<p>It is easy enough to do the calculations, formulae for the in and circum radii of the Platonic solids can be found <a href="http://whistleralley.com/polyhedra/platonic.htm">here</a> which give ratios of circ to in radii of (note formulea for the radii have dropped common factor of the side length, which we don't need as we are interested in the ratios):</p>

<pre><code>&gt;ri4=sqrt(6)/12,rc4=sqrt(6)/4
     0.204124 
     0.612372 
&gt;
&gt;ri6=1/2,rc6=sqrt(3)/2
          0.5 
     0.866025 
&gt;
&gt;ri8=sqrt(6)/6,rc8=sqrt(2)/2
     0.408248 
     0.707107 
&gt;
&gt;ri12=sqrt(250+110*sqrt(5))/20,rc12=(sqrt(15)+sqrt(3))/4
      1.11352 
      1.40126 
&gt;
&gt;ri20=(3*sqrt(3)+sqrt(15))/12,rc20=sqrt(10+2*sqrt(5))/4
     0.755761 
     0.951057 
&gt;
&gt;rho4=rc4/ri4
            3 
&gt;rho6=rc6/ri6
      1.73205 
&gt;rho8=rc8/ri8
      1.73205 
&gt;rho12=rc12/ri12
      1.25841 
&gt;rho20=rc20/ri20
      1.25841 
</code></pre>

<p>Which may be compared with the orbital radius ratios from <a href="https://solarsystem.nasa.gov/planets/charchart.cfm">here</a> (radii in km)</p>

<pre><code>&gt;RMecury=57.9e6;
&gt;RVenus=108.2e6;
&gt;REarth=149.6e6;
&gt;RMars=227.9e6;
&gt;RJupiter=778.3e6;
&gt;RSaturn=1426.7e6;
</code></pre>

<p>Now we can compare the corresponding radii ratios:</p>

<pre><code>&gt;[RVenus/RMecury,rho8]
      1.86874       1.73205 
&gt;[REarth/RVenus,rho20]
      1.38262       1.25841 
&gt;[RMars/REarth,rho12]
       1.5234       1.25841 
&gt;[RJupiter/RMars,rho4]
      3.41509             3 
&gt;[RSaturn/RJupiter,rho6]
       1.8331       1.73205 
</code></pre>

<p>Which, as these things go, is not bad.</p>

CC BY-SA 3.0
2015-05-31T04:30:11.447Z
I think you in the end wrote the value for rho20 in the place of rho12 which should be 1.4725 (to be compared with true 1.5234). So the errors were 3% to 12%, that's a good guess without stuff like physics or telescopes. But is was hardly helpful for his later work that it fits better for the most eccentric planet orbits.
LocalFluff
CC BY-SA 3.0
2015-05-31T05:07:50.530Z
I would like to see your calculations as repeating mine using independent method gives the same result as the main post, unfortunately this is not a suitable medium to sort out such a numerical disagreement.
Conrad Turner
CC BY-SA 3.0
2015-05-31T05:13:24.260Z
I use your calculations. Look at your last box, you copy/pasted 1.25841 twice. rho12 should be 1.40126 as you write higher up.
LocalFluff
CC BY-SA 3.0
2015-05-31T05:26:24.963Z
No I did not copy and past the same value twice, that is what the calculation gave for rho12. I have also repeated the rho12 calculation using Maxima and both the formulae from the reference above and from [here](https://en.wikipedia.org/wiki/Platonic_solid#Radii.2C_area.2C_and_volume) and get the same result. (I'm quite prepared to believe there is an arithmetical error somewhere as that is the nature of the beast, but I can't reproduce the one you are reporting)
Conrad Turner
CC BY-SA 3.0
2015-05-31T05:34:26.167Z
Indeed, rho12 and rho20 doesn't differ until the 6th decimal. Sorry for my confusion.
LocalFluff
CC BY-SA 3.0
2015-05-30T17:38:29.683Z
2015-05-30T18:27:57.433Z
2015-05-30T18:27:57.433Z
Conrad Turner
<p>In <a href="https://en.wikipedia.org/wiki/Mysterium_Cosmographicum#Shapes_and_the_planets" rel="nofollow">Mysterium Cosmographicum (1596)</a> Johannes Kepler proposed that the relative distances between the orbits of the six ancient planets (six because heliocentrism had recently added Earth as one of the planet) correspond to the geometry of the five Platonic solids. Each planetary orbit was assumed to be the grand circle of a sphere. Each Platonic solid would circumscribe an inner planet's orbitally defined sphere, and in same position inscribe the next outer planet's "orbital sphere".</p>

<ul>
<li>How well does it fit?</li>
<li>Is it somehow mathematically related to the equally spurious Bode's law?</li>
<li>I would appreciate some hints or links to the geometric calculation itself, for me to compare with my own exercise trying to solve part of it.</li>
</ul>

<p>Johannes Kepler proposed this order of matching planets with Platonics (I suppose because this order gives the best fit):</p>

<p>Mercury &lt;-- <em>planet</em></p>

<p><strong>octahedron</strong> &lt;-- <em>Platonic solid</em></p>

<p>Venus</p>

<p><strong>icosahedron</strong></p>

<p>Earth</p>

<p><strong>dodecahedron</strong></p>

<p>Mars</p>

<p><strong>tetrahedron</strong></p>

<p>Jupiter</p>

<p><strong>cube</strong></p>

<p>Saturn</p>

<solar-system><history><johannes-kepler>
How well do planetary orbits fit with Johannes Kepler's in- & circumscribed Platonic solids?
LocalFluff
# How well do planetary orbits fit with Johannes Kepler's in- & circumscribed Platonic solids?

