# Core Pressure {© 08/11/17}

The term "core pressure" refers to the pressure within any mass from its centre to its surface and includes that of any planet or star.

## Calculations

Isaac Newton's well-known formula can be modified to determine gravitational acceleration at the surface of a sphere (Fig 1) thus:

g = G.m₁ / Rp²

where 'Rp' is a radius form the centre of any mass at which you require the internal pressure

Force at Rp: F = m₂.g

therefore F = G.m₁.m₂ / Rp²

and p = F/A = G.m₁.m₂ / 4.π.Rp⁴

The difference between the application of Newton's formula for:

**two separate bodies**: Rp is the distance between the centres of mass

**within a single body**: Rp is the radius where the force (or pressure) is required

i.e. Newton's formula is indeed universal

Fig 1. Internal Pressure Model (Sphere)

### Atomic Spacing & Density

Where constant density is assumed in the following Tables, this is not strictly correct as density varies with pressure and will be significant in such large masses as planets and stars e.g.:

The repulsion force (F) between two adjacent iron atoms an be calculated as follows:

F = k.(Z.Q)² / ℓ² = 8.19881E-06 N

where:

Z = atomic number of iron (26)

Q is the elementary charge unit

ℓ = separation distance between adjacent atoms that can be calculated at earth-ambient pressure thus:

ℓ = ³√[ 3.m / 4.π.ρ) ] = 1.379205E-10 m

where:

m is the mass of an iron atom (8.648683E-26 kg)

ρ is the density of iron (7870 kg/m³)

Note: The radius of the 3rd electron shell in an iron atom is 4.76259E-10 m, so the density of a single iron atom is 191.13128. We may conclude therefore, that adjacent iron atoms mesh in their lattice structure at ambient pressure.

If the force on an atom generated by the pressure at a radius is greater than 8.19881E-06 N,

ℓ will reduce with radius and density will increase accordingly.

This density modification will be applied to this page after completion of a following paper on Atoms in Lattice Structures.

## Calculation Results

The following Tables comprise the results for various test cases.

### Milky Way Force-Centre

In a sphere of uniform density (7870 kg/m³), e.g. a collapsed star such as that in the centre of the Milky Way, the above calculation produces the following pressure profile:

ρ⁽¹⁾ | Rp | p | p | |

% | (kg/m³) | (m) | (N/m²) | (bar) |

0% | 7870 | 5412.471041 | 1.691E+26 | 1.691E+21 |

5% | 7870 | 2706235521 | 3.381E+20 | 3.381E+15 |

10% | 7870 | 5412471041 | 1.689E+20 | 1.689E+15 |

15% | 7870 | 8118706562 | 1.123E+20 | 1.123E+15 |

20% | 7870 | 10824942083 | 8.386E+19 | 8.386E+14 |

25% | 7870 | 13531177603 | 6.657E+19 | 6.657E+14 |

30% | 7870 | 16237413124 | 5.484E+19 | 5.484E+14 |

35% | 7870 | 18943648645 | 4.623E+19 | 4.623E+14 |

40% | 7870 | 21649884166 | 3.956E+19 | 3.956E+14 |

45% | 7870 | 24356119686 | 3.415E+19 | 3.415E+14 |

50% | 7870 | 27062355207 | 2.959E+19 | 2.959E+14 |

55% | 7870 | 29768590728 | 2.563E+19 | 2.563E+14 |

60% | 7870 | 32474826248 | 2.209E+19 | 2.209E+14 |

65% | 7870 | 35181061769 | 1.887E+19 | 1.887E+14 |

70% | 7870 | 37887297290 | 1.587E+19 | 1.587E+14 |

75% | 7870 | 40593532810 | 1.303E+19 | 1.303E+14 |

80% | 7870 | 43299768331 | 1.031E+19 | 1.031E+14 |

85% | 7870 | 46006003852 | 7.675E+18 | 7.675E+13 |

90% | 7870 | 48712239373 | 5.091E+18 | 5.091E+13 |

95% | 7870 | 51418474893 | 2.538E+18 | 2.538E+13 |

100% | 7870 | 54124705001 | 5.072E+12 | 5.072E+07 |

Table 11) see Atomic Spacing & Density above |

The force-centre at the heart of our Milky Way is a large, slowly rotating ball of iron.

