Kepler's third law or periods affirms that:

"The squares of the times that the planets use to cover their orbits are proportional to the cube of their average distances from the Sun".

font from as an example https://it.wikipedia.org/wiki/Leggi_di_Keplero

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(the first definition) and

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from the English book PHYSICS, James Walker, 5^ edition

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I write $r=\mathrm{d}(\text{Planet,Sun})$ and $r_i$ for $i=1,\ldots n$, are the radius vectors of the planet when it moves during its period of revolution around the Sun. I have written only $r_1, r_2$ and $r_3$. Considering that in the starting definition we speak of average distances, is it possible to write

$$\frac{T^2}{\langle r\rangle^3}=\text{constant}\tag 1$$

where I indicate the arithmetic average of the distances of a planet from the Sun when it travels through its elliptical orbit?

For example we have an equation of an canonical ellipse,

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ where $a$ is the major semi-axis, $b$ the minor semi-axis with $a>b>0$. Supposing to keep the numerator constant in the $(1)$ if I take just three distances $r_1$, $r_2$ and $r_3$ and I consider, using for example, Geogebra with a drawing

$$\langle r \rangle=\frac{r_1+r_2+r_3}{3}\approx a \tag 2$$

If this approach is meaningful then I can also write, with good approximation, that

$$\frac{T^2}{a^3}=\text{constant}\tag 3$$

So the $(3)$ is justified by the $(1)$. But in almost all books in Italian language of an high school books, the first definition is not given, but it is written that

The ratio between the square of the revolution period and the cube of the semi-axis major of the orbit is the same for all planets.

My question is:

Is there a correlation of average distances $\langle r \rangle$ with the $a$ or $\langle r \rangle\equiv a$?

Any answer is welcome and I hope with a lot of serenity.

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    $\begingroup$ Your statement of Kepler's 3rd law is incorrect. Instead of average distances from the Sun it should be semi-major axis of its orbit. $\endgroup$ – Bill N Oct 4 '20 at 16:43
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    $\begingroup$ There are multiple ways to define the average of a continuously varying quantity. The two most obvious for $\langle r\rangle$ are an average over the angle around the ellipse or an average over time. Neither of these gives the semimajor axis $a$. Without stating what average you are usng, your first statement is meaningless. $\endgroup$ – G. Smith Oct 4 '20 at 16:57
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    $\begingroup$ One “average” which does work is averaging just two specific distances: that at aphelion and that at perihelion. $\endgroup$ – G. Smith Oct 4 '20 at 17:03
  • $\begingroup$ @G.Smith "One “average” which does work is averaging just two specific distances: that at aphelion and that at perihelion". yes it is correct the $\langle r \rangle=a$ without approximation. Do you, please, give an answer to delete my doubts. What is the reason for continue downvotes and I trust you with such sincerity that it is humiliating. When someone is looking for help and is doing everything possible to be as clear as possible can not always be clubbed. $\endgroup$ – Sebastiano Oct 4 '20 at 18:45

Is there a correlation of average distances $\langle r \rangle$ with the $a$ or $\langle r \rangle\equiv a$?

Since $r$ changes continuously, most people would assume that $\langle r \rangle$ means either a continuous average over all angles $\theta$ around the ellipse,

$$\langle r \rangle_\theta\equiv\frac{1}{2\pi}\int_0^{2\pi}r(\theta)d\theta,\tag1$$

or a continuous time average over one period $T$ of the orbit,

$$\langle r \rangle_t\equiv\frac{1}{T}\int_0^T r(t)dt.\tag2$$

Let's calculate these two averages.

The elliptical orbit is given by


where $a$ is the semimajor axis and $e$ the eccentricity. Substituting this into (1) and doing the integral gives

$$\langle r \rangle_\theta\equiv\frac{a(1-e^2)}{2\pi}\int_0^{2\pi}\frac{d\theta}{1-e\cos\theta}=\frac{a(1-e^2)}{2\pi}\frac{2\pi}{\sqrt{1-e^2}}=a\sqrt{1-e^2}.\tag4$$

So the angular average is not equal to $a$; it is less than $a$.

