Core Pressure

{Keith Dixon-Roche © 08/11/17}

A mathematical explanation of the pressure inside any viscous mass, including celstial bodies.

Sources: Solar System Orbits; Laws of Motion; Physical Constants; Newton's 'G'
Related Books: Philosophiæ Naturalis Principia Mathematica Rev. IV; The Theory of Spin; The Mathematical Laws of Natural Science
Related Calculators: Orbital Motion; Physics; Planetary Spin
Online Calculator: Core-Pressure
CalQlata comment: This is the only study we (at CalQlata) know on this subject that manifestly works.

Introduction

The purpose of this study is to answer the following questions:
1) What is the source of core pressure?
2) What is the relationship between core pressure and atomic fusion?

Conclusions

Core pressure can be calculated using Isaac Newton's force law (Fig 4).
Natural atomic fusion occurs in massive bodies; galactic force-centres, the Great Attractor and the Ultimate Body due to core pressure.
Together with spin theory, core pressure may be used to define the internal structure of a planet, star, galactic force-centre, or in fact any celestial body.

Calculations

Isaac Newton's well-known formula can be modified to determine gravitational force (F) at the surface of a sphere thus:
F = G.m₁.m₂ / R²
where 'R' is the radius of the spherical pressure plane, 'm₁' is the mass of the substance inside the pressure plane (Fig 1) and 'm₂' is the mass of substance outside the pressure plane.

Gravitational ...
... acceleration at 'R'; g = G.m₁/R²
... force on spherical surface at 'R'; F = m₂.g
... pressure at 'R'; p = F/A = G.m₁.m₂ / 4.π.R⁴
where A is the surface area of the pressure plane at 'R'
moreover, these properties apply to the inside of a block of wood, a lump of iron or a piece of plastic.

An active planetary body will comprise a number of layers each of varying density that can be accommodated using the polar moment of inertia for a thin shell of varying density:
J = 8.532041001 x mn x (ρ₂/ρ₁) / (r₁⁵/r₂³)
where 'mn' is the mass of the shell, ρ₂ is its density at its outer radius (r₂) and ρ₁ is its density at its inner radius (r₁)

We already know the polar moment of inertia (J) for any planetary body, which can be determined using spin theory.
All we need to do is calculate the 'best-fit' for each layer from the information we know. For example; the inner-core of the earth is said to be 13,000 kg/m³ and the density of the outer surface is seawater (slightly higher than this actually) but following calculator shows a compatible construction for the earth based upon such data (Fig 1).

Calculation for a the earth as it is today

Fig 1. Internal Pressure Model of the Earth as it is Today

The earth comprises a number of layers similar to those shown in Fig 1 that match its calculated mass and polar moment of inertia.

Given that Newton's laws of motion are known to be correct, and that planetary spin theory accords exactly with Newton's laws of motion as does the above Core Pressure, it can be concluded that Fig 1 may be regarded as representative of the earth's construction

The above calculation achieves 1.0 for both factors; Fm and FJ (for mass and polar moment of inertia) and reveals the reason why the substructure of mountain ranges falls away from its continental mass; the density of the continental crust is higher than the upper mantle, and as such once melted will fall into the mantle.

It is claimed that the density of the earth's core is around 13,000 kg/m³ (Fig 1) and that about 90% of it is iron. However, we also know that the earth's core pressure is insufficient to compress iron owing to the coupling ratio.
Therefore, most of the core elements must be much heavier than iron.

An earth of constant density (5506.35 kg/m³) would have a polar moment of inertia of 9.68391E+37 kg.s² as shown in Fig 2 in which Fm & FJ are both equal to 1.0

Calculation for a the earth if it had an homogeneous internal structure

Fig 2. The Earth with an Homogeneous structure

Core Fusion

In order to find out if fusion is possible at the earth's core, which is said to be about 5000K:
The nucleic radius of an iron atom; R ≈7E-15 m
The [repulsive] electrical charge in a proton at 5000K is; e′ 7.8142246480E-19 C
There are eight neighbouring atoms in an iron lattice, therefore
the force between adjacent atoms is; F = k.e′² / 8.R² = 9.1 N/m²
The spherical surface area at R; A = 4πR² = 6E-28 m²
The pressure needed to fuse iron atoms at 5000K; p = F/A = 1.51054E+28 N/m²
The highest possible pressure at the earth's core is 1.9561E+18 N/m² (Fig 1; pₒ)

It can be concluded therefore, that fusion cannot occur naturally in the earth's core. Moreover, this same calculation has been performed for our sun with the same results.
However, fusion will occur naturally within the core elements of Hades, because it is sufficiently massive and its core elements are cold.

Claims

Claim 1: The pressure at the centre of the earth and probably any other planet (not star) is less than that required to generate Fusion in it atoms

Claim 2: The density of the earth's upper mantle is significantly less than the continental crust, but far hotter (upper mantle plumes) and sufficient to melt the substructure of mountains

Claim 3: The pressure at the centre of the earth's inner core is 1.93E+13 ≈bar