Question

A double star is a composite system of two stars rotating about their centre of mass under their mutual gravitational attraction. Let us consider such a double star which has two stars of masses m and 2m at a separation l. If T is the time period of rotation about their centre of mass then
A. $T=2\pi \sqrt{\dfrac{{{l}^{3}}}{mG}}$
B. $T=2\pi \sqrt{\dfrac{{{l}^{3}}}{2mG}}$
C. $T=2\pi \sqrt{\dfrac{{{l}^{3}}}{3mG}}$
D. $T=2\pi \sqrt{\dfrac{{{l}^{3}}}{4mG}}$

Answer 1

Hint: First find an expression for the gravitational force between the two stars. Then calculate the position of the centre of mass. Later find the angular velocity of anyone star and with this you can find the time period of rotation.

Formula used:
$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$,
where $F$ is the gravitational force between two masses ${{m}_{1}},{{m}_{2}}$, separated by distance d. $G$ is the gravitational constant.
${{x}_{com}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}$,
where ${{x}_{com}}$ is the location of centre of mass, ${{x}_{1}}$ and ${{x}_{2}}$ are the positions of the two masses.
$a={{\omega }^{2}}r$,
where $a$ is centripetal acceleration of a body moving in circular path of radius $r$ and $\omega $ is its angular velocity.
 $T=\dfrac{2\pi }{\omega }$,
where $T$ is time period.

Complete step by step answer:
The gravitational force on the smaller start is equal to
$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}\\
\Rightarrow F=\dfrac{Gm(2m)}{{{l}^{2}}}$
It is said that two stars of masses m and 2m are rotating about the centre of mass of the two. So, let us first find the location of the centre of mass of the system with respect to the star of mass m.
Let us assume that the centre of mass of the system is at rest and the two stars are always in a straight line. Now, let us consider that the star of mass m is at origin (${{x}_{1}}=0$) and the other start is on x-axis at ${{x}_{2}}=l$.Therefore, the location of the centre of mass is,
${{x}_{com}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\\
\Rightarrow{{x}_{com}}=\dfrac{m(0)+2m(l)}{m+2m}$
$\Rightarrow {{x}_{com}}=\dfrac{2ml}{3m}\\
\Rightarrow {{x}_{com}} =\dfrac{2l}{3}$
This means that the centre of mass of the given system at a distance $\dfrac{2l}{3}$ from the star of mass m. Now, we can consider that the smaller star is into uniform rotational motion about the centre of mass. This means that $r=\dfrac{2l}{3}$Therefore, the star will have a centripetal acceleration equal to,
$a={{\omega }^{2}}r={{\omega }^{2}}\left( \dfrac{2l}{3} \right)$.
And we know that $F=ma$.Therefore,
$\dfrac{Gm(2m)}{{{l}^{2}}}=m{{\omega }^{2}}\left( \dfrac{2l}{3} \right)$
$\Rightarrow {{\omega }^{2}}=\dfrac{3Gm}{{{l}^{3}}}$
$\Rightarrow \omega =\sqrt{\dfrac{3Gm}{{{l}^{3}}}}$.
This means that the time period of rotation of the stars is equal to,
$T=\dfrac{2\pi }{\omega }\\
\therefore T=2\pi \sqrt{\dfrac{{{l}^{3}}}{3mG}}$

Hence, the correct option is C.

Note:The magnitude of the gravitational force on the start of mass $2m$ is equal to that of the star of mass $m$. In addition, both of them will have the same angular velocity. Therefore, we can also analyse the motion of the star of mass $2m$ and find the time period of rotation. Both the stars will have equal periods of rotation although they have different radius of rotation.