Question

Two bodies 'P' and 'Q' having masses m1 and m2 , are separated by a distance d1 exert a force 'F' on each other .What happens(a) masses of both the objects are doubled(b) distance between the two bodies is reduced to half(c) the space between the two objects has no air and is complete vacuum

Answer 1

Force of Gravitation = (G x m1 x m2)/(d)^2

Force after addition of mass and reduction of distance

= {G x [(2 x M1) x (2 x M2)]}/[(1/2) x d]^2

= 4  x (G x m1 x m2)/ {[(d)^2]/ 4}

= 4 x 4 x (G x m1 x m2)/(d)^2

= 16 x (G x m1 x m2)/(d)^2

Therefore the force will be sixteen times the initial force

Answer 2

Answer:

Hey dear,

◆ Solution-

Gravitational force acting on the bodies is given by -

F = Gm1m2 / r^2

a) When masses are doubled,

F' = G(2m1)(2m2) / r^2

F' = 4Gm1m2 / r^2

F' = 4F

b) When distance is halved -

F' = Gm1m2 / (r/2)^2

F' = 4Gm1m2 / r^2

F' = 4F

c) When medium is complete vacuum,

As gravitational force is independent of intervening medium.

F' = F

Hope this helps you...