A satellite of mass m moving around the earth of mass `m` in a circular orbit of radius R has angular momentum L. The rate of the area swept by the line joining the centre of the earth and satellite is

Options

A. `2Lm`
B. `L//2m`
C. `sqrt(Lm)`
D. `L//P`

Answer

To solve the problem of finding the rate at which the area is swept by the line joining the center of the Earth and a satellite in circular orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the rate at which the area is swept by the line joining the center of the Earth and the satellite.

2. **Identify the Variables**:
- Let the mass of the satellite be \( m \).
- Let the mass of the Earth also be \( m \) (though typically, the mass of the Earth is much larger).
- Let the radius of the orbit be \( R \).
- Let the angular momentum of the satellite be \( L \).

3. **Angular Speed**:
The angular speed \( \omega \) of the satellite can be defined in terms of the angle \( d\theta \) it sweeps in a small time interval \( dt \):
\[ d\theta = \omega dt \]

4. **Area Swept in a Small Time Interval**:
The area \( dA \) swept by the radius vector in the time interval \( dt \) can be expressed as:
\[ dA = \frac{1}{2} R^2 d\theta \]
Substituting \( d\theta \):
\[ dA = \frac{1}{2} R^2 (\omega dt) \]

5. **Rate of Area Swept**:
To find the rate of area swept \( \frac{dA}{dt} \), we differentiate \( dA \) with respect to \( dt \):
\[ \frac{dA}{dt} = \frac{1}{2} R^2 \omega \]

6. **Relate Angular Momentum to Angular Speed**:
The angular momentum \( L \) of the satellite can be expressed as:
\[ L = I \omega \]
where \( I \) is the moment of inertia. For a point mass in circular motion, \( I = m R^2 \):
\[ L = m R^2 \omega \]
Rearranging gives:
\[ \omega = \frac{L}{m R^2} \]

7. **Substituting Angular Speed into Area Rate**:
Substitute \( \omega \) back into the equation for the rate of area swept:
\[ \frac{dA}{dt} = \frac{1}{2} R^2 \left(\frac{L}{m R^2}\right) \]
Simplifying this gives:
\[ \frac{dA}{dt} = \frac{L}{2m} \]

8. **Final Answer**:
Thus, the rate at which the area is swept by the line joining the center of the Earth and the satellite is:
\[ \frac{dA}{dt} = \frac{L}{2m} \]