A planet revolves round the sun in an elliptical orbit of minor and major axes x and y respectively . Then the time period of revolution is proportional to

Options

A. `(x+y)^(3//2)`
B. `(y-x)^(3//2)`
C. `x^(3//2)`
D. `y^(3//2)`

Answer

To solve the problem, we need to determine how the time period of revolution of a planet around the sun is related to the lengths of the semi-major and semi-minor axes of its elliptical orbit.

### Step-by-Step Solution:

1. **Understanding the Elliptical Orbit**:
- The orbit of the planet is elliptical, characterized by two axes: the semi-major axis (y) and the semi-minor axis (x). The semi-major axis is the longer axis of the ellipse, while the semi-minor axis is the shorter one.

2. **Kepler's Third Law**:
- According to Kepler's Third Law of planetary motion, the square of the time period (T) of a planet's orbit is directly proportional to the cube of the semi-major axis (a) of its orbit. This can be mathematically expressed as:
\[ T^2 \propto a^3 \]
- In our case, the semi-major axis is represented by \( \frac{y}{2} \) (since the semi-major axis is half of the major axis).

3. **Applying Kepler's Law**:
- Therefore, we can write:
\[ T^2 \propto \left(\frac{y}{2}\right)^3 \]
- Simplifying this, we have:
\[ T^2 \propto \frac{y^3}{8} \]
- This shows that \( T^2 \) is proportional to \( y^3 \).

4. **Finding the Proportionality**:
- Since we are interested in the relationship between the time period and the semi-major axis, we can express the time period \( T \) as:
\[ T \propto y^{3/2} \]
- This indicates that the time period of revolution is proportional to the semi-major axis raised to the power of \( \frac{3}{2} \).

5. **Conclusion**:
- The final answer is that the time period of revolution \( T \) is proportional to \( y^{3/2} \).

### Final Answer:
The time period of revolution is proportional to \( y^{3/2} \).