Chapter 7, Problem 55PQ

(a)

To determine

Moon’s gravitational field at the side of Earth which is facing Moon.

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing Moon is $\underset{\_}{3.44\times {10}^{-5}\text{m}/\text{s}}$.

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance minus radius of Earth.

$g=\frac{G{M}_{\text{moon}}}{r^2}$ (I)

Here, $g$ is the gravitational field, $G$ is the gravitational constant, $M_{moon}$ is the mass of Moon, $d$ is the Earth-Moon distance, $R_e$ is the radius of Earth.

Write the expression to find $r$.

  $r=d-R_e$

Here, $d$ is the Earth-Moon distance, and $R_e$ is the radius of Earth.

Use the expression for $r$ in equation (I).

$g=\frac{G{M}_{\text{moon}}}{{\left(d-R_e\right)}^2}$ (II)

Conclusion

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$ ,$7.35\times {10}^{22}\text{kg}$ for $M_{\text{moon}}$,$3.84\times {10}^8\text{m}$ for $d$, and $6.37\times {10}^6\text{m}$ for $R_e$ in equation (II) to get $g$

  $\begin{array}{c}g=\frac{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(7.35\times {10}^{22}\text{kg}\right)}{{\left(3.84\times {10}^8\text{m}-6.37\times {10}^6\text{m}\right)}^2}\\ =3.44\times {10}^{-5}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Therefore, Moon’s gravitational field at the side of Earth facing Moon is $\underset{\_}{3.44\times {10}^{-5}\text{m}/\text{s}}$.

(b)

To determine

Moon’s gravitational field at the side of Earth facing away from Moon.

Answer to Problem 55PQ

Moon’s gravitational field at the side of Earth facing away from Moon is $\underset{\_}{3.22\times {10}^{-5}\text{m}/\text{s}}$.

Explanation of Solution

Write the equation to find the gravitational field due to Moon at a distance of Moon-Earth distance plus radius of Earth.

$g=\frac{G{M}_{\text{moon}}}{r^2}$ (III)

Write the expression for $r$.

  $r=d+R_e$

Substitute the expression for $r$ in equation (III).

$g=\frac{G{M}_{\text{moon}}}{{\left(d+R_e\right)}^2}$ (IV)

Conclusion:

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $7.35\times {10}^{22}\text{kg}$ for $M_{\text{moon}}$, $3.84\times {10}^8\text{m}$ for $d$, and $6.37\times {10}^6\text{m}$ for $R_e$ in equation (IV) to get $g$.

  $\begin{array}{c}g=\frac{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(7.35\times {10}^{22}\text{kg}\right)}{{\left(3.84\times {10}^8\text{m}+6.37\times {10}^6\text{m}\right)}^2}\\ =3.22\times {10}^{-5}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Therefore, Moon’s gravitational field at the side of Earth facing away Moon is $\underset{\_}{3.22\times {10}^{-5}\text{m}/\text{s}}$.

(c)

To determine

The gravitational field of Moon at the center of Earth.

Answer to Problem 55PQ

Moon’s gravitational field at the center of Earth is ${\underset{\_}{3.32\times {10}^{-5}\text{m}/\text{s}}}^2$.

Explanation of Solution

Write the equation to find the gravitational field due to Moon.

$g=\frac{G{M}_{moon}}{d^2}$ (IV)

Conclusion:

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $7.35\times {10}^{22}\text{kg}$ for $M_{moon}$,and $3.84\times {10}^8\text{m}$ for $d$ in equation (IV) to get $g$

  $\begin{array}{c}g=\frac{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(7.35\times {10}^{22}\text{kg}\right)}{{\left(3.84\times {10}^8\text{m}\right)}^2}\\ =3.32\times {10}^{-5}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Therefore, Moon’s gravitational field at center of Earth is $\underset{\_}{3.32\times {10}^{-5}\text{m}/\text{s}}$.

(d)

To determine

Sketch Earth and include the three vectors from parts (a) through (c).

Answer to Problem 55PQ

Sketch of Earth and the three vectors from parts (a) through (c) is shown in Figure 1.

Explanation of Solution

Figure 1 shows the sketch of the Earth and the magnitude and direction of the gravitational field vectors found in part (a), (b) and (c).

Conclusion:

Therefore, Sketch of Earth and the three vectors from parts a through c is shown in Figure 1.

(e)

To determine

The reason why there are two tides a day on most places on Earth due to Moon.

Answer to Problem 55PQ

There are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.

Explanation of Solution

Figure below shows the Earth and Moon. High tides and low tides on either side of Earth are due to the lunar activity on Earth. The gravitational pull by Moon on Earth causes high tide and low tide.

The force of Moon is large at water bodies which are close to Moon and lowest on water bodies which are far away. Thus the amplitude or strength of tides is dependent on the distance of the water body and Moon. This is the reason for two types of tides on Earth.

Conclusion:

Therefore, there are two tides a day on most places on Earth due to Moon because the force is larger on bodies of water closer to Moon and smaller on bodies of water on far side of Earth.