*

THE COMPLETE COSMOS & THE MATTER  /  THE SELF-REGULATING UNIVERSE & THE LIFE

Thoughts from a new start and about a beginning... Earth

***                                                                              comet

* HOME |     | ABOUT |     | FIND THE EXIT! |     | COSMOS & LIFE |     | COSMOLOGY

*

KNOWLEDGE &THINKING     |      MAN & SOCIETY     |     THE BOOKS     |      PHILOSOPHERS

?

DECEPTION

 

 

For creatively thinking!


 

 

 

CIRCULAR TIME – COMPLETE & SIMULTANEOUS UNIVERSE
Theory about a completed time and the relativity of energy
(An unified theory about time, space and matter)

 

 The close relation between the mass and speed of electromagnetic waves (5)

2010 ISBN978-960-93-2431-1  |  2012 ISBN978-960-93-4040-3

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 observed and Evangelos Karamicha's theory

 

For each speed Vg that comes out from the type 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 x10-8 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 x10-8 kg. 

In order to comes out in the result the speed of the light from the type V = √GM/r the mass should is equal with called in the physics Planck's mass  Mpl =√(h c /G). That is to say c = √(G Mpl / λpl ).

According to the relation :

 

* Under the term that in the denominator of the type 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 type in the 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 types of Newtonian physics in the sub atomic world can to reveal us very important relations and unexpected information. However, the types 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:

 

 

≈ DATA OF EARTH AND THE SUN ≈

 

Earthy mass   Mer : 5,973 x 1024 kg

Solar mass Ms :  1,99 x1030 kg

 

Middle distance Sun-Earth rers :

1,495978 x 1011 m

 

Middle orbit speed of Earth:

2,9784 x 104 m/sec (in 365,2564 middle sun days).

 

Length orbit: 9,4 x 1011 m

 

Diameter of Earth: 12,7574 x 106 m

Diameter of Sun: 13,92 x 108 m

 

G = 6,6725 x10-11 m3/kg sec2  

c = 2,997924 x108 m/sec

 

Fers = G Mer  Ms / r2    (Attraction Force between 2 spherical bodies).

 

Fers = (6,6725 x10-11 ) (5,973 x 1024 ) (1,99 x1030 ) / (1,495978 x 1011 m)2 

 

Fers = G (11,886 x 1054 ) / 2,23795 x 1022

 

Fers = 79,3093 x 1043 /  2,23795 x 1022

 

Fers = 35,4383 x 1021 Newton 

 

 

35,4383 x 1021 x r2 = 79,3093 x 1043

 

We observe Fers r2 = G Mer Ms

 

 

From the above types of Newtonian Physics are result and we should look the following relations :

 

F r^2 = M1 M2 G

 

F r^2 / G = M1 M2

 

F r^2 / G = M1 M2

 

G M1 M2 / r = F r^2 / r = E | F r^2 = M1 M2 G

 


 

 

► The known type G M / r2 gives us the acceleration g where result from the force Fg of the 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 x10-11 ) x (5,973 x1024 ) / (6,3787 x106 )2 = 39,8548 x1013 / 40,68781 x1012 = 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 x 1011 m and with mass of Earth Mer = The mass of Earth:

g = G Mer / rers2 (6,6725 x10-11 ) x (5,973 x1024 ) / (1,495978 x 1011 )2 = 17,8086 x10-9 m/s2

 

We find the same acceleration as from the type Fers / Ms when we put the mass of Sun in the denominator and - attention- not the mass of Earth. That is to say

g = G Mer / rers2 = Fers / Ms

 

g = Fers / Ms 35,4383 x 1021 / 1,99 x1030 = 17,808 x10-9 m/s2

 

 

Now, with mass Ms = the mass of Sun :

g = G Ms / rers2 (6,6725 x10-11 ) x (1,99 x1030 ) / (1,495978 x 1011 )2 = 5,9332 x10-3 m/s2

 

We find the same acceleration as from the type Fers / Mer when we put the mass of Earth in the denominator and - attention- not the mass of Sun. That is to say

g = G Ms / rers2 = Fers / Mer

 

g = Fers / Mer 35,4383 x 1021 / 5,973 x1024 = 5,933 x10-3 m/s2

 

We observe: G M1 / r2 = F / M2

F = M1 g2 = M2 g1


 

► If the type 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 type V = √GM/r we find the speed of Earth orbit around the Sun at the middle radius : r = 1,495978 x 1011 m

V = (G Msun / r ) =( 13,2782 x1019 / 1,495978 x1011 )= √8,875982 x108

V = 2,979 x104 m/sec  (This is the middle speed of Earth around the Sun).

 

This same speed V=2,979 x104 of the type V = √G Msun / r we find from the type 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 the time T of  one complete rotation.

 

sq root G M1/r = sq root F r / M2 = V = r 2pi / T

 

 

The type 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 the physics "Escape velocity". For example, from mass of Earth is result :

V = √ (2 G 5,973 x1024 / 6,3787 x106 ) = √(79,709685 x1013 / 6,3787 x106 )

V = 12,496227 x107 = 11,178 x 103 m/sec

 

 

► According to Newtonian physics, we found the force F between the mass of the Earth and Sun and the speed of the Earth in the orbit (centripetal speed V=2,979 x104 m/sec) around of the Sun. We know their middle distance and the length of the orbit, where the Earth follow r x 2π = 9,4 x 1011 m

The known types, allow us to find the period of one complete orbit of the Earth around of the Sun, the time until come one rotation. This time, we find from the type T = r x 2π / Vκ

V = 2pi r / T | T = 2pi r / V | r = T V / 2pi

T = r x 2π / Vκ → 9,4 x 1011 / 2,979 x104 = 3,155421 x 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 x 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 type :

F = M V^2 / r = G M M / r^2

 

F = M V2 / r → F = 5,973 x1024 (2,979 x104 )2 / 1,495978 x 1011 = 35,43303 x1021 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 the 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 type 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 :

M1 = F r^2 / G M2

M2 = F r2 / G M135,43303 x1021 x (1,495978 x 1011 )2 / G 5,973 x1024 = 1,9896 x1030 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 types we can find any other second mass, when we imagine any mass in orbit, with any speed and radius.

 

 

 

 

www.kosmologia.gr  1st publication on Earth

 

 

 

The simpler relations of physics, which use the most capable researchers in order to they solve the most tangled mathematic problems and in order to find solutions in the impasses of modern physics, these relations should they had been supplemented and they had been delimited by the professionals physicists.

 

 

 

back

next page

 

 

 

 

 

 

COSMOLOGICAL THEORY

ABOUT A COMPLETED TIME OR UNIVERSE AND THE RELATIVITY OF ENERGY

 

 

 

e-mail

2004-2013 Konstantinos G. Nikoloudakis - All rights reserved

 

THE THEOLOGY        http://www.kosmologia.gr             SCIENCE            

OF

 HOME PAGE

 

ABOUT

clock

Best view at 1024x768px | screen 17"minimum | IE v.6.0 +