Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space.

Q: Why do they rotate about the centre of mass and not the heavier star?

In a closed system with no external forces, the centre of mass remains at rest or moves with constant velocity. For the stars to stay together and orbit, they must revolve around this stationary centre of mass to conserve linear momentum.

Q: Do both stars have the same time period?

Yes! If they had different time periods, the distance between them would constantly change, and they wouldn't remain on opposite sides of the centre of mass. They are locked in a synchronous dance.

Q: Can I use the lighter star to write the force equation?

Absolutely. The gravitational force is the same for both. If you use the lighter star $m$, its orbital radius would be $d - x = \frac{2d}{3}$. The equation $\frac{G(2m)(m)}{d^2} = m\omega^2\left(\frac{2d}{3}\right)$ will give you the exact same $\omega$.

Q: Why is the distance in the denominator of the gravitational force $d^2$ and not $x^2$?

Newton's Law of Universal Gravitation states that the force depends on the *total separation* between the two masses, which is $d$. The radius $x$ is only used for the centripetal acceleration of the individual star.

Q: What is 'reduced mass' and can it be used here?

Yes! The two-body problem can be reduced to a single body of 'reduced mass' $\mu = \frac{m_1 m_2}{m_1 + m_2}$ orbiting a fixed centre at distance $d$. The equation becomes $\frac{G(2m)(m)}{d^2} = \mu \omega^2 d$. Try it, you'll get the same result!

Question

Two stars of masses $m$ and $2m$ at a distance $d$ rotate about their common centre of mass in free space. The period of revolution is

Options:

(A) $\frac{1}{2\pi}\sqrt{\frac{3Gm}{d^3}}$

(B) $2\pi\sqrt{\frac{d^3}{3Gm}}$

(C) $2\pi\sqrt{\frac{3Gm}{d^3}}$

(D) $\frac{1}{2\pi}\sqrt{\frac{d^3}{3Gm}}$

Sigma Percentile · Accepted Answer

### The Cosmic Dance of Binary Stars

Imagine you are floating in the deep, silent expanse of free space.

Far away from any other celestial bodies, you spot two stars locked in an eternal embrace.

One star has a mass of $m$, and its heavier companion boasts a mass of $2m$.

They are separated by a vast distance, $d$.

Because there are no external forces acting on this isolated system, they don't just sit there; they orbit each other.

But what exactly are they orbiting?

They are dancing around a shared, invisible pivot point known as the **centre of mass**.

### Finding the Invisible Pivot: The Centre of Mass

To understand their motion, we first need to pinpoint this centre of mass.

Let's set up a coordinate system to make things easy.

Imagine placing the heavier star, $2m$, right at the origin $(0,0)$.

This means the lighter star, $m$, is located at a distance $d$ along the x-axis.

The formula to find the position of the centre of mass, $x$, is a simple weighted average of their positions:

$$x = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

Let's plug in our specific values.

For the heavier star, the mass is $2m$ and its position is $0$.

For the lighter star, the mass is $m$ and its position is $d$.

Substituting these into our master equation gives:

$$x = \frac{2m(0) + m(d)}{2m + m}$$

This simplifies beautifully.

The numerator becomes $md$, and the denominator becomes the total mass, $3m$.

$$x = \frac{md}{3m} = \frac{d}{3}$$

So, the centre of mass is located at a distance of $\frac{d}{3}$ from the heavier star.

As expected, the balance point is much closer to the more massive object!

### The Gravitational Tether

Now, let's look at the dynamics of their movement.

Both stars revolve in perfectly concentric circles around this centre of mass.

To ensure they always stay on opposite sides of this pivot, they must complete one full revolution in the exact same amount of time.

This means they share the exact same angular velocity, $\omega$.

But what invisible string keeps them from flying off into the void?

It is the universal force of gravity.

The mutual gravitational attraction between the two stars acts as the **centripetal force** required to keep them in their circular paths.

We can express this fundamental physical truth as:

$$F_G = F_C$$

### Unlocking the Orbital Speed

Let's focus our attention on the heavier star, $2m$, to build our equation.

According to Newton's Law of Universal Gravitation, the force pulling it towards the lighter star is:

$$F_G = \frac{G(2m)(m)}{d^2}$$

Notice that the denominator uses $d^2$, the total distance between the stars, because gravity acts across the entire gap.

Now, for the centripetal force keeping it in its circular orbit of radius $x$:

$$F_C = (2m)\omega^2 x$$

Equating these two forces gives us the master dynamic equation for our system:

$$\frac{G(2m)(m)}{d^2} = (2m)\omega^2 x$$

We already did the hard work of finding $x$, which is $\frac{d}{3}$.

Let's substitute that into our equation:

$$\frac{2Gm^2}{d^2} = 2m\omega^2 \left(\frac{d}{3}\right)$$

Now, the magic of algebra takes over.

We can cancel the common factor of $2m$ from both sides of the equation.

Rearranging the remaining terms to isolate $\omega^2$, we get:

$$\omega^2 = \frac{3Gm}{d^3}$$

Taking the square root of both sides reveals the angular velocity of our binary system:

$$\omega = \sqrt{\frac{3Gm}{d^3}}$$

### The Final Countdown: Period of Revolution

We have unlocked the angular velocity, but the question asks for the period of revolution, $T$.

The time period is simply the time taken to complete one full circle, which is $2\pi$ radians.

The relationship between time period and angular velocity is:

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

All that is left is to substitute our expression for $\omega$ into this formula.

When we divide by a fraction, we simply multiply by its reciprocal.

Flipping the fraction inside the square root gives us our final, elegant result:

$$T = 2\pi \sqrt{\frac{d^3}{3Gm}}$$

This perfectly matches option (b).

You have successfully decoded the orbital mechanics of a binary star system!

Animated Solution for Physics - Gravitation: Two stars of masses $m$ and $2 m$ at a distance $d$ rotate about their common centre of mass in free space. The period of revolution is

Select Answer:

Visualized Solution

$Binary Star System$

$Centre of Mass (CM)$

$Locating the CM$

$Calculating x$

$Distance of CM from 2 m$

$Circular Orbits$

$Dynamics of the System$

$Force Equation for 2 m$

$Substituting x$

$Solving for ω^{2}$

$Angular Velocity ω$

$Time Period T$

$Final Answer$

$The Way Forward$

The Sigma Insight: Gravitational Force

The Cosmic Dance of Binary Stars

Finding the Invisible Pivot

The Gravitational Tether

Unlocking the Orbital Speed

The Final Countdown

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Animated Solution for Physics - Gravitation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Binary Star System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Centre of Mass (CM)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Locating the CM

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance of CM from 2m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circular Orbits

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dynamics of the System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Force Equation for 2m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for ω2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Velocity ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward