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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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Dark Matter {© 05/11/17}
(does it really exist!)

This paper, which was released on the 5th of November 2017, represents yet another important discovery by Keith Dixon-Roche (one of CalQlata's Contributors) in that it not only blows away any need for the unknown dark matter but also ...

... identifies the properties and behaviour of the force-centre at the heart of our Milky Way, the total mass of the Milky Way and confirms his own predictions for collapsible stars.

Note: All the theories are provided by CalQlata's Laws of Motion and Planetary Spin
All the calculations are the sole copyright priority of Keith Dixon-Roche © 2017
Keith Dixon-Roche is also responsible for all the other web pages on this site related to planetary motions
A 'pdf' version of this paper can be found at: Dark Matter - The Paper


The purpose of this paper is to determine an accurate description of the behaviour of our sun in the Milky Way by applying Isaac Newton’s laws of motion and Keith Dixon-Roche's planetary spin theory and thereby discount the need for universal dark matter.


Isaac Newton's laws of motion together with the author's planetary spin theory provide an accurate prediction of the Sun’s movement within our Milky Way.

CalQlata can now confirm that the need for dark matter is unnecessary and therefore does not exist.

CalQlata can also announce that contrary to popular belief, our Milky Way’s force-centre is not a neutron star spinning at light-speed but comprises 2.6E+06 solar masses of iron rotating at less than 1E-08 radians per second, and there are at most; 9.4179883E+10 equivalent suns in our Milky Way.

The Milky Way System

The only part of the Milky Way required to establish the behaviour of the sun within it is the sun itself along with its own orbiting bodies and its force-centre. This has been demonstrated by Isaac Newton and the author’s own planetary spin theory.


1) Use Newton’s theories to replicate the Sun's orbit in the Milky Way:
     a) Alter the angular velocity of the force-centre until both independently
     calculated values for Ř are identical (Table 1)
b) Alter the mass of the force-centre until the eccentricity of the sun's orbit is correct

2) Use Planetary Spin Theory to:
     a) Alter the density of the force-centre until the Sun’s angular velocity is correct
     b) Calculate the number of equivalent Sun's in the Milky Way

Calculation Results

The following Table provides the sun’s orbital parameters according to Newton:

Sym. (units)FormulaResultDescription
G (m³/kg/s²)Constants6.67359232E-11gravitational constant
m₁ (kg)Input5.1701000E+36 ⁽¹⁾mass
Orbiting Body:
m₂ (kg)Input1.9885E+30mass
R₂ (m)Input6.9571E+08radius (of body)
J (kg.m²)⅖.m₂.R₂²3.849834662E+39polar moment of inertia
Orbit Shape:
T (s)Input7.25825E+15orbit period
a (m)³√[G.m₁ / (2.π/T)²]7.721841E+18major semi-axis
b (m)√[a².(1-e²)]3.800484E+18minor semi-axis
e[-R̂ + √(R̂² - 4.a.{R̂-a})]0.870497196⁽²⁾eccentricity
p (m)a.(1-e²)1.870497E+18half-parameter
ƒ (m)a.(1-e)1.000000E+18focus distance from Perigee
x' (m)a-ƒ6.721841E+18focus distance from ellipse centre
L (m)π . √[ 2.(a²+b²) - (a-b)² / 2.2 ]3.732437E+19ellipse circumference
K (s²/m³)(2.π)² / G.m₁1.144198E-25factor
A (m²)π.a.b9.219548E+37orbit total area
Body Properties at Perihelion or Perigee:
R̂ (m)Input1.0000000E+18⁽³⁾distance from force centre to body
F̌ (N)G.m₁.m₂ / R̂²6.86095E+20centripetal force on orbiting body
g (m/s²)-G.m₁ / R̂²-3.45031E-10gravitational acceleration on body
v̌ (m/s)h / R̂25404.33545body velocity
h (m²/s)√[F.p.R̂² / m₂]2.54043E+22Newton's motion constant
PE (J)m₂.g.R̂-6.86095E+38potential energy
KE (J)½.m₂.v̌² + ½.J.(2π/t)².ω₁6.41669E+38kinetic energy
E (J)PE+KE-4.442561E+37total energy
Body Properties at Aphelion or Apogee:
Ř (m)x' + a1.44436811E+19 ⁽⁴⁾distance from force centre to body
Ř (m)(-b-(b^2-4.a.c)⁰˙⁵) / 2.a1.44436811E+19⁽⁴,⁵⁾distance from force centre to body
F̂ (N)G.m₁.m₂ / Ř²3.28873160E+18centripetal force on orbiting body
g (m/s²)-G.m₁ / Ř²-1.65388E-12gravitational acceleration on body
v̂ (m/s)h / Ř1758.854639body velocity
h (m²/s)h2.54043E+22Newton's motion constant
PE (J)m₂.g.Ř-4.75014E+37potential energy
KE (J)E-PE3.07578E+36kinetic energy
E (J)E-4.442561E+37total energy
A½.m₂.h²6.416693233E+74part formula
B½.I.G².m₁²2.291549879E+92part formula
CG.m₁.m₂6.86094932E+56part formula
Table 1: Calculations for the Sun’s orbit

