The average radius of an ellipse

Hints:

The polar equation of the ellipse with one focus at the origin and the other at the positive real axis ($e = $ eccentricity) is:

$$r = \frac{a(1 - e^2)}{1 - e\cos(\theta)}.$$

And the average radius can be calculated using the integral

$$\frac{1}{2\pi}\int_0^{2\pi}r(\theta)d\theta.$$

Can you continue?

Update: already asked at Astronomy SE. The important fact:

It's the semi-major axis that defines the period, not the average distance.

Bottom line: the "simple average" $\ne$ the integral average. Also interesting (and different): the time average.

Update 2: doing the cov $z = \tan(\theta/2)$, $$\int\frac{1}{1 - e\cos(\theta)}d\theta = \int\frac{1}{1 - e\frac{1 - z^2}{1 + z^2}}\frac{2}{1 + z^2}dz = \int\frac{2}{(1 + e)z^2 + (1 - e)}dz =$$ $$ = \frac{2}{\sqrt{1 - e^2}}\arctan\left(\frac{\sqrt{1 + e}}{\sqrt{1 - e}}z\right) = \frac{2}{\sqrt{1 - e^2}}\arctan\left(\frac{\sqrt{1 + e}}{\sqrt{1 - e}}\tan(\theta/2)\right).$$


@Martin-Blas has already suggested two possible notions of average

  1. Take an average with respect to $\theta$, the central angle

  2. Take an average with respect to time, i.e.,$$ \frac{1}{T}\int_0^T \| u(t) \| dt, $$ where $T$ is the period, and $u(t)$ is the position of the planet at time $t$, and the sun is located at the origin of the coordinate system.

There's a third notion, which is

  1. An average with respect to arclength, i.e., $$ \int_0^L \| u(t) \| ds$$ where $L$ is the total arclength of the ellipse, and $ds = \|u'(t)\| dt$ is arclength integrand.

I'm pretty certain that one could even hoke up some others, but the key thing here is the one mentioned by others: this question doesn't have an answer until you know the measure with respect to which the "average" is being computed.

Tags:

Geometry

Integration

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