Central Orbit Problem: Time Taken at Perihelion Revealed

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The discussion centers on determining the time taken by a planet in an elliptical orbit to travel from the end of the minor axis to perihelion, given its eccentricity \( e \) and time period \( T \). Options for the time duration include various mathematical expressions involving \( T \) and \( e \). The participants eliminate options (a) and (d) based on the condition \( 0 < e < 1 \) and lean towards option (b) as it suggests a time less than one-fourth of \( T \). A special case of a circular orbit (where \( e = 0 \)) is referenced to support the reasoning for option (b) being the most plausible answer. The conversation highlights the complexity of deriving a solution while relying on elimination and special cases for clarity.
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Homework Statement



A planet revolves around the sun in an elliptical orbit of eccentricity \ e and time period \ T.The time taken by the planet between the end of the minor axis and at perihelion should be:

(a) \ T(\frac{e}{2\pi}-\frac{1}{2})

(b) \ T(\frac{1}{4}-\frac{e}{2\pi})

(c) \ T(\frac{e}{2\pi})

(d) \ T(\frac{2\pi}{e})


Homework Equations





The Attempt at a Solution



Since \ 0&lt;e&lt;1, we must have (a) and (d) eliminated.

The anser is likely (b) as it correctly predicts that it should take time less than one fourth of T...But,I am not sure...

Any better reasoning?and what should be the answer?
 
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neelakash said:

Homework Statement



A planet revolves around the sun in an elliptical orbit of eccentricity \ e and time period \ T.The time taken by the planet between the end of the minor axis and at perihelion should be:

(a) \ T(\frac{e}{2\pi}-\frac{1}{2})

(b) \ T(\frac{1}{4}-\frac{e}{2\pi})

(c) \ T(\frac{e}{2\pi})

(d) \ T(\frac{2\pi}{e})


Homework Equations





The Attempt at a Solution



Since \ 0&lt;e&lt;1, we must have (a) and (d) eliminated.

The anser is likely (b) as it correctly predicts that it should take time less than one fourth of T...But,I am not sure...

Any better reasoning?and what should be the answer?

Other than solving completely the problem (which would be quite complicated), this is the only way to do it (by elimination)

Maybe a better way to see that it is the correct answer is to consider the special case of a circular orbit (e=0). Then the correct answer is obvious.
 
exactly...
 
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