The purpose of this web-page is to see if we can dispense with the force and energy constants derived by Newton (G) and Coulomb (k), and calculate all potential forces and energies between bodies in the usual fashion; 'm.a' & 'm.a.d' respectively.
Two forces control the behaviour of atoms; magnetic and electrical.
These forces are responsible not only for the state of matter, but also the 'Big-Bang', and therefore all universal energy.
The question is; what are they?
Listed below are the important constants used in the following calculations:
a unit mass of ultimate density; mᵤ = 7.1266079635045E+16 kg
the number of electrons in mᵤ; ϵₑ = 7.82336489952175E+46
the number of protons in mᵤ; ϵₚ = 4.26074122343073E+43
Rydberg created an orbital radius (aₒ), that gave us a factor (γ), that eventually gave us the dynamic ratio (ξᵥ):
Rydberg's factor; γ = 18778.8808461551 {no units}
Rydberg's orbital radius; aₒ = Rₙ.γ = 5.2917721067E-11 {m}
Rydberg's wave number; R∞ = 1 / ξᵥ.aₒ = 10973726.9561359 {/m}
Newton's & Kepler's constant of proportionality for a proton-electron pair; Kₐ = 0.15587874533403 s²/m³
Any two bodies are attracted to each other through magnetic charge (gravity).
Isaac Newton gave us the mathematical formula for the potential force (and potential energy) between them:
PFₘ = G.m₁.m₂/R²
PEₘ = G.m₁.m₂/R
Because magnetic charges are accrued, 'm₁' and 'm₂' must be the total mass of each body in these formulas. This is the mechanism that holds celestial bodies in orbit.
Bodies are also attracted or repulsed via electrical charge. Charles Augustine de Coulomb gave us the mathematical formula for the potential force (and potential energy) between them:
PFₑ = k.e²/R²
PEₑ = k.e²/R
Because electrical charges are shared, both charges 'e' in these formulas must be equal to the lowest magnitude of the two bodies. This is the mechanism that holds atomic particles in orbit.
Whilst magnetic charge forces also apply to atomic particles, they are miniscule by comparison to electrical charge forces between two particles.
However, because magnetic charge forces accrue and electrical charge forces are shared; magnetic charge forces in atoms, and electrical charge forces between celestial bodies, are both insignificant.
The ratio of these two forces is the coupling ratio; 'φ'
We currently refer to the potential force between two bodies as gravity, but it is actually the collective field force between the magnetic charges in all their atomic particles.
Potential force and energy is normally calculated like this:
PF = m.g (magnetic) = m.a (electrical)
PE = m.g.d (magnetic) = m.a.d (electrical)
where d and R are interchangeable.
So, why can't we calculate these forces and energies using g or a? Why do we need Newton's and Coulomb's constants?
Let's begin with Isaac Newton's gravitational constant:
which is calculated like this; G = aₒ.c²/mᵤ {m³ / kg.s²}
Magnetic [potential] acceleration;
g = G.m₁/R² = aₒ.c²/mᵤ . m₁/R² = γ.Rₙ.c²/mᵤ . m₁/R² {m/s²}
but we can use an alternative calculation method:
proportionality factor: χ = φ . (2π)²/Kₐ = 1.11623949165606E-37 m³/s²
where; Kₐ is Newton's constant of proportionality for a proton-electron pair & φ converts the atomic (electrical) representation to magnetism (gravity).
now we convert the force-centre mass to its atomic particles (protons), and thereby calculate its potential (gravitational) acceleration;
g = χ.m₁ / mₚ.R² {m/s²} (quasi-static spacing)
g = v.vᴬ / R.(1-e) (elliptical orbits)
where; v is satellite (m₂) orbital velocity where you want to find the potential acceleration, vᴬ is satellite orbital velocity at the orbital apogee, e is the orbital eccentricity
Potential [magnetic] force and energy can now be defined thus:
PFₘ = m₂.g {N}
PEₘ = m₂.g.R {J}
neither of which requires a gravitational constant.
