Chapter 10, Problem 66QAP

To determine

The gravitational force exerted by the bowling balls on a ping pong ball

Answer to Problem 66QAP

Total gravitational force $=-8.023\times {10}^{-12}N$

Explanation of Solution

Given info:

Mass of a bowling ball = $8.0kg$

Radius = $11.0cm$

Distance from point P = $1.0m$

Mass of ping pong ball = $2.70g$

Formula used:

  $F=-\frac{GMm}{R^2}$

  $F=$ Gravitational force

  $G=$ Gravitational constant

  $M=$ Mass of bowling ball

  $R=$ Radius

  $m=$ Mass of ping pong ball

Calculation:

1. Radius of each bowling bowl,
Using Pythagorean theorem,

  $x1=\sqrt{{(1.0m)}^2+{(0.11*2m)}^2}=1.02m$

  $x2=\sqrt{{(1.0m)}^2+{(0.11*4m)}^2}=1.09m$

  $x3=\sqrt{{(1.0m)}^2+{(0.11*6m)}^2}=1.20m$

Gravitational force of each bowling bowl,

  $F{1}_y=-\frac{GMm}{x{1}^2}=-\frac{(6.67*{10}^{-11}N{m}^2/k{g}^2)*8.0kg*(2.70*{10}^{-3}kg)}{{(1.00m)}^2}=-1.44*{10}^{-12}N$

  $F2=F5=-\frac{GMm}{x{1}^2}=-\frac{(6.67*{10}^{-11}N{m}^2/k{g}^2)*8.0kg*(2.70*{10}^{-3}kg)}{{(1.02m)}^2}=-1.38*{10}^{-12}N$

  $F{2}_y=F{5}_y=\frac{F2}{x1}=-\frac{-1.38*{10}^{-12}N}{1.02m}=-1.35*{10}^{-12}N$

  $F3=F6=-\frac{GMm}{x{1}^2}=-\frac{(6.67*{10}^{-11}N{m}^2/k{g}^2)*8.0kg*(2.70*{10}^{-3}kg)}{{(1.09m)}^2}=-1.21*{10}^{-12}N$

  $F{3}_y=F{6}_y=\frac{F3}{x2}=-\frac{-1.21*{10}^{-12}N}{1.09m}=-1.11*{10}^{-12}N$

  $F4=F7=-\frac{GMm}{x{1}^2}=-\frac{(6.67*{10}^{-11}N{m}^2/k{g}^2)*8.0kg*(2.70*{10}^{-3}kg)}{{(1.20m)}^2}=-1.00*{10}^{-12}N$

  $F{4}_y=F{7}_y=\frac{F3}{x3}=-\frac{-1.00*{10}^{-12}N}{1.20m}=-8.33*{10}^{-13}N$

Total gravitational force,

  $\begin{array}{l}{F}_{net}=F{1}_y+F{2}_y+F{3}_y+F{4}_y+F{5}_y+F{6}_y+F{7}_y=-1.44*{10}^{-12}N+((-1.35*{10}^{-12}N-1.11*{10}^{-12}N-8.33*{10}^{-13}N)*2)\\ F_{net}=-8.023*{10}^{-12}N\end{array}$

Conclusion:

Total gravitational force $=-8.023\times {10}^{-12}N$