Geometry of the Solar System, part III
by Frank Hoogerbeets

16 January 2017

Orbital Period

Just as in part 2, we define the orbital period as 1 (t = 1). Next we want to know how the orbital period of a given planet n relates to the orbital time of planet 1. We do this in the same way as with the orbital velocity (see formula (2.9) in part 2):

${v}_{n}^{2}=\frac{4\cdot {\mathrm{\pi }}^{2}\cdot {r}_{1}^{3}}{{r}_{n}}$
formula (3.1)

Because we want to integrate time into the equation, we use formula (2.2) of part II but squared:

${v}^{2}=\frac{{S}^{2}}{{t}^{2}}$
formula (3.2)

We know that the distance is the circumference, so we substitute S. For a given planet n the formula will then be:

${v}_{n}^{2}=\frac{4\cdot {\mathrm{\pi }}^{2}\cdot {r}_{n}^{2}}{{t}_{n}^{2}}$
formula (3.3)

We can now set up the matrix so that the orbital period of each planet n will relate to the orbital period of planet 1:

$\frac{4\cdot {\mathrm{\pi }}^{2}\cdot {r}_{n}^{2}}{{t}_{n}^{2}}=\frac{4\cdot {\mathrm{\pi }}^{2}\cdot {r}_{1}^{3}}{{r}_{n}}⇔{t}_{n}^{2}\cdot {r}_{1}^{3}={r}_{n}^{3}⇔{t}_{n}^{2}=\frac{{r}_{n}^{3}}{{r}_{1}^{3}}⇔{t}_{n}=\sqrt{\frac{{r}_{n}^{3}}{{r}_{1}^{3}}}$
formula (3.4)

Finally, we use our key number 15 for the orbital radius of the first planet:

${t}_{n}=\sqrt{\frac{{r}_{n}^{3}}{{15}^{3}}}$
formula (3.5)

Thus we know the orbital period of each planet n, by entering its orbital radius. The values are shown in the table below.

 planet math.orb. radius astr.orb. radius math.orb. velocity astr.orb. velocity math.orb. time astr.orb. time ? 15 - 94.25 - 1.00 - Mercury 60 57.9 47.12 47.87 8.00 7.59 Venus 105 108.2 35.62 35.02 18.52 19.41 Earth 150 149.6 29.80 29.78 31.62 31.56 Mars 240 227.9 23.56 24.13 64.00 59.34 Ceres 420 414.0 17.81 17.94 148.16 145.00 Jupiter 780 778.4 13.07 13.06 374.98 374.49 Saturn 1500 1426.7 9.42 9.64 1000.00 929.90 Uranus 2940 2871.0 6.73 6.79 2744.00 2656.70 Neptune 4380 4498.3 5.52 5.43 4989.70 5205.09 Pluto 5820 5906.2 4.78 4.67 7642.71 7829.72 Eris 11580 10123.4 3.39 3.44 21449.93 17578.92
(table 3.1)

So far we have only worked with numbers and values based on the mathematical model. We have not translated them to any measurement system, like meters and hours. We could use any measurement system in use by any culture as it will not affect how the values relate. Again we use a matrix to do this.

For example, if we want to know how many earth-days it takes for Venus to complete one revolution, we take the orbital periods of these planets and relate them to the 365 days of one earth year:

$\frac{31.62}{18.52}=\frac{365}{x}⇔x=\frac{365\pi 18.52}{31.62}⇔x=213.78$

This is of course based on the mathematical model. The real values as measured by astronomers always differ slightly. In the case of Venus, the measured number of earth-days is about 225, which is slighly longer. This is because the measured orbital radius is a bit longer (108.2) than the methematical number (105).

In the next part, which is currently being revised, I will touch upon gravity and mass and how number 15 gives us a unique insight.

Copyright © 2006-2017 Frank Hoogerbeets, Ditrianum Media Center. You have my permission to copy and distribute this article as long as you do not change its content including this copyright notice.