Part 5. Gravitational Forces among masses. Check of the New Expression of the Newton's Universal Gravitation Law. New Expressions of the Gravitational Kepler's Equations.
In this part we are going to use the previously obtained first correction of the Newton's Gravitation Law in Part 4, in which a fixed mass attracted another moving mass (with zero rest mass), and we will see that this procedure is incomplete to arrive at the correct Kepler's equation when applied to masses with nonzero mass rest. We know that in an interaction of two masses, both rotate around the center of mass of the system (which will be considered fixed).
Two masses rotating around their Center of Mass.
As before, we will assume that we should arrive at a differential equation similar to that of Kepler, which, in the case of a moving photon attracted by a fixed massive body, we saw that this differential equation does not directly depend upon the photon mass m = p/c. We are going to establish unknown the expression of the magnitude of the gravitational force, and we will call it F. Our problem reduces to find out what is the actual expression of F, for this case. We will try to encounter in this Part 5 the correction of the Newton's definition of gravitational force for the case of a mass, rotating around another fixed mass with mass M.
Let's denote with the scalar F the unknown magnitude of the gravitational force exerted by the mass, considered fixed with mass M, onto the moving mass m. By making some necessary previous relationships:
So, dp = -(F/v).dr ==> (1/p).dp/dt = -[F/(p.v)].dr/dt
On the other hand,
dp = v.dm + m.dv ==> dm/m = dp/p - dv/v
Starting from the begining, with the second condition obtained in Part 4, but developing it with the mass m instead of the momentum p, for deriving the expression of the unknown F:
Which in radial movement, w = 0, dr/dt = v and force does not depend on w, originates the following identity:
d²r/dt² - dv/(v.dt).(dr/dt) = 0.
At this moment, it is necessary to do some reflections for non radial or curvilinear movement. In Part 4, our main assumption was to arrive at the modified Kepler's equation
d²r/dt² - r.w² + (mo/m).G.M/r² = 0
If we do the same in this development, we have to make the last two terms of the previous equation, F.(w².r²)/(m.v²) - dv/(v.dt).(dr/dt), be equal to (mo/m).G.M/r². My first assumption was, precisely, do that thing, but I arrived at inconsistences or contradictions in precession analysis. If you have seen my page before February 18 '99, you can realize that I put the phrase "NOT CONCLUDED" at the begining of this Part 5, and observe the changes done after, in order to solve the inconsistences encountered in the procedure.
The other option I chose was to make the last two terms of the previous equation, F.(w².r²)/(m.v²) - dv/(v.dt).(dr/dt), be equal to (mo/m).G.M/r² - dv/(v.dt).(dr/dt). In this way, disappears the term dv/(v.dt).(dr/dt), and the expression of the unknown force F becomes:
F.(w².r²)/(m.v²) = (mo/m).G.M/r² ==> F = (mo/m).(G.M/r²).m.v²/(w².r²)
F = (G.M/r²).mo.v²/(w².r²)
In which F arrives at a expression similar to that encountered in photon's analysis (see Part 4). Observe that the force F depends on the initial mass mo and not on the dynamic mass m. Remember that in Part 4 this came from the analysis of velocities at perihelio and aphelio and the criterion of respecting the Law of the angular moment conservation.
In order to circumvent the imposibility of having a fixed mass and other moving one in the real world, let's define an equivalent mass, or fictitious mass, located at the center of mass of the system of two masses, Me, fixed, around which masses rotate.
At this same point let's locate an observer making measurements of the movement (remember for those thinking in proper time, time, etc: in this case there is only one observer's reference system). Then, both masses move around this fictitious mass, and in this way, they could be studied separately. This situation, with a fixed mass and another moving one, resembles exactly the configuration studied in Part 4 and we can apply similar relationships.
The measured forces, according to the previous configuration, from Me over each one of the masses m1 and m2, would be given by:
F1 = [(m01/m1).G.Me/r1²].m1.v1²/(w².r1²)
F2 = [(m02/m2).G.Me/r2²].m2.v2²/(w².r2²)
where, m1 and m2 are the generic masses of the moving bodies and m10, m20 the masses at their minimum distance, r1 and r2 the distances from the masses to the center of mass of the system. Given that the masses are rotating around the center of mass in elliptical motion, they are allways opposite and the angular velocity must be equal. This is the reason for putting the same angular velocity for both masses. Because the center of mass of the system is charged to the greater mass, forces F1 and F2 are not equal. It seems that equillibrium is ensured by the equality of the angular momenta.
Let's assume it, as a logical assumption. Then:
m1.r1².w = m2.r2².w ==> m1.r1² = m2.r2²
So, the masses describe proportional ellipses around Me (let's assume this simple movement). Let's try to find a relationship between both forces. Let's say that they are related by a factors, K1 and K2, to be determined, in the following way:
Because this equation must be met in any situation, let's apply it for the simpler case of circular motion, where v2² = (w².r2²) and v1² = (w².r1²), in order to calculate K1 and K2:
For K1 = (r1²)² and K2 = (r2²)², this equation fulfils the condition of equal angular momenta (m1.r1² = m2.r2²), because, as we will see later, (m01/m1) = (m02/m2).
Then, F1 = (r2²/r1²)².F2
Substituting these results, and that of m1 = m2.r2²/r1², we obtain:
[G.Me].v1²/r1² = [G.Me].v2²/r2²
This gives us that v1/r1 = v2/r2, which is true.
