Question

The centers of a 10 kg lead ball and a 100 g lead ball are separated by 10 cm.
a. What gravitational force does each exert on the other?
b. What is the ratio of this gravitational force to the gravitational force of the earth on the 100 g ball?

Short answer

a) The force exerted is 6.67 x 10^-7 N

b) The ratio of gravitational force is 6.80 x 10^-7

Step-by-step solution

Part(a) Step 1: Given information

Mass of ball 1 is m₁=10 kg
Mass of ball 2 is m₂= 100 g =0.10 kg
Distance between balls, r=10 cm=0.10 m

Part(a) Step 2: Explanation

Gravitational force can be calculated between two object by

$F=\frac{G{m}_1{m}_2}{r^2}$

Using this calculate the force between objects

$$\begin{gathered} F=\frac{G{m}_1{m}_2}{r^2} \\ =\frac{\left(6.67\times {10}^{-11}{\mathrm{m}}^3.{\mathrm{kg}}^{-1}.s^{-2}\right)(10\mathrm{kg})(0.100\mathrm{kg})}{(0.010\mathrm{m}{)}^2} \\ =6.67\times {10}^{-7}\mathrm{N} \end{gathered}$$

Part(b) Step 1: Given information

Mass of ball 1 is m₁=10 kg
Mass of ball 2 is m₂= 100 g =0.10 kg
Distance between balls, r=10 cm=0.10 m

Part(b) : Step2 : Explanation

Gravitational force by earth is

W = mg = 0.1 x 9.8 m/s²= 0.98 N

Now use the gravitational force calculated in part(a) to get the ratio

$\frac{F}{W}=\frac{6.67\times {10}^7\mathrm{N}}{0.98\mathrm{N}}=6.80\times {10}^{-7}$