Research Article Open Access
Masses of celestial bodies
A. S. Orlov, S. A. Orlov
SPGU, PGU
*Corresponding author: S. A. Orlov, Petrozavodski State University, Petrozavodsk, Russia, Tel: 79535496127
E-mail: @
Received: Jan 16,2019; Accepted: Jan 20,2019 ; Published: March 4,2019
Citation: Orlov SA, (2019) Masses of celestial bodies. Int J Mol Theor Phy 3(1):1-5
DOI: 10.15226/2576-4934/3/1/00116
Abstract
The calculation of the masses of celestial bodies based on the theory of vortex gravity, cosmology and cosmogony is proposed. Vortex gravity and cosmology is a new model of the emergence and existence of the universe and celestial objects. Based on this theory, the interaction of natural forces can be explained using ordinary physical patterns. The article calculates the actual values of the masses of celestial bodies. The values obtained are two orders of magnitude greater than the generally accepted values.

Keywords: Vortex gravity; cosmology; cosmogony; Heavenly mechanics.
Introduction
The masses of all celestial bodies are determined on the basis of the theory world gravity theories. The author of this theory I. Newton [1]. Newton presented the gravitational interaction of two bodies in the equation:
F π = G · m 1 . m 2 r 2 ( 1 )     Where MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaauaabeqabiaaaeaacaWGgbWaaSbaaSqaaiabec8aWbqabaGccqGH 9aqpaeaacaWGhbWaaSbaaSqaaiabl+y6NbqabaaaaOWaaSaaaeaaca WGTbWaaSbaaSqaaiaaigdaaeqaaOGaaiOlaiaad2gadaWgaaWcbaGa aGOmaaqabaaakeaacaWGYbWaaWbaaSqabeaacaaIYaaaaaaaaOqaam aabmaabaGaaGymaaGaayjkaiaawMcaaaaacaqGGaGaaeiiaiaabcca caqGGaaeaaaaaaaaa8qacaqGxbGaaeiAaiaabwgacaqGYbGaaeyzaa aa@4D32@
m1, m2 - the masses of bodies 1 and 2, respectively, G = 6.672 10-11 N ∙ m2 / kg2 is the gravitational constant, r is the distance between the bodies.

On the surface of the earth, this equation has the form:
F n =gm ( 2 )  Where MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaaiaadAeadaWgaaWcbaGaamOBaaqabaGccqGH9aqpcaWGNbGaamyB aaqaamaabmaabaGaaGOmaaGaayjkaiaawMcaaaaacaqGGaaeaaaaaa aaa8qacaqGxbGaaeiAaiaabwgacaqGYbGaaeyzaaaa@426C@ M - The mass of the body, on the surface of the Earth
g - Free fall acceleration.
Fn - the force of gravity.

From the known values of the acceleration of free fall g and G, the Earth’s mass Me = 5.97 x 1024 kg was determined [2]. The average density of its substance is 5500 kg / m3, which varies from values in the earth’s crust — 2200 kg / m3 to 13100 kg / m3 in the core of the Earth. [3]

In 1915, 1916 A. Einstein proposed the general theory of relativity [4]. In this theory, Newton’s law was considered as a special case.

In these theories, the general principle was the hypothesis about the property of a substance (body) to create the force of gravity.

The masses of other celestial bodies were determined based on the third Kepler law [5] taking into account equation (1).

The law of the world Newton (Einstein) has no experimental or theoretical evidence. Therefore, using this law in research, one can get contradictory results. In particular, according to the above method, the masses and densities of other planets of our solar system, including the Sun, were determined.

The sun has a density of 1410 kg / m3, the force of gravity on the surface F = 273.1 m [6]

Saturn - density of 687 kg / m3, gravity force F = 10.44 m. [7]

Planet Earth has gravity on the surface less than the indicated celestial bodies, but the average density of the Earth greatly exceeds the density of both the Sun and Saturn. Sealing of any bodies, including celestial bodies, occurs under the influence of external forces. These forces can only be attributed to the forces of its own gravity. Consequently, neither the Sun nor Saturn can have a density several times less than that of the Earth. In addition, the average density of Saturn is less than the density of water. Then on the surface the density of these celestial bodies must be equal to the density of the gas. This is an absurd conclusion. Such contradictions appeared within the framework of the theory of universal gravity. Three hundred years ago, Newton’s hypothesis was imposed on the world of scientists that bodies create the force of gravity. All scientists, in their calculations, were based on this erroneous concept. Therefore, the results of these studies were erroneous. Newton himself was not sure about the gravitational properties of bodies (matter). He expressed that the cause of attraction may be a change in density in the space environment. But he could not find a scientific justification for this assumption.

