Author Topic: Calculation of the Earth's absolute radius to optimize the orbits of spaceships  (Read 4215 times)

Offline Hamster

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Hello, all. Developing my model of the hydrogen atom, I have unexpectedly derived a formula, which proved to be suitable for calculating the Earth's radius:

where is the geocentric gravitational constant ≈ 3.986∙1014 m3/s2, me is the electron mass, ħ is the reduced Planck constant, c is the speed of light. (More details is here: https://drive.google.com/open?id=0B90mGmUYbDopMXlTaWVMVTB5LVU
and here: https://sites.google.com/site/snvspace22/science/earthradius )

Since this formula is obtained using fundamental physical constants, the calculated radius can be called the absolute radius of the Earth. So, I have a question: Can this formula be used to optimize the orbits of spaceships?

Online meberbs

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Since it is obvious that the radius of the Earth has nothing whatsoever to do with fundamental constants (There are billions of planets in the galaxy and they all have different radii, the Earth is not special) It can only be concluded that what you have found is pure coincidence. Many examples of formula that produce results that happen to line up with something else are known. It isn't intuitive just how many ways formula can be rearranged, so such coincidences are much more common than most people's intuition would suggest.


https://xkcd.com/687/

Offline MATTBLAK

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In a world where flat earthism is now some sort of mainstream thing (insert rage emoji) I'm actually glad somebody is doing work like this.
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Online Phil Stooke

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Equatorial radius (km)           6378.137   
Polar radius (km)               6356.752         
Volumetric mean radius (km)     6371.008

source: NASA GSFC

Back to the drawing board!

Online AnalogMan

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, the geocentric gravitational constant, depends upon the earth's mass  ( = G.Mearth) so is not a fundamental constant.

Offline Bob012345

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Hello, all. Developing my model of the hydrogen atom, I have unexpectedly derived a formula, which proved to be suitable for calculating the Earth's radius:

where is the geocentric gravitational constant ≈ 3.986∙1014 m3/s2, me is the electron mass, ħ is the reduced Planck constant, c is the speed of light. (More details is here: https://drive.google.com/open?id=0B90mGmUYbDopMXlTaWVMVTB5LVU
and here: https://sites.google.com/site/snvspace22/science/earthradius )

Since this formula is obtained using fundamental physical constants, the calculated radius can be called the absolute radius of the Earth. So, I have a question: Can this formula be used to optimize the orbits of spaceships?

So why would it not include the mass of all the protons and neutrons? Also, in what way are orbits not optimized now?

Offline Jim Davis

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Hello, all. Developing my model of the hydrogen atom, I have unexpectedly derived a formula, which proved to be suitable for calculating the Earth's radius:

where is the geocentric gravitational constant ≈ 3.986∙1014 m3/s2, me is the electron mass, ħ is the reduced Planck constant, c is the speed of light.

If you use the appropriate gravitational constants for other planets/stars/bodies do you similarly calculate their radii? If so, you might be on to something. If not, just a coincidence, like the sun and moon having the same apparent size from earth.

Offline Hamster

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, the geocentric gravitational constant, depends upon the earth's mass  ( = G.Mearth) so is not a fundamental constant.

Well, it's almost a fundamental constant. :) G and M have no direct relation to this formula because can be obtained from Kepler's third law.

So why would it not include the mass of all the protons and neutrons?

That is unnecessary. It already works well.

Also, in what way are orbits not optimized now?

There is no limit to perfection. :)

If you use the appropriate gravitational constants for other planets/stars/bodies do you similarly calculate their radii?

No, I could not obtain a dependence of the radii of other planets on their gravitational constants.

If so, you might be on to something. If not, just a coincidence, like the sun and moon having the same apparent size from earth.

Thank's. But it's an interesting coincidence, is not it?
« Last Edit: 08/05/2017 09:22 AM by Hamster »

Online meberbs

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Thank's. But it's an interesting coincidence, is not it?
No, it is basically useless.

Offline Jim

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So, I have a question: Can this formula be used to optimize the orbits of spaceships?

What does it mean to "optimize the orbits of spaceships"? And why does it need "perfection"?


Offline Hamster

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What does it mean to "optimize the orbits of spaceships"? And why does it need "perfection"?

I thought that an accurate knowledge of the center of the Earth can help calculate the optimal trajectory for launching a spaceship into orbit and saving its fuel. But I can be wrong. That's why I asked my question.

Online meberbs

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What does it mean to "optimize the orbits of spaceships"? And why does it need "perfection"?

I thought that an accurate knowledge of the center of the Earth can help calculate the optimal trajectory for launching a spaceship into orbit and saving its fuel. But I can be wrong. That's why I asked my question.
More useful is accurate knowledge of the actual equatorial and polar radii, rather than some spherical estimate without any real meaning that is off from any meaningful measurement by kilometers. Or more directly an accurate non-spherical geopotential model
(Note: the model I linked is excessive for almost any normal satellite.)

Offline Jim

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What does it mean to "optimize the orbits of spaceships"? And why does it need "perfection"?

I thought that an accurate knowledge of the center of the Earth can help calculate the optimal trajectory for launching a spaceship into orbit and saving its fuel. But I can be wrong. That's why I asked my question.

That has to be known by measurement and not derived

Offline tdperk

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In a world where flat earthism is now some sort of mainstream thing

No evidence of that.

Offline MATTBLAK

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Okay... So you've never heard of Facebook, YouTube and the comments sections of nearly every science news website then. Got it...  ::)
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Offline vladimirph

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The radii of the planets and the parameters of the stationary orbits are described by the formula with a golden cross section

The graph of the dependence of the radii of the planets looks like this

The dependence of the orbits of the planets in the solar system looks like this

In the formula there is also a correlating function

Offline vladimirph

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Quote
Hamster
I have unexpectedly derived a formula, which proved to be suitable for calculating the Earth's radius
You are moving in the right direction. From Kepler's aunt's law follows:

Y1=A/X2
Y2=B/Z3

where  Y1=Y2=Y  is performed at special points and under special conditions for cases of stationary orbits or body radii. These conditions are universal for both the micro-world and the macro-world. World constants for the micro-world and macro-world are general and differ by the conditions of quantization between worlds. In the formula with a golden section, the parameter R0 takes these conditions into account.
By the way, the planet Nubir, about which there is so much talk now, should be sought in the areas of values following from the formula with a golden section.

Online meberbs

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Golden ratio is not relevant.

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

Offline vladimirph

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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.

Offline vladimirph

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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

Online meberbs

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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

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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

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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

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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

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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



Online meberbs

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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

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The result of taking a derivative does not contain "dx"


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

Offline vladimirph

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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

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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.

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