# GIsaac Newton's Gravitational Constant {© 15/05/17}

Isaac Newton's gravitational constant has remained an enigma for more than 300 years. But it has now (at last) been solved:
G = aₒ.c² / mN = 6.67359232004334E-11 m³ / kg.s²
This webpage is its mathematical description.

It is now self-evident, given that Newton's gravitational constant is based upon the properties of Quanta, that his laws of orbital motion also apply to the atom and that gravity is indeed magnetism:
thereby destroying the need for Quantum Theory and the associated unification theory!

## The Mathematical Explanation

You will today see the units for this constant (G) written as: N.m²/kg², which were units of convenience originally assigned to imitate Newton's formula:
F = G.m₁.m₂ / R² → G = F.R² / m₁.m₂ {N.m²/kg²}
This was because the formula for 'G' was unknown, hence its value and its units were also unknown.

From the relationship between Newton's Atom and Planck's Atom:
Fᴺ/Fᴾ ≡ Vₚ.Vₑ = 3E-91 (exact)
where:
Fᴺ = G.mₚ.mₑ/aₒ²
Fᴾ = c⁴/G
G = √[aₒ².c⁴ . Σ/(mₚ.mₑ)] = 6.67359232004334E-11 {√[m².m⁴ / s⁴.kg²] = m³ / s².kg}
giving us a value and units for G as a constant
However;
Vₚ = mₚ/ρᵤ
Vₑ = mₑ/ρᵤ
G.mₚ.mₑ/aₒ² ÷ c⁴/G = mₚ/ρᵤ . mₑ/ρᵤ = mₚ.mₑ/ρᵤ²
G² / aₒ².c⁴ = 1/ρᵤ²
G² = aₒ².c⁴/ρᵤ²
G = aₒ.c² / ρᵤ {m⁶/kg/s²}
giving us its value and units as a factor (see 'The origin of factor 1.5' below).
where:
Fᴺ = Newtonian force between a proton and an electron
Fᴾ = Planck's force
Vₚ = volume of a proton
Vₑ = volume of an electron
mₚ = mass of a proton
mₑ = mass of an electron

From its units (m³ / kg.s²), we can now define the purpose of Newton's gravitational constant, i.e.;
G = aₒ.c² / mN {m³/kg/s²}
G = 5.2917721067E-11 x 299792459² / 7.12660796350450E+16
= 6.67359232004334E-11 m³ / s².kg
where:
mN = mᵤ = unit mass = 1m³ of substance of ultimate density (ρᵤ)
G = gravitational acceleration multiplied by spherical area (at radius R) per unit mass; 'g.A/m'

If G = g.A/m, the following calculations show that, contrary to popular belief, the gravitational force of a mass does not vary with distance from its source.
G = 1·5.c² / A.ρᵤ = 6.67359232004334E-11 {m²/s² / (m².kg/m³) = m³ / s².kg}
A = 1·5.c² / G.ρᵤ = 2.83458918818674E+10 {m²/s² / (m³ / s².kg . kg/m³) = m²} #
R = √[A / 4.π] = 47494.1512680647 {m} ##
V = ⁴/₃.π.R³ = 4.48754692288540E+14 {m³}
Rs = 2.G.m / c² = 47494.1512680647 {m³ / s².kg . kg / (m²/s²) = m} ##
Because R = Rs;
m = R.c² / 2.G = 3.19809876372352E+31 {m . m²/s² / (m³ / s².kg) = kg}
ρ = m/V = 7.12660796350450E+16 {kg/m³}
If ρ = ρᵤ;
G = Rs.c² / 2.m = 6.67359232004334E-11 {m . m²/s² / kg = m³ / s².kg}
Rs = 2.G.m / c² = 47494.1512680647 {m³ / s².kg . kg / (m²/s²) = m} ##
g = G.m / R² = 9.46174592804013E+11 {m³ / s².kg . kg / m² = m/s²}
Fᴺ = G.m² / R² = 3.02595979551312E+43 {m³ / s².kg . kg² / m² = kg.m/s² = N}
Fᴾ = c⁴ / G = 1.21038391820525E+44 {m⁴/s⁴ / (m³ / s².kg) = kg.m/s² = N}
Fᴾ/Fᴺ = 4
R = ⁴√[9.Fᴾ / 16.π².G.ρ².4]
= 47494.1512680647 # {⁴√[kg.m/s² / (m³ / s².kg) / (kg²/m⁶]) = m}
Because G = g.A/m
aₒ.c² / mₚ = G.mₚ/R² . 4.π.R² / mₚ = G.4.π = 8.38628344228055E-10 {m³ / kg.s²}
which means that Newton's force law should really be; F = G.m.m/A,
Whilst this changes nothing with respect to Newton's law, it does confirm that gravitational force does not diminish with increasing distance, it is simply spread over a larger [spherical] area.
This discovery complies with the conservation of energy.

