Question

Three particles, each of mass $M$ are placed at the three corners of an equilateral triangle of side $l$.What is the force due to this system of particles on another particle of mass $m$ placed at the midpoint of any side?

• A. $\frac{3GMm}{{4l}^2}$
• B. $\frac{4GMm}{{3l}^2}$
• C. $\frac{GMm}{{4l}^2}$
• D. $\frac{4GMm}{l^2}$

Answer 1

The correct option is B $\frac{4GMm}{{3l}^2}$

Forces from the two nearer masses would be same in magnitude and opposite in direction. Hence they would cancel out. The only remaining force would be due to the third mass.

Distance between that mass and the mass kept on the side= $l\sin {60}^{\circ }=\frac{\sqrt{3}}{2}l$

Hence the gravitational force would be $\frac{GMm}{(\frac{\sqrt{3}}{2}l{)}^2}=\frac{4GMm}{3l^2}$.