Question

Three identical particles each of mass m are arranged at the corners of an equilateral triangle of side $"L"$. If they are to be in equilibrium, the speed with which they must revolve under the influence of one anothers gravity in a circular orbit circumscribing the triangle is:

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

Answer 1

The correct option is B $\sqrt{\frac{Gm}{L}}$

$r=\frac{L}{2\cos {30}^{\circ }}$

From Newton's law of universal gravitation, the force of attraction between two stars is $F_1=\frac{G{M}^2}{L^2}cos{30}^{\circ }$

The net force of gravity on one star due to the other two stars = $F=2F_1=\frac{2G{M}^2}{L^2}cos{30}^{\circ }$

The centripetal force acting on the stars to orbit around the triangle is $F_c=\frac{M{v}^2}{r}$

Here, speed of stars is v.

Substitute $\frac{L}{2cos{30}^{\circ }}$ for $r$ in above equation.

$F_c=\frac{M{v}^2}{r}=\frac{M{v}^2}{\frac{L}{2cos{30}^{\circ }}}=\frac{2M{v}^{2}cos{30}^{\circ }}{L}$

The necessary centipetal acceleration is provided by the net forcedue to the three stars.

$F=F_c$

$\frac{2G{M}^{2}cos{30}^{\circ }}{L^2}=\frac{2M{v}^{2}cos{30}^{\circ }}{L}$

$v=\sqrt{\frac{GM}{L}}$

Therefore, the speed with which the stars will revolve so as to preserve the equilateral triangle is $\sqrt{\frac{GM}{L}}$