Question

A double star system of two stars moving around the centre of inertia of the system due to gravitation. Find the distance between the components of the double star, if its total mass equals M and the period of revolution T.

Answer 1

Hint: To solve this question, express the double star system as a single star system with reduced mass moving around their centre of mass. Obtain the expressions for their gravitational attraction and the centripetal force. Equate these two forces to find the period of revolution T.

Complete step by step solution:
Consider a double star system with mass of the stars ${{m}_{1}}$ and ${{m}_{2}}$ .
The double star system can be replaced by a single star of reduced mass $\dfrac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}$ .
This reduced mass star system is moving about the centre of mass subjected to the force,
$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Where r is the distance between the two stars and G is the universal gravitational constant.
We can also define the above force in terms of centripetal force as,
$F=\dfrac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{\omega }^{2}}r$
Equating the above two, we get that,
$\begin{align}
  & \dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}=\dfrac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{\omega }^{2}}r \\
 & {{r}^{3}}=\dfrac{G\left( {{m}_{1}}+{{m}_{2}} \right)}{{{\omega }^{2}}} \\
\end{align}$
Time period of revolution is given as T. So, we can write,
$\begin{align}
  & T=\dfrac{2\pi }{\omega } \\
 & \omega =\dfrac{2\pi }{T} \\
\end{align}$
Putting this value on the above equation,
$\begin{align}
  & {{r}^{3}}=\dfrac{G\left( {{m}_{1}}+{{m}_{2}} \right)}{{{\left( \dfrac{2\pi }{T} \right)}^{2}}} \\
 & {{r}^{3}}=G\left( {{m}_{1}}+{{m}_{2}} \right){{\left( \dfrac{T}{2\pi } \right)}^{2}} \\
\end{align}$
Now the total mass of the two stars will be,
$M={{m}_{1}}+{{m}_{2}}$
So, from the above equation, we can write,
$\begin{align}
  & {{r}^{3}}=GM{{\left( \dfrac{T}{2\pi } \right)}^{2}} \\
 & r={{\left\{ GM{{\left( \dfrac{T}{2\pi } \right)}^{2}} \right\}}^{\dfrac{1}{3}}} \\
\end{align}$
So, the distance between the stars in a double start system of total mass M is given as ${{\left\{ GM{{\left( \dfrac{T}{2\pi } \right)}^{2}} \right\}}^{\dfrac{1}{3}}}$.

Note: Reduced mass can be defined as the effective inertial mass in a two-body system about the centre of mass of the system. It allows us to solve the two-body problem as a one body problem which is easier for us.
In this question we considered the double star system as a single mass orbiting the centre of mass of the system. Then we considered the gravitational force and the centripetal force of the one body system to find our answer.