(Asked by: username_0 on 2015-05-30)

In [Mysterium Cosmographicum (1596)](https://en.wikipedia.org/wiki/Mysterium_Cosmographicum#Shapes_and_the_planets) Johannes Kepler proposed that the relative distances between the orbits of the six ancient planets (six because heliocentrism had recently added Earth as one of the planet) correspond to the geometry of the five Platonic solids. Each planetary orbit was assumed to be the grand circle of a sphere. Each Platonic solid would circumscribe an inner planet's orbitally defined sphere, and in same position inscribe the next outer planet's "orbital sphere".

* How well does it fit?
* Is it somehow mathematically related to the equally spurious Bode's law?
* I would appreciate some hints or links to the geometric calculation itself, for me to compare with my own exercise trying to solve part of it.

Johannes Kepler proposed this order of matching planets with Platonics (I suppose because this order gives the best fit):

Mercury <-- *planet*

**octahedron** <-- *Platonic solid*

Venus

**icosahedron**

Earth

**dodecahedron**

Mars

**tetrahedron**

Jupiter

**cube**

Saturn

**Tags:** solar-system, history, johannes-kepler

---

## 1 answers

### Answer by username_1 on 2015-05-30. Score: 5

It is easy enough to do the calculations, formulae for the in and circum radii of the Platonic solids can be found here which give ratios of circ to in radii of (note formulea for the radii have dropped common factor of the side length, which we don't need as we are interested in the ratios):
>ri4=sqrt(6)/12,rc4=sqrt(6)/4
     0.204124 
     0.612372 
>
>ri6=1/2,rc6=sqrt(3)/2
          0.5 
     0.866025 
>
>ri8=sqrt(6)/6,rc8=sqrt(2)/2
     0.408248 
     0.707107 
>
>ri12=sqrt(250+110*sqrt(5))/20,rc12=(sqrt(15)+sqrt(3))/4
      1.11352 
      1.40126 
>
>ri20=(3*sqrt(3)+sqrt(15))/12,rc20=sqrt(10+2*sqrt(5))/4
     0.755761 
     0.951057 
>
>rho4=rc4/ri4
            3 
>rho6=rc6/ri6
      1.73205 
>rho8=rc8/ri8
      1.73205 
>rho12=rc12/ri12
      1.25841 
>rho20=rc20/ri20
      1.25841 

Which may be compared with the orbital radius ratios from here (radii in km)
>RMecury=57.9e6;
>RVenus=108.2e6;
>REarth=149.6e6;
>RMars=227.9e6;
>RJupiter=778.3e6;
>RSaturn=1426.7e6;

Now we can compare the corresponding radii ratios:
>[RVenus/RMecury,rho8]
      1.86874       1.73205 
>[REarth/RVenus,rho20]
      1.38262       1.25841 
>[RMars/REarth,rho12]
       1.5234       1.25841 
>[RJupiter/RMars,rho4]
      3.41509             3 
>[RSaturn/RJupiter,rho6]
       1.8331       1.73205 

Which, as these things go, is not bad.

> **Comments:**
> **username_0**: I think you in the end wrote the value for rho20 in the place of rho12 which should be 1.4725 (to be compared with true 1.5234). So the errors were 3% to 12%, that's a good guess without stuff like physics or telescopes. But is was hardly helpful for his later work that it fits better for the most eccentric planet orbits.  
> **username_1**: I would like to see your calculations as repeating mine using independent method gives the same result as the main post, unfortunately this is not a suitable medium to sort out such a numerical disagreement.  
> **username_0**: I use your calculations. Look at your last box, you copy/pasted 1.25841 twice. rho12 should be 1.40126 as you write higher up.  
> **username_1**: No I did not copy and past the same value twice, that is what the calculation gave for rho12. I have also repeated the rho12 calculation using Maxima and both the formulae from the reference above and from [here](https://en.wikipedia.org/wiki/Platonic_solid#Radii.2C_area.2C_and_volume) and get the same result. (I'm quite prepared to believe there is an arithmetical error somewhere as that is the nature of the beast, but I can't reproduce the one you are reporting)  
> **username_0**: Indeed, rho12 and rho20 doesn't differ until the 6th decimal. Sorry for my confusion.  

---

