What happens to the gravitational force between two objects if the mass of both the objects is doubled and the distance between objects is tripled?

Hint:We will use Newton’s law of gravitation to solve the above problem. As we can clearly see, only a few variables are changed, that too are mass and distance between the two objects. Therefore we can easily calculate the resulting gravitational force between the two objects in terms of their gravitational force in the first case.

Complete answer:
Let the masses of the two objects be ${{m}_{1}}$ and ${{m}_{2}}$ respectively .
And let the distance between their centers be given by $r$ .
Then, using Newton’s law of gravitation, the force of attraction (say $F$ ) between the two objects can be given as:
$\Rightarrow F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Now, in the second case it is given in the problem that, the masses of the two objects have been doubled and the distance between them has been tripled. Therefore,
$\begin{align}
  & \Rightarrow m_{1}^{'}=2{{m}_{1}} \\
 & \Rightarrow m_{2}^{'}=2{{m}_{2}} \\
\end{align}$
Also,
$\Rightarrow {{r}^{'}}=3r$
Putting the value of these new masses of the two object and their new distance between the centers in Newton’s law of gravitation, we get the new force of attraction (say ${{F}^{'}}$ ):
$\begin{align}
  & \Rightarrow {{F}^{'}}=G\dfrac{m_{1}^{'}m_{2}^{'}}{{{({{r}^{'}})}^{2}}} \\
 & \Rightarrow {{F}^{'}}=G\dfrac{(2{{m}_{1}})(2{{m}_{2}})}{{{(3r)}^{2}}} \\
 & \Rightarrow {{F}^{'}}=\dfrac{4}{9}G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}} \\
 & \Rightarrow {{F}^{'}}=\dfrac{4}{9}F \\
\end{align}$
Hence, the gravitational force becomes ${{\left( \dfrac{4}{9} \right)}^{th}}$ of the original.

Note:
These types of questions are simple questions of ratio and proportionality. The parameters are changed by a certain value so we need not worry if the absolute value of any parameter is known or unknown to us. We should just move ahead in our calculation with variables in our equations and at the end, all the unknown variables will disappear from our equation giving us the final result.