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We've all heard $$F_g=\frac{gm_1m_2}{r^2}.$$ However, since I took physics, we've discovered "dark energy," which if I have any concept of the current thinking is caused by space being produced out of whole cloth in the intergalactic void, rather than a classical energy source.

However, if this is happening in the intergalactic void, it doesn't seem reasonable to think this doesn't happen everywhere. In proximity to significant mass, the effect could easily be overwhelmed by gravity to the point that it's difficult to distinguish independently, but certainly two small masses with sufficient distance between them would drift apart rather than attract. That would indicate that the Y-intercept of the gravitational function is nonzero, or $$F_g=\frac{gm_1m_2}{r^2}-K.$$

While this error (if extant) would probably be negligible when dealing with masses and densities that we encounter in this neighborhood, it might make a difference when we project it to exotic phenomena like neutron stars or black holes.

I know this sounds a bit crackpot, but I'm an EE with a few semesters of physics from the 1970s. Kindly take this into account when responding and try to make it simple and intuitive for me. :-)

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    $\begingroup$ I'm not sure I understand what your question is. $\endgroup$
    – BowlOfRed
    Commented May 21 at 19:42
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    $\begingroup$ Unfortunate that this question is closed - I would like to point out that the "anti-gravity" produced by dark energy and matter scales with distance, so things farther away are pushed more. Hence, the units of the Hubble constant. It's not that there's another force acting on everything, it's that space is expanding, and what's observed is that the farther something away is, the more this force accelerates it. It doesn't have anything to do with the mass of any particular objects. $\endgroup$
    – controlgroup
    Commented May 21 at 23:15
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    $\begingroup$ We've all heard $F_g=\frac{gm_1m_2}{r^2}$. Actually, no one writes $F_g=\frac{gm_1m_2}{r^2}$. Do you understand the difference between $g$ and $G$? $\endgroup$
    – Ghoster
    Commented May 21 at 23:31
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    $\begingroup$ If you are talking about dark energy, then you need to leave the known oversimplification of the Newtonian formulation of gravity and use the Einstein field equations of GR instead, because in Newtonian gravity, energy doesn't gravitate. See en.wikipedia.org/wiki/Einstein_field_equations The G in Einstein's field equations is the same as the G in the Newtonian formulation of gravity (if you write out EFE constant as 8πG/c^4), but you add a second free parameter, Lambda a.k.a. the cosmological constant, which is the way that the leading paradigm in cosmology explains dark energy. $\endgroup$
    – ohwilleke
    Commented May 22 at 1:24
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    $\begingroup$ Newton's constant G is known to about six significant digits, while Lambda, the cosmological constant, is known to about four significant digits. en.wikipedia.org/wiki/Gravitational_constant en.wikipedia.org/wiki/Cosmological_constant Both are free parameters in EFE. You can't derive Lambda from G. By way of comparison, these constants are known much less precisely than the fine structure constant or electron or muon masses, but comparable to the tau mass and weak force constants. G is known to more precision than the strong force coupling constant or the quark masses. $\endgroup$
    – ohwilleke
    Commented May 22 at 1:34

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Force is a vector, so your law as stated is not well-formed. I am going to go through some possible interpretations of your force law and point out some issues, although I'm not doing a definitive analysis -- there are likely other issues to consider beyond the few I'm mentioning.

One interpretation is that you are suggesting a modification of Newton's law for the gravitational force on particle 1 due to particle 2 should have the form $$ \vec{F}_{12} = -G\frac{m_1 m_2}{r^2} \hat{e}_r+ \vec{K} $$ where $\hat{e}_r$ is a unit vector pointing from mass 2 to mass 1. This form will violate Newton's third law, and thus momentum conservation, unless the sign of $\vec{K}$ is flipped when considering the force on particle $2$. Furthermore, if we consider multiple particles, do the $K$ vectors add up? Since $K$ does not depend on the distance to the particle, this sum will scale like $N$, the number of particles in the Universe, so will become enormous.

Another interpretation is that every particle has a constant force $\vec{K}$ on them at all times. One question would be, what direction does $\vec{K}$ point in? We do not observe any special directions in the Universe on cosmological scales, and rotational invariance leads to consequences like conservation of angular momentum that do seem to hold in our Universe.

A final interpretation is that $$ \vec{F}_{12} = \left(-G\frac{m_1 m_2}{r^2} + K\right)\hat{e}_r $$ If we have $N$ particles in the Universe more or less isotropically distributed (which is what we observe on cosmological scales), the $K$ term above will tend to average to zero, because we'll be adding together the contributions of many particles -- contributions which do not decay with distance so are all of equal strength -- and if their positions are random then the vector sum will tend to zero.

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