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Stars - Active and Dead incl. Black-Holes {© 03/12/17}

Note: apart from the theories proposed by Isaac Newton and Henri Poincaré, all the formulas and proposals provided below are exclusively from CalQlata.

Mathematical proof for the following theories is provided in:

All celestial bodies; stars, galactic force-centres, planets, moons, comets, etc. originally comprised the same matter and were ejected from the ultimate-body during the 'Big Bang'.

An Active Star

After a star has collected sufficient satellite mass (planets) to create the internal frictional heat through spin energy necessary to generate atomic fission it will have become active (see Stars & The Gas Planets). In this way, active stars create the universe's energy-bank together with the lighter elements and ultimately, hydrogen (proton-electron pairs).

All the heavy elements are created in the ultimate-body; the only place with sufficient gravitational (magnetic charge) energy to generate the fusion required to create them.

During its life, an active star continually emits (radiates) the energy generated by its nuclear fission in the form of alpha and beta particles, and electro-magnetic radiation.

Apart from a little core heat, the only energy a star will possess at the end of its life will be gravitational, but it will have retained all the mass it had when originally ejected from the ultimate-body.

A star's mass, and therefore its gravitational energy, does not [significantly] alter throughout its life. However, as it grows older, having converted most of its matter to hydrogen, gravitational attraction at a star's surface decreases as it grows in size.
i.e. the greater a star's density, the greater its gravitational attraction at its surface
and the greater a star's mass, the greater its gravitational energy

A Dead Star

A dead star is one that has converted all its matter to lighter elements and hydrogen atoms, and therefore generates insufficient internal frictional heat for fission to occur.

Black Holes

A black-hole is simply a star with no satllelites of its own, or no force-centre, and therefore generates no internal heat through planetary spin. I.e. it cannot emit electro-magnetic radiation; making it impossible to observe visually/directly.

As a theoretical exercise (that has no place in reality) it is possible to predict today's misconception of a black-hole as follows:

It is currently believed that a black-hole is body with sufficient density to prevent photons from escaping its surface.

For example; VY Canis Majoris has a radius (9.879082E+11 m) almost 20 times larger than that of the minimum iron star that could be expected to collapse to an instant black-hole (5.3218365E+10 m) according to the Calculations below⁽¹⁾


Whilst CalQlata has adopted 'U' to symbolise energy, with due respect to Henri Poincaré we have adopted his symbol 'E' for this page

From Isaac Newton's gravitational force: F = G.m₁.m₂ / R² → m.g = G.m₁.m₂ / R² → g = G.m / R²

The following Table contains the principle formulas used in the Calculations (below) to determine the properties of collapsed and active stars and black-holes.

