Two bodies of mass m and 9 m are placed at a distance R . The gravitational potential on the line joining…

Let the gravitational field at point $Q$ is zero. Therefore, we can write

$$\begin{gathered} \left(\frac{-Gm}{x^2}\right)+\left(\frac{-G\times 9m}{{\left(R-x\right)}^2}\right)=0 \\ \Rightarrow \frac{{\left(R-x\right)}^2}{x^2}=9 \\ \Rightarrow x=\frac{R}{4} \end{gathered}$$

Now, net gravitational potential at point $Q$ will be

$$\begin{gathered} V_{net}=\left(\frac{-Gm}{x}\right)+\left(\frac{-G\times 9m}{R-x}\right) \\ =\left(\frac{-Gm}{{\displaystyle \frac{R}{4}}}\right)+\left(\frac{-G\times 9m}{{\displaystyle \frac{3R}{4}}}\right) \\ =\frac{-16Gm}{R} \end{gathered}$$$$