Keith Dixon-Roche (one of CalQlata's Contributors) asked himself the question; "does dark matter really exist?", given that a celestial *atmosphere* would have caused all celestial satellites to stop orbiting long ago. And he discovered that Isaac Newton actually told us that it doesn't 300 years ago. And in doing so, Keith also ...

... identified 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.

The dark matter referred to today by the scientific community is nothing more than a galactic force-centre. Since having discovered the mass and spin-rate of our own galactic force-centre, Keith Dixon-Roche has given it the name 'Hades' for easier reference.
A calculator is now available for you to prove it for yourself: 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 1.76572E+41kg of iron, 3.4993E+12m diameter and rotating at 3.6465E-07ᶜ/s; based upon 100 billion equivalent suns in the Milky Way.

CalQlata can now confirm that dark matter does not exist simply by proving that there is no need for it.

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

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

Sym. (units) | Formula | Result | Description |
---|---|---|---|

Force-Centre: | |||

G (m³/kg/s²) | Constants | 6.67359232E-11 | gravitational constant |

m₁ (kg) | Input | 1.76572019E+41 ⁽¹⁾ | mass |

Orbiting Body: | |||

m₂ (kg) | Input | 1.9885E+30 | mass |

R₂ (m) | Input | 6.9571E+08 | radius (of body) |

J (kg.m²) | ⅖.m₂.R₂² | 3.900081153490E+46 | polar moment of inertia |

Orbit Shape: | |||

T (s) | Input | 7.258248E+15 ⁽²⁾ | orbit period |

a (m) | ³√[G.m₁ / (2.π/T)²] | 2.505311941E+20 | major semi-axis |

b (m) | √[a².(1-e²)] | 2.504993572E+20 | minor semi-axis |

e | [-R̂ + √(R̂² - 4.a.{R̂-a})] | 0.015941744 | eccentricity |

p (m) | a.(1-e²) | 2.504675243E+20 | half-parameter |

ƒ (m) | a.(1-e) | 2.465372900E+20 | focus distance from Perigee |

x' (m) | a-ƒ | 3.993904106E+18 | focus distance from ellipse centre |

L (m) | π . √[ 2.(a²+b²) - (a-b)² / 2.2 ] | 1.574033901E+21 | ellipse circumference |

K (s²/m³) | (2.π)² / G.m₁ | 3.350257446E-30 | factor |

A (m²) | π.a.b | 1.971597673E+41 | orbit total area |

Body Properties at Perihelion or Perigee: | |||

R̂ (m) | Input | 2.465372900E+20 ⁽³⁾ | distance from force centre to body |

F̌ (N) | G.m₁.m₂ / R̂² | 3.8551556343E+20 ⁽⁴⁾ | centripetal force on orbiting body |

F̌c (N) | m₂.v̌²/R̂ . ƒ/p © | 3.9166135377E+20 ⁽⁴⁾ | centrifugal force on orbiting body |

g (m/s²) | -G.m₁ / R̂² | -1.9387254887E-10 | gravitational acceleration on body |

v̌ (m/s) | h / R̂ | 2.2036056214E+05 | body velocity |

h (m²/s) | √[F.p.R̂² / m₂] | 5.4327095813E+25 | Newton's motion constant |

PE (J) | m₂.g.R̂ | -9.5043962261E+40 | potential energy |

KE (J) | ½.m₂.v̌² | 4.8279564378E+40 | kinetic energy |

E (J) | PE+KE | -4.6764397883E+40 | total energy |

Body Properties at Aphelion or Apogee: | |||

Ř (m) | x' + a | 2.54525E+20 ⁽⁵⁾ | distance from force centre to body |

Ř (m) | (E - ½.m₂.v̌²) / m₂.g | 2.54525E+20 ⁽⁵⁾ | distance from force centre to body |

F̂ (N) | G.m₁.m₂ / Ř² | 3.616978470E+20 ⁽⁶⁾ | centripetal force on orbiting body |

F̂c (N) | m₂.v̂²/Ř . p/ƒ © | 3.616059254E+20 ⁽⁶⁾ | centrifugal force on orbiting body |

g (m/s²) | -G.m₁ / Ř² | -1.81895E-10 | gravitational acceleration on body |

v̂ (m/s) | h / Ř | 213444.9459 | body velocity |

h (m²/s) | h | 5.4327095813E+25 | Newton's motion constant |

PE (J) | m₂.g.Ř | -9.2061180022E+40 | potential energy |

KE (J) | E-PE | 4.5296782139E+40 | kinetic energy |

E (J) | E | -4.6764397883E+40 | total energy |

Table 1: Calculations for the Sun's orbit1) 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 |

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) | Input | 7870⁽¹⁾ | FC density |
---|---|---|---|

JFC (kg.m²) | ⅖.m₁.(3.m₁ / 4.π.ρ₁)²/³ | 2.1621587048E+65 | Polar moment of inertia of the FC |

KE (J) | ½.(KETa + KETp) | 1.989651235897E+35 | Average kinetic energy of the sun's orbitals |

ωᵢ⁽⁵⁾ (ᶜ/s) | (2.KE / J)⁰˙⁵ | 2.865329084572E-06 | Angular velocity of the Sun due to orbitals |

J (kg.m²) | ⅖.m₂.(Δ.R)² | 3.900081153490E+46 | polar moment of inertia of the Sun |

ES (J) | ½.J.ωₒ² | 1.461301327544E+16 | Spin energy generated by the orbiting Sun |

ωₒ (ᶜ/s) | 2.π / T | 8.656614250684E-16 | Angular velocity of orbiting Sun |

EFC (J) | ½.(PETa + PETp) | -1.490043768819E+42 | FC ave. energy that induces spin in the Sun |

ωᵣ (ᶜ/s) | (2.EFC / JFC)⁰˙⁵ | 1.166416422318E-13 | FC induced angular velocity |

ω (ᶜ/s) | ωᵢ + ωₒ + ωᵣ | 2.865329084572E-06 | Calculated angular velocity of the Sun |

ωₐ (ᶜ/s) | 2.865329084572E-06 | Actual angular velocity of the Sun | |

error | 1 - ω/ωₐ | 0.0000000 ⁽⁵⁾ | |

Table 2: Calculations for the Sun's angular velocity |

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

m = 1.7657E+41 kg (the mass of Milky Way's force-centre)

ρ = 7870 kg/m³ (density of Milky Way's force-centre ⁽¹⁾)

R = (3.m / 4.π.ρ)⅓ = 1.7496567118E+12 m (radius of Milky Way's force-centre)

J = ⅖.m.R² = 2.1621587048E+65 kg.m² (moment of angular inertia of Milky Way's force-centre)

KEp = 4.8279564378E+40 J (kinetic energy of the sun at its perigee)

PEₐ = -9.2061180022E+40 J (gravitational energy of the sun at its apogee)

KEsun = KEₐ - PEp = 1.4034074440E+41 J (KE of the sun used to rotate Milky Way's force-centre)

KEFC = ½.J.ω₁² = 1.4375071338E+52 J (rotational KE in Milky Way's force-centre)

N = KEFC / KEsun = 1.02429E+11

(number of {our} solar systems needed to rotate Milky Way's force-centre)

ωFC = √[2.KEFC / J] = 2.0124E-07 ᶜ/s (actual angular velocity of the Milky Way's force-centre)

- Refer to Table 1. The density of iron is used for this calculation as an example. It is expected that the actual density of the Milky Way's force-centre is somewhere between the the earth's density and less than that of iron.

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