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

Answer 1

Answer: (B)

Gravitational force on particle at
$C$ due to particle at $A$ is
$F_1=\frac{Gmm}{r^2}=\frac{G{m}^2}{r^2}$ along $CA$

Gravitational force on particle at $C$ due to particle at $B$ is
$F_2=\frac{G{m}^2}{r^2}$ along $CB$
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
$F=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos 60^{\circ }}$
$=\sqrt{F_1^2+F_1^2+2F_1\times F_1\times \frac{1}{2}}$
$=\sqrt{3}{F}_1=\frac{\sqrt{3}G{m}^2}{r^2}$
$\left(\because F_1=F_2\right)$