Question

Three identical bodies of mass $M$ are located at the vertices of an equilateral triangle of side $L$. They revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbital velocity is

• A. $\sqrt{\frac{G M}{L}}$
• B. $\sqrt{\frac{3 G M}{2 L}}$
• C. $\sqrt{\frac{3 G M}{L}}$
• D. $\sqrt{\frac{2 G M}{3 L}}$

Answer 1

Answer: (A)

From figure, $\frac{L}{2}=r \cos 30^{\circ}=\frac{\sqrt{3}}{2} r$ or $r=\frac{L}{\sqrt{3}}$
The net force of attraction on mass at
$A$ due to masses at $B$ and $C$ is
$F_{A}=2\left[\frac{G M^{2}}{L^{2}}\right] \cos 30^{\circ}=\left[\frac{G M^{2}}{L^{2}} \sqrt{3}\right]$
As this force provides the necessary centripetal force,
$\therefore \quad \frac{M v^{2}}{r}=\frac{G M^{2}}{L^{2}} \sqrt{3}$ or $\frac{M v^{2} \sqrt{3}}{L}=\frac{G M^{2} \sqrt{3}}{L^{2}} \therefore v=\sqrt{\frac{G M}{L}}$