Question

If three 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 a side and (ii) at the centre of the triangle are respectively.

Answer 1

force on the given mass by the two masses on the same line will cancel out

the net force is due to the mass placed on the 3rd corner.

the distance between these two masses is equal to the median of the triangle which is $\frac{\sqrt{3}a}{2}$

now$F=\frac{GMM}{r^2}=\frac{G{M}^2}{{\left(\frac{\sqrt{3}a}{2}\right)}^2}=\frac{4G{M}^2}{3a^2}$

now at the center of the triangle distance of separation is $\frac{a}{2}$

now the attraction between corner and the center of mass is$F=\frac{GXM}{k^2}$

from three sides the force is$F=\frac{3GXM}{k^2}whenX=MthenF=\frac{3G{M}^2}{K^2}$