Question

A particle of mass m1 is kept at x=0 and another of mass m2 at x=d. When a third particle is kept at x=d/4, it experience no net gravitational force due to the two particles. Find m2/m1.

Answer 1

The third particle experiences no net gravitational force so the force on it due to m1 is equal to the force on it due to m2.
Let the mass of the third body be k.
=> (G*m1*k)/(d/4)²= (G*m2*k)/(3d/4)²
=> 16*G*m1*k/d²= 16*G*m2*k/9d²
=> m2/m1 = {(16*G*k)*9d²}/{(16*G*k)*d²}
=> m2/m1= 9/1
So, m2:m1 = 9:1