Chapter 7, Problem 67A

a.

To determine

The mass of each satellite.

Answer to Problem 67A

  $1.642\times {10}^3\text{kg}$

Explanation of Solution

Given:

The distance between two satellites is $30.0\text{m}$ .

The gravitational force between them is $2.0\times {10}^{-7}\text{N}$ .

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 the gravitational force and the distance, the mass of each satellite will be,

  $\begin{array}{c}F=\frac{Gmm}{r^2}\\ 2.0\times {10}^{-7}=\frac{6.67\times {10}^{-11}\times m^2}{{\left(30\right)}^2}\\ =1.642\times {10}^3\text{kg}\end{array}$

Conclusion:

Thus, the mass of each satellite is $1.642\times {10}^3\text{kg}$ .

b.

To determine

The initial acceleration given to each satellite by the gravitational force.

Answer to Problem 67A

  $1.21\times {10}^{-10}\text{}{\text{m/s}}^{\text{2}}$

Explanation of Solution

Given:

The gravitational force between the satellites is $2.0\times {10}^{-7}\text{N}$ .

From part (a), the mass of each satellite is $1.642\times {10}^3\text{}\text{kg}$ .

Formula used:

Newton’s second law of motion,

  $F=ma$

Where $F$ is the force, m is the mass, and a is the acceleration of the mass.

Calculation:

Now, the initial acceleration of the satellite due to the gravitational force will be,

  $\begin{array}{c}F=ma\\ 2.0\times {10}^{-7}=\left(1.642\times {10}^3\right)a\\ a=1.21\times {10}^{-10}\text{}{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the initial acceleration of the satellite due to the gravitational force is $1.21\times {10}^{-10}\text{}{\text{m/s}}^{\text{2}}$ .