If there particles, each of mass `M`, are placed at the three corners of an equilateral triangle of side, a the force exerted by this system on another particle of mass `M` placed (i) at the midpoint of side and (ii) at the centre of the triangle are, respectively.

Options

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

Answer

B

Explanation

(i) Gravitational force on the particle placed at mid point D of side BC of length a is
`F=F_(1)+F_(2)+F_(3)`
here `F_(2)=-F_(3)`
`rarr F_(2)+F_(3)=0`
`therefore F=F_(1)+0=F_(1)`
or `F=F_(1)=(GMM)/(AD)6(2)=(GM)^(2)/(3a^(2)//4)`
`=(4 GM^(2))/(3a^(2))`