Question

The particles each of mass $M$, are located at the vertices of an equilateral triangle of side $a$. At what speed must they move, if they revolve under the influence of their gravitational force of attraction in a circular orbit, circumscribing the triangle while still preserving the equilateral triangle?

• A. $\sqrt{\frac{\text{MG}}{a}}$
• B. $\sqrt{\frac{\text{MG}}{3a}}$
• C. $\sqrt{\frac{\text{MG}}{2a}}$
• D. $\sqrt{\frac{3\text{MG}}{a}}$

Answer 1

The correct option is A

$\sqrt{\frac{\text{MG}}{a}}$

Force on particle A due to particle B is $\overrightarrow{F_{AB}}=\frac{{\text{GM}}^2}{a^2}$
Force on particle A due to particle C is $\overrightarrow{F_{AC}}=\frac{{\text{GM}}^2}{a^2}$
Net Force on particle A be $\overrightarrow{F_A}$
$\overrightarrow{F_A}=\overrightarrow{F_{AB}}+\overrightarrow{F_{AC}}=2\left[\frac{{\text{GM}}^2}{a^2}\right]\text{cos}{30}^0=[\frac{{\text{GM}}^2}{a^2}\times \sqrt{3}]$

Now, Net force corresponds to centripetal acceleration
$\therefore \frac{M{v}^2}{r}=F_{AB}$ ( $r=\frac{a}{\sqrt{3}}$).

$v=\sqrt{\frac{\text{MG}}{\text{a}}}$