Three mass points each of mass ‘m’ are placed at the vertices of an equilateral ∆ of side l. What is the gravitational field & potential due to three masses at the centroid of  ∆ ?

Let ABC be the equilateral Δ of each side equal to l. Three mass points, each of mass ‘m’ lie at the three vertices as shown in figure.

If G is the centroid of Δ , then AG = BG = CG.

Now, the centroid G divides the normal AP in ratio 2 : 1.

& there4;  AG = 2/3 AP = 2/3 AB sin 60° = 2/3 l × √3/2 = l/√3

The gravitational field at point G is equal to the force on a unit mass placed at the point G due to three equal mass points lying at distances AG = BG = CG = l/√3 form it. The three forces will be equal in magnitude & will be directed as shown. The symmetry shows that the resultant force i.e., the gravitational field at point. G will be zero.

Now, the gravitational potential at pt G

= −Gm/AG+ (−Gm/BG) + (−Gm/BG) = − 3Gm/l/√3== −3√3 Gm/l