
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 (9)
©2010 ISBN9789609324311
 ©2012 ISBN9789609340403

Attention! Exist translational errors
THE GRAVITATIONAL CONSTANT G
(Some of the
thoughts that became in the start of investigation, was comprehended the
role about the constant G of gravity).
Observe how many expresses the things the known formula for the gravitational force:
Fgrav
= G M1 M2 / r^{2}
We say that the force is equal with the
product of two spherical masses M and reversely proportionally to the square of their distance r^{2}
(from their centers). If however, we do not insert the gravitational constant G, then
the result is not correct in units of force (Newton). The masses from alone them and the
distance that separates be taken like static sizes. Is not expressed the phenomenon of
attraction except the presence of two masses in some distance from each other. We separately
insert the constant G, which (in its dimensional content) includes time t
(sec) or acceleration (m/sec^{2}), while without time it could not this constant to express the phenomenon
of motion and acceleration, that the force between of masses can causes. We
can and insert
the constant G, because it has been observed that the sizes (M, r, F
and g) are increased or altered with the same rate and proportion. When we insert the constant G then
the numerator in the fraction minimizes, after this is a quantity smaller
than the unit (10^{11}). Thus, for masses 1kg and distance 1m we have
the result Fgrav=(6,6725 × 10^{11} ) × 1kg × 1kg / 1m^{2} = 6,6725 × 10^{11} N
Even if the formula gives a result in force of attraction, still we observe in the denominator a distance r^{2} that separates two (bodies) masses M that are attracted. This distance r^{2}
in the known formula is expressed as a straight line that links the centers of two masses, while the masses are considered spherical and we know that they
attract from all the radius of their ball. That is to say, the gravitational force exists to each radius in the overall surface of masses and in opposite
direction to the interior of bodies.
The straight line r^{2} that links two overall masses constitutes the conceivable straight
line that links only one from the radius of each ball with the second ball. We could consider that the gravitational force F in the Newton's formula is only one percentage of real gravitational force of each mass, the percentage
that it corresponds in their one only radius and only in straight line. The constant G
does expresses this percentage for the force of attraction in straight line distance?
We could still think, that the force of attraction from the opposite
directions of a same body possibly causes a distortion in its effect toward
an other separate body. Also, we easily observe that we speak for one
conceivable straight line that links the centres of two bodies, but this straight line does not coincide without fail with the straight line for the
application of force or from the shortest length between bodies, after
their attraction does not cause a rectilinear motion up to the conflict
between of
bodies.
With the units that
are found in the dimensional content in the constant G, the result in the formulas
is in agreement with
physics very simply, without it needs we make more numerical calculations. Actually, we could
not calculate this constant, if the phenomena of mass, gravitational force and distance they
were not altered according to certain laws, so as the one depends from other in
such way, that
they do not force certain limits.
We will watch their
relation with an
example of mass M and radius r of our planet Earth.
.
* Attention to the decimal point. This is a different mark for many countries
g = G M / r^{2}
= 9,795 m / sec^{2}

Mass (kg) 
Radius (m) 
Acceleration (g) 
Ratio g / M 
Mass x g
(F) 
M 
5,973000 × 10^{24}

(6,3787 × 10^{6} )^{2}

9,79527 
1,6399 × 10^{24}

58,50714 ×10^{24}

√M 
2,443972 × 10^{12}

(6,3787 × 10^{6} )^{2}

0,40079 ×10^{11}

1,6399 × 10^{24}

9,7952 
_{2}√M 
1,563320 × 10^{6}

(6,3787 × 10^{6} )^{2}

0,256372 × 10^{17}

1,6399 × 10^{24}

0,40079 ×10^{11}

_{3}√M 
1,250327 × 10^{3}

(6,3787 × 10^{6} )^{2}

0,205044 × 10^{20}

1,6399 × 10^{24}

0,256372 ×10^{17}

_{4}√M 
3,535996 × 10 
(6,3787 × 10^{6} )^{2}

0,579877 × 10^{22}

1,6399 × 10^{24}

0,205044 ×10^{20}


1 
(6,3787 × 10^{6} )^{2}

0,163992 × 10^{23}

1,6399 × 10^{24}

0,163992 ×10^{23}

We observe that the acceleration of gravity g
increases depending on mass M (for the same radius).
When mass M increases in square M^{2}
then the acceleration g of the M increases on x M, that is to say g × M
If mass M
decreases in square root √M, then the radius r^{2} should downsize in r^{2
}/ √M so that results the same acceleration g.
Mass of Earth in the square root √ Μ_{1} :
2,443972 ×10^{12}
F = G √M_{1} √M_{1} / r^{2}
→
F = G (2,443972 ×10^{12} ) (2,443972
×10^{12} ) / (6,3787 ×10^{6} )^{2} = 39,8548 ×10^{13} / 40,6878
×10^{12} →
F = 9,795 Newton
g = G M_{1} / r^{2} → G ×
5,973 ×10^{24} / (6,3787 ×10^{6} )^{2} = 9,795 m / sec^{2}
The gravitational force F between two same spherical
bodies is equal with gravitational accelerator g that cause the product of these two masses at
the same radius r. Or the gravitational accelerator g is equal with force F where they would
exert the 2 smaller masses between them, but smaller in the square root of the initial
mass √M but in the same distance r.
► From the equations
resolved for the constant G and with the example of Sun  Earth
still we observe :
From the equation
V^{2}earth rers Mearth = M1 M2 G we can find mass of the Sun.
We find the same, from the equation
V^{2}earth Mearth /
rers =
F
Observe, how in one formula of equation the information is hidden, that us presents from the other
formula :
F rers / Mearth = G Msun / rers
► The gravitational
constant G in the Newton's formula shows, that the force between two spherical bodies in certain relation with their distance and mass can be altered, but with such proportion so
as an immutable relation results from all together, that we determine it with the physical constant G. In the constant G is found already the answer that we ask.
We do
not make something else from dissolve the constant G mathematically in order to remains force F
or the acceleration a.
IT WILL BE CONTINUED…
Simple relations of
physics, which the most capable researchers use them in order to they 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 the professionals physicists. A lot of decades were
lost and now, a philosopher reveals, that
simple formulas of physics could be supplemented of a student of medium education! Unless
deliberately the simple observations have not been propagated because
they lead to new technologies... 
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