Three particles each of mass m are kept at the vertices of an equilateral triangle of side L . What is the gravitational potential at the centroid of the triangle?

Options

A. zero
B. `-3sqrt3(Gm)/(a)`
C. `-2sqrt3(Gm)/(a)`
D. `-sqrt3(Gm)/(a)`

Answer

b

Explanation

Given `AB=BC=AC=a` (See figure). The perpendiculars from A, B and C on opposite side meet at the centroid O, which bisect the sides AB, BC and AC. Let `r=AO=BO=CO.` Centroid alos divides the lines AD, BE and CF in the ratio `2:1`, i.e.,
`AO=(2)/(3)AD, BO=(2)/(3)BE, CO=(2)/(3)CF`

In triangle ABD,
`AD=a sin 60^(@)=(sqrt3a)/(2)`
Similarly,
`BE=CF=(sqrt3a)/(2)`
`therefore" "r=AO=OB=OC=(2)/(3)xx(sqrt3a)/(2)=(a)/(sqrt3)`
`V=V_(1)+V_(2)+V_(3)`
`V=V_(1)+V_(2)+V_(3)`
`V=-(Gm)/(r)-(Gm)/(r)-(Gm)/(r)`
`=(3Gm)/(r)=-(3Gm)/(a//sqrt3)=-3sqrt3(Gm)/(a)`