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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 = 4 π 2 r 1 3 r n v_{n}^{2} = {4 cdot %pi^{2} cdot r_{1}^{3}} over {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 = S 2 t 2 v^{ 2 }= {S^{2}} over {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 = 4 π 2 r n 2 t n 2 v_{n}^{2}={4 cdot %pi^{2} cdot r_{n}^{2}} over {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:

4 π 2 r n 2 t n 2 = 4 π 2 r 1 3 r n t n 2 r 1 3 = r n 3 t n 2 = r n 3 r 1 3 t n = r n 3 r 1 3 {4 cdot %pi^{2} cdot r_{n}^{2}} over {t_{n}^{2}} = {4 cdot %pi^{2} cdot r_{1}^{3}} over {r_{n}} dlrarrow t_{n}^{2} cdot r_{1}^{3}=r_{n}^{3} dlrarrow t_{n}^{2}= {r_{n}^{3}} over {r_{1}^{3}} dlrarrow t_{n}= sqrt{{r_{n}^{3}} over {r_{1}^{3}}}
formula (3.4)

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

t n = r n 3 15 3 t_{n}= sqrt{{r_{n}^{3}} over {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:

31.62 18.52 = 365 x x = 365 π 18.52 31.62 x = 213.78 {31.62} over {18.52}= {365} over {x} dlrarrow x= {365 cdot 18.52} over {31.62} drarrow 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.

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