Question

Find the force of attraction between a 60 kg student and:A. Another student of mass 80 kg, 1.4 m away.B. A 130,000 kg blue whale, 10 m away.C. The Great Pyramid in Egypt, with an estimated mass of 5.22 x 109 kg, 1.0 km away.D. A 45 g golf ball, 95 cm away.

Answer 1

According to Newton's Law of Universal Gravitation, the force between to bodies of masses m and M separated by a distance r is:

$F=G\frac{Mm}{r^2}$

Where G is the gravitational constant:

$G=6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2}$

In all cases, use M=60kg, and replace the values of the second mass m and the distance r accordingly.

A) m=80kg, r=1.4m

$\begin{gathered} F=(6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2})\times\frac{(60\operatorname{kg})(80\operatorname{kg})}{(1.4m)^2} \\ =1.6\times10^{-7}N \end{gathered}$

B) m=130,000kg, r=10m

$\begin{gathered} F=(6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2})\times\frac{(60\operatorname{kg})(130,000\operatorname{kg})}{(10m)^2} \\ =5.2\times10^{-6}N \end{gathered}$

C) m=5.22*10^9kg, r=1000m

$\begin{gathered} F=(6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2})\times\frac{(60\operatorname{kg})(5.22\times10^9\operatorname{kg})}{(1000m)^2} \\ =2.1\times10^{-5}N \end{gathered}$

D) m=0.045kg, r=0.95m

$\begin{gathered} F=(6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2})\times\frac{(60\operatorname{kg})(0.045\operatorname{kg})}{(0.95m)^2} \\ =2.0\times10^{-10}N \end{gathered}$