An artificial satellite relolves around a planet for which gravitational force (F) varies with distance r from its centres as `F prop r^(2)` If `v_(0)` its orbital speed , then

Options

A. `v_(0) prop r^(-1//2)`
B. `v_(0)prop r^(3//2)`
C. `v_(0) prop r^(3//2)`
D. `v_(0) prop r`

Answer

To solve the problem step by step, we will analyze the relationship between the gravitational force acting on the satellite and its orbital speed.

### Step-by-Step Solution:

1. **Understanding the Gravitational Force**:
The problem states that the gravitational force \( F \) varies with distance \( r \) from the center of the planet as:
\[ F \propto r^2 \]
This means we can express the gravitational force as:
\[ F = k \cdot r^2 \]
where \( k \) is a constant of proportionality.

2. **Centripetal Force Requirement**:
For a satellite in circular motion, the gravitational force provides the necessary centripetal force to keep the satellite in orbit. The centripetal force \( F_c \) is given by:
\[ F_c = \frac{m v_0^2}{r} \]
where \( m \) is the mass of the satellite and \( v_0 \) is its orbital speed.

3. **Equating Forces**:
Since the gravitational force is providing the centripetal force, we can set the two forces equal to each other:
\[ k \cdot r^2 = \frac{m v_0^2}{r} \]

4. **Rearranging the Equation**:
To find an expression for \( v_0^2 \), we can rearrange the equation:
\[ k \cdot r^3 = m v_0^2 \]
Thus, we can express \( v_0^2 \) as:
\[ v_0^2 = \frac{k}{m} \cdot r^3 \]

5. **Finding Orbital Speed**:
Taking the square root of both sides gives us the expression for the orbital speed \( v_0 \):
\[ v_0 = \sqrt{\frac{k}{m}} \cdot r^{3/2} \]

6. **Identifying the Relationship**:
Since \( \frac{k}{m} \) is a constant, we can express \( v_0 \) in terms of \( r \):
\[ v_0 \propto r^{3/2} \]
This indicates that the orbital speed \( v_0 \) varies with the distance \( r \) from the center of the planet as \( r^{3/2} \).

### Conclusion:
The correct relationship is:
\[ v_0 \propto r^{3/2} \]