Notes from Table 1:
1) This is the only value (2.6E+06 solar masses) that provides the eccentricity in Note 2 below
2) Taken from Wikipedia
3) Taken from NASA
4) Geometric calculation that must be equal to the Note 5) below
5) Energy calculation that must be equal to the Note 4) above
6) The angular velocity of force-centre that will ensure correct matching of Notes 4) & 5) above is; ω₁ ≤ 1E-08ᶜ/s (see Table 2)
7) The accuracy of these calculations is dependent upon the accuracy of the information in Notes 2) and 3) above

Milky Way Force-Centre

The following Table provides the sun’s angular velocity according to the author’s planetary spin theory {FC stands for the force-centre of the Milky Way}:

ρ₁ (kg/m³)Input7870⁽¹⁾FC density
JFC (kg.m²)⅖.m₁.(3.m₁ / 4.π.ρ₁)²/³6.014266510E+57Polar moment of inertia of the FC
KE (J)½.(KETa + KETp)3.7918553E+34⁽²⁾Average kinetic energy of the sun’s orbitals
ωᵢ⁽⁵⁾ (ᶜ/s)(2.KE / J)⁰˙⁵2.8653291E-06Angular velocity of the Sun due to orbitals
R (m)Input6.9571E+08Sun radius
m₂ (kg)Input1.9885E+30Sun mass
Δ⁽⁵⁾Input0.15489783⁽³⁾factor for sun's radial centre of mass
J (kg.m²)⅖.m₂.(Δ.R)²9.2370381E+45polar moment of inertia of the Sun
ES (J)½.J.ωₒ²3.4609783E+15Spin energy generated by the orbiting Sun
ωₒ (ᶜ/s)2.π / T8.6566143E-16Angular velocity of orbiting Sun
EFC (J)½.(PETa + PETp)3.6679816E+38⁽⁴⁾FC ave. energy that induces spin in the Sun
ωᵣ (ᶜ/s)(2.EFC / JFC)⁰˙⁵3.4925066E-10FC induced angular velocity
ω (ᶜ/s)ωᵢ + ωₒ + ωᵣ2.86568E-06Calculated angular velocity of the Sun
ωₐ (ᶜ/s)2.86533E-06Actual angular velocity of the Sun
error1 - ω/ωₐ0.0001219 ⁽⁵⁾
Table 2: Calculations for the Sun’s angular velocity

Notes from Table 2:
1) The density of iron (7870 kg/m³) is the only value that provides the correct spin.
    If the force-centre was a neutron star ωᵢ would have been 1.01449E-05 ᶜ/s which is incorrect
2) Taken from the author’s analysis of the Solar System
3) Taken from the author’s planetary spin paper
4) Taken from the author’s planetary spin paper
5) This minor error can be corrected by altering 'ωᵢ' and 'Δ'

The number of Stars in the Milky Way (Planetary Spin Theory)

m = 5.1701E+36kg (the mass of Milky Way’s force-centre {Table 1})

ρ = 7870 kg/m³ (density of Milky Way’s force-centre {Table 1})

KEp = 6.41669E+38 J (kinetic energy of the sun at its perigee {Table 1})

KEₐ = 3.0757816E+36 J (kinetic energy of the sun at its apogee {Table 1})

R = (3.m / 4.π.ρ) (radius of Milky Way’s force-centre)

J = ⅖.m.R² = 6.01427E+57 kg.m² (moment of angular inertia of Milky Way’s force-centre)

KEsun = KEₐ - KEp = 6.3859354E+38 J (KE of the sun used to rotate Milky Way’s force-centre)

KEFC = J.ω₁ = 6.0142665E+49 J (rotational KE in Milky Way’s force-centre)

N = KEFC / KEsun = 9.4179883E+10
(number of {our} solar systems needed to rotate Milky Way’s force-centre)

Further Reading

You will find further reading on this subject in reference publications(55, 60, 61, 62, 63 & 64)

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