Central Orbit Problem: Time Taken at Perihelion Revealed

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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<e<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<e<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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The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
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