It is clear that the gravitational pull diminishes with increasing distance, but I can't find the answer to whether there is a point where it becomes "exactly zero". Is there such distance? Is it calculable?
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$\begingroup$ Do you know how to calculate the gravitational force from the distance between two masses? Set it to zero, solve for $r$ and tell us what you found out. $\endgroup$– DK2AXCommented Jan 28, 2020 at 14:14
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1$\begingroup$ Related: physics.stackexchange.com/q/200635/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jan 28, 2020 at 14:35
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1$\begingroup$ What is meant by "exactly zero", and why is it in quotes? $\endgroup$– JMacCommented Jan 28, 2020 at 15:00
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1$\begingroup$ $1/r^2$ is never zero. It decreases toward zero as $r$ increases toward $\infty$. $\endgroup$– G. SmithCommented Jan 28, 2020 at 18:53
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$\begingroup$ JMac I was trying to emphasize that I don't mean approaching to zero but being exactly zero. @G. Smith Well, I agree with you but I was not sure about the limit case. $\endgroup$– gunakkocCommented Jan 28, 2020 at 20:41
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3 Answers
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Is there such distance?
No. Gravity is a fundamental force whose range is infinite. The same applies to the electromagnetic force. For more info see
http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html
Hope this helps.
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$\begingroup$ Might be worth adding that this is, essentially, „only“ a mathematical model. Concerning the „only“ it is again worth mentioning that no discrepancies between the model and reality have been measured (leaving MOND aside). I think that OPs question is exactly pointing at the interface between model and reality. $\endgroup$ Commented Jan 28, 2020 at 14:33
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$\begingroup$ @HartmutBraun It's worth noting that the question is tagged "Newtonian-gravity", so it seems fair to assume he's asking about the model and not reality. $\endgroup$– JMacCommented Jan 28, 2020 at 14:43
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$\begingroup$ @JMac I'm not very conversant in GR, but is not the range of gravity infinite under GR as well? $\endgroup$– Bob DCommented Jan 28, 2020 at 14:54
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$\begingroup$ @JMac that’s exactly my point (and no offence intended). The math is simple enough to assume OP can do the calculation him/herself but has difficulty to understand the relationship of mathematical model (with infinities and limits) to reality. But I shouldn’t speculate: if OP reads our comments, he/she can restate the question. $\endgroup$ Commented Jan 28, 2020 at 14:55
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$\begingroup$ @BobD Me neither, but I was actually just discussing this with someone in chat. I think you could make the case that the edge of the observable universe is actually the maximum distance that gravitational "force" acts; since the field is limited to travel at the speed of light. The issue is that in GR, I don't think it's really considered a "force", so discussing that in an answer (especially with the Newtonian gravity tag), would probably just confuse things. $\endgroup$– JMacCommented Jan 28, 2020 at 14:59
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Yes, in a way. Gravitational force decreases inside a planet. At the center it is 0 because all parts of the planet pull outward symmetrically. The formula you used only applies outside.
Also at an infinite distance, the force would be 0. That is of course a bit of a cheat. The limit is 0, but you can't find a point where the value is the same as the limit.
answered Jan 28, 2020 at 17:26
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Not in Newtonian gravity, but in relativistic cosmology. The most suitable metric to study this problem is the Schwarzschild De Sitter metric. By setting $\dot{r}=\dot{\theta}=\dot{\phi}=0$ we get
$$\ddot{r}=\frac{\lambda c^2 r}{3}-\frac{G M}{r^2}$$
where $\lambda$ is the cosmological constant. By setting $\ddot{r}=0$ and solve for $r$ we get
$$r=\sqrt[3]{3 G M / \lambda / c^2}$$
which is the coordinate radius at which the test particle is neither attracted nor repelled and keeps a constant distance to the dominant mass $M$.
Another approach is to set the recessional velocity equal to the Newtonian orbital velocity
$$H r = \sqrt{G M/r}$$
and solve for $r$, then we get
$$r=\sqrt[3]{G M / H^2}$$
Since in a De Sitter universe the Hubble and the cosmological constants are related by
$$H=\sqrt{\lambda c^2/3}$$
this is the same expression for $r$ as obtained above.
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$\begingroup$ You can have cosmological constant in the Newtonian gravity. $\endgroup$– A.V.S.Commented Jan 28, 2020 at 16:22
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$\begingroup$ In that case it should also work then $\endgroup$– YukterezCommented Jan 28, 2020 at 16:23
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$\begingroup$ What "gravitational force" is there in GR? $\endgroup$– ProfRobCommented Feb 6, 2020 at 7:15
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$\begingroup$ If you are in free fall there is none, but you can still use the concept of a force when you consider the force required to stay stationary. In this case, the required force to keep a constant distance to the dominant mass is zero when you are at the right distance, otherwise you need a rocket or something else to keep your position constant. $\endgroup$– YukterezCommented Feb 6, 2020 at 7:19