Chapter 2, Problem 60P

(a)

To determine

The net gravitational force on the moon during a lunar eclipse.

Answer to Problem 60P

The net gravitational force on the moon during a lunar eclipse is $4.77\times {10}^{20}\text{N}$.

Explanation of Solution

Write the expression for gravitational force acts on the Moon by the Earth.

  $F_{ME}=\frac{G{M}_E{M}_M}{r_{EM}^2}$        (I)

Here, $F_{EM}$ is the gravitational force acts on the Moon by the Earth, $G$ is the gravitational constant, $M_E$ is the mass of the Earth, $M_M$ is the mass of the Moon and $r_{EM}$ is the distance between the moon and the earth.

Write the expression for gravitational force acts on the Moon by the Sun.

  $F_{SM}=\frac{G{M}_S{M}_M}{\left(r_{SE}-r_{EM}\right)}$        (II)

Here, $F_{SM}$ is the gravitational force acts on the Moon by the Sun, $G$ is the gravitational constant, $M_S$ is the mass of the Sun, $M_M$ is the mass of the Moon, $r_{EM}$ is the distance between the moon and the Earth and $r_{SE}$ is the distance between the sun and the Earth.

The net gravitational force on the moon is,

  $F=\sqrt{F_{SM}^2+F_{EM}^2}$        (III)

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $7.35\times {10}^{22}\text{kg}$ for $M_M$, and $\text{3}\text{.85}\times {\text{10}}^8\text{m}$ for $r_{EM}$ in the equation (I) to find $F_{EM}$.

  $\begin{array}{c}{F}_{EM}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)\left(7.35\times {10}^{22}\text{kg}\right)}{{\left(\text{3}\text{.85}\times {\text{10}}^8\text{m}\right)}^2}\\ =1.98\times {10}^{20}\text{N}\end{array}$

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2$ for $G$, $1.99\times {10}^{30}\text{kg}$ for $M_S$, $7.35\times {10}^{22}\text{kg}$ for $M_M$, $3.85\times {10}^8\text{m}$ for $r_{EM}$, and $1.50\times {\text{10}}^{11}\text{m}$ for $r_{SE}$ in the equation (II) to find $F_{SM}$.

  $\begin{array}{c}{F}_{SM}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2\right)\left(1.99\times {10}^{24}\text{kg}\right)\left(7.35\times {10}^{22}\text{kg}\right)}{{\left(1.50\times {\text{10}}^{11}\text{m}-3.85\times {10}^8\text{m}\right)}^2}\\ =4.34\times {10}^{20}\text{N}\end{array}$

Substitute $1.98\times {10}^{20}\text{N}$ for $F_{EM}$ and $4.34\times {10}^{20}\text{N}$ for $F_{SM}$ in the equation (III) to find $F$.

  $\begin{array}{c}F=\sqrt{{\left(4.34\times {10}^{20}\text{N}\right)}^2+{\left(1.98\times {10}^{20}\text{N}\right)}^2}\\ =4.77\times {10}^{20}\text{N}\end{array}$

Therefore, the net gravitational force on the moon during a lunar eclipse is $4.77\times {10}^{20}\text{N}$.

(b)

To determine

The net gravitational force on the moon during a solar eclipse.

Answer to Problem 60P

The net gravitational force on the moon during a solar eclipse is $5.6\times {10}^4\text{N}$.

Explanation of Solution

Write the expression for gravitational force acts on the Sun by the Earth.

  $F_{SE}=\frac{G{M}_S{M}_E}{r_{SE}^2}$        (IV)

Here, $F_{SE}$ is the gravitational force acts on the Sun by the Earth, $G$ is the gravitational constant, $M_E$ is the mass of the Earth, $M_M$ is the mass of the Moon and $r_{EM}$ is the distance between the moon and the earth.

Write the expression for gravitational force acts on the Moon by the Sun.

  $F_{SM}=\frac{G{M}_S{M}_M}{r_{SM}^2}$        (V)

Here, $F_{SM}$ is the gravitational force acts on the Moon by the Sun, $G$ is the gravitational constant, $M_S$ is the mass of the Sun, $M_M$ is the mass of the Moon and $r_{SM}$ is the distance between the moon and the sun.

The net gravitational force on the moon is,

  $F=\sqrt{F_{SM}^2+F_{SE}^2}$        (VI)

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_E$, $1.99\times {10}^{30}\text{kg}$ for $M_S$, and $1.50\times {\text{10}}^{11}\text{m}$ for $r_{SM}$ in the equation (I) to find $F_{SE}$.

  $\begin{array}{c}{F}_{SE}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)\left(1.99\times {10}^{22}\text{kg}\right)}{{\left(1.50\times {\text{10}}^{11}\text{m}\right)}^2}\\ =\text{3}\text{.53}\times {\text{10}}^{14}\text{N}\end{array}$

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2$ for $G$, $1.99\times {10}^{30}\text{kg}$ for $M_S$, $7.35\times {10}^{22}\text{kg}$ for $M_M$, and $1.50\times {\text{10}}^{11}\text{m}$ for $r_{SM}$ in the equation (II) to find $F_{SM}$.

  $\begin{array}{c}{F}_{SM}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2\right)\left(1.99\times {10}^{24}\text{kg}\right)\left(7.35\times {10}^{22}\text{kg}\right)}{{\left(1.50\times {\text{10}}^{11}\text{m}\right)}^2}\\ =4.34\times {10}^{14}\text{N}\end{array}$

Substitute $3.53\times {10}^{14}\text{N}$ for $F_{SE}$ and $4.34\times {10}^{14}\text{N}$ for $F_{SM}$ in the equation (III) to find $F$.

  $\begin{array}{c}F=\sqrt{{\left(3.53\times {10}^{14}\text{N}\right)}^2+{\left(4.34\times {10}^{14}\text{N}\right)}^2}\\ =5.6\times {10}^4\text{N}\end{array}$

Therefore, the net gravitational force on the moon during a solar eclipse is $5.6\times {10}^4\text{N}$.