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 experiences not net gravitational force due to the particle. Find m2/m1 in the ratio form?

Answer 1

Gravitational attractive force b/w m1 and m is given by

F1 = G × (m1 × m )/(d/4)^2

= G × ( 16m × m1 )/ d^2

Gravitational attractive force b/w m2 and m is given by

F2 = G × ( m × m2 )/( 3/4 × d) ^2

= G × ( 16m × m2 )/ 9d^2

Since , F1 = F2 , we have

G × (16m × m1)/d^2 = G × (16m × m2 )/ 9d^2

=> m2/m1 = 9/1 Ans...

Hope it helps you....