Variable density (due to compression) will be dealt with in a future revision.

### Earth with Uniform Density

If the Earth had a uniform density of 5506.351327 kg/m³, the above calculation produces the following pressure profile:

ρ⁽¹⁾ | Rp | p | p | |

% | (kg/m³) | (m) | (N/m²) | (bar) |

0% | 5506.35 | 0.637100069 | 1.14675E+18 | 1.14675E+13 |

5% | 5506.35 | 318550.0343 | 2.29322E+12 | 2.29322E+07 |

10% | 5506.35 | 637100.0685 | 1.14561E+12 | 1.14561E+07 |

15% | 5506.35 | 955650.1028 | 7.61922E+11 | 7.61922E+06 |

20% | 5506.35 | 1274200.137 | 5.68790E+11 | 5.68790E+06 |

25% | 5506.35 | 1592750.171 | 4.51534E+11 | 4.51534E+06 |

30% | 5506.35 | 1911300.206 | 3.71930E+11 | 3.71930E+06 |

35% | 5506.35 | 2229850.24 | 3.13596E+11 | 3.13596E+06 |

40% | 5506.35 | 2548400.274 | 2.68340E+11 | 2.68340E+06 |

45% | 5506.35 | 2866950.308 | 2.31612E+11 | 2.31612E+06 |

50% | 5506.35 | 3185500.343 | 2.00682E+11 | 2.00682E+06 |

55% | 5506.35 | 3504050.377 | 1.73811E+11 | 1.73811E+06 |

60% | 5506.35 | 3822600.411 | 1.49842E+11 | 1.49842E+06 |

65% | 5506.35 | 4141150.445 | 1.27973E+11 | 1.27973E+06 |

70% | 5506.35 | 4459700.48 | 1.07631E+11 | 1.07631E+06 |

75% | 5506.35 | 4778250.514 | 8.83956E+10 | 8.83956E+05 |

80% | 5506.35 | 5096800.548 | 6.99520E+10 | 6.99520E+05 |

85% | 5506.35 | 5415350.582 | 5.20592E+10 | 5.20592E+05 |

90% | 5506.35 | 5733900.617 | 3.45300E+10 | 3.45300E+05 |

95% | 5506.35 | 6052450.651 | 1.72164E+10 | 1.72164E+05 |

100% | 5506.35 | 6371000.685 | -7.639E-05 | 7.639E-10 |

Table 21) see Atomic Spacing & Density above |

The above pressure profile is exactly as expected given the profile for Milky Way's force-centre as would be the case for any spherical solid, irrespective of the size or density.

Variable density (due to compression) will be dealt with in a future revision.

### Earth with Variable Density

The Earth, however, does not have a uniform density. And whilst we have an indication of its boundary layers and and a good idea of its total mass, the density variation must represent the total planet mass & a 'Δ' value of 0.1938197 and accord with Newton's laws of gravitational attraction.

In order to achieve the above, the layer-densities were iterated until all the Spin Theory requirements were met (**Table 3**), including the fact that the density at the very centre of the iron core is assumed to be that of iron
under compression (13000 kg/m³).

