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.

At the vertices of an equilateral triangle of side a, three particles

1.	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.
Answer» SOLUTION :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.