Deduction of the Formula of the Force of Gravity that a body A exerts on a body B

Influence of the speed of the energetrons on the quantity Q₁ of mini-energetrons from a line of gravity which reaches a body
A body A is always connected to a body B by lines of gravity.
A_n observer in the referential of body A, sees one line of gravity that connects body A to body B

The energetrons are emitted with the speed c by the bodies. Mini-energetrons are emitted by the energetrons with the speed c. Thus, one observer at the referential of an energetrons thinks that he is standing still and sees the mini-energetrons with the speed c.
Now, an observer at the body referential thinks he is standing still and sees the energetrons with the speed c and the mini-energetrons with the speed c+c.

From the figure above, we may conclude:

1º) Gravity lines emitted by body A are composed of energetrons that are emitted at the speed c in the unit of time.

The quantity P of energetrons of a line that reaches body B is proportional to the speed of emission c, that is P a c.

2º) In turn, these energetrons emit mini-energetrons in the unit of time, with the speed c, which upon arriving at body B exert a force of gravity f in the direction of arrival of the mini-energetrons.

The quantity J of mini-energetrons that arrive at body B is proportional to the speed c of emission of these, namely J a c.

3º) Thus, we may state that the quantity Q₁ of mini-energetrons emitted by body A that arrive at body B is proportional to the product P x J, namely, Q₁ a c x c, or Q₁ a c².

Influence of the number of lines of gravity emitted by a body A of mass M, which reach a body B due to the distance between them.

Lines of gravity emitted by a body of mass M

4º) A Body A of mass M_a emits a quantity M of lines of gravity that reach a body B. M is proportional to the number of massives of unit mass contained in such body A, that is M a M_a.

Lines of gravity that arrive at a body of mass m

5º) The quantity of lines of gravity emitted by a body A, which reach a body B of mass M_b is proportional to the quantity N of unit massives contained in that body of mass M_b, that is, N a M_b.

Influence of the distance between two bodies A and B over the quantity of lines of gravity that are emitted by body A in the direction of B

Quantity L of lines emitted by a body A that reach a surface S of a sphere of radius R, which involve body A is the same that reach a surface S₁ of a sphere of radius R₁.
One surface DS of the sphere S₁ of radius R₁ will receive a quantity K₁ of lines of gravity, where , or DS L = S1K₁.

The surface Ds of the sphere of surface S and radius R will receive the quantity K
of lines of gravity, where , or DS L = SK.
Thus,

The area of the surface
The area of the surface
Therefore, we have ,
where:

6º) The quantity of lines K of energetrons emitted by a body that arrive at another body is inversely proportional to the square of the distance R between them, namely K µ .

Quantity of energetrons

7º) Base don conclusions 3, 4, 5 and 6, the quantity Q of mini-energetrons arriving at the body B, coming from A, is equal to:
.

Force of Gravity

Final conclusion:
The force of gravity F_g that a body A of mass Ma exerts on a body B of mass M_b is the sumo f all unit forces f of each mini-energetron from body A that arrive at body B.
Q being the quantity of such mini-energetrons, we have:

or else, having in view conclusion no. 7, we obtain

where:
F_g → force of gravity that body A exerts on body B
M_a → mass of A;
M_b → mass of B;
c → speed of the energetrons on one line of gravity, to an observer at the referential of body A and of the mini-energetrons to an observer at the referential of an energetron;
c + c → speed of the mini-energetrons to an observer at the referential of body A;
R → distance between body A and body B;
f → gravitational constant
By making we have:

And by making G´ = , where G = universal constant of gravity
We have:

where, G = 6,6725985 x 10^-11 Nam²/Rg², which is the universal gravitational constant.
Whence we conclude that the force of gravity F_g that a body A exerts on a body B standing still in the gravitational field of body A is identical to the Newton’s gravitational force.

In the formula ,
Since R = c x t, where t = time that the energetron takes to move from body A to body B, we will have:
,
or else,

where

Thus, we may express the force of gravity that a body A exerts over body B as a function of time t that an energetron takes to travel from body A to body B at a speed c.
We have just defined the force that the lines of gravity emitted by a body of mass Ma exerts over another body of mass Mb. Therefore, we may assume:
F₁ = gravitational force that a body of mass Ma exerts over the body of mass M_b;
F₂ = gravitational force that a body of mass M_a exerts over a body of mass M_a;
Since the modules of such forces are equal, we have,

Camila Effect