The author’s theory of vortex gravity, cosmology and cosmogony [8] is free from this contradiction. On this basis, it is possible to determine the true masses of celestial bodies.

The next chapter proposes the basic principles of the theory of vortex gravity.
Theory Of Vortex Gravitation, Cosmology And Cosmogony
The theory of vortex gravity is based on the well-known astronomical fact - all celestial objects rotate. In the vortex gravity model, the condition is accepted that this rotation is inertial and was caused by the rotation of the ether. Ether is a cosmic, gaseous, extremely low dense substance. Ether forms in the world space a system of interconnected vortices. The orbital velocities of the ether in each vortex (torsion) decrease in the direction from the center to the periphery. Velocities decrease in accordance with the law of the inverse square of this deletion. According to the laws of aerodynamics - the lower the flow velocity, the greater the pressure in it. The pressure gradient generates a pushing force towards the zones with the lowest pressure, that is, towards the center of this torsion. Thereby, in the center of the torsion, cosmic substance is accumulated or created, from which a celestial body is generated.

Let us consider the vortex gravity equation obtained in the theory [8].

The forces are shown acting on body 2: Fc – the centrifugal force, Fп – the force of attraction of body 2 from body 1; v2 – linear velocity of body 2 at the orbit, R – the radius of the orbit, r1 – the radius of body 1, r2 – the radius of body 2, w1 – angular velocity of ether rotation at the surface of body 1, and m2 are the mass of body 2.

Next we consider the appearance of the attraction force in more detail and derive a formula describing it. As was said above, a pressure gradient arises as the result of the vortex motion. Let’s find the radial distribution of the pressure and the ether velocity. For this purpose, we write the Navier-Stokes equation for the motion of a viscous liquid (gas).
ρ[ t + v .grad ] v = F gradP+η v ( 3 ) MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaaiabeg8aYnaadmaabaWaaSaaaeaacqGHciITaeaacqGHciITcaWG 0baaaiabgUcaRmaaFiaabaGaamODaaGaay51GaGaaiOlaiaadEgaca WGYbGaamyyaiaadsgaaiaawUfacaGLDbaadaWhcaqaaiaadAhaaiaa wEniaiabg2da9maaFiaabaGaamOraaGaay51GaGaeyOeI0Iaam4zai aadkhacaWGHbGaamizaiaadcfacqGHRaWkcqaH3oaAcqWIZwIvdaWh caqaaiaadAhaaiaawEniaaqaamaabmaabaGaaG4maaGaayjkaiaawM caaaaaaaa@5A68@ Where ρ is the ether density, v MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabiODayaala aaaa@3701@ and P are, respectively, its velocity and pressure, and η - the ether viscosity. In cylindrical coordinates, taking into account the radial symmetry vr=vz=0, vϕ=v(r), P=P(r), the equation can be written as the system:
{ v (r) 2 r = 1 ρ dP dr η.( 2 v( r ) r 2 + v( r ) rr v( r ) r 2 ) }=0 ( 4 ) MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaamaacmaaeaqabeaacqGHsisldaWcaaqaaiaadAhacaGGOaGaamOC aiaacMcadaahaaWcbeqaaiaaikdaaaaakeaacaWGYbaaaiabg2da9i abgkHiTmaalaaabaGaaGymaaqaaiabeg8aYbaadaWcaaqaaiaadsga caWGqbaabaGaamizaiaadkhaaaaabaGaeq4TdGMaaiOlamaabmaaba WaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWG2bWaaeWa aeaacaWGYbaacaGLOaGaayzkaaaabaGaeyOaIyRaamOCamaaCaaale qabaGaaGOmaaaaaaGccqGHRaWkdaWcaaqaaiabgkGi2kaadAhadaqa daqaaiaadkhaaiaawIcacaGLPaaaaeaacaWGYbGaeyOaIyRaamOCaa aacqGHsisldaWcaaqaaiaadAhadaqadaqaaiaadkhaaiaawIcacaGL PaaaaeaacaWGYbWaaWbaaSqabeaacaaIYaaaaaaaaOGaayjkaiaawM caaaaacaGL7bGaayzFaaGaeyypa0JaaGimaaqaamaabmaabaGaaGin aaGaayjkaiaawMcaaaaaaaa@6743@
After transformations, an equation is obtained to determine the gravity forces in the ether’s vortex:
F= v n ×ρ× v e 2 r ( 5 ) MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaauaabeqabiaaaeaacaWGgbGaeyypa0dabaGaamODamaaBaaaleaa daWgaaadbaGaamOBaaqabaaaleqaaOGaey41aqRaeqyWdiNaey41aq 7aaSaaaeaacaWG2bWaaSbaaSqaaiaadwgaaeqaaOWaaWbaaSqabeaa caaIYaaaaaGcbaGaamOCaaaaaaaabaWaaeWaaeaacaaI1aaacaGLOa Gaayzkaaaaaaaa@4687@ With the following relationship v e ~ 1 r MathType@MTEF@5@5@+= feaagGart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 qacaqG2bWdamaaBaaaleaapeGaaeyzaaWdaeqaaOWdbiaab6hadaWc aaWdaeaapeGaaGymaaWdaeaapeWaaOaaa8aabaWdbiaabkhaaSqaba aaaaaa@3BA4@ where

