What happens to the gravitational 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?
Give reasons in each case.

(i)
According to universal law of gravitation, the force between 2 objects $({\mathrm{m}}_1$ and ${\mathrm{m}}_2$) is proportional to their plenty and reciprocally proportional to the sq. of the distance(R) between them.
$F=\frac{\left(G2m_1{m}_2\right)}{R^2}$
If the mass is doubled for one object. $\mathrm{F}=2\mathrm{F}$, so force is also doubled.
(ii)
If the distance between the objects is doubled and tripled If it's doubled
Hence,
$\begin{array}{rl}\mathrm{F}& =\left({\mathrm{Gm}}_1{\mathrm{m}}_2\right)/(2\mathrm{R}{)}^2\\ \mathrm{F}& =\left({\mathrm{Gm}}_1{\mathrm{m}}_2\right)/4{\mathrm{R}}^2\\ & =\mathrm{F}/4\end{array}$
$F=F/4$ hence force thus becomes one-fourth of its initial force.
Now, if it's tripled
Hence,
$\begin{array}{rl}& \mathrm{F}=\left({\mathrm{Gm}}_1{\mathrm{m}}_2\right)/(3\mathrm{R}{)}^2\\ & \mathrm{F}=\left({\mathrm{Gm}}_1{\mathrm{m}}_2\right)/9{\mathrm{R}}^2\end{array}$
$\mathrm{F}=\mathrm{F}/9$ hence, force becomes one-ninth of its initial force.
(iii)
If masses of both the objects are doubled, then
$F=\frac{\left(G2m_{1}2m_2\right)}{R^2}$
$\mathrm{F}=4\mathrm{F}$, Force will therefore be four times greater than its actual value.