Three equal masses m are placed at the three corners of an equilateral tringle of side a. find the force exerted by this system on another particle of mass m placed at a. the mid point of a side b. at the centre of the triangle.

Options

A. `0, (4GM^(2))/(3a^(2))`
B. `(4GM^(2))/(3a^(2)),0`
C. `(3GM^(2))/(a^(2)),(GM^(2))/(a^(2))`
D. `0,0`

Answer

B

Explanation

i. Gravitational force on the particle placed at the midpoint `D` of side `AB` of length `a` is `=vecF_(1)+vecF_(2)+vecF_(3)`. But `vecF_(1)` and `vecF_(3)` are equal in magnitude and opposite in direction.
`F=F_(3)(4GMm)/(3a^(2))=(4GM^(2))/(3a^(2)) [:'CD^(2)=(3a^(2))/4]`
ii. Gravitational force on the particle placed at the point of intersection of three medians `=vecF_(1)+vecF_(2)+vecF_(3)=0`, since the resultant of `vecF_(1)` and `vecF_(2)` is equal and opposite to `vecF_(3)`.