 6/10 CYCLICAL TIME – COMPLETE & SIMULTANEOUS UNIVERSE
Theory on a complete time and the relativity of energy
(A unified theory about time, space and matter)

#### The close relation between mass and speed of electro­magnetic waves (6)

Attention! Exist translational errors

Now we look from the start and briefly how unravel the big conundrums of modern physics, as it would be supposed they had made since a lot of decades ago.

 ► Reminder. What Evangelos Karamicha's observed and his theory   For each speed Vg that comes out from the formula V= √GM/r can calculate theoretically a particle of certain mass M. So, reversely, for each particle mass M it can be calculated a certain speed Vg. This speed results faster c for particle of biggest mass Plank Mpl =√(h c /G) = 5,45624 ×10-8 kg. If however the speed of light c is the superior speed in to the Universe, then how much mass will have the particle, which we would find speed of c? This speed results maximum c for a particle with a maximum Plank's mass  Mpl =√(h c /G) = 5,45624 ×10-8 kg.  In order to comes out in the result the speed of light from the formula V = √GM/r mass should is equal with called in physics Planck's mass  Mpl =√(h c /G). That is to say c = √(G Mpl / λpl ). According to the relation : * Under this term that in the denominator of the formula GM/r we put the Compton's length for Planck's Energy λ=h/Μpl c and not the radius r that is in effect for big masses (celestial bodies). That is to say c = √(G Mpl / λpl ). The unit m (meter) remains same.   From the relation are result with the simple method of three following relations: the same as: Mplanck = √(h c /G)    |     c = √(G Mpl / λpl ). The above relations agree with the known formula in physics V = √GM/r, after Vm =m c / Mplanck = √(G m / λm) where λm the length Compton of particle M from the relation λ=h/M c and not radius r that is in effect for big masses. The unit m (meter) remains same. In other formulation, Evangelos Karamichas says to us :  The application of the formulas of Newtonian physics in the sub atomic world can to reveal us very important relations and unexpected information. However, the formulas and the equations of Newtonian physics result with the observations in the visible world and are applied successfully in a multitude of cases. With their description on the usual and visible world, we know well what they describe, what phenomena and what relations are connect the phenomena between them. So, it's useful to remember them so that we avoid childish errors with their application in the microscopic world and in order to reveal potential relations and phenomena, respectively with the known relations and phenomena of the visible our world.

From Newton and his famous laws we know the following relations: * Attention to the decimal point. This is a different mark for many countries

 ≈ DATA OF EARTH AND THE SUN ≈   Earthy mass   Mer : 5,973 × 1024 kg Solar mass Ms :  1,99 × 1030 kg   Middle distance Sun-Earth rers : 1,495978 × 1011 m   Middle orbit speed of Earth: 2,9784 × 104 m/sec (in 365,2564 middle sun days).   Length orbit: 9,4 × 1011 m   Diameter of Earth: 12,7574 × 106 m Diameter of Sun: 13,92 × 108 m   G = 6,6725 × 10-11 m3/kg sec2   c = 2,997924 × 108 m/sec Fers = G Mer  Ms / r2    (Attraction Force between 2 spherical bodies).   Fers = (6,6725 ×10-11 ) (5,973 × 1024 ) (1,99 × 1030 ) / (1,495978 × 1011 m)2    Fers = G (11,886 × 1054 ) / 2,23795 × 1022   Fers = 79,3093 × 1043 /  2,23795 × 1022   Fers = 35,4383 × 1021 Newton      35,4383 × 1021 × r2 = 79,3093 × 1043   We observe Fers r2 = G Mer Ms

