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**New Theories / Re: How gravity works in spiral galaxy?**

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**Yesterday**at 05:56:01 »

OKQuoteIn other words - Gravity only contributes rotation velocity - But that is good enough to create new energy in our Universe.

No it doesn't. Conservation of energy won't let you. Give it up already. What part of "energy cannot be created or destroyed" do you not understand?

My job is to prove that the SMBH's gravity force contributes more energy than the energy that is needed to create those new particle pairs.

Let's set the calculation and verify the real contribution of the mighty SMBH's Gravity force to the total energy.

Let's start with the Virtual particle pair that orbit at almost the speed of light.

Before the creation they have no mass and therefore they have no Kinetic Energy.

At the moment of the creation, they actually get all their mass energy from the SMBH (by the magnetic field).

However, they also get kinetic energy from the SMBH's gravity force.

In other words-

The SMBH's internal energy is transformed into real particle and antiparticle, while the Kinetic energy for those particles are contributed by the SMBH's gravity force.

So, the calculation should be as follow:

M - Represents the mass of the particle or Antiparticle

Eout - The total energy that the SMBH have lost in order to generate those particle and Antiparticle air

Eout = 2M

Ek - The kinetic energy of each particle at the moment of creation. As their orbital velocity is almost the speed of light then:

Ek= 1/2 M c^2

Ep - The potential energy of a particle at the moment of creation

Er = The Total rotation energy of the particle at the moment of creation:

Er = Ek + Ep

The Ep is transformed into kinetic energy as the antiparticle is falling into the SMBH.

Therefore, if the orbital velocity of the antiparticle at the moment of creation is the speed of light, than as it drifts inwards to the SMBH than theoretically, its orbital velocity should be higher than the speed of light.

However, as it gets below the event horizon, even at that Ultra high velocity, it can't go out any more.

However, at the moment of the collision between the antiparticle to the SMBH, all the total rotation energy is transformed into new Energy in the SMBH.

The SMBH creates two particles (Particle + Antiparticle). However, only the antiparticle is falling in.

So, we need to verify if its total rotation energy is greater that the energy that was needed to create the pair.

Therefore, we need to prove that:

Er is greater than Eout

Er = Ek + Ep

For this simple calculation lets ignore the impact of the Ep

So, let's assume that:

Er = Ek = 1/2 M c^2

Eout = 2M

Er = n * Eout

n = represents how bigger is Er from Eout

Er = 1/2 M c^2 = n 2M

n = 1/4 c^2

Conclusions:

For any energy that the SMBH contributes to create the Pair particle, it gets back at least 1/4c^2 times the "invested" energy.

This does not include:

1. The extra mass of the in falling antiparticle

2. It's Potential energy

3. The particle that is drifted outwards to the accretion disc

4. The transformation of the energy from the accretion disc back to the SMBH.

Therefore, I have proved that only the Kinetic energy of the in falling antiparticle contributes much more energy that was needed to set the whole creation activity.

You realize that would make me right, not wrong, don't you?It is totally a diffrent senario from the one that you have used.

Attraction between the Earth and Sun:

F = (GMm)/r2

F = ((6.674 x 10−11(5.97237 x 1024)(1.9885 x 1030))(149,598,023,000)2)

F = (7.926080939013 x 1044)/(2.2379568485508529 x 1022)

F = 3.5418308070902173002872574484922 x 1022 newtons

Attraction between the Moon and Sun:

F = (GMm)/r2

F = ((6.674 x 10−11(7.342 x 1022)(1.9885 x 1030))(149,982,422,000)2)

F = (9.7437510158 x 1042)/(2.2494726908986084 x 1022

F = 4.3315711523075275820924842308266 x 1020

Sum of the two forces: 3.5851465186132925761081822908005 x 1022 newtons

Attraction between the Sun and a hypothetical Earth-Moon combination object:

F = (GMm)/r2

F = ((6.674 x 10−11(6.04579 x 1024)(1.9885 x 1030))(149,602,691,137)2)

F = (8.023518449171 x 1044)/(2.2380965195362677778850495607477 x 1022)

F = 3.5849742757445814209639045480079 x 1022 newtons.

In this case, you have set all the objects at the same radius of 149,602,691,137.

However, in your first example you have set one object at minimal distance of 9,000,000 m

The lower sphere is at an altitude of 9,000,000 meters and has a mass of 50 kilograms.While the other at 11,000,000 m:

The higher sphere is at an altitude of 11,000,000 meters and has a mass of 50 kilograms.After merging the two objects, their center of mass had been set at 10,000,000 m:

Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres. What was once two objects has now become a single object with a mass equal to the single, 100 kilogram sphere with a center of gravity located at the same altitude (10,000,000 meters).I can promise you that if in your example you have used the same radius for all objects - you would probably get fully balanced forces.

Therefore, please use the minimal and maximal distance/radius from the Earth Moon to the Sun (maximal tidal) and then set your recalculation.

Mathematical fiction is a direct outcome if we use the error/delta in the Taylor series to prove something which is totally based on that error/delta.QuoteTaylor series is an excellent tool to get close to the solution.Then what was that nonsense you were saying when you called it "mathematical fiction"?