Four point masses each of mass $\text{'}m\text{'}$ are placed at four vertices$A,B,C$and $D$ of a regular hexagon of side$\text{'} a \text{'}$as shown in figure. Find gravitational potential and field strength at the centre $O$ of the hexagon. Given E=$\frac{Gm}{a^2} and \theta ={60}^0$

 Gravitational potential is a scalar quantity. Therefore,

$V_o=$scalar sum of gravitational potentials produced by four point masses at $A,B,C$and $D$.

$=-\frac{Gm}{a}-\frac{Gm}{a}-\frac{Gm}{a}-\frac{Gm}{a}$

$=-\frac{4Gm}{a}$

Gravitational field strength is a vector quantity. So, it is a vector sum of four vectors of equal magnitudes.

$E=\frac{Gm}{a^2}=E_A=E_B=E_C=E_D$

$E_A$ and $E_D$are cancelled. So, net field strength is a vector sum of $E_B$ and $E_C$ at angle $60°$.

$E_{\text{net }}=\sqrt{E^2+E^2+2\left(E\right)\left(E\right)\cos 60°}=\sqrt{3}E$

$=\frac{\sqrt{3}Gm}{a^2}$

If $E_1=E_2=E$and $\theta$ be the angle between them, then $E_{\text{net }}=2E\cos \left(\frac{\theta }{2}\right)$

This net field strength is along the bisector line of $\angle COB$, away from $O$, between $E_C$ and $E$