Chapter 8.2, Problem 14P

To determine

To calculate: Number of times the gravitational force of earth is greater on each sphere than between the spheres.

Answer to Problem 14P

The gravitational force of earth on each sphere is $4.1\times {10}^{14}$ times than between the spheres.

Explanation of Solution

Given information:

The following variation:

  $F=k\frac{m_1{m}_2}{r^2}$

Radius of the Earth is $\text{6400 km}$ and mass of the Earth is $6\times {10}^{24}\text{kg}$ .

Distance between the spheres is $\text{3 m}$ . Mass of each sphere is $3200\text{kg}$ and radius Is $\text{1 m}$ .

Calculation:

Let mass of earth be M, mass of each sphere be m and R be the distance between their centers,

Therefore, gravitational force of Earth on each sphere is,

  $F_1=k\frac{Mm}{R^2}$

Let distance between the two spheres be r. Therefore, Gravitational force of Earth between the spheres is,

  $F_2=k\frac{mm}{r^2}$

Ratio between the two forces is:

  $\begin{array}{l}\frac{F_1}{F_2}=\frac{k\frac{Mm}{R^2}}{k\frac{mm}{r^2}}[\text{Ratio]}\\ \frac{F_1}{F_2}=\frac{\cancel{k}M\cancel{m}}{R^2}\times \frac{r^2}{\cancel{k}m\cancel{m}}[\text{Multiply by reciprocal]}\\ \frac{F_1}{F_2}=\frac{M{r}^2}{m{R}^2}[\text{Simplify]}\\ \frac{F_1}{F_2}=\frac{6\times {10}^{24}\times {\left(3\right)}^2}{3200{\left(6400\right)}^2}[\text{Substitute]}\\ \frac{F_1}{F_2}=\frac{6\times {10}^{24}\times 9}{3200\times 4096\times {10}^4}[\text{Find square]}\\ \frac{F_1}{F_2}=4.1\times {10}^{14}[\text{Simplify]}\end{array}$