FormulaVariables & ConstantsComments
V = 4/3 π.R³ V = volume of a sphere
R = radius of a sphere
ρ = m/V V = volume of a sphere
ρ = density
m = mass
R = ³√[3.m / 4π.ρ] R = radius of a sphere
ρ = density
m = mass
Calculating a star's radius based upon its known density and mass
F = G.m₁.m₂ / R² F = centripetal force between a force centre and an orbiting body
G = Newton's gravitational constant
m₁ = mass of the force centre
m₂ = mass of the orbiting body
R = the straight-line distance between the centres of the force centre and the orbiting body
The attracting (gravitational) force between a force centre (a sun) and an orbiting body (a planet)
F = m.a = m.g F = force
m = mass
a = acceleration
g = gravitational acceleration
As described by Isaac Newton
g = G.m / R² g = gravitational acceleration at 'R'
G = Newton's gravitational constant
m = mass of force centre (star)
R = radius of the force centre (star)
Used to determine the gravitational acceleration of a mass at radial distance 'R' from its centre
p = ρ.R p = core mass-pressure of the hydrogen gas
ρ = average planet density
R = planet radius
Mass-pressure at planet core
p = g/G p = mass-pressure
g = gravitational acceleration
G = Newton's gravitational constant
special case: unknown
m = p.R² m = mass of body
p = mass-pressure at interface of iron core and hydrogen gas
R = radius of star
Used to determine the mass of a star of known radius from its core pressure⁽³⁾
p₁/R₁ = p₂/R₂ R₁ = radius of star₁
R₂ = radius of star₂
p₁ = mass-pressure at core of R₁
p₂ = mass-pressure at core of R₂
A relationship between any two masses of the same average density⁽³⁾
R₁/R₂ = g₁/g₂ R₁ = radius of star₁
R₂ = radius of star₂
g₁ = gravitational acceleration₁
g₂ = gravitational acceleration₂
A relationship between any two masses of the same average density⁽³⁾
R₁/ρ₁ = R₂/ρ₂ R₁ = radius of star₁
R₂ = radius of star₂
ρ₁ = density of star₁
ρ₂ = density of star₂
A relationship between any two masses of the same material (e.g. hydrogen)⁽³⁾
R₂ =
(3.m₂.R₁ / 4.π.ρ₁)⁰˙²⁵
R₁ = radius of star₁
R₂ = radius of star₂
ρ₁ = density of star₁
m₂ = mass of star₂
Used to find the radius of a star of known mass of the same material (e.g. hydrogen)⁽³⁾
E = m.c² E = relativistic momentum (energy) in 'm' at 'c'
m = mass of star
c = speed of light in a vacuum
Limiting momentum (energy) in a body of mass (m) when moving at the speed of light (c)
E = m.g.R E = gravitational energy in 'm'
m = mass of body
R = radius of 'm'
g = gravitational acceleration at 'R'
Gravitational energy in a body of mass (m) at radius 'R'
E = G.m²/R G = Newton's gravitational constant
E = gravitational energy of star
m = mass of star
R = radius of star
Gravitational energy
v² = 2.a.R c = speed of light
g = gravitational acceleration at 'R'
R = radius of body (star)
Relationship between velocity and acceleration
c² = 2.g.R c = speed of light
g = gravitational acceleration at 'R'
R = radius of body (star)
Used to determine the equivalent limiting radius 'R' below which light cannot escape a given mass
v² = 2.G.m/R G = Newton's gravitational constant
m = mass of star
R = radius of star
v = velocity
Newton's escape velocity for a planetary body
m = ¾(E/G)⁰˙⁶ / ρ⁰˙² m = mass of body (star)
E = gravitational energy
G = Newton's gravitational constant
ρ = density of body
Alternative formula for the mass of a body
m =
(3.c⁶ / 32.π.ρ.G³)⁰˙⁵
m = mass of body (star)
c = speed of light
G = Newton's gravitational constant
ρ = density of body
Minimum (limiting) mass of a star that can be expected to collapse directly to form a black hole

Calculations {© 15/02/17}

The following comprises a series of calculations that predict the expected ultimate destiny of three stars according to the above formulas and data developed in CalQlata's 'Laws of Motion' web page.

The Earth's Sun

Known Properties (active):
mass (m) = 1.98850E+30 kg
radius (R) = 6.95710E+08 m
Calculated Properties (active):
Volume (V) = 4/3 . π.R³ = 1.4105E+27 m³
average density (ρ) = m/V = 1409.7829 kg/m³
gravitational acceleration (g) = G.m/R² = 274.1755834 m/s²
gravitational energy (E) = m.g.R = 3.793E+41 J
Known Properties (dead) assuming the density of iron:
average density (ρ) = 7870 kg/m³ ⁽⁴⁾
mass (m) ≈1.98850E+30 kg
Calculated Properties (dead):
volume (V) ≈ 4/3 . π.R³ ≈ 2.52668E+26 m³
radius (R) ≈ ³√[3.m / 4π.ρ] ≈ 3.921818E+08 m
gravitational acceleration (g) ≈ G.m/R² ≈ 862.8 m/s²
2.g.R ⁽⁵⁾ ≈ 6.767493E+11 m²/s² (c² = 8.987552E+16)

Given that 2.g.R is so much less than c², it is unlikely that the earth’s sun will become a black-hole immediately after collapse

VY Canis Majoris

Known Properties (active):
Radius (R) = 9.87908E+11 m
mass (m) = 5.967E+31 kg
Calculated Properties (active):
volume (V) = 4/3 . π.R³ = 4.03867E+36 m³
average density (ρ) = m/V = 1.477467E-05 kg/m³
gravitational acceleration (g) = G.m/R² = 4.0802E-03 m/s²
gravitational energy (E) = m.g.R = 2.4052224E+41 J
Known Properties (dead): (assuming the density of iron)
mass (m) = 5.967E+31 kg
average density (ρ) = 7870 kg/m³ ⁽⁴⁾
Calculated Properties (dead):
volume (V) ≈ 4/3 . π.R³ ≈ 7.58196E+27 m³
radius (R) ≈ ³√[3.m / 4π.ρ] ≈ 1.21870E+09 m
gravitational acceleration (g) ≈ G.m/R² ≈ 2681.156 m/s²
2.g.R ⁽⁵⁾ ≈ 6.53505E+12 m²/s² (c² = 8.9875518E+16)

Whilst VY Canis Majoris is very large, it has a mass little larger than our sun and therefore is unlikely to collapse into an instant black hole

