Question

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

• A. $0,\frac{4G{M}^2}{{3a}^2}$
• B. $\frac{4G{M}^2}{{3a}^2},0$
• C. $\frac{3G{M}^2}{a^2},\frac{G{M}^2}{a^2}$
• D. $0,0$

Answer 1

Answer: (B)

$\frac{4G{M}^2}{{3a}^2},0$

When the body is at the center of the triangle, forces from the three corners would be exactly equal and all at an angle of ${120}^{\circ }$ to each other. Hence the resultant of the forces would be zero.

When the object would be at the mid point point of a side, the case would be as shown in the figure.

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=$a\sin {60}^{\circ }=\frac{\sqrt{3}}{2}a$

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