A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius R `(R lt lt L)` A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the time period of star is T and its distance from the galaxy's axis is r, then-

Options

A. `Tpropsqrt(r)`
B. `Tpropr`
C. `Tpropr^(2)`
D. `T^(2)propr^(3)`

Explanation

To solve the problem, we need to analyze the gravitational field produced by the cylindrical galaxy and how it affects the motion of a star orbiting around it. Here's a step-by-step solution:

### Step 1: Understand the Gravitational Field of the Cylinder
The gravitational field (E_g) produced by a long cylindrical mass distribution can be derived from the analogy with the electric field of a charged cylinder. For a cylinder of radius \( R \) and linear mass density \( \lambda \), the gravitational field at a distance \( r \) from the axis of the cylinder is given by:

\[ E_g = \frac{2G\lambda}{r} \]

where \( G \) is the universal gravitational constant.

### Step 2: Relate the Gravitational Field to the Centripetal Force
The star is moving in a circular orbit due to the gravitational attraction from the galaxy. The gravitational force acting on the star provides the necessary centripetal force for circular motion. Thus, we can equate the gravitational force to the centripetal force:

\[ F_{gravity} = F_{centripetal} \]

This can be expressed as:

\[ m \cdot E_g = m \cdot \frac{v^2}{r} \]

where \( m \) is the mass of the star, \( v \) is the orbital speed, and \( r \) is the distance from the axis of the galaxy.

### Step 3: Substitute the Gravitational Field into the Equation
Substituting the expression for the gravitational field into the equation gives:

\[ m \cdot \frac{2G\lambda}{r} = m \cdot \frac{v^2}{r} \]

We can cancel \( m \) and \( r \) (assuming \( r \neq 0 \)):

\[ 2G\lambda = v^2 \]

### Step 4: Relate Orbital Speed to Angular Velocity
The orbital speed \( v \) can be related to the angular velocity \( \omega \) by the equation:

\[ v = r \cdot \omega \]

Substituting this into the previous equation gives:

\[ 2G\lambda = (r \cdot \omega)^2 \]

### Step 5: Solve for Angular Velocity
Rearranging the equation to solve for \( \omega \):

\[ \omega^2 = \frac{2G\lambda}{r^2} \]

Taking the square root:

\[ \omega = \sqrt{\frac{2G\lambda}{r^2}} \]

### Step 6: Calculate the Time Period
The time period \( T \) of the star's orbit is related to the angular velocity by:

\[ T = \frac{2\pi}{\omega} \]

Substituting the expression for \( \omega \):

\[ T = \frac{2\pi}{\sqrt{\frac{2G\lambda}{r^2}}} \]

This simplifies to:

\[ T = 2\pi \cdot \frac{r}{\sqrt{2G\lambda}} \]

### Step 7: Determine the Proportionality
From the final expression for \( T \), we can see that:

\[ T \propto r \]

This means that the time period \( T \) is directly proportional to the distance \( r \) from the axis of the galaxy.

### Final Answer
Thus, the time period \( T \) of the star is directly proportional to its distance \( r \) from the axis of the galaxy.

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