From the above formulas of Newtonian Physics are result and we should look the following relations :    ► The known type G M / r2 gives us the acceleration g where result from the force Fg of gravity at one body with mass Μ, that is to say g = G M / r2 and in radius r. Well. if we put the data of our planets we will find approximate the acceleration that result the gravitational field in the radius r. g = G M / r2 → (6,6725 ×10-11 ) × (5,973 ×1024 ) / (6,3787 ×106 )2 = 39,8548 ×1013 / 40,68781 ×1012 = 9,795  m / sec2     ► Now, we will find what acceleration result from the same type g = G M / r2 when we put in the denominator the middle radius of the Earth until the Sun rers = 1,495978 × 1011 m and with mass of Earth Mer = The mass of Earth: g = G Mer / rers2 → (6,6725 ×10-11 ) × (5,973 ×1024 ) / (1,495978 × 1011 )2 = 17,8086 ×10-9 m/s2   We find the same acceleration as from the formula Fers / Ms when we put mass of Sun in the denominator and - attention- not mass of Earth. That is to say g = G Mer / rers2 = Fers / Ms   g = Fers / Ms → 35,4383 × 1021 / 1,99 ×1030 = 17,808 ×10-9 m/s2     ► Now, with mass Ms = mass of Sun : g = G Ms / rers2 → (6,6725 ×10-11 ) × (1,99 ×1030 ) / (1,495978 × 1011 )2 = 5,9332 ×10-3 m/s2   We find the same acceleration as from the formula Fers / Mer when we put mass of Earth in the denominator and - attention- not mass of Sun. That is to say g = G Ms / rers2 = Fers / Mer   g = Fers / Mer → 35,4383 × 1021 / 5,973 ×1024 = 5,933 ×10-3 m/s2   We observe: G M1 / r2 = F / M2 ► If the formula change in GM/r its result in units is a speed at square V2. With root on the result we find the clear speed V,  V = √GM/r. The speed of rotation in orbit for a satellite around of earth is given from this relation. From the same formula V = √GM/r we find the speed of Earth orbit around the Sun at the middle radius : r = 1,495978 × 1011 m V = √(G Msun / r ) = √( 13,2782 ×1019 / 1,495978 ×1011 )= √8,875982 ×108 → V = 2,979 ×104 m/sec  (This is the middle speed of Earth around the Sun).   This same speed V=2,979 ×104 of the formula V = √G Msun / r we find from the formula √ Fers x rers / Mer  when we put the other mass (of Earth Μer and not of Sun). That is to say :   √(G Msun / rers ) = √(Fers x rers / Mer ) = V   (because G Mer Ms / r = Fers r = E )   ► Also, we can find  the speed of the Earth orbit from the simple type V = r 2π / Τ when we know the distance r 2π from the central mass and time T of  one complete rotation. ► The formula V = √ (2 G M / r ) give the speed of a body that is require so as to escape from the gravitational field of one body mass Μ and with radius r and this speed called in physics "Escape velocity". For example, from mass of Earth is result : V = √ (2 G 5,973 ×1024 / 6,3787 ×106 ) = √(79,709685 ×1013 / 6,3787 ×106 )→ V = √ 12,496227 ×107 = 11,178 × 103 m/sec     ► According to Newtonian physics, we found the force F between mass of the Earth and Sun and the speed of the Earth in the orbit (centripetal speed V=2,979 ×104 m/sec) around of the Sun. We know their middle distance and the length of the orbit, where the Earth follow r × 2π = 9,4 × 1011 m The known formulas, allow us to find the period of one complete orbit of the Earth around of the Sun, time until come one rotation. This time, we find from the formula T = r × 2π / Vκ T = r x 2π / Vκ → 9,4 × 1011 / 2,979 ×104 = 3,155421 × 107 sec The result is in seconds, because we use a speed in meters per second and the length of the orbit in meter, again. If we convert the seconds in days, then we will see the known number of the days of one earthly year. 3,155421 × 107 sec / 60 / 60 / 24 = 365,21 days (in 24hours). The small divergence from the accurate result, here do not engross us.   ► The force F between the Sun and the Earth is a centripetal force and this is expressed from the formula : F = M V2 / r → F = 5,973 ×1024 (2,979 ×104 )2 / 1,495978 × 1011 = 35,43303 ×1021 N We find the same force in N as from the Newton's type with 2 masses. Observe that in type F = M V2 / r  we put mass of Earth M that finds in orbit of radius r with speed V and not the central mass of the Sun. The relation  V2 / r = a =g  as the  F / Mer We can write F = M V2 / r = M g   With force F that we find from the formula F = M V2 / r we can find the second mass M2 , where that is applied or would be applied bilaterally the above force F. It will need the Newton's type : M2 = F r2 / G M1 → 35,43303 ×1021 × (1,495978 × 1011 )2 / G 5,973 ×1024 = 1,9896 ×1030 kg (The mass of Sun) We put the numbers that we know about mass M of the Earth, speed V in the orbit and the distance r from the sun and with the last type, we found the central mass of Sun. With the same formulas we can find any other second mass, when we imagine any mass in orbit, with any speed and radius.  Simple relations of physics, which the most capable researchers use them in order to solve the most tangled mathematic problems and find solutions in the impasses of modern physics, these relations should they had been supplemented and delimited by professionals physicists.

 COSMOLOGICAL THEORY ON A COMPLETE TIME OR UNIVERSE AND THE RELATIVITY OF ENERGY

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