What happens to the force between two objects, if

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?

(i) We know,

F = G $\dfrac{\text{m}$1\text{m}2}{\text{d}^2}

Where,

F = force of attraction

G = constant of proportionality

m₁ = mass of first object

m₂ = mass of second object

d = distance between the two objects.

When mass of one object is doubled.

m' = 2m₁

So, substituting we get,

F' = G $\dfrac{\text{2m}$1\text{m}2}{\text{d}^2} = 2G $\dfrac{\text{m}$1\text{m}2}{\text{d}^2} = 2F

Hence, when the mass of one object is doubled, the gravitational force increases two times.

(ii) When distance is doubled then, d' = 2r

So, substituting we get,

F' = G $\dfrac{\text{m}$1\text{m}2}{(2d)^2} = G $\dfrac{\text{m}$1\text{m}2}{4d^2} = $\dfrac{\text{F}}{4}$

Hence, when the distance between the objects is doubled, the gravitational force becomes one fourth of its original force.

Now, if it's tripled

d' = 3r

So, substituting we get,

F' = G $\dfrac{\text{m}$1\text{m}2}{(3d)^2} =G $\dfrac{\text{m}$1\text{m}2}{9d^2} = $\dfrac{\text{F}}{9}$

Hence, when the distance between the objects is tripled, the gravitational force becomes one ninth of its original force.

(iii) If masses of both the objects are doubled, then

m₁' = 2m₁

and

m₂' = 2m₂

So, substituting we get,

F' = G $\dfrac{\text{2m}$1\text{2m}2}{\text{d}^2} = 4G $\dfrac{\text{m}$1\text{m}2}{\text{d}^2} = 4F

Hence, when the masses of both the objects are doubled, the gravitational force becomes four times the original force.