Chapter 7, Problem 68A

To determine

The force of attraction between the spheres.

Answer to Problem 68A

  $8.33\times {10}^{-9}\text{N}$

Explanation of Solution

Given:

The weight of the first sphere is $W_1=9.8\times {10}^2\text{N}$ .

The weight of the second sphere is $W_2=1.96\times {10}^2\text{N}$ .

The distance between the spheres is $r=4\text{m}$ .

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.

The weight of an object of mass m on the planet is:

  $W=mg$

Where, g is the gravitational strength on the planet.

Calculation:

Using the above formula and the known value of g on Earth, the mass of the spheres will be,

  $\begin{array}{l}{m}_1=\frac{W_1}{g}=\frac{9.8\times {10}^2}{9.8}=100\text{kg}\\ m_2=\frac{W_2}{g}=\frac{1.96\times {10}^2}{9.8}=20\text{kg}\end{array}$

The value of gravitational constant is,

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

Then, the gravitational force between the spheres is,

  $\begin{array}{c}F=\frac{G{m}_1{m}_2}{r^2}\\ =\frac{6.67\times {10}^{-11}\times 100\times 20}{{\left(4\right)}^2}\\ =8.33\times {10}^{-9}\text{N}\end{array}$

Conclusion:

Thus, the gravitational force between the spheres is $8.33\times {10}^{-9}\text{N}$ .