Chapter 10, Problem 22Q

(a)

To determine

The tidal force that Earth exerts on $1\text{ kg}$ rock located on the Moon when at perigee by referring table 10-1.

Answer to Problem 22Q

Solution:

$5.778\times {10}^{-5}\text{ N}$.

Explanation of Solution

Given data:

The mass of the rock present on the Moon is $1\text{ kg}$.

The Moon is at perigee position from Earth.

Formula used:

The tidal force that Earth exerts on the Moon is expressed as:

$F_{\text{tidal}}=\frac{2G{M}_{\text{Earth}}md}{r^3}$

Here, $G$ represents the universal gravitational constant, $M_{\text{Earth}}$ represents the mass of Earth, $m$ represents the mass of the rock, $d$ represents the diameter of the Moon, and $r$ represents the average distance between Earth and the Moon.

Explanation:

Recall the expression for the tidal force that Earth exerts on the rock present on the Moon when it is at perigee:

$F_1=\frac{2G{M}_{\text{Earth}}md}{r_1^3}$

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $5.974\times {10}^{24}\text{ kg}$ for $M_{\text{Earth}}$, $1\text{ kg}$ for $m$, $3.476\times {10}^6\text{ m}$ for $d$ and $3.633\times {10}^8\text{ m}$ for $r_1$:

$\begin{array}{c}{F}_1=\frac{2\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(5.974\times {10}^{24}\text{ kg}\right)\left(1\text{ kg}\right)\left(3.476\times {10}^6\text{ m}\right)}{{\left(3.633\times {10}^8\text{ m}\right)}^3}\\ =5.778\times {10}^{-5}\text{ N}\end{array}$

Conclusion:

Hence, the tidal force experienced by $1\text{ kg}$ rock at the Moon, when it is at perigee, is $5.778\times {10}^{-5}\text{ N}$.

(b)

To determine

The tidal force experienced by a $1\text{ kg}$ rock on the Moon when it is at apogee by referring table 10-1.

Answer to Problem 22Q

Solution:

$4.15\times {10}^{-5}\text{ N}$.

Explanation of Solution

Given data:

The mass of the rock present on the Moon is $1\text{ kg}$.

The Moon is at apogee position from Earth.

Formula used:

The tidal force that Earth exerts on the Moon is expressed as:

$F_{\text{tidal}}=\frac{2G{M}_{\text{Earth}}md}{r^3}$

Here, $G$ represents the universal gravitational constant, $M_{\text{Earth}}$ represents the mass of Earth, $m$ represents the mass of the rock, $d$ represents the diameter of the Moon, and $r$ represents the average distance between Earth and the Moon.

Explanation:

Recall the expression for the tidal force that Earth exerts on the rock present on the Moon when it is at apogee:

$F_2=\frac{2G{M}_{\text{Earth}}md}{r_2^3}$

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $5.974\times {10}^{24}\text{ kg}$ for $M_{\text{Earth}}$, $1\text{ kg}$ for $m$, $3.476\times {10}^6\text{ m}$ for $d$ and $4.055\times {10}^8\text{ m}$ for $r_2$:

$\begin{array}{c}{F}_2=\frac{2\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(5.974\times {10}^{24}\text{ kg}\right)\left(1\text{ kg}\right)\left(3.476\times {10}^6\text{ m}\right)}{{\left(4.055\times {10}^8\text{ m}\right)}^3}\\ =4.154\times {10}^{-5}\text{ N}\end{array}$

Conclusion:

Hence, the tidal force experienced by a $1\text{ kg}$ rock on the Moon when it is at apogee is $4.154\times {10}^{-5}\text{ N}$.

(c)

To determine

The ratio of the tidal forces experienced by a $1\text{ kg}$ rock at the Moon at perigee to the tidal force on the same rock on the Moon when it is at apogee position.

Answer to Problem 22Q

Solution:

$1.4$.

Explanation of Solution

Introduction:

The Moon and Earth, both exerts tidal forces, which are pulling in nature and are the direct results of the gravitational forces exerted by them on each other. This force is inversely proportional to the cube of the distance between the centers of Earth and Moon.

Due to the tidal force by the Moon on Earth, tidal waves are generated whereas the Moon recedes away from Earth due to the tidal force of Earth.

Explanation:

The ratio of tidal force experienced by the $1\text{ kg}$ rock when the Moon is at perigee to the instant when it is at apogee, is expressed as:

$\frac{F_{\text{tidal}\left(\text{perigee}\right)}}{F_{\text{tidal}\left(\text{apogee}\right)}}=\frac{F_1}{F_2}$

Refer to the sub-parts (a) and (b) of the problem and substitute $5.778\times {10}^{-5}\text{ N}$ for $F_1$ and $4.154\times {10}^{-5}\text{ N}$ for $F_2$:

$\begin{array}{c}\frac{F_{\text{tidal}\left(\text{perigee}\right)}}{F_{\text{tidal}\left(\text{apogee}\right)}}=\frac{5.778\times {10}^{-5}\text{ N}}{4.154\times {10}^{-5}\text{ N}}\\ =1.391\\ \approx 1.4\end{array}$

Conclusion:

Hence, the rock on the Moon experiences around $40\%$ more force when the Moon is at perigee compared to when it is at apogee position.