Author Topic: Calculation of the Earth's absolute radius to optimize the orbits of spaceships  (Read 4895 times)

Offline Hamster

  • Member
  • Posts: 3
  • Liked: 0
  • Likes Given: 5
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?

Offline meberbs

  • Full Member
  • ****
  • Posts: 1384
  • Liked: 1252
  • Likes Given: 317
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/

Online MATTBLAK

  • Elite Veteran & 'J.A.F.A'
  • Senior Member
  • *****
  • Posts: 3851
  • 'Space Cadets' Let us; UNITE!! (crickets chirping)
  • New Zealand
  • Liked: 765
  • Likes Given: 1444
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.
"Those who can't, Blog".   'Space Cadets' of the World - Let us UNITE!! (crickets chirping)

Offline Phil Stooke

  • Full Member
  • ***
  • Posts: 362
  • Canada
  • Liked: 204
  • Likes Given: 0
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

  • Member
  • Senior Member
  • *****
  • Posts: 2943
  • Cambridge, UK
  • Liked: 661
  • Likes Given: 22
, the geocentric gravitational constant, depends upon the earth's mass  ( = G.Mearth) so is not a fundamental constant.

Offline Bob012345

  • Full Member
  • ****
  • Posts: 563
  • Liked: 118
  • Likes Given: 209
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?

Online Jim Davis

  • Full Member
  • ****
  • Posts: 495
  • Liked: 78
  • Likes Given: 0
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

  • Member
  • Posts: 3
  • Liked: 0
  • Likes Given: 5
, 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 »

Offline meberbs

  • Full Member
  • ****
  • Posts: 1384
  • Liked: 1252
  • Likes Given: 317
Thank's. But it's an interesting coincidence, is not it?
No, it is basically useless.

Offline Jim

  • Night Gator
  • Senior Member
  • *****
  • Posts: 31351
  • Cape Canaveral Spaceport
  • Liked: 9628
  • Likes Given: 299
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

  • Member
  • Posts: 3
  • Liked: 0
  • Likes Given: 5
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.

Offline meberbs

  • Full Member
  • ****
  • Posts: 1384
  • Liked: 1252
  • Likes Given: 317
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

  • Night Gator
  • Senior Member
  • *****
  • Posts: 31351
  • Cape Canaveral Spaceport
  • Liked: 9628
  • Likes Given: 299
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

  • Full Member
  • **
  • Posts: 243
  • Liked: 73
  • Likes Given: 24
In a world where flat earthism is now some sort of mainstream thing

No evidence of that.

Online MATTBLAK

  • Elite Veteran & 'J.A.F.A'
  • Senior Member
  • *****
  • Posts: 3851
  • 'Space Cadets' Let us; UNITE!! (crickets chirping)
  • New Zealand
  • Liked: 765
  • Likes Given: 1444
Okay... So you've never heard of Facebook, YouTube and the comments sections of nearly every science news website then. Got it...  ::)
"Those who can't, Blog".   'Space Cadets' of the World - Let us UNITE!! (crickets chirping)

Offline vladimirph

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0
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

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0
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.

Offline meberbs

  • Full Member
  • ****
  • Posts: 1384
  • Liked: 1252
  • Likes Given: 317
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

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0
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.

Offline vladimirph

  • Member
  • Posts: 15
  • Тьмутаракань
  • Liked: 3
  • Likes Given: 0

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

Tags: