Two particles of masses m and 3m are moving under their mutual gravitational force, around their centre of mass, in circular orbits. The separation between the masses is r. The gravitational attraction the two provides nessary centripetal force for circular motion
If `v_(1)` and `v_(2)` be the linear speeds of m and 3m respectively, then the value of `(v_(1))/(v_(2))` is

Options

A. 1
B. 2
C. 3
D. `(1)/(3)`

Answer

To solve the problem, we need to find the ratio of the linear speeds \( v_1 \) and \( v_2 \) of two masses \( m \) and \( 3m \) moving in circular orbits around their center of mass under the influence of their mutual gravitational force.

### Step 1: Determine the position of the center of mass
The center of mass (COM) of the two masses can be found using the formula:
\[ R_{cm} = \frac{m_1 \cdot r_1 + m_2 \cdot r_2}{m_1 + m_2} \]
where \( m_1 = m \) and \( m_2 = 3m \). The total separation between the two masses is \( r \).

Let \( r_1 \) be the distance from mass \( m \) to the center of mass, and \( r_2 \) be the distance from mass \( 3m \) to the center of mass. We know:
\[ r_1 + r_2 = r \]
Substituting the values into the center of mass equation:
\[ r_1 = \frac{3m \cdot r_2}{m + 3m} = \frac{3m \cdot r_2}{4m} = \frac{3}{4} r_2 \]
From \( r_1 + r_2 = r \), we can express \( r_2 \):
\[ r_2 = \frac{r}{4}, \quad r_1 = \frac{3r}{4} \]

### Step 2: Write the centripetal force equations
For mass \( m \):
The centripetal force required for mass \( m \) is provided by the gravitational force between the two masses:
\[ \frac{m v_1^2}{r_1} = \frac{G \cdot m \cdot 3m}{r^2} \]
Substituting \( r_1 = \frac{3r}{4} \):
\[ \frac{m v_1^2}{\frac{3r}{4}} = \frac{3Gm^2}{r^2} \]
This simplifies to:
\[ \frac{4 v_1^2}{3r} = \frac{3Gm}{r^2} \]
Thus:
\[ v_1^2 = \frac{9Gm r}{4} \]

For mass \( 3m \):
The centripetal force for mass \( 3m \) is:
\[ \frac{3m v_2^2}{r_2} = \frac{G \cdot m \cdot 3m}{r^2} \]
Substituting \( r_2 = \frac{r}{4} \):
\[ \frac{3m v_2^2}{\frac{r}{4}} = \frac{3Gm^2}{r^2} \]
This simplifies to:
\[ 12 v_2^2 = \frac{3Gm}{r} \]
Thus:
\[ v_2^2 = \frac{Gm r}{4} \]

### Step 3: Find the ratio of the speeds
Now, we can find the ratio \( \frac{v_1}{v_2} \):
\[ \frac{v_1^2}{v_2^2} = \frac{\frac{9Gm r}{4}}{\frac{Gm r}{4}} = 9 \]
Taking the square root gives:
\[ \frac{v_1}{v_2} = 3 \]

### Final Answer
The ratio \( \frac{v_1}{v_2} \) is \( 3 \).

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