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.
(This question has multiple correct options)
$\text{A}\text{. }{{r}^{\dfrac{3}{2}}}$
B. r
$\text{C}\text{. }{{\text{m}}^{\dfrac{1}{2}}}$
$\text{D}\text{. }{{\text{m}}^{-\dfrac{1}{2}}}$

Answer 1

Hint: Use the formula for finding the position of centre of mass of the two masses, i.e. ${{x}_{com}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$. Time period in terms of angular frequency ($\omega$) is given as $T=\dfrac{2\pi }{\omega }$. Find $\omega$ with the help of centripetal acceleration of the masses. The centripetal acceleration is given as $a=r{{\omega }^{2}}$.
Formula used:
${{x}_{com}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$
$F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
$a=r{{\omega }^{2}}$
F=ma
$T=\dfrac{2\pi }{\omega }$

Complete step-by-step answer:
Let us first calculate the position of the centre of mass of the system of two stars. Consider a Cartesian plane whose origin lies on the mass m. Therefore, the position of mass m is x=0 and position of mass 2m will be r.
The position of centre of mass of a system of two masses is given as ${{x}_{com}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$.
Therefore, the position of centre of mass of the given system is ${{x}_{com}}=\dfrac{m(0)+2m(r)}{m+2m}$.
$\Rightarrow {{x}_{com}}=\dfrac{2m(r)}{3m}=\dfrac{2r}{3}$.
Hence, the position of centre of mass of the system is at distance of $\dfrac{2r}{3}$ from the mass m on the line joining the two masses.
It is given that both masses are rotating about the centre of mass under the gravitational force.
The gravitational force between two masses ${{m}_{1}}$and ${{m}_{2}}$is given as $F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$, where G is proportional constant and r is the distance between the centres of the two masses.
Consider the motion of mass m.
The force acting on this mass will be equal to $F=G\dfrac{(2m)m}{{{r}^{2}}}$ ………(i).
This the mass m is rotation due to force F, it will have a centripetal acceleration equal to $a=r{{\omega }^{2}}$, where $\omega$ is the angular velocity of the mass and r is the distance of the mas from the axis of rotation.
By the second law of motion F=ma. Hence, $F=m\left( \dfrac{2r}{3} \right){{\omega }^{2}}$ …….(ii).
Equate (i) and (ii).
Hence, we get
$G\dfrac{(2m)m}{{{r}^{2}}}=\dfrac{2mr{{\omega }^{2}}}{3}$
$\Rightarrow G\dfrac{3m}{{{r}^{3}}}={{\omega }^{2}}$
$\Rightarrow \omega =\sqrt{G\dfrac{3m}{{{r}^{3}}}}$
Time period (T) of the system is given as $T=\dfrac{2\pi }{\omega }$.
Substitute the value of $\omega$ in the above equation.
$T=\dfrac{2\pi }{\sqrt{G\dfrac{3m}{{{r}^{3}}}}}=\dfrac{2\pi }{{{\left( G\dfrac{3m}{{{r}^{3}}} \right)}^{\dfrac{1}{2}}}}=\dfrac{2\pi {{r}^{\dfrac{3}{2}}}}{{{\left( 3mG \right)}^{\dfrac{1}{2}}}}$
Therefore, the time period of the system is proportional to ${{r}^{\dfrac{3}{2}}}$ and ${{\text{m}}^{-\dfrac{1}{2}}}$.
Hence, the correct options are A and D.

Note: It is not a compulsion to consider only mass m for calculating angular velocity of the system. Calculate the distance of the centre of mass of the system from the mass 2m and proceed in the method considering the motion of mass 2m.
Whichever mass we may choose the answer will not change.