Gravity Spheres theory in terms of equations
In the beginning of our presentation we compared the gravitational force field to magnetic force field. We did that in order to make visualization of gravity spheres easier. To define the gravity spheres theory in terms of equations, we, in a similar manner, pair it with electric field. We compare gravitational density of a unit of volume in gravitational field to a charge density of a unit of volume in electrical field. Gauss' law for vector electric field in vacuum is
∇⋄E = ρ / ε
where: ρ - charge density per unit volume ε - electric constant
In a similar manner, Gauss' law for vector gravitational field would be
∇⋄g = d / γ
where: d - gravity density per unit volume γ - gravity constant
That is the Maxwell equation number five. Let's find the value of the gravity constant.
In the movie Contact a woman-scientist made a journey to Andromeda galaxy. To all observers, her trip appeared to be instantaneous, though to her it lasted about eighteen hours. During the public inquiry, she was asked by the panel, "Do you expect us to take what you said on faith?!" Developing our theory, we took several well known equations on faith. That saved us a bucket of time. These well known equations were developed by you and scientists like you. At the present time there are no tools to measure gravity in all its details. The only tools we have is math and logic. If you have anything to contribute to this theory, please do so.
Gravitational field is a force field similar to any other force field. Perhaps it is more rarefied and weak, but it is still a force field. We already compared gravitational force field to magnetic and electric force fields. We will continue this process in parallel. Gravitational density (gilottis: gi) is similar to electric charge (coulombs). Gravitational intensity (in) is similar to electric current (amperes). Other units of measurement are the same: newtons, kilograms, meters, seconds, etc. We'll use the SI scale.
The Ampere force law gives us
F = μ ∗ I∗∗2 ⁄ 2∗ π ∗ r (1)
The force is 2 ∗ 10∗∗-7 newtons per meter. With current of 1 A and distance 1 m, we compute the field's permeability, or magnetic constant μ
μ = 4 ∗ π ∗ 10∗∗-7 = 1.256637061 ∗ 10∗∗-6
Gravitational force is 10∗∗39 times weaker than magnetic or electric force. Its value then is 2 ∗ 10∗∗-46 newtons per meter. Using the equation (1) with distance of 1 m and current of 1 A (or distance of 1 m and gravitational intensity of 1 in), we obtain gravitational permeability
μ = 4 ∗ π ∗ 10∗∗-46 = 1.256637061 ∗ 10∗∗-45 kg ∗ s∗∗4 ⁄ m
Electric constant is
ε = 1 ⁄ μ ∗ C∗∗2 (2)
where μ is magnetic constant and C is the speed of light.
Using gravitational permeability in equation (2) we obtain gravitational permittivity, or gravity constant γ = 1 ⁄ 1.256637061 ∗ 10∗∗-45 x 8.987551787 ∗ 10∗∗16 = 8.854187823 ∗ 10∗∗27 kg∗∗-1 ∗ m∗∗-1 ∗ s∗∗-2
Let's check whether the new constant fits. Define C from (2), use γ
C = 1 ⁄ √μ ∗ γ (3) √μ ∗ γ = 3.335640952 ∗ 10∗∗-9 C = 299792458 m/s
The equation (3) validates the gravity constant.
We must verify the process of finding gravitational constants in terms of units. In our MATH pages we equated 1 gi (field density) to 1 m ∗ s∗∗-2 (acceleration) -- a striking convenience! Again, we will go in parallel: electric field units and gravitational field units.
C and gi (coulomb and gilotti) A and in (ampere and intensity)
gi = m ∗ s∗∗-2 in = gi ⁄ s = m ∗ s∗∗-3
μ = kg ∗ m ∗ s∗∗-2 ∗ in∗∗-2 = kg ∗ m ∗ s∗∗-2 ∗ m∗∗-2 ∗ s∗∗6 = kg ∗ m∗∗-1 ∗ s∗∗4 = kg ∗ s∗∗4 ⁄ m (gravitational permeability unit)
γ = kg∗∗-1 ∗ m∗∗-3 ∗ in∗∗2 ∗ s∗∗4 = kg∗∗-1 ∗ m∗∗-3 ∗ m∗∗2 ∗ s∗∗-6 ∗ s∗∗4 = kg∗∗-1 ∗ m∗∗-1 ∗ s∗∗-2 (gravitational permittivity unit)
Use the equation (3) to verify the constants units
C = 1 ⁄ √μ ∗ γ μ ∗ γ = kg ∗ m∗∗-1 ∗ s∗∗4 ∗ kg∗∗-1 ∗ m∗∗-1 ∗ s∗∗-2 = m∗∗-2 ∗ s∗∗2 C = 1 ⁄ √m∗∗-2 ∗ s∗∗2 = m ⁄ s (meters per second -- unit of the speed of light)
The gravity constant stands.
