Chapter 13, Problem 60A

To determine

Factor of changing force of gravity.

Answer to Problem 60A

The resulting force of gravity between the planets is 4 times the initial force.

Explanation of Solution

Given:

Mass of both planets is doubled.

Distance between the planets remains the same.

Formula Used:

From Newton’s law of gravitation,

  $F=\frac{G{m}_1{m}_2}{r^2}$

  $F$ is force of gravity, $G$ is gravitational constant, $m_1$ and $m_2$ are mass of two objects and $r$ is distance between the two objects.

Calculation:

Let the initial force of gravity is,

  $F=\frac{G{m}_1{m}_2}{r^2}$

As, Mass of both planets is doubled.

So, $m_1\text{'}=2m_1$ and $m_2\text{'}=2m_2$

And distance between the planets remains the same.

So, $r\text{'}=r$

Therefore, the resulting force of gravity between the planets is

  $\begin{array}{l}F\text{'}=\frac{G{m}_1\text{'}{m}_2\text{'}}{{\left(r\text{'}\right)}^2}=\frac{G\left(2m_1\right)\left(2m_2\right)}{{\left(r\right)}^2}\\ =\frac{G{m}_1{m}_2}{r^2}\times 4\\ =F\times 4\left(\text{Since,}F=\frac{G{m}_1{m}_2}{r^2}\right)\\ \Rightarrow F\text{'}=4F\end{array}$

It means the force is increased.

Conclusion:

Therefore, the resulting force of gravity between the planets is 4 times the initial force.