In a binary star system one of the stars has mass equal to mass of sun `(M)`. Separation between the stars is four times the average separation between earth and sun. Mass of the other star so that time period of revolution of the stars each other is equal to one earth year is

Options

A. `31M`
B. `63M`
C. `M`
D. `2M`

Explanation

To solve the problem, we need to find the mass of the second star in a binary star system such that the time period of revolution of the stars around each other is equal to one Earth year. Let's break down the solution step by step.

### Step 1: Understand the Given Information
- One star has a mass equal to the mass of the Sun, denoted as \( M \).
- The separation between the two stars is four times the average distance between the Earth and the Sun, denoted as \( r = 4R \), where \( R \) is the average distance between the Earth and the Sun (approximately \( 1.496 \times 10^{11} \) meters).
- We need to find the mass of the second star, denoted as \( M' \), such that the time period of revolution \( T \) is equal to one Earth year.

### Step 2: Use Kepler's Third Law
Kepler's Third Law states that the square of the orbital period \( T \) of two bodies is proportional to the cube of the semi-major axis \( r \) of their orbit divided by the total mass \( M + M' \) of the system:
\[ T^2 = \frac{4\pi^2}{G(M + M')} r^3 \]
Where:
- \( T \) is the time period of revolution,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the first star,
- \( M' \) is the mass of the second star,
- \( r \) is the separation between the stars.

### Step 3: Substitute the Known Values
We know that \( T = 1 \) year (which we convert to seconds for calculations),
\[ T = 1 \text{ year} = 365.25 \times 24 \times 60 \times 60 \text{ seconds} \approx 3.156 \times 10^7 \text{ seconds} \]
And the separation \( r = 4R \).

### Step 4: Set Up the Equation
Substituting \( r = 4R \) into Kepler's law gives:
\[ T^2 = \frac{4\pi^2}{G(M + M')} (4R)^3 \]
This simplifies to:
\[ T^2 = \frac{4\pi^2}{G(M + M')} \cdot 64R^3 \]
\[ T^2 = \frac{256\pi^2 R^3}{G(M + M')} \]

### Step 5: Relate to the Time Period for the Sun-Earth System
For the Earth, the time period \( T \) is given by:
\[ T^2 = \frac{4\pi^2 R^3}{GM} \]

### Step 6: Set the Two Equations Equal
Since both expressions equal \( T^2 \), we can set them equal to each other:
\[ \frac{4\pi^2 R^3}{GM} = \frac{256\pi^2 R^3}{G(M + M')} \]
Cancel \( \frac{4\pi^2 R^3}{G} \) from both sides:
\[ 1 = \frac{64}{M + M'} \]

### Step 7: Solve for \( M' \)
Rearranging gives:
\[ M + M' = 64 \]
Substituting \( M = 1 \) (mass of the Sun):
\[ 1 + M' = 64 \]
\[ M' = 64 - 1 = 63 \]

### Conclusion
The mass of the other star \( M' \) is \( 63M \).

### Final Answer
The mass of the other star is \( 63 \) times the mass of the Sun.

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