Now we can address Charles Augustine de Coulomb's constant:
which is calculated like this; k = μₒ.c² = mₑ.Rₙ.c²/e² {kg.m³ / C².s²}
The acceleration between atomic particles can be calculated using Christiaan Huygens formula for centrifugal (centripetal) force and energy:
orbital velocity: v² = Ṯ/X {m²/s²}
orbital radius: R = Xᴿ/Ṯ {m}
potential (centripetal) acceleration: a = v²/R {m/s²}
a = Ṯ² / X.Xᴿ {m/s²}
PFₑ = mₑ.a {N}
PEₑ = mₑ.a.R {J}
neither of which require Coulomb's constant.
Using the proton-electron pair at the neutronic condition, the ratio of these forces (Fₘ:Fₑ) is the coupling ratio,
which defines nuclear energy; the basis for all universal energy, from elemental half-life to the 'Big-Bang', so it is quite important.
Fₘ = G.ξₘ.mₑ²/Rₙ² = 1.28051247006E-38 {N}
Fₑ = k . e²/Rₙ² = μ . (e.2π/tₙ)² = mₑ.c²/Rₙ = 29.0535538991261 {N}
φ = Fₘ/Fₑ = 4.40742111792334E-40
Now if we break these constants down to their fundamentals;
Newton: G = aₒ.c² / mᵤ = Rₙ.γ.c² / mₑ.ϵₑ = 6.67359232004333E-11 {m³ / kg/s²}
Coulomb: k = μₒ.c² = mₑ.Rₙ.c² / e² = 8.98755184732666E+09 {kg.m³ / C²/s²}
The ratio of these constants is:
Constant ratio: G/k = γ/ϵₑ . RC² = 7.425372819438460E-21 {(C/kg)²}
and the ratio of the charges is:
Force ratio: ξₘ.mₑ²/e² = ξₘ/RC² = 5.935622661781240E-20 {C²/kg²}
φ = G/k . ξₘ/RC² = 4.40742111792334E-40
φ = (γ/ϵₑ . RC²).(ξₘ/RC²) = ξₘ.γ/ϵₑ
and because; ξₘ/ϵₑ = 1/ϵₚ
the coupling ratio is defined thus:
φ = γ/ϵₚ
and because γ is derived from Rydberg's radius, the coupling ratio can also be attributed to his work;
and because the coupling ratio is responsible for the 'Big-Bang', that too can also be attributed to his work.
The following calculations apply the above techniques to the earth's orbit and a proton-electron pair, the results of which have been verified using Newton's laws of orbital motion and Coulomb's force law.
sun's mass; m₁ = 1.9885E+30 kg
earth's mass; m₂ = 5.95786303763713E+24 kg
orbital distance @ 180°; Rᴾ = 1.4709500000E+11 m
orbital distance @ 45°; R = 1.51340948066215E+11 m
orbital distance @ 0°; Rᴬ = 152094196155.084 m
Earth's orbit @ 45°:
v = 2.A / t.R = 2.94363192490184E+04 m/s
g = -v.vᴬ / R.(1-e) = -5.7939183081124E-03 m/s²
PF = m₂.g = -3.4519371730992E+22 N
PE = PF.R = -5.22419444441841E+33 J
Earth's orbit @ 0°:
vᴬ = 2.A / t.Rᴬ = 2.92905355716777E+04 m/s
g = -vᴬ.vᴬ / Rᴬ.(1-e) = -5.73667153622504E-03 m/s²
PF = m₂.g = -3.41783033047402E+22 N
PE = PF.Rᴬ = -5.19832156707912E+33 J
Earth @ 180° (perigee; quasi-static):
g = -χ.m₁ / mₚ.Rᴾ² = 6.13323276137771E-03 m/s²
PF = m₂.g = -3.65409607702373E+22 N
PE = PF.Rᴾ = -5.37499262449805E+33 J
proton-electron pair @ 300K:
m₂ = mₑ = 9.1093897E-31 kg
R = Ṯ/Xᴿ = 5.85488721693444E-09 m
a = -Ṯ² / X.Xᴿ = 7.38815108322507E+18 m/s²
PF = mₑ.a = -6.73015473795743E-12 N
PE = PF.R = -3.94042969432577E-20 J
You will find further reading on this subject in reference publications(68, 69, & 70)