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}, m_{e} 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?

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}, m_{e} is the electron mass, ħ is the reduced Planck constant, c is the speed of light.

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

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

Also, in what way are orbits not optimized now?

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.

Thank's. But it's an interesting coincidence, is not it?

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"?

Quote from: Jim on 08/05/2017 01:12 PMWhat 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.

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

Hamster I have unexpectedly derived a formula, which proved to be suitable for calculating the Earth's radius

Golden ratio is not relevant.

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

QuoteGolden 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.

QuoteThe 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

From Kepler's aunt's law follows:

Quote from: vladimirph on 08/11/2017 07:23 AMFrom 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.

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

Stable orbits exist at any distance from the sun.