-
Calculation of the Earth's absolute radius to optimize the orbits of spaceships
by
Hamster
on 04 Aug, 2017 05:19
-
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
µ is the geocentric gravitational constant ≈ 3.986∙10
14 m
3/s
2,
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=0B90mGmUYbDopMXlTaWVMVTB5LVUand 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?
-
#1
by
meberbs
on 04 Aug, 2017 05:56
-
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/
-
#2
by
MATTBLAK
on 04 Aug, 2017 06:55
-
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.
-
#3
by
Phil Stooke
on 04 Aug, 2017 07:19
-
Equatorial radius (km) 6378.137
Polar radius (km) 6356.752
Volumetric mean radius (km) 6371.008
source: NASA GSFC
Back to the drawing board!
-
#4
by
AnalogMan
on 04 Aug, 2017 11:16
-
µ, the geocentric gravitational constant, depends upon the earth's mass (µ = G.Mearth) so is not a fundamental constant.
-
#5
by
Bob012345
on 04 Aug, 2017 17:14
-
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 µ 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?
-
#6
by
Jim Davis
on 04 Aug, 2017 17:54
-
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 µ 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.
-
#7
by
Hamster
on 05 Aug, 2017 08:56
-
µ, the geocentric gravitational constant, depends upon the earth's mass (µ = 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 µ 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?
-
#8
by
meberbs
on 05 Aug, 2017 13:01
-
Thank's. But it's an interesting coincidence, is not it?
No, it is basically useless.
-
#9
by
Jim
on 05 Aug, 2017 13:12
-
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"?
-
#10
by
Hamster
on 05 Aug, 2017 14:19
-
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.
-
#11
by
meberbs
on 05 Aug, 2017 14:51
-
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.)
-
#12
by
Jim
on 05 Aug, 2017 15:36
-
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
-
#13
by
tdperk
on 06 Aug, 2017 17:48
-
In a world where flat earthism is now some sort of mainstream thing
No evidence of that.
-
#14
by
MATTBLAK
on 07 Aug, 2017 18:09
-
Okay... So you've never heard of Facebook, YouTube and the comments sections of nearly every science news website then. Got it...
-
#15
by
vladimirph
on 10 Aug, 2017 05:03
-
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
-
#16
by
vladimirph
on 11 Aug, 2017 07:23
-
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:
Y
1=A/X
2Y
2=B/Z
3where Y
1=Y
2=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 R
0 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.
-
#17
by
meberbs
on 11 Aug, 2017 07:52
-
Golden ratio is not relevant.
The graphs don't even have their axes labelled, preventing a proper explanation of why they are nonsensical.
-
#18
by
vladimirph
on 11 Aug, 2017 08:25
-
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.
-
#19
by
vladimirph
on 11 Aug, 2017 08:27
-
-
#20
by
meberbs
on 11 Aug, 2017 13:21
-
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.
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.
-
#21
by
vladimirph
on 11 Aug, 2017 15:09
-
Consider Kepler's third law:
GMT
2=4π
2a
3Where: 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:
a
13/T
12=a
23/T
22or
a
13/a
23=T
12/T
22Fix a
1 and T
1We get:
Y
1=A/X
2Y
2=B/Z
3Further 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.
-
#22
by
as58
on 11 Aug, 2017 17:30
-
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.
-
#23
by
vladimirph
on 12 Aug, 2017 02:14
-
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.
-
#24
by
vladimirph
on 12 Aug, 2017 03:52
-
And now I will prove two very important facts that follow from equations
Y
1=A/X
2Y
2=B/Z
3Y
1=Y
21) stationary orbits (and radii of planets) exist
2) the gravitational constant is an oscillating quantity.
We rewrite the equations in the form
Y
1X
2=A
Y
2Z
3=B
and differentiate them:
dY
1X
2 + Y
12XdX=0
dY
2Z
3 + Y
23Z
2dZ=0
The first equation is shortened by X, and the second is shortened by Z
2. But such a reduction in these differential equations can not be directly done, because they are related by the condition:
Y
1=Y
2dY
1=dY
2Therefore, there must be a function that "swallows" X and Z
2. 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
-
#25
by
meberbs
on 12 Aug, 2017 06:21
-
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.
-
#26
by
vladimirph
on 12 Aug, 2017 07:05
-
The result of taking a derivative does not contain "dx"
I did not take the derivative, but calculated the differential.
-
#27
by
vladimirph
on 12 Aug, 2017 07:08
-
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?
-
#28
by
vladimirph
on 12 Aug, 2017 07:40
-
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
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.
G and M have no direct relation to this formula because µ can be obtained from Kepler's third law.
n = 7 - 10.7 au. - Saturn