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I remember hearing back in school that each object is attracted to every other object in the universe. Of course this has to be taken with the understanding that most of those objects have very very very*10^very little attraction.

Isn't there a limit though, where the attraction hits a planck barrier that once it gets past it becomes non-existent?

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  • $\begingroup$ Related (possible duplicate?): physics.stackexchange.com/questions/8688 $\endgroup$ – Kyle Kanos May 27 '15 at 0:31
  • $\begingroup$ Even in the light cone limits , the question will not have a definitive answer before a quantum gravity theory. GR doesn't postulate curvatures quantization. According to it , under Planck barrier curvatures might be added. $\endgroup$ – user46925 May 27 '15 at 1:22
  • $\begingroup$ I guess to be more specific, if mass is "stretching" space-time there has to be a point where it stretches it less then one planck length which will essentially mean zero curvature on space time. Am I wrong with this assumption? $\endgroup$ – Joe May 28 '15 at 0:18
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Yes! It is true that your mass and, in fact, every single mass in the universe affects every single other mass in the universe gravitationally. This can be seen through Newton's equation for gravitational force: $$\vec{F}_g=G\frac{m_1 m_2}{r^2} $$ where G is the gravitational constant: $$G \approx 6.67 *10^{-11} \tfrac{m^3}{kgs^2}$$ $m1$ is the mass of one object (let's say your mass), $m2$ is the mass of the other object and $r$ is the vector denoting the distance between your two objects. As you might notice from this equation, the only way to have $F_g = 0$ is if one or both of your masses are equal to $0$. As long as the two objects you are measuring the force between have some measurable mass, they will experience a gravitational force, no matter how small.

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    $\begingroup$ This addresses part of the question, but not what I would consider the most fundamental part -- if things like velocities, energies, etc. are discrete, is there an $r < \infty$ such that $F = 0$ due to, in essence, "truncation" of discrete values. $\endgroup$ – tpg2114 May 27 '15 at 0:55
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    $\begingroup$ if outside the light cone ? $\endgroup$ – user46925 May 27 '15 at 1:08
  • $\begingroup$ You're forgetting of the consequences of relativity $\endgroup$ – Jimmy360 May 27 '15 at 4:48

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