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The common distance of the stars
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The common distance of the stars
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aetzbar
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The common distance of the stars
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on:
18/03/2015 17:52:31 »
The common distance of the stars , by Aetzbar ………………………………..……………………………1
Countlees stars moving infinite space.
The stars are combined according to the idea of a central star , and ather stars revolving
around it. Each star is a central star and ather star, at the same time.
Kepler discovered the univerce formula D^3 : T^2 = P ( P is number )
D – The distance of ather star from central star.
T – Time that ather star rotates around a central star.
Newton : Each star has its material quantity ( M – view Newton)
M determines P that appesrs in Kepler's universe formula.
If M will increase 5 times,then either P increase 5 times.
How to get the P of earth ? with Kepler's universe formula. D^3 : T^2 = P
The earth is a central star for the moon.
Moon is far from earth 384000 km, and time it orbited the earth 672 hours.
P = 384000^3 : 672^2 = 1.25*10^11
P of earth is determined by M of earth.
There are an infinite number of combination of D and T ,if selected D ,the formula tell us T. If you want to send a satellite to earth, a height of 88000 km ,time it will circle the earth,
predetermined. T = root of ( 88000^3 : 1.25*10^11) = 74 hours
Finding p of the Sun.
The Sun is a central star for the earth (with the moon moves around)
P is obtained by the distance ( Earth – Sun) and time of 1 year in hours.
P of Sun = 429*10^14 P of the Sun determined by M of the Sun.
The common distance of the stars , by Aetzbar ………………………………..……………………………2
Galileo introduced an imaginary experiment , in which a ball falls to the entrance of a
Tunnel that reaches the other side of the earth. This ball will illustrate a pendulum
Movement , between the two openings of the tunnel.
The cycle time will be 1.4 hours. Marked ]T[
P of earth and ]T[ of earth ,determines by M of earth.
The common distance of the stars
As we move away from a star, lap time around it will grow. ( T increases) Therefore ,the
Equality T = ]T[ must appear each star, at some distance from the center of the star.
So that all the stars in the universe will be in harmony, need only a single distance which appears the equality T = ]T[
How do we find the common distance of the stars ?
Since M of star determines its P and ]T[ , must be a connection between P and ]T[
The connection is P* ]T[^2 = number
This connection will reveal the common distance of the stars.
P of earth = 1.3*10^11 and ]T[ of earth = 1.4 hours
1.3 * 10^11)*1.4^2 = 2.5 * 10^11 )
( D^3 : T^2) * [T[^2 = 2.5 * 10^11
T = ]T[
D^3 = 2.5*10^11
D = 6300 kilometers
The common distance of the stars = 6300 km
Satellite close to the ground, makes a complete revolution in 1.4 hours.
]T[ of earth is 1.4 hours.
Earth space are within 6300 km from the center of the earth.
The common distance of the stars , by Aetzbar ………………………………..……………………………3
6300 km is a Natural Physical Constant.
The equality of times T = ]T[ appears at 6300 km from the center of each star.
Each star has its unique M and unique ]T[
M* ]T[^2 = constant
We live in a place of equality times
]T[ of Sun is 8.7 seconds
In distance of 6300 km from the center of the Sun, T is 8.7 seconds
]T[ of Earth is 5050 seconds
In distance of 6300 km from the center of the Earth, T is 5050 seconds
6300 km is common distance of all stars
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aetzbar
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Re: The common distance of the stars
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Reply #1 on:
20/03/2015 12:09:08 »
5 steps Creating a dynamic universe.
First step: Each star is moving and never rests.
Second step: Each star is a central star and side star, at the same time.
The Sun is a central star for the Earth ,and Earth is a central star for the Moon.
The Earth is a side star for the Sun, and the Moon is a side star for the Earth.
Side star revolves around central stare, with lap time T.
Third step: Each star has mass M and Cycle time ]T[
Fourth step: M determines ]T[ by the formula M* ]T[^2 = constant
Fifth step: The equality………. Lap time = Cycle time ….( or ]T[ = T ) will appear at
D= 6300 km, from the center of each star.(6300 km is a Natural Physical Constant)
Thus we have obtained the known universe.
Each star has a unique P resulting from the formula D^3 : T^2
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aetzbar
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Re: The common distance of the stars
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Reply #2 on:
21/03/2015 14:26:55 »
Dynamic universe without gravity, by Aetzbar --------------------------------------------------1 .
Each star has its Material quantity ( M – Newtonian view)
The stars are moving and never stop.
Stars move like a screw – turn and advanced.
Each star is a central star and side star, at the same time.(Copernican view)
The Sun is a central star for the Earth ,and Earth is a central star for the Moon.
The Earth is a side star for the Sun, and the Moon is a side star for the Earth.
Side star revolves around central stare, with lap time T
Let's say that the sun is a point and earth as a point
If we move away from the sun,the distance D will increase from zero km to infinity km
lap time T will increase (from zero hours to infinity hours)
If we move away from the Earth,( D will increase from zero km to infinity km )
lap time T will increase (from zero hours to infinity hours)
The Sun has a unique fit between D and T, and the Earth has a unique fit of D and T.
Each star has its unique fit between D and T.
Kepler fined the formula of the stars D^3 : T^2 = P
P is a unique number of star, that represents a unique fit between D and T.
There is a link between M of star to P of star ( M1 : M2 = P1 : P2 )
How to get the P of earth ?
The earth is a central star for the moon.
Moon is far from earth D = 384000 km, and time it orbited the earth T = 672 hours.
P of Earth = 384000^3 : 672^2 = 1.25*10^11
Dynamic universe without gravity, by Aetzbar ------------------------------------------------ 2
Finding p of the Sun.
The Sun is a central star for the earth (with the moon moves around)
P is obtained by the distance ( Earth – Sun) and time of 1 year in hours.
P of Sun = 429*10^14
Each star has also its Cycle time ( marked ]T[ )
Galileo introduced an imaginary experiment , in which a ball falls to the entrance of a
Tunnel that reaches the other side of the earth. This ball will illustrated a pendulum
Movement ( of 1.4 hours ) between the two openings of the tunnel.
The cycle time of the Earth will be 1.4 hours.
each star has its Cycle time.
M of star determines its ]T[ , by the formula M* ]T[^2 = K (constant)
Because M1 : M2 = P1 : P2
We have the formula P* ]T[^2 = K
According to P of Earth and ]T[ of Earth , K = 2.5*10^11
With the formula P* ]T[^2 = 2.5 *10^11 can calculate ]T[ of Sun = 8.7 seconds
Natural Physical Constant
The formula P* ]T[^2 = 2.5*10^11
is the formula (D^3 : T^2)* ]T[^2 = 2.5 * 10^11
The formula shows that the situation of T = ]T[ appears at the distance of 6300 km, from
the center of each star.
D^3 = 2.5*10^11
D = 6300 km
Thererfor , 6300 km is a Natural Physical Constant.
Dynamic universe without gravity, by Aetzbar ------------------------------------------------ 3
The equality of times law
The equality………. Lap time = Cycle time ….( or ]T[ = T ) will appear only at
D= 6300 km, from the center of each star.
6300 km is a Natural Physical Constant.
The equality of times law determines that each star has a unique fit between D and T
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