When I first heard that there is no easy exact formula for the circumference of the standard ellipse, I found it hard to believe. I had already learned that a planet traces out an ellipse with the Sun at one of the foci. And that the orbital period is proportional to the three-halves power of the semi-major axis of the orbit,![](https://upload.wikimedia.org/wikipedia/commons/thumb/4/4d/Ellipse-param.svg/440px-Ellipse-param.svg.png)
where:
- a: semi-major axis,
- b: semi-minor axis.
This is Kepler's third law of planetary motion. It gives this easy equation for the period of the orbit:
![](https://wikimedia.org/api/rest_v1/media/math/render/svg/84a962a4e5da8183f6a38ef23d0a5a9452667148)
where
This ignores relativistic effects, of course, so the orbit is and exact ellipse.
If the elliptical period is so simple, then why is the circumference of the orbit a problem? After all, Kepler's second orbital law also states that the planet sweeps out a constant area per unit time, as observed in Sun-centered coordinates.
This is the consequence of the constancy (conservation) of angular momentum around an orbit, so for each position on the ellipse perimeter, it is trivial to calculate the orbital velocity. So we have velocity and we have time - the distance that the planet travel should also be trivial. Or not?
The precise value for the perimeter length (circumference) is given by the following infinite series: (Wikipedia Ellipse Circumference)
![](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e225de865dc476c19e57cf9278cecae8fff3998)
where !! is the double factorial.
This series converges, but by expanding in terms of
h = ( a − b ) 2 / ( a + b ) 2 ,
James Ivory and Bessel derived an expression that converges much more rapidly:
![](https://wikimedia.org/api/rest_v1/media/math/render/svg/be763d0ca06c104f63d304145e65701b57384811)
I leave you with the puzzle: why is the period and the velocity of the elliptical orbit trivial, but the precise perimeter length so complex?
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