Binary star motion about the center of mass

In a binary star system, both stars revolve about their common center of mass.

Let their angular speed be $\omega$.

Since they are always on opposite sides of the line joining them and complete one revolution together, both stars have the same angular speed and hence the same time period.

Relation between time period and angular speed

For any circular motion, $T = \frac{2\pi}{\omega}$

Therefore, for the two stars, $T_A = \frac{2\pi}{\omega}, \qquad T_B = \frac{2\pi}{\omega}$

Hence, $T_A = T_B$

Why this is true physically

If one star completed a revolution in a different time than the other, the line joining the two stars would not remain intact as a rigid two-body configuration. In a bound circular binary system, both must orbit the common center with the same angular frequency.

Checking the options

• Option A: $\frac{T_A}{T_B} = \left(\frac{r_A}{r_B}\right)^{3/2}$ This is false for the two stars in the same binary system, because actually $\frac{T_A}{T_B} = 1$

• Option B: $T_A = T_B$ This is correct.

• Option C: $T_A > T_B \quad (\text{if } m_A > m_B)$ False, because time periods are equal regardless of masses.

• Option D: $T_A > T_B \quad (\text{if } r_A > r_B)$ False, because even if orbital radii differ, the angular speed is the same, so periods are equal.

Final answer

The correct option is: $\boxed{B}$

Comparison with stored answer

Stored correct answer: $B$

My derived answer matches the stored answer.