Two particles of mass M each are placed at the vertices of an equilateral triangle of side ‘a’, the gravitational force on the particle of mass ‘m’ placed at the third vertex is:

force of attraction between two masses=F1=F2=GMma2 ; θ=600 here universal gravitation constant=G; mass=m; distance=a The resultant force on ‘m’ is   F=F12+F22+2F1F2cosθ  from parallelogram law of addition of vectors F=F12+F22+2F12×12 F=3F1 F=3×GMma2=net force on mass m