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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 collapsed 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.
A calculator is now available to prove this claim: Newton's Laws of Motion


Isaac Newton's laws of motion together with the author's planetary spin theory provides an accurate prediction of the Sun’s movement within our Milky Way. All calculations correlate without the need for dark matter.

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 5.647E+12 solar masses of iron rotating at about 3.7E-10 radians per second; if there are 250 billion equivalent suns in the Milky Way.

CalQlata can now confirm that dark matter does not exist.

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) Verify the density of the force-centre
b) Correlate Newton's gravitational force with the Sun's centrfugal force

2) Use Planetary Spin Theory to:
a) Calculate the angular velocity of the Milky Way's force centre for 250 billion orbiting solar masses

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)Input1.1228E+43 ⁽¹⁾mass
Orbiting Body:
m₂ (kg)Input1.9885E+30mass
R₂ (m)Input6.9571E+08radius (of body)
J (kg.m²)⅖.m₂.R₂²3.900081153E+46polar moment of inertia
Orbit Shape:
T (s)Input7.258248E+15 ⁽²⁾orbit period
a (m)³√[G.m₁ / (2.π/T)²]9.999738631E+20major semi-axis
b (m)√[a².(1-e²)]9.999738603E+20minor semi-axis
e[-R̂ + √(R̂² - 4.a.{R̂-a})]7.38261E-05eccentricity
p (m)a.(1-e²)9.999738576E+20half-parameter
ƒ (m)a.(1-e)9.9990003890E+20focus distance from Perigee
x' (m)a-ƒ7.382416009E+16focus distance from ellipse centre
L (m)π . √[ 2.(a²+b²) - (a-b)² / 2.2 ]6.283021075E+21ellipse circumference
K (s²/m³)(2.π)² / G.m₁5.268629510E-32factor
A (m²)π.a.b3.141428424E+42orbit total area
Body Properties at Perihelion or Perigee:
R̂ (m)Input9.999000389E+20 ⁽³⁾distance from force centre to body
F̌ (N)G.m₁.m₂ / R̂²1.4903027452E+21 ⁽⁴⁾centripetal force on orbiting body
c (N)m₂.v̌²/R̂ . ƒ/p ©1.4904127684E+21 ⁽⁴⁾centrifugal force on orbiting body
g (m/s²)-G.m₁ / R̂²-7.4946077205E-10gravitational acceleration on body
v̌ (m/s)h / R̂8.6570270841E+05body velocity
h (m²/s)√[F.p.R̂² / m₂]8.6561617182E+26Newton's motion constant
PE (J)m₂.g.R̂-1.4901537729E+42potential energy
KE (J)½.m₂.v̌²7.4513189257E+41kinetic energy
E (J)PE+KE-7.4502188035E+41total energy
Body Properties at Aphelion or Apogee:
Ř (m)x' + a1.00005E+21 ⁽⁵⁾distance from force centre to body
Ř (m)(E - ½.m₂.v̌²) / m₂.g1.00005E+21 ⁽⁵⁾distance from force centre to body
F̂ (N)G.m₁.m₂ / Ř²1.489862717E+21 ⁽⁶⁾centripetal force on orbiting body
c (N)m₂.v̂²/Ř . p/ƒ ©1.489862709E+21 ⁽⁶⁾centrifugal force on orbiting body
g (m/s²)-G.m₁ / Ř²-7.49239E-10gravitational acceleration on body
v̂ (m/s)h / Ř8.65575E+05body velocity
h (m²/s)h8.65616E+26Newton's motion constant
PE (J)m₂.g.Ř-1.48993E+42potential energy
KE (J)E-PE7.44912E+41kinetic energy
E (J)E-7.45022E+41total energy
Table 1: Calculations for the Sun’s orbit

Notes from Table 1:
1) The mass necessary for the orbital period of the sun and its distance from its force-centre at its perigee
2) Taken from NASA
3) This value gives zero error, NASA specifies 1.0E+21
4) Must be equal to each other for calculations to be correct
5) Must be equal to each other for calculations to be correct
6) Must be equal to each other for calculations to be correct

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)²/³2.190391624E+68Polar moment of inertia of the FC
KE (J)½.(KETa + KETp)3.759372853802E+34Average kinetic energy of the sun’s orbitals
ωᵢ⁽⁵⁾ (ᶜ/s)(2.KE / J)⁰˙⁵2.865329202071E-06Angular velocity of the Sun due to orbitals
J (kg.m²)⅖.m₂.(Δ.R)²9.157910197249E+45polar moment of inertia of the Sun
ES (J)½.J.ωₒ²3.431330221626E+15Spin energy generated by the orbiting Sun
ωₒ (ᶜ/s)2.π / T8.656614250684E-16Angular velocity of orbiting Sun
EFC (J)½.(PETa + PETp)-1.490181473005E+42FC ave. energy that induces spin in the Sun
ωᵣ (ᶜ/s)(2.EFC / JFC)⁰˙⁵1.166335572038E-13FC 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

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

m = 1.1228E+43 kg (the mass of Milky Way’s force-centre {Table 1})

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

R = (3.m / 4.π.ρ) = 6.9836053247E+12 m (radius of Milky Way’s force-centre)

J = ⅖.m.R² = 2.1903916245E+68 kg.m² (moment of angular inertia of Milky Way’s force-centre)

KEp = 7.4513189257E+41 J (kinetic energy of the sun at its perigee {Table 1})

PEₐ = -1.48993E+42 J (gravitational energy of the sun at its apogee {Table 1})

KEsun = KEₐ - PEp = 2.2350656573E+42 J (KE of the sun used to rotate Milky Way’s force-centre)

KEFC = ½.J.ω₁² = 5.5989147911E+53 J (rotational KE in Milky Way’s force-centre)

N = KEFC / KEsun = 2.5050336991E+11
(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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