ρ⁽¹⁾ | Rp | p | p | ||||

% | (kg/m³) | (m) | (N/m²) | (bar) | |||

0% | 13000 | 0.637100069 | 6.39189E+18 | 6.39189E+13 | |||

5% | 12978.8 | 318550.0343 | 1.27406E+13 | 1.27406E+08 | |||

10% | 12957.7 | 637100.0685 | 6.34398E+12 | 6.34398E+07 | |||

15% | 12936.5 | 955650.1028 | 4.20550E+12 | 4.20550E+07 | |||

19% | 12918.0 | 1234845 | 3.23260E+12 | 3.23260E+07 | |||

25% | 12294.2 | 1592750.171 | 2.25093E+12 | 2.25093E+07 | |||

30% | 11739.0 | 1911300.206 | 1.69042E+12 | 1.69042E+07 | |||

35% | 11183.8 | 2229850.24 | 1.29366E+12 | 1.29366E+07 | |||

40% | 10628.6 | 2548400.274 | 9.99790E+11 | 9.99790E+06 | |||

45% | 10073.4 | 2866950.308 | 7.75148E+11 | 7.75148E+06 | |||

50% | 9518.2 | 3185500.343 | 5.99637E+11 | 5.99637E+06 | |||

54% | 9083.6 | 3434844.999 | 4.88130E+11 | 4.88130E+06 | |||

60% | 8082.1 | 3822600.411 | 3.22816E+11 | 3.22816E+06 | |||

65% | 7259.3 | 4141150.445 | 2.22426E+11 | 2.22426E+06 | |||

70% | 6436.6 | 4459700.48 | 1.47068E+11 | 1.47068E+06 | |||

75% | 5613.8 | 4778250.514 | 9.18794E+10 | 9.18794E+05 | |||

80% | 4791.0 | 5096800.548 | 5.29582E+10 | 5.29582E+05 | |||

85% | 3968.3 | 5415350.582 | 2.70381E+10 | 2.70381E+05 | |||

90% | 3145.5 | 5733900.617 | 1.12682E+10 | 1.12682E+05 | |||

95% | 2322.8 | 6052450.651 | 3.06354E+09 | 3.06354E+04 | |||

100% | 1500 | 6371000.685 | -5.66874E-06 | -5.66874E-11 | |||

Table 3: Pressure Profile for the Earth with a Variable Density1) The densities were selected for a smooth profile based upon the values defined in Planetary Spin Theory |

Despite variable densities, the pressure profile remains pretty much as would be expected for a sphere of uniform density.

The following Table represents the rotational model of the Earth according to Newton's laws of motion that correctly predicts its Δ value along with the densities, layer dimensions and angular velocities according to planetary spin theory.

ρ | Rp | Boundary Details | m | J | |||

% | kg/m³ | (m) | (kg) | (kg.m²) | |||

0% | 13000 | 0.637100069 | ω = 7.22399E-05 ᶜ/s Rp = 1234845 m | 1.408E+04 | 2.286E+03 | ||

5% | 12978.8 | 318550.0343 | 1.759E+21 | 7.139E+31 | |||

10% | 12957.7 | 637100.0685 | 1.229E+22 | 2.210E+33 | |||

15% | 12936.5 | 955650.1028 | 3.331E+22 | 1.502E+34 | |||

19% | 12918.0 | 1234845 | 5.470E+22 | 4.495E+34 | |||

25% | 12294.2 | 1592750.171 | ω = 3.3142E-05 ᶜ/s Rp = 3434845 m | 1.139E+23 | 1.569E+35 | ||