In fact, by putting the equality of angular momenta in the following way, where swept angles should be equal:
m1.v1.r1.sin a = m2.v2.r2.sin a ==> (m2.r2²/r1²).v1.r1 = m2.v2.r2 ==> v1/r1 = v2/r2
Because, r1 + r2 = r, is the distance from one mass to the other, and v1 + v2 = v, the velocity of one of the masses measured relative to the other, we can put:
v1/r1 = v2/r2 = (v1 + v2)/(r1 + r2) = v/r
Because of the proportionality of the masses' movements, dr1/r1 = dr2/r2 ( this also comes out from v1/r1 = v2/r2 and from the velocity definition: v² = (dr/dt)² + w².r² ) and dv1/v1 = dv2/v2 (this can be also derived from the previous referred relationship, which implies that r1/r2 = v1/v2 = Constant, and that d²r1 = (r1/r2).d²r2 )
Also, it can be observed that m01 = m02.(r02²/r01²) and that m1 = m2.(r2²/r1²). This relationship implies that m01/m1 = m02/m2, because, as it was suggested before, r02²/r01² = (r2²/r1²) = constant.
In order to obtain other relationships and verify if there are not any other factors that have to be taken in account, let's allow the fixed observer located in the center of mass of the system (CMS) to calculate the force that a moving mass exerts over the other, taking as reference the first moving mass (as it were fixed) and viceversa. Such forces obviously are equal, because the distance between the masses is unique and the velocity of each mass relative to the other is the same (remember that they are being measured in the same reference system, by only one observer). With this considerations in mind, let's work. Then:
F = [(m02/m2).G.m1/r²].m2.v²/(r.w)² = [(m01/m1).G.m2/r²].m1.v²/(r.w)²
Which is true, because, as we have established before, m01/m1 = m02/m2.
where, v = v1 + v2 and r = r1 + r2.
Let's try to relate the forces between Me over m1, m2, (unequal forces) and the forces between m1 and m2 (equal forces), in order to check the used expression for the force between the moving masses. Let's establish that one of the unequal forces, F1, is directly proportional to the equal force F, through K, and the other unequal force F2 will be then, inversely proportional to the equal force F, Through 1/K, in the following way:
leads to the very important relationship: m1.r1² = m2.r2² = Me.r²
where, in some way, the conservation of angular momentum is extended to the equivalent mass, Me. We can imagine this, as that the total system, with an equivalent mass Me, revolves with a rotational angular velocity w, around its center of mass.
Substituting, we obtain: F = (r1/r2)².F1 = (r2/r1)².F2 ==> F1 = (r2²/r1²)².F2
We already had been obtained this last result, through the factors K1 and K2, which ratifies and confirms the relation done before and the used procedure, through K.
In this way the equations fulfill, and make the obtained expressions of any kind of forces to be compatible, in all cases: Through K, or through K1 and K2.
This set of expressions for the equal forces between masses m1 and m2, together with the set of equations of unequal forces between Me and m1, m2, ratify the correction to Newton´s definition of gravitational force which leads to obtain the correspondent Kepler's equation for analyzing completely moving masses (according to this development).
With these considerations the equations of movement for masses m1 and m2 (similar to that of Kepler, using the set of equations from Me over m1 and m2) are modified to:
In this relationship, the equivalent mass Me only depends on the value of the constituent masses of the system, m1 and m2, independently of their radiuses to the center of mass of the system (CMS).
Three or more attracting masses in motion.
For three or more masses moving around their center of mass, for the special case in that they have the same angular velocity w, it is apparent that they meet :
But, this is not so obvious. Let's start the discussion about this theme, because it is very interesting and with surprising results (at least for me!).
Let m1, m2 and m3 be the three masses with equal angular velocity w, revolving around their center of mass of the system, CMS. The first idea that comes out is to consider two of them, m2 and m3, as only one body opposing its movement to the third one, m1. In this way we reduce the analysis to only two masses, work that we already had done. At this time, there are two possibilities: EITHER they constitute a unique body whose mass is the addition of the two masses, m = m2 + m3 located at its private center of mass, PCM, and whose PCM of this "body" should become closer to the CMS than the closest mass, because its mass concentrates such masses in only one body; OR, masses m2 and m3 can be seen as another body whose inverse of the square root of the mass is the addition of the inverses of the square root of the component masses: 1/(m')¹/² = 1/(m2)¹/² + 1/(m3)¹/², resulting a body whose mass is smaller than the smallest of the two masses, located farther than the farthest of the two masses.
If you take the first option, and make the analysis of the equal and unequal forces, previously presented, you arrive at the following result for the equivalent mass of the three masses, Me (make the exercise):
but such expressions of 1/(Me)¹/², obviously, are not the same. Then, there is an inconsistence in this choosing.
This means that the correct option is the second one, ratifying our initial presentation at the beginning of this Part.
It was surprising for me because the first option was the "logical" one when we treat with bodies connected in a fixed shape, as we used to, in where equillibrium is ensured by the acting forces. As we have observed before, equillibrium of bodies under their gravitational forces, without acting external forces, is ensured by the conservation of the angular momentum.
In this way, by making the analysis of the equal and unequal forces we arrived at the previously presented results for three or more masses revolving around their CMS.
These results indicate that for three or more masses, the equivalent mass Me, can be calculated as:
independently of how they move: All moving around the CMS, or two (or more) masses moving joined (as the case of earth-moon) also moving around the CMS in an opposite movement to the other mass.
As you see, these results make the analysis of the gravitation problem for many masses very simple: after calculation the mass Me, we can do the analysis of each mass separately. The velocity v contained inside the relationship are simply the linear summation of the velocities of the constituent masses, and the same ocurrs for radius r. That is luck, isn't it?
Get your own Free Home Page