Consider that the Earth is revolving around the Sun in a circular orbit with a period T. The area of the circular orbit is directly proportional to

Options

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

Answer

To solve the problem, we need to determine how the area of the circular orbit of the Earth around the Sun is related to the time period \( T \) of the orbit.

### Step-by-Step Solution:

1. **Understand the Circular Orbit**:
- The Earth revolves around the Sun in a circular orbit. The radius of this orbit can be denoted as \( r \).

2. **Use Kepler's Third Law**:
- According to Kepler's Third Law, the square of the period \( T \) of a planet's orbit is directly proportional to the cube of the semi-major axis \( r \) of its orbit:
\[ T^2 \propto r^3 \]
- This can be expressed as:
\[ T^2 = k \cdot r^3 \]
where \( k \) is a constant.

3. **Express \( r \) in terms of \( T \)**:
- Rearranging the equation gives:
\[ r \propto T^{2/3} \]
- This indicates that the radius \( r \) of the orbit is proportional to the time period \( T \) raised to the power of \( \frac{2}{3} \).

4. **Calculate the Area of the Circular Orbit**:
- The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
- Substituting the expression for \( r \):
\[ A \propto r^2 \propto (T^{2/3})^2 \]
- This simplifies to:
\[ A \propto T^{4/3} \]

5. **Conclusion**:
- Therefore, the area of the circular orbit is directly proportional to the time period \( T \) raised to the power of \( \frac{4}{3} \):
\[ A \propto T^{4/3} \]

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