Question

Question 3
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is $6\times {10}^{24}kg$ and radius of the earth is $6.4\times {10}^{6}m$)

Answer 1

Given that,
Mass of the body, m = 1 kg
Mass of the Earth, $M=6\times {10}^{24}kg$
Radius of the Earth, $R=6.4\times {10}^{6}m$

From the universal law of gravitation, the magnitude of the gravitational force (F) between the earth and the body is given by,

$F=G\frac{M\times m}{R^2}$

where $G=6.67\times {10}^{-11}N{m}^2/k{g}^2$ is the universal gravitational constant.

Hence, magnitude of the gravitational force (F) between the earth and the body is given by,

$F=G\frac{M\times m}{R^2}=\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times 1}{(6.4\times {10}^6{)}^2}$ = 9.77 N = 9.8 N (approx).