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In an atomic and particle physics model, the gravitational force is ignored or neglected. Physicists might have freedom of ignoring such minute scale value. But can we really avoid it in calculations? The field exists the phenomenon exists.

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  • $\begingroup$ This is probably a question for physics.se. The short answer is, yes, we have freedom to ignore it on a minute scale value; particle physics models do just fine when gravity is weak. When it becomes strong, as you say, they break down, but because the other forces are so much stronger, gravity only becomes relevant in really exotic situations like black holes, the cores of massive stars, the beginning of the universe, etc. $\endgroup$ – Neal Aug 28 '16 at 16:45
  • $\begingroup$ Yes it can be at low energy scales. Gravity is a minuscule fraction of even the next weakest force. $\endgroup$ – Cameron Williams Aug 28 '16 at 16:45
  • $\begingroup$ @OmarNagib imagine A SPACE constrained by forces. the particles flow according to curvature of that space. for instance how can we think of solar system. space curved by gravitational field of Sun . so what is the geometry of this space. how sharp is that curvature. this must be question of differential geometry. and how planets revolve the sun, their speed as the function of the curvature; this must be differrential equation. $\endgroup$ – SK1987 Aug 28 '16 at 17:04
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    $\begingroup$ Have you tried including it in a calculation? What effect did it have? $\endgroup$ – sammy gerbil Aug 28 '16 at 21:45
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If you take gravity as 1 unit, the electromagnetic force is 1000,000,000,000,000,000,000,000,000,000,000,000 times stronger, ($10^{36}$).

We can ignore gravitional force at atomic scales, because for example, when we use an extremely high precision measurement of the quantized energies of the cyclotron orbits, or Landau levels, of the electron, compared to the quantized energies of the electron's two possible spin orientations, gives a value for the electron's spin g-factor, electomagnetic related:

g/2 = 1.001 159 652 180 85 (76),

a precision of better than one part in a trillion, that is 1,000,000,000,000 ($ 10^{12} $.)

So the gravitational force at this scale is not measurable.

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