Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orb

Q: Why can't the two stars have different time periods?

If they had different time periods, they would not remain diametrically opposite. This would cause their center of mass to move in a circle, which violates the law of conservation of momentum since there are no external forces acting on the system.

Q: Does the heavier star move faster?

The heavier star is closer to the center of mass (smaller radius). Since both have the same angular velocity $\omega$, the linear speed $v = \omega r$ of the heavier star is actually smaller than that of the lighter star.

Q: What provides the centripetal force for their circular motion?

The mutual gravitational force of attraction between the two stars provides the necessary centripetal force for both of them to maintain their circular orbits.

Q: Can we apply Kepler's third law directly here?

Standard Kepler's third law assumes the central body is stationary and much more massive. For a binary system, both stars orbit the center of mass, so the formula is modified to $T^2 = \frac{4\pi^2}{G(m_A + m_B)} (r_A + r_B)^3$.

Q: Why is the distance between the stars $r_A + r_B$?

Because the stars are always diametrically opposite to each other with the center of mass in between them. The total distance separating them is the sum of their individual orbital radii.

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 $r_A$ and $r_B$, respectively. If $T_A$ and $T_B$ are the time period of star A and star B respectively, then

Options:

(A) $T_A = T_B \left(\frac{r_A}{r_B}\right)^{\frac{3}{2}}$

(B) $T_A = T_B$

(C) $T_A > T_B$ (if $m_A > m_B$)

(D) $T_A > T_B$ (if $r_A > r_B$)

Sigma Percentile · Accepted Answer

### The Cosmic Dance of Binary Stars

Imagine looking up at the night sky and spotting a single, bright point of light, only to realize through a powerful telescope that it is actually two massive stars locked in a cosmic dance. This is a binary star system. In such a system, the two stars do not orbit one another; instead, they both orbit a common, invisible point in space known as the **Center of Mass (CM)**.

Understanding the dynamics of this system is a classic and beautiful application of Newton's laws of motion and gravitation. Let's break down the physics behind their synchronized movement.

### The Anchor: The Center of Mass

In the absence of any external forces, the center of mass of the binary star system must remain perfectly stationary (or move at a constant velocity). For our analysis, we consider the center of mass to be at rest at the origin.

For the center of mass to remain stationary while the two stars, $A$ and $B$, revolve around it, their positions must always perfectly balance each other out. This means that at any given instant, Star A, the Center of Mass, and Star B must lie on a single straight line. They are diametrically opposite to each other.

### The Synchronized Symphony

Because the two stars are perpetually tethered by this invisible straight line passing through the center of mass, they are forced to sweep out the exact same angle in any given time interval.

If Star A completes a $90^\circ$ turn, Star B must also complete a $90^\circ$ turn to remain on the opposite side of the center of mass. Consequently, their rate of angular displacement—their **angular velocity $\omega$**—must be identical.

$$ \omega_A = \omega_B = \omega $$

### The Mathematical Conclusion

We know from the kinematics of circular motion that the time period $T$ (the time taken to complete one full revolution) is inversely proportional to the angular velocity. The relationship is given by the simple formula:

$$ T = \frac{2\pi}{\omega} $$

Since we have already established that both stars share the exact same angular velocity $\omega$, it mathematically guarantees that their time periods must also be exactly the same.

$$ T_A = \frac{2\pi}{\omega_A} $$
$$ T_B = \frac{2\pi}{\omega_B} $$

Therefore, we arrive at our elegant conclusion:

$$ \mathbf{T_A = T_B} $$

### Beyond the Question: Deriving the Time Period

While the question only asks for the relationship between the time periods, a true physics enthusiast might wonder: *What is the actual value of this time period?*

To find it, we equate the mutual gravitational force of attraction between the stars to the centripetal force required to keep either star in its circular orbit. Let's look at Star A:

The gravitational force is provided by Star B, which is at a total distance of $(r_A + r_B)$ from Star A.

$$ F_g = \frac{G m_A m_B}{(r_A + r_B)^2} $$

The centripetal force required for Star A to move in a circle of radius $r_A$ is:

$$ F_c = m_A \omega^2 r_A $$

Equating the two:

$$ \frac{G m_A m_B}{(r_A + r_B)^2} = m_A \omega^2 r_A $$

By substituting $\omega = \frac{2\pi}{T}$ and using the center of mass relation $m_A r_A = m_B r_B$, you can derive the generalized form of Kepler's Third Law for binary systems. Try it out!

Animated Solution for Physics - Laws of Motion: 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 $r_{A}$ and $r_{B}$ , respectively. If $T_{A}$ and $T_{B}$ are the time period of star A and star B respectively, then

Select Answer:

Visualized Solution

$Visualizing the Binary System$

$The Center of Mass Constraint$

$Equal Angular Velocities$

$Time Period Formula$

$Final Conclusion$

$The Way Forward$

The Sigma Insight: Dynamics of Circular Motion

The Cosmic Dance of Binary Stars

The Anchor

The Synchronized Symphony

The Mathematical Conclusion

Beyond the Question

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