At the vertices of an equilateral triangle of side a, three particles each of mass m are kept. Find the gravitational potential and the gravitational field at the centroid of the triangle.

Explanation

Here AB =BC=CA =a [Fig.1.10], perpendiculars drawn from the points A,B and C on respective opposite sides meet at the centroid O and Proceed to bisect the sides BC,CA and AB respectively.

Let AO =BO =CO =r
From `Delta ABD ,AD =asin 60^@=sqrt(3)/(2)a`
Also `AO =2/3AD`
`therefore r=2/3xxsqrt(3)/2a=a/sqrt(3)`.
Hence , the gravitational potential at the centroid O is
`V=-(Gm)/r +(-Gm)/r+(-Gm)/r=-(3Gm)/(a/sqrt(3))=-(3sqrt(3)Gm)/a`
Let the gravitational field intensities at O due to masses at A,B and C be `F_A` (direction from O to A), `F_B` (direction from O to B) and `F_C` (directed from O to C ) respectively [Fig.1.11] .

`therefore F_A=(Gm)/(r^2)=F_B=F_C`
Resolving the intensities along and perpendicular to the direction of AO ,we get,
`F_A-F_B cos 60^@-F_C cos 60^@=F_A-(F_B)/(2)-(F_C)/(2)=0`
`[because F_B=F_C=F_A]`
and `0+F_C sin 60^@-F_Bsin 60^@=(F_Csqrt(3))/2-(F_Bsqrt(3))/2=0`
`[because F_B=F_C]`
Hence the gravitational field intensity at the centroid of the triangle =0.