The small dense stars rotate about their common centre of mass as a binary system, each with a period of `1` year. One star has mass double than that of the other, while mass of the lighter star is one-third the mass of the Sun. The distance between the two stars is `r` and the distance of the earth from the Sun is `R`, find the relation between `r` and `R`.

Explanation

To solve the problem, we need to find the relation between the distance \( r \) between the two stars and the distance \( R \) from the Earth to the Sun. Let's break down the solution step by step.

### Step 1: Define the masses of the stars
Let the mass of the lighter star be \( M_2 \) and the mass of the heavier star be \( M_1 \). According to the problem:
- \( M_2 = \frac{1}{3} M_{Sun} \)
- \( M_1 = 2M_2 = 2 \times \frac{1}{3} M_{Sun} = \frac{2}{3} M_{Sun} \)

### Step 2: Determine the center of mass
The center of mass \( r_{cm} \) of the two-star system can be calculated using the formula:
\[ r_{cm} = \frac{M_1 r_1 + M_2 r_2}{M_1 + M_2} \]
where \( r_1 \) and \( r_2 \) are the distances of the stars from the center of mass. The total distance between the two stars is \( r = r_1 + r_2 \).

Using the relation \( M_1 = 2M_2 \), we can express \( r_1 \) and \( r_2 \) in terms of \( r \):
\[ M_1 r_1 = M_2 r_2 \]
Let \( r_1 = x \) and \( r_2 = r - x \). Then:
\[ \frac{2}{3} M_{Sun} x = \frac{1}{3} M_{Sun} (r - x) \]
Solving this gives:
\[ 2x = r - x \implies 3x = r \implies x = \frac{r}{3} \]
Thus, \( r_1 = \frac{r}{3} \) and \( r_2 = r - r_1 = r - \frac{r}{3} = \frac{2r}{3} \).

### Step 3: Use Kepler's Third Law
For a binary system, the gravitational force provides the necessary centripetal force for the stars to rotate around their common center of mass. The gravitational force \( F \) between the two stars is given by:
\[ F = \frac{G M_1 M_2}{r^2} \]
The centripetal force required for the rotation of each star about the center of mass is given by:
\[ F_{centripetal} = M_1 \omega^2 r_1 \quad \text{and} \quad F_{centripetal} = M_2 \omega^2 r_2 \]
where \( \omega \) is the angular velocity. Since both stars have the same period \( T = 1 \) year, we can relate \( \omega \) to \( T \):
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{1 \text{ year}} \]

### Step 4: Set up the equations
Setting the gravitational force equal to the centripetal force for the heavier star:
\[ \frac{G \left(\frac{2}{3} M_{Sun}\right) \left(\frac{1}{3} M_{Sun}\right)}{r^2} = \left(\frac{2}{3} M_{Sun}\right) \left(\frac{2\pi}{1 \text{ year}}\right)^2 \left(\frac{r}{3}\right) \]
This simplifies to:
\[ \frac{G \cdot \frac{2}{9} M_{Sun}^2}{r^2} = \frac{2}{3} M_{Sun} \cdot \frac{4\pi^2}{1 \text{ year}^2} \cdot \frac{r}{3} \]

### Step 5: Solve for \( r \)
Rearranging and simplifying gives:
\[ \frac{2G M_{Sun}}{9r^2} = \frac{8\pi^2 M_{Sun}}{9 \text{ year}^2} \implies r^2 = \frac{G M_{Sun}}{4\pi^2} \text{ year}^2 \]

### Step 6: Relate \( r \) to \( R \)
For the Earth-Sun system, we have:
\[ R^2 = \frac{G M_{Sun}}{4\pi^2} \text{ year}^2 \]
Thus, we find:
\[ r^2 = R^2 \implies r = R \]

### Final Relation
The relation between the distance \( r \) between the two stars and the distance \( R \) from the Earth to the Sun is:
\[ r = R \]