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  • Proof of the non-exitence of Dark MatterDOES NOT EXIST
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The solution to Newton's G1 The theory controlling planetary spin2 Pressure at the centre of the Earth3 Proof of the non-exitence of Dark Matter4 The atom as Newton describes it5
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G
Isaac Newton's Gravitational Constant

Isaac Newton's gravitational constant has remained unknown and undefined for 300 years.
The following is a mathematical description of its true and accurate values and units
G = aₒ.c² / ρᵤ = 6.67359232004334E-11 m³ / kg.s² per cubic metre
G is therefore a factor

It is now evident that Newton's gravitational constant is based upon the properties of Quanta, making it all the more probable that his laws of orbital motion also apply to the the atom:
destroying the need for Quantum Theory and precluding the need for a unification theory!

The Mathematical Explanation

You will see the units for this constant (G) written as: N.m²/kg², which were units of convenience originally assigned to reflect Newton's formula:
F = G.m₁.m₂ / R²
This was because the formula for 'G' was unknown, hence 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 = √[3E-91.aₒ².c⁴ / mₚ.mₑ] = 6.67359232004334E-11 {√[m².m⁴ / s⁴.kg²] = m³ / s².kg}
giving us a value and units for G, moreover;
Vₚ = mₚ/ρᵤ
Vₑ = mₑ/ρᵤ
G.mₚ.mₑ/aₒ² ÷ c⁴/G = mₚ/ρᵤ . mₑ/ρᵤ = mₚ.mₑ/ρᵤ²
G.mₚ.mₑ/aₒ² ÷ c⁴/G = mₚ.mₑ/ρᵤ²
G² / aₒ².c⁴ = 1/ρᵤ²
G² = aₒ².c⁴/ρᵤ²
G = aₒ.c² / ρᵤ {m⁶/kg/s²}

Because we know the units for G are m³ / kg.s², there must be a per unit volume component in this last formula and given that ρ.V = mass; Newton's gravitational constant must be gravitational acceleration multiplied by an area per unit mass
i.e. G = aₒ.c² / mᵤ {m³/kg/s²}
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
mᵤ = unit mass = 1m³ of substance of ultimate density (ρᵤ)
c = speed of light
aₒ = Bohr's radius
G = Newton's gravitational constant
G = aₒ.c² / mᵤ
G = 5.2917721067E-11 x 299792459² / 7.12660796350450E+16
     = 6.67359232004334E-11 m³ / s².kg

If G = gravitational acceleration times spherical area per unit mass, the following calculations show that, contrary to popular belief, gravitational force does not simply vary with the square of the distance from the centre of its source.
Factors 1.5 & 4 below are exact values and will be explained in due course
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} #
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}
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}
# Schwarzschild radius (Rs) of mass 'm'

From the above formulas:
G = g.A / 4.π.m = 6.67359232004334E-11 {m³ / s².kg}
4.π.G = g.A / m = 8.38628344228057E-10 {m³ / s².kg}
Indicating that Fᴺ = 4.π.G.m₁.m₂ / 4.π.R²
In other words; Newton’s gravitational force equation should read:
Fᴺ = G.m₁.m₂ / A
in which his gravitational constant would be: G = 8.38628344228057E-10 m³/s²/kg;
i.e. gravitational force is constant irrespective of distance from the centre of its source, but it will vary at distance (R) according to its distribution over the spherical area.
This claim upholds the conservation of energy.
Two aspects of this discovery require further explanation:
1) aₒ = 1·5 / A.V (V = 1m³) where A = 2.83458918818674E+10 m²; what is area 'A'?
2) Fᴾ/Fᴺ = 4 (exact)
Both of which will be addressed later but neither of which alters the final result.

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 probably means his theories can be applied throughout all science, from the largest to the smallest ...
i.e. there should be no need for a unification theory

Further Reading

You will find further reading on this subject in reference publications(68, 69, & 70)

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