Consider a planet moving around a star in an elliptical orbit with period T. Area of elliptical orbit is proportional to

Options

A. `T^(4//3)`
B. `T`
C. `T^(20)`
D. `T^(1//2)`

Answer

To determine how the area of an elliptical orbit is proportional to the time period \( T \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Kepler's Third Law**: According to Kepler's Third Law of planetary motion, the square of the time period \( T \) of a planet is directly proportional to the cube of the semi-major axis \( r \) of its orbit. This can be expressed mathematically as:
\[ T^2 \propto r^3 \]

2. **Expressing Radius in Terms of Time Period**: From the above relationship, we can express the semi-major axis \( r \) in terms of the time period \( T \):
\[ r \propto T^{2/3} \]

3. **Calculating the Area of the Ellipse**: The area \( A \) of an ellipse is given by the formula:
\[ A = \pi r^2 \]
Since we have \( r \) in terms of \( T \), we can substitute \( r \) into this area formula.

4. **Substituting for Radius**: Substituting \( r \propto T^{2/3} \) into the area formula, we have:
\[ A \propto \pi (T^{2/3})^2 \]

5. **Simplifying the Area Expression**: Simplifying the expression gives:
\[ A \propto \pi T^{4/3} \]
Since \( \pi \) is a constant, we can drop it when considering proportionality:
\[ A \propto T^{4/3} \]

6. **Conclusion**: Therefore, we conclude that the area of the elliptical orbit is proportional to the time period raised to the power of \( \frac{4}{3} \):
\[ A \propto T^{4/3} \]

### Final Answer:
The area of the elliptical orbit is proportional to \( T^{4/3} \).