Three mass points each of mass m are placed at the vertices of an equilateral tringale of side l. What is the gravitational field and potential due to three masses at the centroid of the triangle ?

Explanation

From `Delta OAD cos 30^(@) = (AD)/(OA) = (a//2)/(OA)`
`(sqrt(3))/(2) = (a)/(2 OA)`
`OA = (a)/(sqrt(3))`
Gravitational fields at O due to masses at A, B, C
`E_(1) = (Gm)/((OA)^(2)) = (Gm)/((a//sqrt(3))^(2)) = (3Gm)/(a^(2))` along OA
`E_(2) = (Gm)/((OB)^(2)) = (Gm)/((a//sqrt(3))^(2)) = (3Gm)/(a^(2))` along OB
`E_(3) = (Gm)/((OC)^(2)) = (Gm)/((a//sqrt(3))^(2)) = (3Gm)/(a^(2))` OC
As the angle between `E_(1)` and `E_(3)` is `120^(@)`
`:. E = sqrt(E_(1)^(2) + E_(3)^(2) + 2E_(1)E_(3) cos 120^(@))`
`= (3GM)/(a^(2))` (along EM)
As E is equal and opposite to `vec(E)_(2)`, the resultant gravitational field at O is zero.
Total gravitational potential at O, `V = V_(1) + V_(2) + V_(3)`
`= -(Gm)/(OA) -(Gm)/(OB) - (Gm)/(OC)`
`= -(3Gm)/(OA) (because OA = OB = OC = (a)/(sqrt(3)))`
`= (-3Gm)/(a//sqrt(3))`
`= -3sqrt(3)(Gm)/(a)`