Question

Three stars each of mass $m$ revolve in circular path about their common centre of mass. Stars are at distance $a$ from each other

• A. Orbital speed for each mass is $\sqrt{\frac{Cm}{a}}$
• B. Orbital speed for each mass is $\sqrt{\frac{Cm}{3a}}$
• C. Escape speed for a point mass placed at common centre of mass of stars is $\sqrt{\frac{6\sqrt{3}Cm}{a}}$
• D. Escape speed for a point mass placed at common centre of mass of star is $\sqrt{\frac{3Gm}{a}}$

Answer 1

Answer: (A), (C)

A and C

Answer 2

For orbital speed: The net gravitational force on one star from the other two is $\sqrt{3} \frac{Gm^2}{a^2}$. The radius of the circular orbit is $R = \frac{a}{\sqrt{3}}$. Equating the net gravitational force to the centripetal force ($F_c = \frac{mv^2}{R}$), we get $\sqrt{3} \frac{Gm^2}{a^2} = \frac{mv^2}{a/\sqrt{3}}$. Solving for $v$ yields $v = \sqrt{\frac{Gm}{a}}$.

For escape speed: The gravitational potential energy of a test mass $M_{test}$ at the center of mass is $U = 3 \times (-\frac{GmM_{test}}{R})$, where $R = \frac{a}{\sqrt{3}}$. This gives $U = -3\sqrt{3} \frac{GmM_{test}}{a}$. The escape speed $v_e$ is achieved when $\frac{1}{2}M_{test}v_e^2 = -U$. Solving for $v_e$ yields $v_e = \sqrt{\frac{6\sqrt{3}Gm}{a}}$.