Minimum Potential Black-Hole Star

Known Properties (dead):
c² = 2.g.R = 8.9875518E+16 m²/s²
average density (ρ) = 7870 kg/m³ (assuming the density of iron)
Calculated Properties (dead):
mass (m) = √(3.c⁶ / 32π.ρ.G³) = 9.6237854E+37 kg
volume (V) = m/ρ = 1.22284441E+34 m³
radius (R) = ³√(3.V / 4π) = 1.42920392E+11 m ⁽⁵⁾
gravitational acceleration (g) = G.m/R² = 314425.10331 m/s²
Known Properties (active):
mass (m) = 9.62378552E+37 kg
Calculated Properties (active):
radius (R) = ⁴√(R₁⁴ . m₁/m₂) = 5.802742927E+10 m
volume (V) = 4/3π.R³ = 8.1844330782E+32 m³
average density (ρ) = m/V = 117586.46475 kg/m³ ⁽⁴⁾
gravitational acceleration (g) = G.m/R² = 1907387.5622 m/s²
gravitational energy (E) = m.g.R = 1.0651682497E+55 J
2.g.R = 8.9875517874E+16 m²/s² (c² = 8.9875517874E+16)

A black-hole at the time of collapse (see Neutron Star below)

The above results are summarised in the following Table (input data):

UnitsThe Earth’s SunVY Canis MajorisMinimum Potential Black-Hole Star
average densitykg/m³1409.782931.477467E-05117586.46475
gravitational accelerationm/s²274.1755834.08021E-031907387.5622
gravitational energyJ3.79299803E+41 2.405222E+411.06516825E+55
average density ⁽⁴⁾kg/m³787078707870
gravitational accelerationm/s²862.8005882681.14672314425.10331
2.g.R ⁽⁵⁾m²/s²6.76749348E+116.5350384E+128.9875518E+16

Properties of a Black-Hole {© 15/02/17}

From our formulas above:
v² = 2.a.R
Special [limiting] case to trap photons: c² = 2.g.R → g = c² / 2.R
Gravitational (potential) energy: E = m.g.R → E = m.R.c² / 2.R
Limiting gravitational energy that will trap photons: E = m.c² / 2
i.e.: Kinetic energy of the mass at light speed: E = ½.m.c²
If we apply these calculations to the above (iron) limiting black-hole:
E = m.g.R = 9.6287855E+37 x 314425.10331 x 1.4292039174E+11 = 4.32696038E+54 J
KE = ½.m.c² = 9.6287855E+37 x 2997924582² ÷ 2 = 4.32696038E+54 J
CalQlata has therefore shown that the gravitational energy in a planet that will trap photons is equal to its kinetic energy at light speed

Assuming that the limiting mass for a star that can be expected to collapse to form an instant black hole (trap high-energy photons) is governed by the relationship 2.g.R = c², its mass can be calculated thus:
m = √(3.c⁶ / 32.π.ρ.G³) = 9.62378551609E+37
and its radius; R = 5.3218365E+10 m⁽¹⁾

However, it is claimed that all black holes are neutron stars, i.e. they collapse to the density of a neutron. If this is so its properties may be calculated as follows:

Neutron Star

The properties of a neutron:
m = 1.6749272928E-27 kg
R = 1.11328405737E-15 m
V = 5.77971706488E-45 m³
ρ = 2.8979399407E+17 kg/m³ ⁽⁶⁾

The radius of this star is related to its mass thus:
R³ = 3.m / 4.π.ρ
R = ³√(3.m / [4 x π x 2.8979399407E+17])
R = ³√m . ³√(3 / [4 x π x 2.8979399407E+17])
R = 9.37433933214E-07 x ³√m

According to Newton: v² = 2.G.m/R
Limiting condition: c² = 2.G.m/R
R = 2.G.m / c²
2.G.m / c² = 9.37433933214E-07 x ³√m
m/³√m = ¹˙⁵√m = 9.37433933214E-07 x c²/2.G
m = (9.37433933214E-07 x c² ÷ 2.G)¹˙⁵ = 1.58677073182E+31 kg
{check: m = √(3.c⁶ / 32.π.ρ.G³) = 1.58677073182E+31 kg}
which represents the minimum mass that could create a neutron star on collapse
Its radius: R = 2.G.m / c² = 23556.57050854 m
The gravitational acceleration at its surface is:
g = c² / 2.R = 1.90765285297E+12 m/s²
{check: g = G.m/R² = 1.90765285297E+12}
KE = ½.m.c² = 7.13059206345E+47 J
Eg = m.g.R = 7.13059206345E+47 J
confirming CalQlata's claim that the gravitational energy of a black-hole is equal to its kinetic energy at light speed

The problem is, however, whether or not iron can collapse to the density of a neutron with the above mass(es), because in order to do so, there can be no protons or electrons (or photons) within the body.
Moreover, its ultimate density is governed by the compressibility or iron.

