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 of finding the angular momentum \( L \) of a satellite revolving around the Earth in a circular orbit of radius \( r \) with speed \( v \), we can follow these steps:

### Step 1: Understand the definition of angular momentum
The angular momentum \( L \) of an object moving in a circular path is given by the formula:
\[ L = r \times p \]
where \( p \) is the linear momentum of the object, and \( r \) is the radius of the circular path.

### Step 2: Express linear momentum
The linear momentum \( p \) of the satellite can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the satellite and \( v \) is its speed.

### Step 3: Substitute linear momentum into the angular momentum formula
Substituting the expression for linear momentum into the angular momentum formula, we have:
\[ L = r \times mv = mvr \]

### Step 4: Find the expression for speed \( v \)
For a satellite in a circular orbit, the gravitational force provides the necessary centripetal force. The gravitational force acting on the satellite is given by:
\[ F = \frac{GMm}{r^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. The centripetal force required to keep the satellite in circular motion is:
\[ F = \frac{mv^2}{r} \]
Setting these two forces equal gives:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]

### Step 5: Simplify 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 \( r \):
\[ \frac{GM}{r} = v^2 \]
Taking the square root of both sides gives:
\[ v = \sqrt{\frac{GM}{r}} \]

### Step 6: Substitute \( v \) back into the angular momentum equation
Now we substitute \( v \) back into the angular momentum equation:
\[ L = mvr = m \left(\sqrt{\frac{GM}{r}}\right) r \]
This simplifies to:
\[ L = m \sqrt{GM} \sqrt{r} \]

### Step 7: Conclusion
From the final expression, we can conclude that the angular momentum \( L \) is directly proportional to the square root of the radius \( r \):
\[ L \propto \sqrt{r} \]