Question

if the distance b/w 2 point masses decreases by 25%, how does the force of gravitation change?

Answer 1

let two body which mass m and M are
separate in r distance

initially gravitational force = F =GmM/r²

now,
r decrese 25℅

then , r" = r - r×25/100 = r-r/4 =3r/4

now, force F" = GmM/r"²

F" = GmM/(3r/4)² = (16/9)GmM/r²

from above GmM/r² = F

so, F" = 16F/9

change in gravitational force = final -initial = 16F/9 - F = 7F/9

% change of gravitational force = 7F/9/F×100 = 700/9 = 77.777 ℅

Answer 2

Answer:

The decrease in gravitational attraction is not a linear one. The force of gravity, as given by Newton's law, is an 'inverse square' law. That means that the force of gravity is inversely proportional to the square of the distance between the centers of mass. So

F α 1 / r^2

So, if the distance doubles, the force or gravity falls by 1/4.

If the distance trebles, the force fall be 1/9 and so on.