If gravitational attraction between two points masses be given by `F=G(m_(1)m_(2))/(r^(n))`. Then the period of a satellite in a circular orbit will be proportional to

Options

A. r
B. `r^(2)`
C. `r^((1)/(2))`
D. independent of `n`

Explanation

To solve the problem, we need to find the relationship between the period of a satellite in a circular orbit and the distance from the center of the mass it is orbiting around, given the modified gravitational force formula.

### Step-by-Step Solution:

1. **Understanding the Gravitational Force**:
The gravitational force between two point masses is given by:
\[ F = \frac{G m_1 m_2}{r^n} \]
where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the two masses.

2. **Centripetal Force Requirement**:
For a satellite in circular motion, the gravitational force provides the necessary centripetal force. The centripetal force required to keep a mass \( m_2 \) moving in a circle of radius \( r \) with speed \( v \) is:
\[ F_{centripetal} = \frac{m_2 v^2}{r} \]

3. **Setting the Forces Equal**:
Setting the gravitational force equal to the centripetal force gives us:
\[ \frac{G m_1 m_2}{r^n} = \frac{m_2 v^2}{r} \]
We can cancel \( m_2 \) from both sides (assuming \( m_2 \neq 0 \)):
\[ \frac{G m_1}{r^n} = \frac{v^2}{r} \]

4. **Rearranging the Equation**:
Rearranging this equation, we get:
\[ v^2 = \frac{G m_1}{r^{n-1}} \]

5. **Finding the Period (T)**:
The period \( T \) of the satellite is related to its speed \( v \) and the radius \( r \) of the orbit:
\[ T = \frac{2\pi r}{v} \]
Substituting \( v \) from the previous step, we have:
\[ T = \frac{2\pi r}{\sqrt{\frac{G m_1}{r^{n-1}}}} = 2\pi r \sqrt{\frac{r^{n-1}}{G m_1}} = 2\pi \sqrt{\frac{r^n}{G m_1}} \]

6. **Identifying Proportionality**:
From the equation \( T = 2\pi \sqrt{\frac{r^n}{G m_1}} \), we can see that the period \( T \) is proportional to \( r^{n/2} \):
\[ T \propto r^{n/2} \]

### Final Answer:
The period of a satellite in a circular orbit will be proportional to \( r^{n/2} \).