Chapter 7, Problem 63A

a.

To determine

The gravitation attractive force between Earth and the Moon.

Answer to Problem 63A

  $1.921\times {10}^{20}\text{N}$

Explanation of Solution

Given:

The mass of the Earth is $5.97\times {10}^{24}\text{}\text{kg}$ , mass of the Moon is $7.34\times {10}^{22}\text{}\text{kg}$ and the distance between Earth and the Moon is $3.9\times {10}^8\text{}\text{m}$ .

Formula used:

Gravitational force:

  $F=\frac{G{m}_1{m}_2}{r^2}$

Where, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between the masses.

Calculation:

The value of gravitational constant is,

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Then using the given values of masses of Earth and the Moon, and the distance between them, the gravitational force between them will be,

  $\begin{array}{c}F=\frac{G{m}_1{m}_2}{r^2}\\ =\frac{6.67\times {10}^{-11}\times 5.97\times {10}^{24}\text{}\times 7.34\times {10}^{22}}{{\left(3.9\times {10}^8\right)}^2}\\ =1.921\times {10}^{20}\text{N}\end{array}$

Conclusion:

Thus, the gravitational attractive force between the Earth and the Moon is $1.921\times {10}^{20}\text{N}$ .

b.

To determine

The Earth’s gravitational field at the Moon.

Answer to Problem 63A

  $0.00261{\text{m/s}}^2$

Explanation of Solution

Given:

The mass of the Moon is $7.34\times {10}^{22}\text{}\text{kg}$ .

From part (a), the gravitational attractive force between Earth and the Moon is $1.921\times {10}^{20}\text{N}$ .

Formula used:

The weight of an object of mass m is given by,

  $W=mg$

Where g is the strength of the gravitational field.

Calculation:

The weight of the Moon due the Earth’s gravitational field will be equal to the force of attraction between them, thus

  $\begin{array}{c}F=M_{moon}g\\ 1.921\times {10}^{20}=7.34\times {10}^{22}\times g\\ g=0.00261{\text{m/s}}^2\end{array}$

Conclusion:

Thus, the Earth’s gravitational field at the Moonis $0.00261{\text{m/s}}^2$ .