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. (a)`T^(2//3)`
B. (b)`T^(1//3)`
C. (c)`T^(4//3)`
D. (d)`T^(1//2)`

Answer

To solve the problem, we need to establish the relationship between the area of the circular orbit and the period \( T \) of the Earth's revolution around the Sun.

### Step-by-Step Solution:

1. **Understanding the Circular Orbit**:
- The Earth revolves around the Sun in a circular orbit with radius \( r \).
- The area \( A \) of a circular orbit is given by the formula:
\[ A = \pi r^2 \]

2. **Applying 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 (or radius \( r \) for a circular orbit):
\[ T^2 \propto r^3 \]
- This can be expressed as:
\[ T^2 = k r^3 \]
where \( k \) is a constant of proportionality.

3. **Rearranging Kepler's Law**:
- From the equation \( T^2 = k r^3 \), we can express \( r \) in terms of \( T \):
\[ r = \left(\frac{T^2}{k}\right)^{1/3} \]

4. **Finding the Area in Terms of Period**:
- Substitute \( r \) back into the area formula:
\[ A = \pi r^2 = \pi \left(\left(\frac{T^2}{k}\right)^{1/3}\right)^2 \]
- This simplifies to:
\[ A = \pi \left(\frac{T^2}{k}\right)^{2/3} = \frac{\pi}{k^{2/3}} T^{4/3} \]

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

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