Question

Three particles, each of mass $m$, are fixed at the vertices of an equilateral triangle of side length $a$. The only forces acting on the particles are their mutual gravitational forces. Then answer the following question.
Force acting on particle $C$, due to particle $A$ and $B$.

• A. $\frac{\sqrt{3}G{m}^2}{a^2}$
• B. $\frac{\sqrt{2}G{m}^2}{a^2}$
• C. $\frac{2\sqrt{3}G{m}^2}{a^2}$
• D. $\frac{\sqrt{3}G{m}^2}{2a^2}$

Answer 1

The correct option is A $\frac{\sqrt{3}G{m}^2}{a^2}$

Due to symmetry $F_C=F_A=\frac{Gmm}{a^2}$

$R=\sqrt{F_A^2+F_C^2+2F_A{F}_C{\cos 60}^0}$

$R=\sqrt{{\left(\frac{{Gm}^2}{a^2}\right)}^2(1+1+1)}=\frac{\sqrt{3}{Gm}^2}{a^2}$