Vn is the volume of nucleons in the body, which is in a torsion orbit with a radius of - r.

ρ = 8.85 x 10-12 kg / m3 - density of ether [4]

Ve - the speed of the ether in the orbit r

r - The radius of the considered ether-vortex orbit

Replace in equation (5) the volume of nucleons on their mass, using the well-known relationship:

r - The radius of the considered ether-vortex orbit

Replace in equation (5) the volume of nucleons on their mass, using the well-known relationship:
v n =m\ ρ n ( 6 )  Where MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaaiaadAhadaWgaaWcbaGaamOBaaqabaGccqGH9aqpcaWGTbGaaiix aiabeg8aYnaaBaaaleaacaWGUbaabeaaaOqaamaabmaabaGaaGOnaa GaayjkaiaawMcaaaaacaqGGaaeaaaaaaaaa8qacaqGxbGaaeiAaiaa bwgacaqGYbGaaeyzaaaa@457D@
ρ ~1017 kg / m3 - density of nucleons, constant for all atoms.

m - The mass of nucleons in the body

Substituting (6) into (5) we get: F g = m ρ n ×ρ× v e 2 r = 10 28 ×m× v e r 2 ( 7 ) MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaaeaafa qabeqacaaabaGaamOramaaBaaaleaacaWGNbaabeaakiabg2da9maa laaabaGaamyBaaqaaiabeg8aYnaaBaaaleaacaWGUbaabeaaaaGccq GHxdaTcqaHbpGCcqGHxdaTdaWcaaqaaiaadAhadaWgaaWcbaGaamyz aaqabaGcdaahaaWcbeqaaiaaikdaaaaakeaacaWGYbaaaiabg2da9i aaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaGaaGioaaaakiab gEna0kaad2gacqGHxdaTdaWcaaqaaiaadAhadaWgaaWcbaGaamyzaa qabaaakeaacaWGYbaaamaaCaaaleqabaGaaGOmaaaaaOqaamaabmaa baGaaG4naaGaayjkaiaawMcaaaaaaaaa@5786@
In aerodynamics, the dependence of pressure in a gas (ether) P on its velocity w is represented by the equation: P(r)= P 0 +ρ w 1 2 r 1 3 [ 1 r 1 1 r ] ( 8 )        Where MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaaiaaccfacaGGOaGaaiOCaiaacMcacqGH9aqpcaGGqbWaaSbaaSqa aiaaccdaaeqaaOGaey4kaSIaaiyWdiabgwSixlaacEhadaWgaaWcba GaaiymaaqabaGcdaahaaWcbeqaaiaackdaaaGccqGHflY1caGGYbWa aSbaaSqaaiaacgdaaeqaaOWaaWbaaSqabeaacaGGZaaaaOGaeyyXIC 9aamWaaeaadaWcaaqaaiaacgdaaeaacaGGYbWaaSbaaSqaaiaacgda aeqaaaaakiabgkHiTmaalaaabaGaaiymaaqaaiaackhaaaaacaGLBb GaayzxaaaabaWaaeWaaeaacaaI4aaacaGLOaGaayzkaaaaaiaabcca caqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccaqaaaaaaaaaWdbi aabEfacaqGObGaaeyzaiaabkhacaqGLbaaaa@5D3D@
P0 - the ether pressure at the surface Using the boundary condition P( )= P b MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaabm aabaGaeyOhIukacaGLOaGaayzkaaGaeyypa0JaamiuamaaBaaaleaa caWGIbaabeaaaaa@3CB2@ , we find that P 0 = P b ρ. W 1 2 . r 2 MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaBa aaleaacaaIWaaabeaakiabg2da9iaadcfadaWgaaWcbaGaamOyaaqa baGccqGHsislcqaHbpGCcaGGUaGaam4vamaaBaaaleaacaaIXaaabe aakmaaCaaaleqabaGaaGOmaaaakiaac6cacaWGYbWaaWbaaSqabeaa caaIYaaaaaaa@4363@
Figure 1: Two-dimensional model of gravitational interaction of two bodies.
Pb- free ether pressure. In fig. 2 graphically illustrates the pressure distribution according to formula (8).

