A trinary star system which is a system of three orbiting around centre of mass of system has time peroid of 3 years, while the distance between any two stars of the system is 2 astronomical unit. All the three stars are identical and mass of the sun is M and total mass of this multiple star system is `(a)/(b)xxM` . Then find `(b)/(3)`? (a and b are lowest possible integers (I astronomical unit is equal to distance between sun and earth, time period of rotation of earth around sun is 1year)

Explanation

To solve the problem, we need to analyze the trinary star system and apply the principles of gravitational force and centripetal motion. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a trinary star system with three identical stars, each having mass \( m_0 \). The distance between any two stars is given as \( 2 \) astronomical units (AU). The time period of the system's rotation around the center of mass is \( T = 3 \) years.

### Step 2: Determine the Geometry
The stars form an equilateral triangle due to their identical masses. The distance between any two stars is \( 2 \) AU. The radius \( r \) from the center of mass to each star can be derived using geometry.

In an equilateral triangle, if the side length is \( s \), the distance from the center to a vertex (the radius) is given by:
\[ r = \frac{s}{\sqrt{3}} \]
For our case, \( s = 2 \) AU, so:
\[ r = \frac{2 \text{ AU}}{\sqrt{3}} = \frac{2}{\sqrt{3}} \text{ AU} \]

### Step 3: Apply Gravitational Force
The gravitational force between two stars can be expressed as:
\[ F = \frac{G m_0^2}{(2 \text{ AU})^2} \]
However, since we have three stars, we need to consider the net gravitational force acting on one star due to the other two.

### Step 4: Calculate the Net Force
The net force acting on one star due to the other two can be calculated by considering the components of the forces. The angle between the forces from the two other stars is \( 60^\circ \). The resultant force directed towards the center can be calculated as:
\[ F_{\text{net}} = 2F \cos(30^\circ) = 2 \left( \frac{G m_0^2}{(2 \text{ AU})^2} \right) \left( \frac{\sqrt{3}}{2}\right) = \frac{G m_0^2 \sqrt{3}}{2 \text{ AU}^2} \]

### Step 5: Relate to Centripetal Force
This net gravitational force provides the necessary centripetal force for circular motion:
\[ F_{\text{centripetal}} = m_0 \frac{v^2}{r} \]
where \( v \) is the orbital speed of the stars. The angular velocity \( \omega \) can be expressed in terms of the time period \( T \):
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{3 \text{ years}} \]
The relationship between linear velocity and angular velocity is:
\[ v = r \omega \]

### Step 6: Substitute and Solve
Substituting \( v \) into the centripetal force equation:
\[ F_{\text{centripetal}} = m_0 \frac{(r \omega)^2}{r} = m_0 r \omega^2 \]
Setting the gravitational force equal to the centripetal force gives:
\[ \frac{G m_0^2 \sqrt{3}}{2 \text{ AU}^2} = m_0 r \left(\frac{2\pi}{3}\right)^2 \]
Substituting \( r = \frac{2}{\sqrt{3}} \text{ AU} \):
\[ \frac{G m_0^2 \sqrt{3}}{2 \text{ AU}^2} = m_0 \left(\frac{2}{\sqrt{3}}\right) \left(\frac{2\pi}{3}\right)^2 \]

### Step 7: Solve for \( m_0 \)
After simplifying, we can solve for \( m_0 \) in terms of \( M \) (mass of the Sun):
\[ m_0 = \frac{8}{27} M \]

### Step 8: Total Mass of the System
The total mass of the system \( M_{\text{total}} \) is:
\[ M_{\text{total}} = 3 m_0 = 3 \left(\frac{8}{27} M\right) = \frac{24}{27} M = \frac{8}{9} M \]

### Step 9: Find \( \frac{b}{3} \)
From the total mass expression \( \frac{a}{b} M \), we have \( a = 8 \) and \( b = 9 \). Thus:
\[ \frac{b}{3} = \frac{9}{3} = 3 \]

### Final Answer
The value of \( \frac{b}{3} \) is \( \boxed{3} \).