A satellite goes along an elliptical path around earth. The rate of change of area swept by the line joining earth and the satellite is proportional to

Options

A. `r^(1/2)`
B. `r`
C. `r^(3/2)`
D. `r^(2)`

Answer

To solve the problem, we need to determine the relationship between the rate of change of area swept by the line joining the Earth and the satellite and the distance between them.

### Step-by-Step Solution:

1. **Understanding the Concept**:
- A satellite moves in an elliptical orbit around the Earth. The line joining the Earth and the satellite sweeps out an area as the satellite moves.

2. **Area Swept by the Satellite**:
- Consider a small segment of the satellite's path where it moves through a small angle \( d\theta \) over a small time interval \( dt \). The distance from the Earth to the satellite is \( r \).

3. **Calculating the Area**:
- The area \( dA \) swept out by the line joining the Earth and the satellite during this small time interval can be approximated as:
\[ dA = \frac{1}{2} r^2 d\theta \]
- This formula comes from the area of a triangle where the two sides are \( r \) and the angle between them is \( d\theta \).

4. **Rate of Change of Area**:
- To find the rate of change of area with respect to time, we differentiate \( dA \):
\[ \frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} \]
- Here, \( \frac{d\theta}{dt} \) is the angular velocity of the satellite.

5. **Proportionality**:
- From the equation \( \frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} \), we can see that the rate of change of area \( \frac{dA}{dt} \) is directly proportional to \( r^2 \) (the square of the distance from the Earth to the satellite).

6. **Conclusion**:
- Therefore, the rate of change of area swept by the line joining the Earth and the satellite is proportional to \( r^2 \).

### Final Answer:
The rate of change of area swept by the line joining the Earth and the satellite is proportional to \( r^2 \).