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Strong Force is definitely NOT gravity itself
Did refer to the gravity/strong force similarity mentioned by Haramein .
He says that if you model a pair of protons as two black holes, the gravitational force between them turns out to be equal to the strength of the strong force. To be fair, a pair of proton-mass black holes could have an attractive force acting between them that was equal to what we call the strong force if they were sufficiently close together. The problem with this, however, is that the electrical repulsion between those charged black holes is going to increase at the exact same rate as the gravitational attraction is. So changing the distance between those black holes will actually have no affect on how strongly they attract or repel each other (and the repulsion will be much, much higher).One could get around that problem if you posit that the inverse-square law ceases to operate for gravity and/or the electromagnetic force at extremely tiny distances, but evidence for this is currently lacking. Even if you did make a model where that was true, then it would need to explain why it doesn't work for many different particles. Electrons, muons and tau particles all have mass, so they should be able to bind together using the strong force if the strong force is a form of gravity. Yet they don't.Then there are particle experiments that demonstrate the existence of gluon particles, which were predicted in advance in order to explain how the strong force operates. The strong force also has its own conservation laws that do not apply to gravity (hypercharge and color charge).
To be as clear as possible
The gravity attraction between 2 protons doesn't have to overcome their electrical repulsion, because diproton is extremely unstable. https://en.wikipedia.org/wiki/Isotopes_of_helium#Helium-2_(diproton)
Proton also has non-zero electric polarizability, which can reduce repulsion at close distance.https://en.wikipedia.org/wiki/Proton
Then I can use helium-3 as the example instead, which is stable. There is still an enormous repulsive force acting between the two protons in a helium-3 nucleus.
How can strong force overcome electrostatic repulsion in Helium-3, but fails to do so in Helium-2?
If Helium-3 is modeled as 3 protons in equilateral triangle configuration with an electron in the middle (instead of 2 protons and 1 neutron that we usually see), we can calculate that attractive force to each proton by central electron can overcome repulsive forces by the other protons.
Anytime someone postulates a new idea they have to make sure that it does not violate any conservation laws. If the proposer cannot show this then they do not have enough knowledge to be stating such propositions. Go away and read Noether and then you may have a better chance.
The strong force does overcome the repulsion between the two protons in helium-2. The reason that it is unstable is because a deuterium nucleus is more stable than it is, so it decays into deuterium.
Care to show those calculations?That model doesn't work anyway, since we know that neutrons exist.and they aren't simply a proton plus an electron.
How much is the strong force?
why not decay to hydrogen-1?
Let's say an electron is placed in the origin of a coordinate. Three protons are arranged in equilateral triangle 1 length unit away from the electron in a flat plane. Thus, the distance between protons is https://upload.wikimedia.org/wikipedia/commons/1/11/Equilateral-triangle-heights.svgElectrostatic electric force by electron to each proton is attractive. The magnitude is inversely proportional to the square of the distance between electron and proton. In this case it's 1. The repulsive force between protons is thus 1/3.Combined repulsive force by other proton is
The existence of particles other than electron and proton, such as neutron, muon, as well as any other particles doesn't prove nor disprove the argument above, hence it's a non sequitur logical fallacy.
By the way, your model also violates conservation of lepton number. When a tritium nucleus decays into a helium-3 nucleus, it releases an electron and an electron anti-neutrino. If the atomic nucleus contains neutrons, this is not a problem. If, however, there is a proton and electron in the nucleus instead of a neutron, then the net lepton number before and after the decay are different.