In accordance with the laws of ether dynamics, [9] in the free state of ether (at rest) has a pressure of Pb=2.1032.
Note 1
Using the vortex gravity equation (7), it is possible to calculate the gravitational forces that act only in the plane of the torsion or in its center. These equations are reliable for calculating the forces of gravity both above the surface and inside celestial bodies.
Figure 2: Radial distribution of pressure in the ether torsion
To determine the attractive forces in remote orbits of torsions, the author’s article “Gravitation - flat power field” [10] presents the calculation of the gravity equation in a three-dimensional model.
Note 2
Ether consists of super small particles - amers, which freely penetrate any substance, except super dense bodies - nucleons.
Determination Of The Mass Of Heavenly Body
The mass (M) of any is determined by multiplying its volume (V) by its density (ρ). M=V×ρ (9) MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaaiaad2eacqGH9aqpcaWGwbGaey41aqRaeqyWdihabaGaaiikaiaa iMdacaGGPaaaaaaa@3EA8@
The density of a substance is directly proportional to the force of compression that acts on this substance. Celestial bodies are compressed by static pressure under the influence of the forces of their own gravity. Based on the law of world attraction (Newton), it is assumed that in the center of the planets or stars the gravitational force decreases to zero. The density in the upper layers (in the crust) of our planet is known, it is 2200-2900 kg/ m3. When immersed, it increases. In the center, in the core of the Earth, the density was assumed to be 1.31 x 104 kg/m3 [3] [11]. The classical values of the density of the nuclei of celestial bodies raise great doubts. Static (technical) pressure in the center of the Earth in modern science is calculated to be 3.7 x 1010 kgs/m2 [12]. That is, the pressure in the center of the Earth increases a million times, and the density of a substance increases only by one order of magnitude compared with the surface layer of the earth. This discrepancy arises from the fact that the usual physical laws are trying to combine with the unproven, empirical equation of universal gravity. Newton’s equation obliges researchers to accept the Earth’s average density of 5,500 kg/m3. At other density values there will be another mass of the Earth, which does not correspond to Newton’s equation.

According to the theory of vortex gravity, the forces of attraction are not created by bodies. Therefore, the gravitational force of a celestial body does not depend on the mass of this body. The density and mass of celestial bodies can be calculated using the classical, physical laws - the more strongly the substance is compressed, the denser it is. The force of compression of the substance of celestial bodies creates only the force of gravity. Since the force of gravity increases towards the center of the torsion (the planet), the density of matter must also increase in proportion to the force of compression. ρP MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdiNaeS ipIOJaamiuaaaa@39B3@ Based on the theory of vortex gravity, vortex flows rotate not only above the surface of planets or stars, but also inside celestial bodies. Therefore, inside a celestial body, gravitational force increases along the same dependence as above the surface of this body. That is, inversely proportional to the square of the distance to the center of the celestial body.

In the earth’s crust, rocks were formed not on the surface of the planet, but inside the earth’s body. As a result of excessive pressure, the molten earth masses were squeezed, through the vents of volcanoes, onto the surface of the Earth [13]. Hardened lava formed the lithosphere on the outer layers of the Earth. The cooled lithosphere is a hard rock. It has a stable crystalline structure and can withstand the pressure of the overlying layers without compaction, to a certain depth. With an increase in the depth of the earth layers, the pressure on these layers increases. There is a destruction of bonds in crystals and atoms approach each other. In this pressure range, synthesis of new materials is possible. They are stable under normal conditions, with new properties. Example:

- at a depth of 30 kilometers, silicon oxide - a quartz mineral with a density of 2650 kg/m3 under a powerful pressure lying above the thickness turns into a denser modification of silicon oxide - with a density of 2910 kg/m3, - graphite with a density of 2230 kg/m3, under a high pressure of 108 kg/m2, is converted into diamond, with a density of 3510 kg/m3.

