Question

Two massive particle of masses $M$ and $m(M>m)$ are seperated by a distance $l$. They rotate with equal angular velocity under their gravitational attraction about a common axis through centre of mass. The linear speed of the heavier particle is

• A. $\sqrt{\frac{GMm}{(M+m)l}}$
• B. $\sqrt{\frac{G{m}^2}{(M+m)l}}$
• C. $\sqrt{\frac{Gm}{2l}}$
• D. $\sqrt{\frac{2G{m}^2}{(M+m)l}}$

Answer 1

Answer: (B)

$\sqrt{\frac{G{m}^2}{(M+m)l}}$

$\text{Reduced mass of the two bodies},\mu =\frac{Mm}{M+m}$

$\text{Force acting between the two objects}=\frac{GMm}{l^2}$

$\text{The two masses rotate with angular velocity}\omega \text{about the center of mass of the system.}$

$\text{The gravitational force causes centripetal acceleration of the masses.}$

$\text{Distance of mass M from center of mass}=\frac{m}{M+m}l$

$\text{Thus,}\frac{GMm}{l^2}=\mu {\omega }^{2}l$

$\text{Thus,}\frac{GMm}{l^2}=\frac{Mm}{M+m}{\omega }^{2}l$

$⟹\omega =\sqrt{\frac{G(M+m)}{l^3}}$

$\text{Linear velocity of mass M}=\omega \left(\frac{m}{M+m}l\right)$

$v=\sqrt{\frac{G{m}^2}{(M+m)l}}$