A system of binary stars of mass `m_(A)` and `m_(B)` are moving in circular orbits of radii `r_(A)` and `r_(B)` respectively. If `T_(A)` and `T_(B)` are at the time periods of masses `m_(A)` and `m_(B)` respectively then

Options

A. If `T_(A)gtT_(B)`, then `R_(A)gtR_(B)`
B. If `T_(A)gtT_(B)`, then `M_(A)gtM_(B)`
C. `((T_(A))/(T_(B)))^(2)=((R_(A))/(R_(B)))^(3)`
D. `T_(A)=T_(B)`

Explanation

To solve the problem regarding the binary star system, we will follow these steps:

### Step 1: Understand the System
We have two stars, A and B, with masses \( m_A \) and \( m_B \) respectively, moving in circular orbits of radii \( r_A \) and \( r_B \). The time periods of these stars are \( T_A \) and \( T_B \).

### Step 2: Center of Mass Condition
The center of mass (COM) of the system can be defined as:
\[ m_A r_A = m_B r_B \]
This equation indicates that the moments about the center of mass are equal. This is our **Equation 1**.

### Step 3: Forces Acting on the Stars
The gravitational force between the two stars provides the necessary centripetal force for their circular motion. The gravitational force \( F_g \) is given by:
\[ F_g = \frac{G m_A m_B}{(r_A + r_B)^2} \]
The centripetal force for star A is:
\[ F_{cA} = \frac{m_A V_A^2}{r_A} \]
And for star B:
\[ F_{cB} = \frac{m_B V_B^2}{r_B} \]
Since the gravitational force provides the centripetal force, we can set up the following equations:
\[ \frac{G m_A m_B}{(r_A + r_B)^2} = \frac{m_A V_A^2}{r_A} \quad (2) \]
\[ \frac{G m_A m_B}{(r_A + r_B)^2} = \frac{m_B V_B^2}{r_B} \quad (3) \]

### Step 4: Expressing Velocities
The velocities \( V_A \) and \( V_B \) can be expressed in terms of the time periods:
\[ V_A = \frac{2 \pi r_A}{T_A} \]
\[ V_B = \frac{2 \pi r_B}{T_B} \]

### Step 5: Substitute Velocities into the Force Equations
Substituting \( V_A \) and \( V_B \) into equations (2) and (3):
\[ \frac{G m_A m_B}{(r_A + r_B)^2} = \frac{m_A \left(\frac{2 \pi r_A}{T_A}\right)^2}{r_A} \]
\[ \frac{G m_A m_B}{(r_A + r_B)^2} = \frac{m_B \left(\frac{2 \pi r_B}{T_B}\right)^2}{r_B} \]

### Step 6: Simplifying the Equations
From the first equation:
\[ \frac{G m_A m_B}{(r_A + r_B)^2} = \frac{4 \pi^2 m_A r_A}{T_A^2} \]
From the second equation:
\[ \frac{G m_A m_B}{(r_A + r_B)^2} = \frac{4 \pi^2 m_B r_B}{T_B^2} \]

### Step 7: Equating the Two Expressions
Since both expressions are equal to the gravitational force, we can set them equal to each other:
\[ \frac{4 \pi^2 m_A r_A}{T_A^2} = \frac{4 \pi^2 m_B r_B}{T_B^2} \]
This simplifies to:
\[ \frac{m_A r_A}{T_A^2} = \frac{m_B r_B}{T_B^2} \]

### Step 8: Relating Time Periods
From the center of mass condition \( m_A r_A = m_B r_B \), we can substitute \( r_B = \frac{m_A}{m_B} r_A \) into the equation:
\[ \frac{m_A r_A}{T_A^2} = \frac{m_B \left(\frac{m_A}{m_B} r_A\right)}{T_B^2} \]
This leads to:
\[ \frac{1}{T_A^2} = \frac{1}{T_B^2} \]
Thus, we find:
\[ T_A = T_B \]

### Conclusion
The time periods of the two stars are equal:
\[ \boxed{T_A = T_B} \]