Author Topic: Calculation of the Earth's absolute radius to optimize the orbits of spaceships  (Read 2917 times)

Offline meberbs

  • Full Member
  • ****
  • Posts: 1134
  • Liked: 1052
  • Likes Given: 273
Quote
Golden ratio is not relevant.
The formula with a golden section follows from the differential equations compiled on the basis of Kepler's third law.
So it's all fair.
No. Kepler's third law is an algebraic equation, not a differential one, and has nothing to do with the golden ratio.


Quote
The graphs don't even have their axes labelled, preventing a proper explanation of why they are nonsensical.

Look at here http://gravitus.ucoz.ru/blog/formula_s_zolotym_secheniem_i_ee_primenenie/2017-02-17-9
Graph axes still aren't labelled.
If the google translate of the page is accurate, it seems that the x axis of the graphs is just "index" which in itself proves your equation is meaningless, beyond the fact that you are arbitrarily rearranging the planets to make them fit and that you basically are doing a curve fit that could be comparably well done by a parabola, or a generic exponential function, so it has nothing to do with the golden ratio in the end.

Offline vladimirph

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0
Consider Kepler's third law:
GMT2=4π2a3
Where: G - Constant gravity
         M - Mass of the central body
         T - Period of revolution of the planet (satellite)
         a - The semimajor axis of the elliptical orbit.

For different a and T we obtain:
a13/T12=a23/T22
or
a13/a23=T12/T22
Fix a1  and T1
We get:
Y1=A/X2
Y2=B/Z3
Further mathematical transformations begin, as a result of which the field theory of gravitation appears.
For example, stable orbits not only lie in the ecliptic plane, but also at an angle of about 30 degrees, the value of which again sets the number of the golden section.


Offline as58

  • Full Member
  • ****
  • Posts: 707
  • Liked: 232
  • Likes Given: 158
From Kepler's aunt's law follows:

Ok... Well, Kepler's mother was accused of witchcraft so maybe her sister, aunt of Johannes, had access to some sort of hidden knowledge.

Offline vladimirph

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0
From Kepler's aunt's law follows:

Ok... Well, Kepler's mother was accused of witchcraft so maybe her sister, aunt of Johannes, had access to some sort of hidden knowledge.

Thank you. I need to change the translator.

Offline vladimirph

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0
And now I will prove two very important facts that follow from equations
Y1=A/X2
Y2=B/Z3
Y1=Y2
1) stationary orbits (and radii of planets) exist
2) the gravitational constant is an oscillating quantity.
We rewrite the equations in the form 
Y1X2=A
Y2Z3=B
and differentiate them:
dY1X2 + Y12XdX=0
dY2Z3 + Y23Z2dZ=0
The first equation is shortened by X, and the second is shortened by Z2. But such a reduction in these differential equations can not be directly done, because they are related by the condition:
Y1=Y2
dY1=dY2
Therefore, there must be a function that "swallows" X and Z2. But where does such a function come from? There is only one option - the constant of gravity G.
And this is proved by the parameters of the orbits of the planets of the solar system



Offline meberbs

  • Full Member
  • ****
  • Posts: 1134
  • Liked: 1052
  • Likes Given: 273
The result of taking a derivative does not contain "dx"

You never defined basically any of your variables.

Also, nothing you are saying has any correlation with reality. Stable orbits exist at any distance from the sun. Also generally orbits are elliptical and have varying distance.

Offline vladimirph

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0
Quote
The result of taking a derivative does not contain "dx"


I did not take the derivative, but calculated the differential.

Offline vladimirph

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0
Quote
Stable orbits exist at any distance from the sun.
Are you sure about that?
Why do satellites on Earth's orbit need a constant adjustment?
« Last Edit: 08/12/2017 07:12 AM by vladimirph »

Offline vladimirph

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0
According to the formula with a golden section, we calculate the parameters of the orbits of the planets of the solar system:
1) n = 0 orbit of Mercury
2) n = 1 - 0.6 au. - Venus
3) n = 2 - 1a.u. - Earth
4) n = 3 - 1.6a.u. - Mars
5) n = 4 - 2.6 au. - asteroids (beginning)
6) n = 5 - 4.2 au. - asteroids (end)
7) n = 6 to 6.7 au. - Jupiter
8) n = 7 - 10.7 au. - Saturn
9) n = 8 - 17.1 au. - Uranus
10) n = 9 - 27.4 au. - Neptune
11) n = 10 - 43.8 au. - Pluto
A.u. - Is an astronomical unit
We build the graph:

The green line is the resulting graph, and the red line is constructed from tabular data.
It can be seen how the red line "interlaces" the green line. This also confirms the presence of an oscillating factor.

Tags: