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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 (8)

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

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 / r2

We say that the force is equal with the product of two spherical masses M and reversely proportionally to the square of their distance r2 (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/sec2), 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 / 1m2 = 6,6725 x 10-11 N

F = ? M1 M2 / r^2

Even if the type gives a result in force of attraction, still we observe in the denominator the distance r2 that separates two (bodies) masses M that are attracted. This distance r2 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 r2 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. 1.

g = G M / r2 = 9,795  m / sec2
 


 

Mass (kg)

Radius (m)

Acceleration (g)

Ratio g / M

Mass x g

(F)

M

5,973000 x 1024

(6,3787 x 106 )2

9,79527

1,6399 x 10-24

58,50714 x1024

M

2,443972 x 1012

(6,3787 x 106 )2

0,40079 x10-11

1,6399 x 10-24

9,7952

2√M

1,563320 x 106

(6,3787 x 106 )2

0,256372 x 10-17

1,6399 x 10-24

0,40079 x10-11

3√M

1,250327 x 103

(6,3787 x 106 )2

0,205044 x 10-20

1,6399 x 10-24

0,256372 x10-17

4√M

3,535996 x 10

(6,3787 x 106 )2

0,579877 x 10-22

1,6399 x 10-24

0,205044 x10-20


 

1

(6,3787 x 106 )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 M2 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 r2 should downsize in r2 / √M so that results the same acceleration g.


 

Mass of Earth in the square root √ Μ1 : 2,443972 x1012

 

F = G √M1 √M1 / r2

F = G (2,443972 x1012 ) (2,443972 x1012 ) / (6,3787 x106 )2 = 39,8548 x1013 / 40,6878 x1012

F = 9,795 Newton

 

g = G M1 / r2 → G x 5,973 x1024 / (6,3787 x106 )2 = 9,795 m / sec2

 

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 :  

F r^2 = V^2 r M

V^2 = F r / M1 = G M2 / r
 

 

From the equation V2earth rers Mearth = M1 M2 G we can find the mass of the Sun.

We find the same, from the equation V2earth 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

 

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

 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 natural 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.  

G = F r^2 / M1 M2

 

 

 

1 G: 6,6725 x10-11   |   Mearth : 5,973 x1024 kg   |   rearth : 6,3787 x106 m   |

 

 

 

 

www.kosmologia.gr  1st publication on Earth

 

 

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...

 

 

 

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COSMOLOGICAL THEORY

ABOUT A COMPLETED TIME OR UNIVERSE AND THE RELATIVITY OF ENERGY

 

 

 

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