Notes:
# A = the spherical area of the surface of a mass of ultimate density (ρᵤ) at 'Rs'
## Rs = the Schwarzschild radius of a mass of ultimate density 'ρᵤ'

The origin of factor 1.5:
The Schwarzschild radius (Rs) of a mass of ultimate density (ρᵤ);
Rs = c.√[3 / 8.π.G.ρᵤ]
3 = Rs²/c² . 8.π.G.ρᵤ
3 = 2 . 4.π.Rs² . G.ρᵤ/c²
3/2 = 4.π.Rs² . aₒ.c²/mᵤ . ρᵤ/c² {m² . m.m²/s²/kg . kg/m³.s²/m² = no units}
1.5 = 4.π.Rs² . aₒ/mᵤ . ρᵤ {m² . m/kg . kg/m³ = no units}
1.5 = 4.π.Rs² . aₒ per unit volume
confirming that G = aₒ.c²/mᵤ (not aₒ.c²/ρᵤ)
, so; G is a constant (not a factor)

If the relationship between Planck's and Newton's atomic particles is correct:
Vₑ.Vp [Planck] = Vₑ.Vp [Newton] must hold true
where Vₑ & Vp are the force-centre and satellite (e.g. proton & electron) volumes respectively.
If G & ρᵤ are as defined above:
Vₑ.Vp [Newton] = Vₑ.Vp [Planck] = 3E-91 {m⁶} (exactly)
If G & ρᵤ are not as defined above:
Vₑ.Vp [Newton] ≠ Vₑ.Vp [Planck] ≠ 3E-91 {m⁶}

Because Vₑ [Planck] = Vp [Planck]; R = ⁶√[9 x 3E-91 ÷ 16π²] = 5.07563837996471E-16 m as this is the radius of a Planck particle, it may be inferred that all the above relationships are correct.

And finally: according to Planck: G = 2π.λ².c³ / h = λ².c³ / ħ {m³ / s².kg}
confirming the units for 'G'.
In this solution for 'G', Planck's atom and Newton's theories have been fully analysed and complement each other perfectly. The discovery of so many atomic associations with G means that Newton actually did anticipate both Poincaré and Planck, which means his theories can be applied throughout all science, from the largest to the smallest ...
i.e. there is no need for a unification theory

## To Summarise

Planck gave us the means to determine the solution to Newton's gravitational constant 'G' using his formula for 'length' (which is actually aₒ).

### Its Units

λ = √[G.ħ / c3]
G = λ2.c3 / ħ {m3 / s2.kg}

### Its Value

λ = √[G.ħ / c3]
Where: ħ = h / 2π & λP = ao = 5.29177210670E-11 m
G = λ2.c3 / ħ = 6.67359232004334E-11 {m3 / s2.kg}

### Its formula

Vₑ = 4π.λᴾ² / Rₙ.aₒ.ξᵥ (. 1 m³)
ħ = ½.Rₙ.mₑ.c.ξᵥ / 2π = Rₙ.mₑ.c.ξᵥ / 4π
1/ħ = 4π / Rₙ.mₑ.c.ξᵥ = Vₑ . aₒ / 1.mₑ.c.λᴾ²
Given that; λᴾ = √[G.ħ / c³]
G = c³ . λᴾ² . 1/ħ
G = c³ . λᴾ² . Vₑ . aₒ / 1.mₑ.c.rᴾ² = c³/c . aₒ . λᴾ²/λᴾ² . [Vₑ / 1.mₑ]
G = c² . aₒ [Vₑ / 1.mₑ] & [ρᵤ = Vₑ / mₑ] & [mᵤ = 1.ρᵤ]     (note: mN = 1.mₑ / Vₑ)
G = c².aₒ / mN {m³ / s².kg}
again confirming that G is a constant, not a factor.

### Conclusion

Despite all of the above;
why was it necessary to fabricate its units (N.m²/kg²) from Newton's force-law formula?
why is it still necessary to estimate its value?
why has nobody applied his laws to the atom?
when ...

Newton's and Kepler's formulas for the constant of proportionality (K) together gave us the correct units for Newton's gravitational constant (G) 300-years ago;
K = (2π)² / G.m = t²/a³
G = (2π)².a³ / t².m {m³ / s².kg}

Today we have accurate values for Earth's; orbital 'a', orbital period (t), and the mass of its force-centre (the sun), so we also have an accurate value for 'G';
(2π)² x ((1.47095E+11+1.52094196155084E+11)/2)³ ÷ 1.9885E+30 ÷ 31558118.40² = 6.67359232004334E-11

And by a remarkable coincidence:
G = c².aₒ / mN = 6.67359232004334E-11 {m³ / s².kg}
confirming that gravity is not based upon the mass of matter, it is based upon the particles of which it comprises.