Question

Determine the force of gravitational attraction between the earth (5.98 x1024

kg) and a 100 kg football player, when he is at a distance of 6.38 x 106 m from

earth’s center.

Answer 1

Given:

$M_E=\rm 5.98*10^{24}\:\rm kg$

$m=\rm 100\:\rm kg$

$R=\rm 6.38*10^{6}\:\rm kg$

$F-?$

The law of universe gravitation says

$$F=G\frac{M_Em}{R^2}\\ =\rm 6.67*10^{-11}\frac{5.98*10^{24}*100}{(6.38*10^6)^2}=980\: N$$