Three particles each of mass `m` are kept at the vertices of an euilateral triangle of side `L`. The gravitational field at the centre due to these particle is

Options

A. `-(3GM)/(L)`
B. `-(sqrt(3)GM)/(L)`
C. `-(3sqrt(3)GM)/(L)`
D. `-(GM)/(3sqrt(3)L)`

Answer

C

Explanation

`AD=AB sin 60^(@)=(L xx sqrt(3))/(2)`
`because` G is the centroid, `AG=(2)/(3)AD=(2)/(3)(L sqrt(3))/(2)=(L)/(sqrt(3))`
`therefore GA = GB = GC = (L)/(sqrt(3))`
`therefore` The gravitational potential at G due to masses at A, B and C
`=-(Gm)/(GA)-(Gm)/(GB)-(Gm)/(GC)=-(3Gm)/(GA)`
`=-(3GM)/((L)/(sqrt(3)))=-(3sqrt(3)GM)/(L)`