Constants {© 02/02/17}

For all stars (and planets): 2.π.R.G.ρ / g = 1.5 {radians}
Proof: 2.π.R.G.ρ / 1.5 g = 1
4/3.π.R.G.(m/(4/3.π.R²)) / g
G.m / g.R² = 1
g = G.m / R²

We can check the above calculations as follows:

The Earth’s Sun
active: 2.π.G.ρ.R / g = 2.π.G x 1409.782932 x 6.95710E+08 ÷ 274.0806631 = 1.5 {radians}
dead: 2.π.G.ρ.R / g = 2.π.G x 7870 x 3.921817841E+08 ÷ 862.5018838 = 1.5 {radians}

VY Canis Majoris
active: 2.π.G.ρ.R / g = 2.π.G x 1.477467E-05 x 9.87908E+11 ÷ 4.07880E-03 = 1.5 {radians}
dead: 2.π.G.ρ.R / g = 2.π.G x 7870 x 1.21870E+09 ÷ 2680 = 1.5 {radians}

Minimum Potential Black-Hole Star
active: 2.π.G.ρ.R / g = 2.π.G x 117601 x 5.80350E+10 ÷ 1907222 = 1.5 {radians}
dead: 2.π.G.ρ.R / g = 2.π.G x 7870 x 1.42945138E+11 ÷ 314370.67118 = 1.5 {radians}

This constant also applies to planets (e.g. the earth):
2.π.G.ρ.R / g = 2.π.G x 5508.25830105342 x 6371000.685 ÷ 9.80663139027614 = 1.5 {radians}

A useful constant (K), based upon Newton's theories, for all matter is:
K = g / ρ.R = 2.7954278140935E-10 m³/s²/kg
G = K / 4/3π = 6.6735923200433E-11 m³/s²/kg

EHT: Black Hole

Looking at this recently released photograph, we (at CalQlata) are left with a number of concerns:

1) There is no such thing as a singularity. A black hole is simply a large body of matter too cold to emit electro-magnetic energy strong enough for us to detect.

2) And even if there were, by definition you would not be able to see it. Therefore, this cannot be a photograph of a black-hole. It is simply a photograph of a black space.

3) Singularities do not exist because they defy the laws of thermodynamics

4) Even if the dark area in the centre of the photograph is a singularity, why is it so large? The Schwarzschild’s radius for a singularity would be much smaller than depicted in this photograph

5) Einstein's theories of relativity have already been proven incorrect; so there is no such thing as an 'event-horizon'

6) The photograph was taken using radio-telescopes, which cannot detect colour, so how do we know that the coloured regions of the photographs are orange

7) The image looks exactly like a large star orbiting a large galactic force-centre; the bottom-front, 'bright' part of the image is our side of the orbit and the upper dark area looks like the back of the orbit

8) The variable nature of the light could be scattered matter left behind in the orbital trails or simply time-lapse photography. At such an orbital radius, the star's velocity would be very high indeed

9) The scientists presenting this discovery used the terms "I think" and "we think" far too many times to convince us of their interpretation

It is our (here at CalQlata) opinion that the European money would have been better spent discovering for our own galactic force-centre 'Hades'

Dark Matter

There is no such thing as dark matter.


  1. Compared with a radius of 3.26E+12 m for what is claimed to be a super-massive black hole at the centre of Messier 87 (spiral galaxy))
  2. At radii below (inside) the surface of a star (or planet), gravitational attraction is less than at its surface because there is less mass generating it. Gravitational attraction above (outside) the surface of a star is less than at its surface because its mass does not increase with radius.
  3. These relationships are representative at the birth of each star and/or stars of similar age, but are approximate for stars of different ages due to the variability of core size and material
  4. The density of a collapsed star will define its ultimate size, not its gravitational energy. Therefore, whilst the average density of the collapsed star may not be as defined above, any such variation will not affect its ability to trap photons
    The average density of iron (7870 kg/m³) for a dead star is estimated based upon this value appearing approximately half-way through its radial distance with lighter elements above this point and denser iron below it.
  5. Schwarzschild formula R = 2.G.m/c² predicts a radius of:
    R = 2 x 6.671282E-11 x 9.6287854E+37 / 299792458² = 1.42920391736095E+11m
  6. Whilst this calculation is based upon a neutron star comrising compacted neutrons throughout its radial distance, this is highly unlikely. Our Dark Matter? calculations show it to be an iron ball.

Further Reading

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

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