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

Let PQR be the equilateral triangle and O be its centroid. Then from figure
OR = OQ = OP, m = 1 gram = 0.001 kg
From right-angled triangle OQB, we have
`cos 30^(@) = (QB)/(OQ) = (5cm)/(OQ)`
`rArr OQ = (5 cm)/(cos 30^(@)) = (5)/(sqrt(3)//2) = (10)/(sqrt(3)) cm = (0.1)/(sqrt(3)) m`
All the three masses will create gravitational field of equal magnitudes at an angle `120^(@)` with each other at the centroid of the triangle. resultant of these vectors is zero.
`:.` Resultant gravitational field at O is zero.
Total gravitational potential at O is
`V = V_(1) + V_(2) + V_(3) = -(Gm)/(OP) - (Gm)/(OQ) - (Gm)/(OR)`
`because OP = OQ = OR = (0.1)/(sqrt(3)) m`
`:. V = -(3Gm)/(OP) = (-3G xx (0.001))/(0.1//sqrt(3))`
`= -3sqrt(3) G xx 0.01`
`:. G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`
`V = -0.03 xx sqrt(3) xx 6.67 xx 10^(-11)`
`= -3.47 xx 10^(-12) V`