Chapter 4, Problem 58Q

To determine

(a)

The gravitational force exerted by the moon on a 1 kg rock placed at the point on Earth’s surface that is closest to the moon.

Answer to Problem 58Q

The gravitational force exerted by the moon on a 1 kg rock at the closest point on the Earth’s surface is $3.43\times {10}^{-5}N$.

Explanation of Solution

Given:

Distance from the moon to the center of the Earth, $d_1=384400km$.

The diameter of Earth, $D=12756km$

The mass of rock, $m=1kg$

Formula used:

The gravitational force between two objects is given by,

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

Where, G is the universal gravitation constant, $m_1$ and $m_2$ are masses of the two objects, and r is the distance between them.

Calculation:

We take the formula for gravitational force, $F=G\frac{m_1{m}_2}{r^2}$

Here, $m_1$ is the mass of the moon, which we know as $m_1$ = $7.349\times {10}^{22}$ kg, and, $m_2$ is the mass of the rock so $m_2$ = 1 kg. Value of G is also known to be $6.67\times {10}^{-11}N{m}^2/k{g}^2$.

First, we calculate the value of $r$, which is the distance between the moon and the closest point on Earth’s surface.

This is given by,

$r=d_1-\frac{D}{2}$

$r=384400-\frac{12756}{2}=378022$ Km = $378022\times {10}^3$ m.

Putting in all the values in the formula for gravitational force, we get,

$\begin{array}{l}F=(6.67\times {10}^{-11})\frac{(7.349\times {10}^{22})(1)}{{\left(378022\times {10}^3\right)}^2}\\ F=3.43\times {10}^{-5}N\end{array}$

Conclusion:

Thus, the gravitational force exerted by the moon on a 1 kg rock at the closest point on the Earth’s surface is $3.43\times {10}^{-5}N$.

To determine

(b)

The gravitational force exerted by the moon on a 1 kg rock placed at the point on Earth’s surface that is farthest to the moon.

Answer to Problem 58Q

The gravitational force exerted by the moon on a 1 kg rock at the farthest point on the Earth’s surface is $3.20\times {10}^{-5}N$

Explanation of Solution

Given:

Distance from the moon to the center of the Earth, $d_1=384400km$.

The diameter of Earth, $D=12756km$ and the mass of rock = 1 kg.

Formula used:

$F=G\frac{m_1{m}_2}{r^2}$, which has been explained in the above section.

Calculation:

Again, we have $m_1$ = $7.349\times {10}^{22}$ kg, $m_2$ = 1 kg and G = $6.67\times {10}^{-11}N{m}^2/k{g}^2$

Here, $r$ is the distance between the moon and the farthest point on the Earth’s surface. This is given by,

$r=d_1+\frac{D}{2}$

$r=384400+\frac{12756}{2}=390778$ km = $390778\times {10}^3$ m.

Substituting all values in $F=G\frac{m_1{m}_2}{r^2}$, we get,

$\begin{array}{l}{F}^{\text{'}}=(6.67\times {10}^{-11})\frac{(7.349\times {10}^{22})(1)}{390778\times {10}^3}\\ F^{\text{'}}=3.20\times {10}^{-5}N\end{array}$

Conclusion:

Thus, the gravitational force exerted by the moon on a 1 kg rock at the farthest point on the Earth’s surface is

$3.20\times {10}^{-5}N$.

To determine

(c)

The difference between the two forces $F$ and $F^{\text{'}}$, calculated in the sections (a) and (b) respectively.

Answer to Problem 58Q

Difference between the two forces $F$ and $F^{\text{'}}$ is $0.23\times {10}^{-5}N$.

Explanation of Solution

Given:

$F=3.43\times {10}^{-5}N$ and $F^{\text{'}}=3.20\times {10}^{-5}N$

Formula used:

The difference between the two forces is calculated by

$\Delta F=F-F^{\text{'}}$

Calculation:

The difference between the two forces is,

$\begin{array}{l}\Delta F=F-F^{\text{'}}\\ \Delta F=3.43\times {10}^{-5}-3.20\times {10}^{-5}\\ \Delta F=0.23\times {10}^{-5}N\end{array}$

The tidal force, i.e., the difference between the two forces $F$ and $F^{\text{'}}$ is $0.23\times {10}^{-5}N$. As we can see, the magnitude of this force is very small. Since the tidal force is of such a small magnitude, so it causes only a small deformation on Earth.

Conclusion:

Thus, the difference between the two forces $F$ and $F^{\text{'}}$ is $0.23\times {10}^{-5}N$.