The distance between two stars of masses 3MS and 6MS is 9R. Here R is the mean distance between the

Q: Why do we use the sum of the masses $(m_1 + m_2)$ in the denominator for binary stars?

In a binary system, both stars orbit their common center of mass. The gravitational force provides the centripetal acceleration for both. When you solve the equations of motion for either star, the effective mass that dictates the orbital period turns out to be the total mass of the system.

Q: Do both stars in a binary system have the same time period?

Yes! They must always remain on opposite sides of their common center of mass to maintain equilibrium. Therefore, they complete one full revolution in the exact same amount of time.

Q: What is the radius of the orbit for the $3M_S$ star?

The distance from the center of mass is inversely proportional to the mass. For the $3M_S$ star, its orbital radius is $r_1 = \frac{6M_S}{3M_S + 6M_S} \times 9R = 6R$.

Q: Why doesn't the Earth's mass appear in the Earth-Sun time period formula?

The Earth's mass is negligibly small compared to the Sun's mass. The exact formula has $(M_S + M_{Earth})$ in the denominator, but since $M_{Earth}$ is so tiny, we approximate it as just $M_S$.

Q: Can we use Kepler's third law directly without deriving the binary star formula?

Yes, Kepler's Third Law in its generalized Newtonian form is exactly what we used: $T^2 = \frac{4\pi^2}{G(M_1 + M_2)} r^3$. It's a fundamental result that applies to any two-body system orbiting under mutual gravity.

Question

The distance between two stars of masses $3M_S$ and $6M_S$ is $9R$. Here $R$ is the mean distance between the centers of the Earth and the Sun, and $M_S$ is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period $nT$, where $T$ is the period of Earth's revolution around the Sun. The value of $n$ is _______.

Sigma Percentile · Accepted Answer

### The Earth-Sun Benchmark

Before we dive into the complexities of a binary star system, let's establish a solid baseline. The problem gives us a reference point: the Earth orbiting the Sun. According to Kepler's Third Law, the time period $T$ of a planet revolving around a central star is determined by the radius of its orbit $R$ and the mass of the central star $M_S$.

Mathematically, this is expressed as:

$$T = 2\pi \sqrt{\frac{R^3}{GM_S}}$$

This equation is our anchor. Whenever we see the expression $2\pi \sqrt{\frac{R^3}{GM_S}}$, we can immediately substitute it with $T$.

### The Mechanics of a Binary Star System

Now, let's shift our gaze to the binary star system. Unlike the Earth-Sun system where one body is overwhelmingly massive and remains essentially stationary, a binary system consists of two stars of comparable masses orbiting their common center of mass.

Despite this complex dance, the generalized form of Kepler's Third Law still applies beautifully. For two stars of masses $m_1$ and $m_2$ separated by a distance $d$, the time period of their mutual revolution is given by:

$$T_{binary} = 2\pi \sqrt{\frac{d^3}{G(m_1 + m_2)}}$$

Notice the key differences: the radius $R$ is replaced by the total separation distance $d$, and the central mass $M_S$ is replaced by the **sum of the masses** of the two stars.

### The Master Equation

We are given specific values for our binary system. The distance between the stars is $d = 9R$, and their masses are $m_1 = 3M_S$ and $m_2 = 6M_S$. The problem states that the time period of this system is $nT$. Let's substitute these values into our binary star formula:

$$nT = 2\pi \sqrt{\frac{(9R)^3}{G(3M_S + 6M_S)}}$$

This is our master equation. The goal now is to simplify the right-hand side until it looks like a multiple of our reference time period $T$.

### The Final Calculation

Let's carefully execute the algebra. First, we expand the numerator and add the masses in the denominator:

$$nT = 2\pi \sqrt{\frac{729 R^3}{G(9M_S)}}$$

Next, we can simplify the fraction inside the square root by dividing $729$ by $9$:

$$nT = 2\pi \sqrt{81 \frac{R^3}{GM_S}}$$

Now, we extract the perfect square $81$ from under the radical:

$$nT = 9 	imes \left( 2\pi \sqrt{\frac{R^3}{GM_S}} ight)$$

Look closely at the expression inside the parentheses. It is exactly the formula for the Earth's time period $T$ that we established at the very beginning! Substituting $T$ back into the equation yields:

$$nT = 9T$$

Therefore, by direct comparison, we find that **$n = 9$**.

Enter Numerical Value:

Visualized Solution

$T_{E a r t h}$

$T_{B ina r y}$

$n T$

$n T$

$n T$

$n T$

$n$

$r_{1}, r_{2}$

The Sigma Insight: Kepler's Laws of Planetary Motion

The Earth-Sun Benchmark

The Mechanics of a Binary Star System

The Master Equation

The Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​,r2​

The Sigma Insight: Kepler's Laws of Planetary Motion

The Earth-Sun Benchmark

The Mechanics of a Binary Star System

The Master Equation

The Final Calculation

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