Chapter 7.1, Problem 8SSC

To determine

The gravitation force between two balls, and its fraction as the weight of one ball.

Answer to Problem 8SSC

  $F=1.22\times {10}^{-11}\text{N}$ and $F\approx \left(8.3\times {10}^{-14}\right)W_{ball}$

Explanation of Solution

Given:

The mass of a ball is 15-kg and they are 35 m apart.

Formula used:

By law of Gravitation:

  $F=\frac{G{m}_1{m}_2}{r^2}$

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

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

Substitute the given values of mass of the balls and the distance between them, the gravitational force between the balls will be,

  $\begin{array}{c}F=\frac{Gmm}{r^2}\\ =\frac{6.67\times {10}^{-11}\times {\left(15\right)}^2}{{\left(35\right)}^2}\\ =1.22\times {10}^{-11}\text{N}\end{array}$

Now, the weight of the ball is,

  $\begin{array}{c}W=mg\\ =15\times 9.8\\ =147\text{N}\end{array}$

Then, the gravitational force between the masses as the fraction of the weight of the ball will be,

  $\begin{array}{c}\frac{F}{W}=\frac{1.22\times {10}^{-11}}{147}\\ F\approx \left(8.3\times {10}^{-14}\right)W\end{array}$

Conclusion:

Thus, the gravitational force between the balls is $1.22\times {10}^{-11}\text{N}$ , and it is $8.3\times {10}^{-14}$ times the weight of the ball.