The GS theory is, obviously, in the realm of spherical geometry, even if its spheres are rarely perfect. Gravity is a spherical force field with influence both inside and outside the mass. At the present moment we will focus our attention on the external part of the field (outside the surface of the mass).
Two new and essential variables in the theory are density and intensity of gravity. Density is the "amount of gravity" in the cubic meter of gravitational force field (gi = in ∗ s) and intensity is how much of gravity "substance" circulates through a cubic meter of the field per second (in = gi / s).
The nature of gravitational force field can be defined for a stationary point-particle with the unit of density from zero at the gravitational limit to maximum at the surface of a mass.
This allows the gravitational field to be dependent on the source distribution alone.
Applying the Coulomb's law to our point-particle
g = (1 ⁄ 4 ∗ π ∗ γ) ∗ (gi ⁄ r∗∗2) ∗ ř
where: gi - gravity density at the point r - distance from the particle with gravity density gi to the g-field evaluation point ř - unit vector pointing from the particle with gravity density gi to the g-field evaluation point γ - gravity constant
The total g-field due to the total n of point-particles is the superposition of the contribution of each individual point-particle
n n g = ∑ gᵢ = ∑ (1 ⁄ 4 ∗ π ∗ γ) ∗ (gi ⁄ rᵢ ∗∗2) ∗ řᵢ i=1 i=1
Well, let's make it even simpler. The total strength of a g-field can be computed by summing up the densities of its spheres. We will take the Earth as an example.
One cubic meter of the gravitational force field is the main unit. Its gravitational density depends on gravitational acceleration at the point (quantitatively, gravitational density equals acceleration). Gravitational density is zero at the Earth's gravitational limit and maximum at the surface of the Earth.
Arbitrarily, we assign the gradient for spheres to be one meter (at this point, the real gradient is not important). This gradient is the depth of density for a sphere. We can now compute the total density for each sphere between the gravitational limit and the surface at one meter intervals, then sum up the spheres rᵥ g = ∑ giᵢ ∗ 4 ∗ π ∗ rᵢ ∗∗2 i=rᵤ
where rᵤ - radius of the outermost sphere rᵥ - radius of the surface sphere giᵢ - density of one cubic meter of the given sphere rᵢ - given radius
Why is it important to know the total strength of the gravitational force field of a mass? Because it contributes to the gravitational force field of the larger "mother" mass, extends and strengthens it. An example? The Earth and the moon. We add both masses together and compute the gravity limit for the sum using one very simple equation
R = km
where m - mass in kg k - galactic gravitational constant in m/kg
And that brings us to the center of our theory.
When we published this website, we received a certain number of responses both positive and negative, about fifty-fifty. One of the questions thrown at us was how did we derive the galactic gravitational constant to compute gravity limits. Here is how.
We had to define the task, adopt some assumptions, and create a working model. What was the task? To find the galactic gravitational constant for computing gravity limits. Assumptions? The assumption that gravitational limits exist and they depend on the mass only. The model? The model of the solar system. What is the algorithm for k? It is a program (too extensive and complex to be included here) which constructs a model of the Solar system, then attempts to fit its workings with the proven parameters. The best fit coefficient is the answer. To the best of our knowledge, there is no interaction between our sun and the closest twin star Alpha Centauri 4.37 light years, or about 41.5 trillion kilometers away.
In order to simplify computations, we did not include any of the moons in the solar system or its belts (we did include Pluto though, it was a planet then). Thus, our model was only an approximation of the system, that is why we call the constant k theoretical.
The gravity limit of the solar system had to be somewhere between the sun and Alpha Centauri. Also, it had to fit within the solar system, in other words, the model of the system had to function flawlessly. Cray's answer was
k = 1.54 ∗ 10∗∗-14 m/kg (meters of distance per kg of mass)
We had to verify the newly created constant. Cray created an equation for computing Mercury's unaccounted perihelion advance based on the new k constant. And we had a perfect fit, better than the conventional one (see the MATH page). The Mercury's perihelion advance was precisely as observed. Then we used the galactic equation to compute the perihelion advance for the other planets, and it worked.
( ... in the process of development )
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