Force of gravity between two bodies A and B when there is a relative speed v between them.

Enviado por Geraldo em ter, 01/22/2008 - 08:33.

Force of gravity between two bodies A and B when there is a relative speed v between them.

To an observer at the referential of A

This observer thinks he is standing still, seeing body B with the speed v.

In this case, the gravity will function in the following manner:

Body B, with its gravitational field, has a speed v in relation to body A. The gravitational field of body B which will exert a force over body A, standing still in the gravitational field of B, will be that B0A that forms an angle b with the direction of v when body B is at position B0.
In this position, the energetron emitted from B on this line will take a time t in order to reach body A, at a speed c, where it will exert a force Fv in the direction of B0, where Fv:


or

Where

Within this same time t, body B moves from position B0 to Bv at a speed v. Thus, when body B is at position Bv, at a distance D from A, body A will be attracted by the force of gravity Fv in the direction of B0, forming an angle a with the axis BvA. For every value of the speed v, body B will occupy different positions Bn in its trajectory and, of the lines of gravity of the gravitational field of B in Bn, just one will hit body A.
The gravitational field of body B has existed since the Big Bang. The line from this field at position B0 which will hit body A will depend upon its speed v. Such position B0 will be defined in such a way that the distance B0Bv will be E, where E = v x t and the distance B0A will be L, where L = c x t.
When v = 0, the field will be standing still at position Bv and angle b will be that formed by vector v with the axis BvA, exerting a gravitational force on A, in the direction of Bv, where


or

Assuming that t0 is the time the energetron takes to go from Bv to A, covering the distance D, where D = c x t0.

To an observer at the referential of B


This observer thinks he is standing still, seeing body A with the speed –v.

In this instance, gravity will function in a similar way as the previous one, assuming that Body B is standing still and body A will have the speed –v.

Thus, the gravity between two bodies, when there is a relative speed v between them will be as in the figure above.

Where:


or
.

Difference between Logical Gravity and Newtonian Gravity

If the relative speed between two bodies is v = 0, it means that they are standing still at the same referential and that the line of gravity of a body that will hit another body will that line BA or AB, of a length D. The energetrons from such lines will cover the trajectories AB and BA at the speed c, within the same time t0 so that, upon arriving on a body, they will exert of gravitational force F0 in the direction of the other, the modules of which will have the same value, where

which is the same formula of Newtonian Gravity.
Whence we conclude that Logical Gravity is the same as that of Newtonian Gravity when two bodies are standing still at the same referential.
The difference between them is when there is a relative speed v between two bodies. In this case, the direction of the logical gravity force will form an angle a with the axis between the two bodies, which does not exist in the Newtonian gravity.

Actual line of gravity and virtual line of gravity

    Observer at the referential of A

This observer thinks he is standing still between A and body B, which has a speed v moves from position B0 to position Bv for each position Bn that the gravitational field of body B may occupy, the gravitational line from this field will exert gravitational force over body A, that will be different with a new value for the angle b, in such a way that, when the energetron from the first line reaches body A the other energetrons from the other lines that will reach body A will be forming a virtual curved line.
For any generic position Bn of body B between B0 and Bv, just one line of gravity from B gravitational field will reach body A, in such a manner that BnBv = v x tn and BnA = c x tn
where:
c = speed of the light;
v = speed of body B;
tn = time in which B moves from position Bn to position Bv and the energetron from this line moves from position Bn to the body A;
Fv = Force of gravity.
Let’s keep in mind that the gravitational force is always exerted by a straight gravitational line from the gravitational field of body B.
Observer at the differential of B
This observer thinks he is standing still at B and body A moves from position A0 to position Av at a speed - v.
All the reasoning applied to the observer standing still at A similarly applies to the observer standing still at B.

    For of Gravity between two bodies A and B when the relative speed between them is perpendicular to the axis AB

    Fv = Force of Gravity between bodies A and B, with v ≠ 0;

    F0 = Force of Gravity between bodies A and B, with v = 0;

    V = relative speed between the bodies

    tang a = .

    In this case, the triangle ABvB0 shall be a rectangle triangle, the sides of which are:

    B0A = L = c x t.

    B0Bv = E = v x t.

    BvA = D = c x t0.

    Resolving this triangle by the Pythagorean Theorem, we will have:

    L2 = E2 + D2

    c2t2 = v2t2 + c2t02

    t2 (c2 – v2) = c2t02

    t2 (1- ) = t02

    [1]

    When we compare the force of Gravity that a body A exerts over a body B when there is a relative speed v between them:

    If v = 0, namely, when they are standing still between themselves, the force of gravity will be F0, where

    [2]

    If v ≠ 0, the force of gravity will be Fv, where

    [3]

    Dividing [2] by [3], we will have:

    [4]

    We have seen that [1]

    From [1] and [4], we will have:

    Hence,

    This is the formula for gravity force between two bodies A and B, when the relative speed between the two is perpendicular to axis AB.

    Since this is a rectangle triangle, the direction of force Fv will form an angle a with axis ABv, where tang a = .

    Ifv = 0

    Fv = F0

    If v c

    Fv ∞