To compute the time average, it is easiest to turn it into another integral over $\theta$ by writing it as

$$\langle r \rangle_t=\frac{1}{T}\int_0^{2\pi}\frac{r(\theta)d\theta}{\dot\theta}.\tag5$$

where the overdot means a time derivative.

To evaluate this, use Kepler's Second Law, which says that

$$\frac{dA}{dt}=\frac12r^2\dot\theta=\text{const}=\frac{A}{T}=\frac{\pi ab}{T}=\frac{\pi a^2\sqrt{1-e^2}}{T}\tag6$$

(here $b=a\sqrt{1-e^2}$ is the semiminor axis)


$$\dot\theta=\frac{2\pi a^2\sqrt{1-e^2}}{T}\frac{1}{r^2}.\tag7$$

Putting (7) into (5), we get

$$\langle r \rangle_t=\frac{1}{2\pi a^2\sqrt{1-e^2}}\int_0^{2\pi}r(\theta)^3d\theta.\tag8$$

Putting (3) into (8) and doing the integral, we get

$$\begin{align}\langle r \rangle_t&=\frac{a(1-e^2)^{5/2}}{2\pi}\int_0^{2\pi}\frac{d\theta}{(1-e\cos\theta)^3}=\frac{a(1-e^2)^{5/2}}{2\pi}\frac{(2+e^2)\pi}{(1-e^2)^{5/2}}\\&=a\left(1+\frac12e^2\right).\tag9\end{align}$$

So the time average of $r$ is not equal to $a$; it is greater than $a$.

Thus neither the continuous angular average of $r$ nor the continuous time average of $r$ is equal to $a$.

The way in which to understand $a$ as an "average" distance is simply as a discrete average of $r$ at two particular points on the orbit, namely aphelion and perihelion:


P.S. I did the two integrals with Mathematica. One way to do them by hand is to turn them into contour integrals around the unit circle in the complex plane and evaluate them using residues.

  • $\begingroup$ Because in each case $\langle r \rangle = af(e)$, and eccentricity varies from one elliptical orbit to another, both $T^2/a^3 = $constant and $T^2/ \langle r \rangle^3 = $constant can't be true, right? $\endgroup$ – Brain Stroke Patient Oct 5 '20 at 6:57
  • $\begingroup$ @BrainStrokePatient Right, but if you define $\langle r \rangle\equiv r_\text{max}+r_\text{min}=a$ then they are equal. $\endgroup$ – G. Smith Oct 5 '20 at 13:25

It is not an approximation to write $T^2 \propto a^3$, it is the actual Kepler's law. There are many ways to show it, but a simple one is to start from his area law to write: $$\mathcal{A} = \pi a b = \frac{LT}{2m}$$

On the other hand, one can show using Newton's third law that the parameter $p=b^2/a$ of the ellipsis is given by: $$p = \frac{L^2}{GMm}$$

Combining these two to eliminate $L$, we obtain: $$\frac{a^3}{T^2} = \frac{GM}{4\pi^2} = \mathrm{const}$$

  • $\begingroup$ In the meantime thank you very very much. Do you start using the conservation of the angular momentum and equals it to the area of the ellipse? What is $p$ and I not rembember the reason $\pi ab=LT/2m$? I should use the law of universal gravitation but Kepler's laws in my book are obviously written before. I really appreciate your effort. $\endgroup$ – Sebastiano Oct 4 '20 at 12:05
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    $\begingroup$ @Sebastiano $p$ is the radius at 90° before/after the perihelion. $\endgroup$ – Thomas Fritsch Oct 4 '20 at 12:20
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    $\begingroup$ I start using Kepler's second law, the area law, which is a consequence of the conservation of angular momentum indeed. In the particular case where we consider a full period $T$, the area covered is the area of the ellipse $\mathcal{A}$, which is equal to $\pi ab$ by geometry $\endgroup$ – Emmy Oct 4 '20 at 18:20

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