Question

Determine the force of gravity of the meteorites when it reaches one quarter of its distance from the center of Earth if the force of gravity of a meteorite approaching Earth is 50,000 N.

Answer 1

$$F_{new}=\dfrac{Gm_mM_E}{(\dfrac{1}{4}d)^2}=\dfrac{Gm_pM_E}{\dfrac{1}{16}d^2}=16F,$$ $$F_{new}=16\times50000\ N=8\times10^5\ N.$$