A planet revolves around the sun in elliptical orbit of eccentricity 'e'. If 'T' is the time period of the planet then the time spent by the planet between the end of the minor axis and close to the sun is

Options

A. `T(1/4-e/(2pi))`
B. `(Te)/(pi)`
C. `(e/(pi)-1)`
D. `(piT)/e`

Explanation

To solve the problem step by step, we will analyze the elliptical orbit of the planet and apply Kepler's laws of planetary motion.

### Step-by-Step Solution:

1. **Understanding the Orbit**:
- The planet revolves around the sun in an elliptical orbit with eccentricity \( e \).
- The semi-major axis is denoted as \( A \) and the semi-minor axis as \( B \).

2. **Kepler's Second Law**:
- According to Kepler's second law, the line joining a planet and the sun sweeps out equal areas in equal times. This means that the area swept out by the planet in a given time is constant.

3. **Total Area of the Ellipse**:
- The total area \( A_{\text{total}} \) of the elliptical orbit is given by:
\[ A_{\text{total}} = \pi \times A \times B \]

4. **Time Period**:
- The time period \( T \) is the time taken for the planet to complete one full revolution around the sun.

5. **Area from Minor Axis to Sun**:
- We need to find the time spent by the planet between the end of the minor axis and the point closest to the sun (the perihelion).
- The area between these two points can be calculated as the area of the sector of the ellipse minus the area of the triangle formed by the minor axis and the line connecting the sun to the perihelion.

6. **Calculating the Area**:
- The area from the end of the minor axis to the perihelion can be expressed as:
\[ A_{\text{desired}} = \text{Area of sector} - \text{Area of triangle} \]
- The area of the sector (which is one-fourth of the total area of the ellipse) is:
\[ A_{\text{sector}} = \frac{1}{4} \times A_{\text{total}} = \frac{1}{4} \times \pi A B \]
- The area of the triangle formed at the perihelion can be calculated using the base and height:
\[ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (A \cdot e) \times B \]

7. **Substituting Values**:
- Now, substituting the areas into the equation:
\[ A_{\text{desired}} = \frac{1}{4} \pi A B - \frac{1}{2} A e B \]

8. **Finding the Time Spent**:
- The time spent \( t \) to cover this area is proportional to the area covered:
\[ \frac{t}{T} = \frac{A_{\text{desired}}}{A_{\text{total}}} \]
- Substituting the areas:
\[ \frac{t}{T} = \frac{\frac{1}{4} \pi A B - \frac{1}{2} A e B}{\pi A B} \]
- Simplifying gives:
\[ t = T \left( \frac{1}{4} - \frac{e}{2} \right) \]

9. **Final Result**:
- Thus, the time spent by the planet between the end of the minor axis and close to the sun is:
\[ t = T \left( \frac{1}{4} - \frac{e}{2} \right) \]