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I have recently been studying the 3rd Kepler´s law which uses the average radius of an orbit. I would love to know how to find the geometric average radius of an ellipse. Would you please show me or at least help me with the process of integration?

Adam Páltik
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    Please refer to another answer of mine here – Ng Chung Tak Feb 10 '19 at 10:37
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    Beware: in the context of books presenting Kepler's laws, by "average radius" of an elliptic orbit they mean the semi-major axis, that is the average between maximum and minimum distance of a planet from the Sun. That is not what I would call "average radius" from a mathematical point of view. – Intelligenti pauca Feb 10 '19 at 11:09
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    @Aretino But I find it interesting that the average distance weighted by arc length is the same as the semi-major axis. – David K Feb 10 '19 at 15:47

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

Community
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Martín-Blas Pérez Pinilla
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@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.

John Hughes
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It depends on your definition of that average radius means in this context. E.g. if we define average radius of an ellipse as the radius $r$ of a circle which has the same area an an ellipse whose length of the semi-major and semi-minor axis are $a$ and $b$ then $$ Area = \pi r^2 = \pi ab $$

which gives $r = \sqrt {ab}$.

Alternative definition is to literally take the average of $a$ and $b$ or $r = \frac{a+b}{2}$.

Nilotpal Sinha
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