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 will use Kepler's Third Law of planetary motion, which relates the time period of revolution of a planet to the dimensions of its orbit.

### Step-by-Step Solution:

1. **Understanding the Orbit**:
The planet revolves around the Sun in an elliptical orbit characterized by its major axis (y) and minor axis (x). For our calculations, we will focus on the major axis.

2. **Applying Kepler's Third Law**:
According to Kepler's Third Law, 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:
\[ T^2 \propto a^3 \]
In the case of an ellipse, the semi-major axis \( a \) is half of the major axis \( y \):
\[ a = \frac{y}{2} \]

3. **Substituting the Semi-Major Axis**:
Substituting \( a \) into Kepler's Third Law gives:
\[ T^2 \propto \left(\frac{y}{2}\right)^3 \]

4. **Simplifying the Equation**:
This can be simplified as:
\[ T^2 \propto \frac{y^3}{2^3} \]
Since \( 2^3 \) is a constant, we can write:
\[ T^2 \propto y^3 \]

5. **Finding the Time Period**:
Taking the square root of both sides, we find:
\[ T \propto y^{3/2} \]

6. **Conclusion**:
Thus, the time period \( T \) of revolution is proportional to \( y^{3/2} \).

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