Consider a binary star system of star A and star B with masses m A and m B revolving in a circular orbit of radii r A an r B , respectively. If T A and T B are the time period of star A and star B, respectively,
Then :

Solution

Physics sir · 4 steps

Both stars circle the same centre of mass and always stay on opposite sides of it, so they share one period.

1. Common centre. Neither star is fixed. Both revolve about their common centre of mass C, which lies on the line joining them, so m_A r_A = m_B r_B.
2. Locked geometry. C always lies between A and B on the line joining them. When A turns through some angle about C, B must turn through the same angle, so both have the same angular velocity ω.
3. Check with forces. The mutual attraction (G m_A m_B)/(r_A + r_B)² supplies the centripetal force for each star. Setting it equal to m_A ω_A² r_A and to m_B ω_B² r_B, and using m_A r_A = m_B r_B, gives ω_A = ω_B.

   T_A = T_B = 2π√((r_A + r_B)³/(G (m_A + m_B)))

4. So. The period does not depend on which star is heavier or which orbit is larger. T_A = T_B, option (b).

The idea to keep. In a binary system, both bodies orbit the centre of mass with the same ω and the same period. The radii split as r ∝ 1/m, and the heavier one moves slower on a smaller circle.

Answer from the NTA final answer key. Solution by PrepWiser.