-
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
-
G and M have no direct relation to this formula because µ can be obtained from Kepler's third law.