Question

A double star is a system of two stars of masses $m$ and $2m$, rotating about their centre of mass only under their mutual gravitational attraction. If $r$ is the separation between these two stars then their time period of rotation about their centre of mass will be proportional to:

• A. $r^{\frac{3}{2}}$
• B. $r$
• C. $m^{\frac{1}{2}}$
• D. $m^{-\frac{1}{2}}$

Answer 1

Answer: (A), (D)

$r^{\frac{3}{2}}$
$m^{-\frac{1}{2}}$

For equilibrium, $F_{gravitational}=F_{centrifugal}$

Distance between the stars is r. So $F_{gravitational}=\frac{Gm\left(2m\right)}{r^2}$

For point O to be the center of mass of the system, then $3m\left(x\right)=0+2m\left(r\right)⟹x=\frac{2r}{3}$

Let the star of mass m revolves around O with angular velocity $w$.

$F_{centrifugal}=m\left(\frac{2r}{3}\right)w^2$ ; where, $w=\frac{2\pi }{T}$

Thus, $\frac{Gm\left(2m\right)}{r^2}=m\left(\frac{2r}{3}\right)\frac{4{\pi }^2}{T^2}$

$⟹T^2=\frac{4{\pi }^2{r}^3}{3Gm}$

Hence, $T\propto r^{\frac{3}{2}}{m}^{\frac{-1}{2}}$