Answer to Question #342748 in Physics for physy

Question

Determine the force of gravitational attraction between the earth (m = 5.98 x 10²⁴ kg) and a 63.0-kg man if the man is standing at sea level, a distance of 6.38 x 10⁶ m from earth's center.

Answer 1

$$F=G\frac{m_1m_2}{r^2}$$

$$F=6.67*10^{-11}*\frac{5.98*10^{24}*63.0}{(6.38*10^6)^2}=617\:\rm N$$