Thus, to change the structure of a substance during compression, the condition must be met - the force applied to it must exceed the strength of the interatomic bonds of the molecules or the strength of the crystal lattice. The result is the complete destruction of crystalline and interatomic bonds and the transition of a substance into an amorphous state. Then the substance will consist of individual atoms. With a further increase in the depth of deposition, the interatomic space in the compressed planetary substance will be reduced to a minimum. Atoms are touching. The substance is converted into plasma. The temperature of the substance rises. The density of a substance grows and reaches such values when not only atoms, but also the nuclei of these atoms get closer. In the center of the planet (stars) they merge into a single nucleus, with a maximum density equal to the density of nucleons ρi= 1017 kg/m3.

Based on the above, for calculating the density of terrestrial rocks should adopt the following scheme:

In the earth’s crust (hl ~ 30 km) the density of the earth’s substance is constant and does not exceed ρ0 ~ 2650 kg / m3,

- The pressure force at the depth hl has the value P = 2650 x 3 x 104 x 9.8 = 7.8 x 108 n/m2

- Below the mark of 30 km the terrestrial substance is in the form of magma without crystalline bonds.

Each underlying layer becomes denser under the influence of:

- Static pressure from the overlying layers (inversely proportional to the distance to the center of the planet),

- Terrestrial, vortex gravity (inversely proportional to the square of the distance to the center of the Earth).

Consequently, the density of terrestrial matter increases in cubic dependence on the depth of the deposit. This dependence can be represented by the equation: ρi= ρ 0 × ( r e r i ) 3 (10)  Where MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaaiabeg8aYjaadMgacqGH9aqpcqaHbpGCdaWgaaWcbaGaaGimaaqa baGccqGHxdaTdaqadaqaamaalaaabaGaamOCamaaBaaaleaacaWGLb aabeaaaOqaaiaadkhadaWgaaWcbaGaamyAaaqabaaaaaGccaGLOaGa ayzkaaWaaWbaaSqabeaacaaIZaaaaaGcbaGaaiikaiaaigdacaaIWa GaaiykaaaacaqGGaaeaaaaaaaaa8qacaqGxbGaaeiAaiaabwgacaqG YbGaaeyzaaaa@4D57@ ρi = is the density at the investigated depth,

ρ0= 2650 kg / m3 - the maximum density of the crust,

re - radius of a celestial body (Earth)

ri - distance from the center of the celestial body to the reservoir under study.

Based on equations (9) and (10), we determine the masses of the Earth, the Sun and the Moon

