Question

Three identical point objects each of mass $m$ are placed at the vertices of an equilateral triangle of side $l$. What is the gravitational potential at the centre of the equilateral triangle due to the point masses?

Answer 1

The distance between the vertices of the equilateral triangle and its centre is equal to $r=\left(l/2\right)\sec {30}^{\circ }=\left(l/\sqrt{3}\right)$. The potential at $P=V_P=V_A+V_B+V_C$, where $V_A,V_B$ and $V_C$ are the potentials due to the masses placed at $A,B$ and $C$ at the centre, respectively.
Since, $V_A=V_B=V_C=-\frac{Gm}{r}$
$V_P=\frac{-3Gm}{r}$ and $r=\frac{l}{\sqrt{3}}$
$\Rightarrow V_P=-\frac{3\sqrt{3}Gm}{l}$.