Chapter 10, Problem 23Q

(a)

To determine

If Earth’s tidal force would have been able to lift rocks off the lunar surface, when the Moon first coalesced by referring table 10-1.

Answer to Problem 23Q

Solution:

No, the tidal force of Earth on the rock would not have been sufficient to lift it off the Moon’s surface.

Explanation of Solution

Introduction:

According to the Collision ejection theory, the Moon was coalesced from the debris of Earth, when a Mars sized protoplanet collided with the proto-earth.

The tidal force exerted by Earth on the Moon is the result of the gravitational pull that it is exerting on the Moon. But, due to the tidal bulge on Earth, the Moon keeps receding away from it and this is the reason when the Moon first coalesced, it was at one tenth distance from Earth compared to present time.

The gravitational force between two bodies is given as:

$F=G\frac{M_1{m}_2}{r^2}$

Here, $G$ is the universal gravitational constant, $M_1$ and $m_2$ are the two masses, respectively and $r$ represent the distance between the center of mass of two masses.

Explanation:

Recall the expression for the gravitational pull exerted by Earth on a $1\text{ kg}$ rock on the Moon, when it first coalesced as:

$F_1=G\frac{M_{\text{Earth}}{m}_{\text{rock}}}{r_1^2}$

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $5.972\times {10}^{24}\text{ kg}$ for $M_{\text{Earth}}$, $1\text{ kg}$ for $m_{\text{rock}}$ and $3.844\times {10}^7\text{ m}$ for $r_1$:

$\begin{array}{c}{F}_1=\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/\text{kg}\right)\frac{\left(5.972\times {10}^{24}\text{ kg}\right)\left(1\text{ kg}\right)}{{\left(3.844\times {10}^7\text{ m}\right)}^2}\\ =0.27\text{ N}\end{array}$

Similarly, the gravitational force exerted by the Moon on the rock to keep it on its surface is given as:

$F_2=G\frac{M_{\text{Moon}}{m}_{\text{rock}}}{r_2^2}$

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $7.349\times {10}^{22}\text{ kg}$ for $M_{\text{Moon}}$, $1\text{ kg}$ for $m_{\text{rock}}$ and $1.738\times {10}^6\text{ m}$ for $r_2$:

$\begin{array}{c}{F}_2=\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/\text{kg}\right)\frac{\left(7.349\times {10}^{22}\text{ kg}\right)\left(1\text{ kg}\right)}{{\left(1.738\times {10}^6\text{ m}\right)}^2}\\ =1.60\text{ N}\end{array}$

Conclusion:

Since, the force exerted by the Moon on the rock is greater than the force exerted by Earth on the same rock. Therefore, the rock would not get off the Moon’s surface.

(b)

To determine

The comparison between the tidal forces that the Moon would experience due to Earth nowadays with the one that it experienced when it first coalesced.

Answer to Problem 23Q

Solution:

The tidal force on the Moon would have been 1000 times greater compared to the one which is experienced by it today.

Explanation of Solution

Introduction:

The tidal force experienced between two bodies is inversely proportional to the cube of the distance between the center of mass of the bodies. The force is expressed as:

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

Here, $G$ represents the universal gravitational constant, $M_{\text{Earth}}$ and $M_{\text{Moon}}$ represents the mass of Earth and the Moon, respectively, $d$ represents the diameter of the Moon whereas $r$ represents the center to center distance between Earth and the Moon.

Explanation:

Recall the expression for the tidal force experienced by the present day Moon due to Earth as:

$F_{\text{tidal}\left(\text{now}\right)}=\frac{2G{M}_{\text{Earth}}{M}_{\text{Moon}}d}{r_{\text{now}}^3}$

Similarly, the expression, for the tidal force experienced by the Moon due to Earth when it first coalesced, is given as:

$F_{\text{tidal}\left(\text{then}\right)}=\frac{2G{M}_{\text{Earth}}{M}_{\text{Moon}}d}{r_{\text{then}}^3}$

Take the ratio of the two expressions of forces as:

$\begin{array}{c}\frac{F_{\text{tidal}\left(\text{then}\right)}}{F_{\text{tidal}\left(\text{now}\right)}}=\frac{2G{M}_{\text{Earth}}{M}_{\text{Moon}}d}{r_{\text{then}}^3}\times \frac{r_{\text{now}}^3}{2G{M}_{\text{Earth}}{M}_{\text{Moon}}d}\\ ={\left(\frac{r_{\text{now}}}{r_{\text{then}}}\right)}^3\end{array}$

Refer the sub-part (a) of the problem and observe that the distance between Earth and the Moon was one-tenth of the present day and its value can be expressed as:

$\begin{array}{l}{r}_{\text{then}}=\frac{r_{\text{now}}}{10}\\ \frac{r_{\text{now}}}{r_{\text{then}}}=10\end{array}$

Hence, refer to the above calculation and write the expression for the ratio of the tidal forces as:

$\begin{array}{c}\frac{F_{\text{tidal}\left(\text{then}\right)}}{F_{\text{tidal}\left(\text{now}\right)}}={\left(\frac{r_{\text{now}}}{r_{\text{then}}}\right)}^3\\ ={10}^3\\ =1000\end{array}$

Conclusion:

Hence, the tidal force on the Moon when it first coalesced was 1000 times stronger compared to the one experienced by the modern day Moon.