To find the mass, we integrate over the radius of the planet with partial sums in the form: dV=4π r 2 MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadA facqGH9aqpcaaI0aGaeqiWdaNaamOCamaaCaaaleqabaGaaGOmaaaa aaa@3D1A@ dρ= ρ 0 ( r e r ) 3   at MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabeg 8aYjabg2da9iabeg8aYnaaBaaaleaacaaIWaaabeaakmaabmaabaWa aSaaaeaacaWGYbWaaSbaaSqaaiaadwgaaeqaaaGcbaGaamOCaaaaai aawIcacaGLPaaadaahaaWcbeqaaiaaiodaaaGccaqGGaGaaeiiaiaa bggacaqG0baaaa@4510@ r n <r< r e  where MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCamaaBa aaleaacaWGUbaabeaakiabgYda8iaadkhacqGH8aapcaWGYbWaaSba aSqaaiaadwgaaeqaaOGaaeiiaiaabEhacaqGObGaaeyzaiaabkhaca qGLbaaaa@4278@ rn is the radius of the core of a celestial body Then the mass is calculated by equation (11) the integral will look like this: m v = r e r e ρ 0 ( r e r ) 3 4π r 2 dr+ m n = ρ 0 r e 3 4π r n r e 1 r dr+ m n = ρ 0 r e 3 4π( In( r e r n ) )+ m n (11) MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeqabeGaaa qaaiaad2gadaWgaaWcbaGaamODaaqabaGccqGH9aqpdaWdXaqaaiab eg8aYnaaBaaaleaacaaIWaaabeaakmaabmaabaWaaSaaaeaacaWGYb WaaSbaaSqaaiaadwgaaeqaaaGcbaGaamOCaaaaaiaawIcacaGLPaaa daahaaWcbeqaaiaaiodaaaGccaaI0aGaeqiWdaNaamOCamaaCaaale qabaGaaGOmaaaakiaadsgacaWGYbGaey4kaSIaamyBamaaBaaaleaa caWGUbaabeaakiabg2da9iabeg8aYnaaBaaaleaacaaIWaaabeaaki aadkhadaWgaaWcbaGaamyzaaqabaGcdaahaaWcbeqaaiaaiodaaaGc caaI0aGaeqiWda3aa8qmaeaadaWcaaqaaiaaigdaaeaacaWGYbaaai aadsgacaWGYbGaey4kaSIaamyBamaaBaaaleaacaWGUbaabeaakiab g2da9iabeg8aYnaaBaaaleaacaaIWaaabeaaaeaacaWGYbWaaSbaaW qaaiaad6gaaeqaaaWcbaGaamOCamaaBaaameaacaWGLbaabeaaa0Ga ey4kIipakiaadkhadaWgaaWcbaGaamyzaaqabaGcdaahaaWcbeqaai aaiodaaaGccaaI0aGaeqiWda3aaeWaaeaacaWGjbGaamOBamaabmaa baWaaSaaaeaacaWGYbWaaSbaaSqaaiaadwgaaeqaaaGcbaGaamOCam aaBaaaleaacaWGUbaabeaaaaaakiaawIcacaGLPaaaaiaawIcacaGL PaaacqGHRaWkcaWGTbWaaSbaaSqaaiaad6gaaeqaaaqaaiaadkhada WgaaadbaGaamyzaaqabaaaleaacaWGYbWaaSbaaWqaaiaadwgaaeqa aaqdcqGHRiI8aaGcbaGaaiikaiaaigdacaaIXaGaaiykaaaaaaa@7F48@ In equation (11) mn is the mass of the nucleus of celestial bodies. The density of the nuclei of celestial bodies is constant and it is rn = 1017 kg/m3. Then the masses of these nuclei are determined: m n = v n × ρ n , where MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa aaleaacaWGUbaabeaakiabg2da9iaadAhadaWgaaWcbaGaamOBaaqa baGccqGHxdaTcqaHbpGCdaWgaaWcbaGaamOBaaqabaGccaGGSaGaae iiaiaabEhacaqGObGaaeyzaiaabkhacaqGLbaaaa@4637@ Vn - core volume of a celestial body
The radius of the nuclei of celestial bodies (rn) is determined from equation (10) by substituting into it the values ρi = ρn = 1017 kg / m3, ρ0 = 2.65 x 103 kg/m3 and the radius of the celestial body re. Then the radius of the nucleus will be equal to rn = ri, and the volume of the nucleus: V n =4/3×π×r_ n 3 MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa aaleaacaWGUbaabeaakiabg2da9iaaisdacaGGVaGaaG4maiabgEna 0kabec8aWjabgEna0kaadkhacaGGFbGaamOBamaaCaaaleqabaGaey 4jIKnaaOGaaG4maaaa@4687@ The masses of nuclei are added to the masses of the corresponding celestial bodies in table 1
Table 1: physical characteristics of celestial bodies
mc, ρc - generally accepted masses and densities of celestial bodies,

mv, ρv - mass and density of bodies according to the calculation (equation 11).

The density of terrestrial matter at a depth equal to half the radius of the Earth with ri = 3,2x106 m is equal to: ρ i =21200kT/ M 3 MathType@MTEF@5@5@+= feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdi3aaS baaSqaaiaadMgaaeqaaOGaeyypa0JaaGOmaiaaigdacaaIYaGaaGim aiaaicdacaWGRbGaamivaiaac+cacaWGnbWaaWbaaSqabeaacaaIZa aaaaaa@41BE@
Conclusion
Based on the theory of vortex gravity and cosmology, the masses of celestial bodies are determined by 1 - 2 orders of magnitude more than the classical values. The obtained values of the masses of celestial bodies fully correspond not only to physical laws, but also to common sense. In astronomy, researchers are forced to operate with absurd physical quantities. A huge planet cannot, having a huge force of gravity, have a density equal to the density of gas.

According to the proposed methodology, any specialist can calculate the masses of any celestial bodies.

Using the laws of vortex gravity and cosmology, many paradoxes can be explained in various fields of natural science. Contradictions and paradoxes in science give rise to an erroneous understanding of the origin and action of the forces of gravity, which are imposed on the scientific world by outdated physical theories. These theories should include Newton’s theory of universal gravity and Einstein’s theory of relativity.
ReferencesTop
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