
CIRCULAR TIME – COMPLETE & SIMULTANEOUS
UNIVERSE
Theory about a completed time and the relativity of energy
(A unified theory about time,
space and matter)
The close
relation between the mass and speed of electromagnetic
waves (8)
©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 type 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 centres). If however, we do not import 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
import the constant G, which (in its dimensional content) includes time t
(sec) or acceleration (m/sec^{2}), while without the 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 import
the constant G, because it has been observed that the sizes (M, r, F
and g) are increased or altered with the same rythm and proportion. When we import 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
x 10^{11} ) x 1kg x 1kg / 1m^{2} = 6,6725 x 10^{11} N
Even if the type gives a result in force
of attraction, still we observe in the denominator the distance r^{2} that separates two (bodies) masses M that are attracted. This distance r^{2} in the known type is expressed like a
straight line that links the centres 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 the 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 type 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 the 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 types
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 the one depends from the other in
such way, that
they do not force certain limits.
We will watch their
relation with an
example of the mass M and radius r of our planet Earth.
.

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 x 10^{24} 
(6,3787 x 10^{6} )^{2}

9,79527 
1,6399 x 10^{24} 
58,50714 x10^{24} 
√M 
2,443972 x 10^{12} 
(6,3787 x 10^{6} )^{2}

0,40079 x10^{11} 
1,6399 x 10^{24} 
9,7952 
_{2}√M 
1,563320 x 10^{6} 
(6,3787 x 10^{6} )^{2}

0,256372 x 10^{17} 
1,6399 x 10^{24} 
0,40079 x10^{11} 
_{3}√M 
1,250327 x 10^{3} 
(6,3787 x 10^{6} )^{2}

0,205044 x 10^{20} 
1,6399 x 10^{24} 
0,256372 x10^{17} 
_{4}√M 
3,535996 x 10 
(6,3787 x 10^{6} )^{2}

0,579877 x 10^{22} 
1,6399 x 10^{24} 
0,205044 x10^{20} 

1 
(6,3787 x 10^{6} )^{2}

0,163992 x 10^{23} 
1,6399 x 10^{24} 
0,163992 x10^{23} 
We observe that the acceleration of gravity g
increases depending on mass M (for the same radius).
When mass M increases in the square M^{2}
then the acceleration g of the M increases on x M, that is to say g x 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 x10^{12}
F = G √M_{1} √M_{1} / r^{2}
→
F = G (2,443972 x10^{12} ) (2,443972
x10^{12} ) / (6,3787 x10^{6} )^{2} = 39,8548 x10^{13} / 40,6878
x10^{12} →
F = 9,795 Newton
g = G M_{1} / r^{2} → G x
5,973 x10^{24} / (6,3787 x10^{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 the mass of the Sun.
We find the same, from the equation
V^{2}earth
Mearth
/ rers
=
F
Observe, how " they are hidden " the
information in one type of equation, that us presents from the other type :
F rers
/ Mearth
=
G Msun
/ rers
►
The gravitational
constant G in the Newton's type shows, that the force between two
spherical bodies in certain relation
with their distance and mass, they 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…
The simpler relations of
physics, which the most capable researchers use them 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. A lot of decades were
lost and now, a philosopher reveals, that the
simpler types 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...


