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A planet revolves around the sun in elli...

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 ends of the minor axis and major axis close to the sun is

A

1. `(T pi)/(2e)`

B

2. `T((2e)/(pi)-1)`

C

2. `(Te)/(2pi)`

D

4. `T((1)/(4)-( e)/(2pi))`

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To solve the problem of finding the time spent by a planet between the ends of the minor axis and the major axis close to the Sun in its elliptical orbit, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Elliptical Orbit**: - An elliptical orbit has a major axis and a minor axis. The Sun is located at one of the foci of the ellipse. 2. **Define Points**: - Let \( O \) be the center of the ellipse, \( P \) be the point on the major axis closest to the Sun, and \( L \) be the end of the minor axis. 3. **Kepler's Second Law**: - According to Kepler's second law, the line segment joining a planet to 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. 4. **Calculate Total Area of the Ellipse**: - The area \( A \) of the ellipse is given by the formula: \[ A = \pi a b \] where \( a \) is the semi-major axis and \( b \) is the semi-minor axis. 5. **Area Covered in One Revolution**: - The planet takes a time period \( T \) to cover the entire area \( A \). 6. **Area of Interest**: - We need to find the area between points \( P \) and \( L \). This area can be determined by calculating the area of triangle \( SOL \) and subtracting it from the area \( OPL \). 7. **Area of Triangle \( SOL \)**: - The area of triangle \( SOL \) can be calculated using the formula: \[ \text{Area}_{SOL} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (ae) \times b \] where \( e \) is the eccentricity of the ellipse. 8. **Area of Sector \( OPL \)**: - The area of sector \( OPL \) is one-fourth of the total area of the ellipse: \[ \text{Area}_{OPL} = \frac{1}{4} \times \pi a b \] 9. **Calculate the Area Between \( P \) and \( L \)**: - The area between \( P \) and \( L \) is: \[ \text{Area}_{PL} = \text{Area}_{OPL} - \text{Area}_{SOL} \] Substituting the values: \[ \text{Area}_{PL} = \frac{1}{4} \pi ab - \frac{1}{2} ae b \] 10. **Relate Area to Time**: - Since the area swept out is proportional to time, we can set up the relationship: \[ \frac{\text{Area}_{PL}}{T} = \frac{\text{Area}_{OPL}}{T} \] 11. **Calculate Time \( t \)**: - The time \( t \) spent between points \( P \) and \( L \) can be expressed as: \[ t = T \cdot \frac{\text{Area}_{PL}}{\text{Total Area}} = T \cdot \frac{\frac{1}{4} \pi ab - \frac{1}{2} ae b}{\pi ab} \] - Simplifying this gives: \[ t = T \cdot \left( \frac{1}{4} - \frac{e}{2} \right) \] ### Final Result: The time spent by the planet between the ends of the minor axis and major axis close to the Sun is: \[ t = \frac{T}{4} - \frac{Te}{2} \]

To solve the problem of finding the time spent by a planet between the ends of the minor axis and the major axis close to the Sun in its elliptical orbit, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Elliptical Orbit**: - An elliptical orbit has a major axis and a minor axis. The Sun is located at one of the foci of the ellipse. 2. **Define Points**: ...
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