Question:
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 :
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 :
Then :
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Both stars orbit their common centre of mass and always stay on opposite sides of it, so they must have the same angular velocity and the same period.
Updated On: Nov 4, 2026
- \(\dfrac{T_A}{T_B} = \left(\dfrac{r_A}{r_B}\right)^{\frac{3}{2}}\)
- \(T_A = T_B\)
- \(T_A \gt T_B\) (if \(m_A \gt m_B\))
- \(T_A \gt T_B\) (if \(r_A \gt r_B\))
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The Correct Option is B
Solution and Explanation
In a binary system the two stars revolve about their common centre of mass, and the line joining them always passes through it.
Hence in any time interval both stars sweep the same angle, i.e. \(\omega_A = \omega_B\).
Therefore \(T_A = \dfrac{2\pi}{\omega_A} = \dfrac{2\pi}{\omega_B} = T_B\), regardless of their masses or orbital radii.
(Kepler's \(T^2\propto r^3\) law applies to the separation \(r_A + r_B\) of the stars, not to each orbit individually, so option (1) is wrong.)
Answer: (2) \(T_A = T_B\)
Hence in any time interval both stars sweep the same angle, i.e. \(\omega_A = \omega_B\).
Therefore \(T_A = \dfrac{2\pi}{\omega_A} = \dfrac{2\pi}{\omega_B} = T_B\), regardless of their masses or orbital radii.
(Kepler's \(T^2\propto r^3\) law applies to the separation \(r_A + r_B\) of the stars, not to each orbit individually, so option (1) is wrong.)
Answer: (2) \(T_A = T_B\)
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