Chapter 7, Problem 39A

To determine

The force of attraction between the small and large lead sphere.

Answer to Problem 39A

  $6.11\times {10}^{-9}\text{N}$

Explanation of Solution

Given:

The mass of the large lead sphere is $m_1=5.9\text{kg}$ .

The mass of the small lead sphere is $m_2=0.047\text{kg}$ .

The distance between the spheres is $r=0.055\text{m}$ .

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,

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

Now, the gravitational force between the spheres is,

  $\begin{array}{c}F=\frac{G{m}_S{m}_J}{r^2}\\ =\frac{6.67\times {10}^{-11}\times 5.9\times 0.047}{{\left(0.055\right)}^2}\\ =6.11\times {10}^{-9}\text{N}\end{array}$

Conclusion:

Thus, the gravitational force between the spheres is $6.11\times {10}^{-9}\text{N}$ .