If L is the angular momentum of a satellite revolving around earth is a circular orbit of radius r with speed v , then

Options

A. `L prop v`
B. ` L prop r`
C. `L prop sqrt(r )`
D. `L prop sqrt(v)`

Answer

To solve the problem, we need to derive the relationship between the angular momentum \( L \) of a satellite in a circular orbit and the radius \( r \) of that orbit.

### Step-by-Step Solution:

1. **Understanding Angular Momentum:**
The angular momentum \( L \) of a satellite moving in a circular orbit can be expressed as:
\[ L = mvr \]
where \( m \) is the mass of the satellite, \( v \) is its orbital speed, and \( r \) is the radius of the orbit.

2. **Equating Gravitational Force and Centripetal Force:**
For a satellite in a circular orbit, the gravitational force provides the necessary centripetal force to keep the satellite in orbit. Thus, we have:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

3. **Simplifying the Equation:**
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r^2 \) gives:
\[ GM = v^2 r \]
Rearranging this gives:
\[ v^2 = \frac{GM}{r} \]

4. **Finding the Relationship Between \( v \) and \( r \):**
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{GM}{r}} \]
This shows that the speed \( v \) is inversely proportional to the square root of the radius \( r \):
\[ v \propto \frac{1}{\sqrt{r}} \]

5. **Substituting \( v \) Back into the Angular Momentum Equation:**
Now, substituting this expression for \( v \) back into the angular momentum equation:
\[ L = mvr = m \left(\sqrt{\frac{GM}{r}}\right) r \]
This simplifies to:
\[ L = m \sqrt{GM} \sqrt{r} \]

6. **Final Relationship:**
From this equation, we can see that:
\[ L \propto \sqrt{r} \]
Thus, the angular momentum \( L \) is directly proportional to the square root of the radius \( r \) of the circular orbit.

### Conclusion:
The correct relationship derived is:
\[ L \propto \sqrt{r} \]
This means that as the radius of the orbit increases, the angular momentum of the satellite also increases proportionally to the square root of that radius.