{ = }\frac{\text{(6}\text{.67}\times {\text{10}}^{-11}{\text{ Nm}}^2{\text{kg}}^{-2})(7.348\times {10}^{22}\text{ kg)(1 kg)}}{{(d\text{)}}^2}$

Substitute $3.91\times {10}^8\text{ m}$ for $d$.

$\begin{array}{c}{F}^{\prime }\text{ = }\frac{\text{(6}\text{.67}\times {\text{10}}^{-11}{\text{ Nm}}^2{\text{kg}}^{-2})(7.348\times {10}^{22}\text{ kg)(1 kg)}}{{(3.91\times {10}^8\text{ m)}}^2}\\ =3.21\times {10}^{-5}\text{ N}\end{array}$

Conclusion:

Hence, the gravitational force that the Moon would exert on a $1\text{-kg}$ rock at a point on the Earth’s surface farthest from the Moon is $3.21\times {10}^{-5}\text{ N}$.

(c)

To determine

The difference between the gravitational force that the Moon would exert on a $1\text{-kg}$ rock at a point on the Earth’s surface closest to the Moon and and at a point on the Earth’s surface farthest to the moon.

Answer to Problem 58Q

Solution:

$0.22\times {10}^{-5}\text{ N}$.

Explanation of Solution

Given data:

Average distance from the Moon to the center of Earth is $384,400\text{km}$.

Diameter of Earth is $12,756\text{ km}$.

Mass of the rock is $1\text{-kg}\text{.}$

Explanation:

The gravitational force of attraction $\left(F\right)$ between the Moon and a $1\text{-kg}$ rock, at a point on the Earth’s surface closest to the Moon, is $3.43\times {10}^{-5}\text{ N}$. [From part(a)].

The gravitational force of attraction $\left(F^{\prime }\right)$ between the Moon and a $1\text{-kg}$ rock, at a point on the Earth’s surface farthest to the Moon, is $3.21\times {10}^{-5}\text{ N}$ [From part(b)].

Calculate the difference (say $D$) between the gravitational force that the Moon would exert on a $1\text{-kg}$ rock at a point on the Earth’s surface closest to the Moon and at a point on the Earth’s surface farthest to the moon.

$\begin{array}{c}D=F-F^{\prime }\\ =\left(3.43\times {10}^{-5}\text{ N}\right)-\left(3.21\times {10}^{-5}\text{ N}\right)\\ =0.22\times {10}^{-5}\text{ N}\end{array}$

As the difference between the gravitational force exerted by Moon on a 1 kg stone closest to the Earth’s surface and on a 1 kg stone farthest to the Earth’s surface is $0.22\times {10}^{-5}\text{ N}$. The tidal energy generated by this will be very small as compared to Earth’s gravity which tries to maintain the spherical surface.

Conclusion:

Hence, the difference between the gravitational force that the Moon would exert on a $1\text{-kg}$ rock at a point on the Earth’s surface closest to the Moon and at a point on the Earth’s surface farthest to the moon is $0.22\times {10}^{-5}\text{ N}$.