# Central Orbit problem

## 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) $$\ T(\frac{2\pi}{e})$$

## 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?

## Answers and Replies

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nrqed
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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) $$\ T(\frac{2\pi}{e})$$

## 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...