Two massive particles of masses `M` & `m` `(M gt m)` are separated by a distance `l`. They rotate with equal angular velocity under their gravitational attraction. The linear speed of the particle of mass `m` is

Options

A. `sqrt((GMm)/((M+m)l))`
B. `sqrt((GM^(2))/((M+m)l))`
C. `sqrt((Gm)/(l))`
D. `sqrt((Gm^(2))/((M+m)l))`

Explanation

To find the linear speed of the particle of mass \( m \) that rotates under the gravitational attraction of another mass \( M \) at a distance \( l \), we can follow these steps:

### Step 1: Understand the System
We have two masses, \( M \) (greater mass) and \( m \) (smaller mass), separated by a distance \( l \). Both masses are rotating around their common center of mass with an angular velocity \( \omega \).

### Step 2: Identify the Center of Mass
The center of mass \( x_{cm} \) of the system can be calculated using the formula:
\[ x_{cm} = \frac{m \cdot 0 + M \cdot l}{m + M} = \frac{M \cdot l}{m + M} \]
Here, we consider the position of mass \( m \) at the origin (0) and mass \( M \) at distance \( l \).

### Step 3: Calculate the Distance from the Center of Mass
The distance \( r \) from the mass \( m \) to the center of mass is:
\[ r = l - x_{cm} = l - \frac{M \cdot l}{m + M} = l \left(1 - \frac{M}{m + M}\right) = \frac{m \cdot l}{m + M} \]

### Step 4: Apply Gravitational Force and Centripetal Force
The gravitational force \( F_g \) between the two masses is given by:
\[ F_g = \frac{G \cdot M \cdot m}{l^2} \]
This gravitational force provides the necessary centripetal force for the mass \( m \) moving in a circular path:
\[ F_c = \frac{m v^2}{r} \]
Setting the gravitational force equal to the centripetal force, we have:
\[ \frac{G \cdot M \cdot m}{l^2} = \frac{m v^2}{r} \]

### Step 5: Substitute for \( r \)
Substituting \( r \) from Step 3 into the centripetal force equation:
\[ \frac{G \cdot M \cdot m}{l^2} = \frac{m v^2}{\frac{m \cdot l}{m + M}} \]
Simplifying this gives:
\[ \frac{G \cdot M \cdot m}{l^2} = \frac{m^2 v^2}{m \cdot l/(m + M)} \]
This simplifies further to:
\[ \frac{G \cdot M}{l^2} = \frac{m v^2}{l/(m + M)} \]

### Step 6: Solve for Linear Speed \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{G \cdot M \cdot l}{(m + M)} \]
Taking the square root gives us the linear speed \( v \):
\[ v = \sqrt{\frac{G \cdot M}{(m + M)} \cdot l} \]

### Final Answer
Thus, the linear speed of the particle of mass \( m \) is:
\[ v = \sqrt{\frac{G \cdot M \cdot l}{m + M}} \]

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