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. $\sqrt{\frac{GM}{L}}$
• B. $\sqrt{\frac{3GM}{2L}}$
• C. $\sqrt{\frac{3GM}{L}}$
• D. $\sqrt{\frac{2GM}{3L}}$

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,
$\therefore \frac{M{v}^2}{r}=\frac{G{M}^2}{L^2}\sqrt{3}$ or $\frac{M{v}^2\sqrt{3}}{L}=\frac{G{M}^2\sqrt{3}}{L^2}\therefore v=\sqrt{\frac{GM}{L}}$