30% | 11739.0 | 1911300.206 | 1.481E+23 | 3.128E+35 | |||

35% | 11183.8 | 2229850.24 | 1.971E+23 | 5.841E+35 | |||

40% | 10628.6 | 2548400.274 | 2.496E+23 | 9.899E+35 | |||

45% | 10073.4 | 2866950.308 | 3.041E+23 | 1.559E+36 | |||

50% | 9518.2 | 3185500.343 | 3.594E+23 | 2.319E+36 | |||

54% | 9083.6 | 3434844.999 | 3.195E+23 | 2.481E+36 | |||

60% | 8082.1 | 3822600.411 | ω = 7.29124E-05 ᶜ/s | 5.512E+23 | 5.241E+36 | ||

65% | 7259.3 | 4141150.445 | 4.871E+23 | 5.721E+36 | |||

70% | 6436.6 | 4459700.48 | 5.072E+23 | 7.139E+36 | |||

75% | 5613.8 | 4778250.514 | 5.148E+23 | 8.632E+36 | |||

80% | 4791.0 | 5096800.548 | 5.079E+23 | 1.013E+37 | |||

85% | 3968.3 | 5415350.582 | 4.845E+23 | 1.153E+37 | |||

90% | 3145.5 | 5733900.617 | 4.426E+23 | 1.271E+37 | |||

95% | 2322.8 | 6052450.651 | 3.802E+23 | 1.351E+37 | |||

100% | 1500 | 6371000.685 | 2.953E+23 | 1.377E+37 | |||

Table 4: Angular Momentum Profile for the Earth as Defined in Table 3 |

**Verification Example**

At a 3 km depth the radius of the earth is 6368000.7 m

If the surface of the earth above this radius was covered with seawater (ρ = 1025 kg/m³), the mass between the surface and this depth (m₂) would be:

m₂ = 4/3.π.(R₂³-R₁³).ρ = 4/3.π.(6371000.7³ - 6368000.7³) x 1025 = 1.56771E+21 kg

and the mass below this radius (m₁) would be:

5.96452E+24 - 1.56771E+21 = 5.96295229E+24 kg

Therefore gravitational acceleration at this depth (3 km):

g = 6.67359232E-11 x 5.96295229E+24 / 6368000.7² = 9.8132938 m/s²

**Pressure**

The force of the water at this radius (6368000.7 m) can be calculated using Newton’s formula:

F = m₂.g = 1.56771E+21 x 9.8132938 = 1.53844E+22 N

and the surface area at this depth (3 km) can be calculated thus:

A = 4.π.R₁² = 4 x π x 6368000.7² = 5.09584335E+14 m²

**Verification**

The pressure at this depth is therefore:

p = F/A = 1.53844E+22 / 5.09584335E+14 = **3.01901E+07** N/m²

A simplified (but less accurate) calculation would be:

p = 1025 x 3000 x 9.8132938 = **3.01759E+07** N/m²

The above calculation was shown simply to prove the procedure.

These calculations confirm that the spherical model for planet Earth accords with Newton's laws, spin theory and the Earth's physical appearance as far as we know it.

The sun is rotating the earth's core through its potential energy and thereby generating its internal heat through friction, i.e. it is not stored heat.

## Fusion Within the Earth

An interesting hypothesis:

If fusion occurs when sufficient gravitational energy (Eg) is available to push two atoms close enough for their nuclei to contact and neutrons are able to bind the nuclei:

This would occur when; Eg = k.(n.Q)² / 2.Rp

where Rp is the radius of a proton

For iron atoms (Z = 26):

Eg = 8.987552E+09 x (26 x 1.6021765E-19) / 2 x 1.7761327E-15 = 4.390393E-11 J

This gravitational energy is achieved in the earth's core at radius

(assuming a maximum core of density 13,000 kg/m³):

R = G.m₂.mₐ / Eg {m³.kg.kg.s² / kg.s².kg.m² = m}

R = 6.67359E-11 x 5.965E+24 x 8.648683E-26 / 4.390393E-11 = 0.78418 m

and a mass of:

m = 4/3.π.R³.ρ = 26259.2076 kg

Which means that 26.26 tonnes of material at the very centre of the earth's core could be generating sufficient fusion energy to create heavier elements than iron.

However, we also know that the density at the very centre of the earth will be considerably higher than 13,000 kg/m³ so this calculation will need to be modified after completion of a following paper on Atoms in Lattice Structures

Refer to the following pages for planetary properties used in the above calculations:

Stars & Planets

Laws of Motion

Solar System Orbits

Planetary Spin

Dark Matter?

### Further Reading

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

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