Question

Three particles each of mass $m$ are placed at the vertices of an equilateral triangle of side $r$ as shown in the figure. The magnitude of the gravitational force on any one particle due to others two is

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

Answer 1

Answer: (B)

Gravitational force on particle at
$C$ due to particle at $A$ is
$F_{1}=\frac{G m m}{r^{2}}=\frac{G m^{2}}{r^{2}}$ along $C A$

Gravitational force on particle at $C$ due to particle at $B$ is
$F_{2}=\frac{G m^{2}}{r^{2}}$ along $C B$
As $\vec{F}_{1}$ and $\vec{F}_{2}$ are inclined at an angle $60^{\circ},$ so the resultant gravitational force on particle at $C$ is
$F=\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos 60^{\circ}}$
$=\sqrt{F_{1}^{2}+F_{1}^{2}+2 F_{1} \times F_{1} \times \frac{1}{2}}$
$=\sqrt{3} F_{1}=\frac{\sqrt{3} G m^{2}}{r^{2}}$
$\left(\because F_{1}=F_{2}\right)$