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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?

Qmechanic
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gunakkoc
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  • 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. – DK2AX Jan 28 '20 at 14:14
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    Related: https://physics.stackexchange.com/q/200635/2451 and links therein. – Qmechanic Jan 28 '20 at 14:35
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    What is meant by "exactly zero", and why is it in quotes? – JMac Jan 28 '20 at 15:00
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    $1/r^2$ is never zero. It decreases toward zero as $r$ increases toward $\infty$. – G. Smith Jan 28 '20 at 18:53
  • 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. – gunakkoc Jan 28 '20 at 20:41

3 Answers3

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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.

Bob D
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  • 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. – Hartmut Braun Jan 28 '20 at 14:33
  • @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. – JMac Jan 28 '20 at 14:43
  • @JMac I'm not very conversant in GR, but is not the range of gravity infinite under GR as well? – Bob D Jan 28 '20 at 14:54
  • @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. – Hartmut Braun Jan 28 '20 at 14:55
  • @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. – JMac Jan 28 '20 at 14:59
  • @JMac Yeah I know under GR gravity is not a force. Maybe we can say that the range of the curvature of space time due to mass and energy is infinite? I don't know. – Bob D Jan 28 '20 at 15:48
  • @JMac That newtonian-gravity tag was added by Qmechanic, not the OP. – PM 2Ring Jan 28 '20 at 16:30
  • @BobD By some reasoning you might be able to argue that it's not infinite; because the information about the mass only travels at the speed of light, the same as our information about the cosmic background. So anything beyond our observable universe may in theory have mass; but the effects of it's gravity wouldn't have reached us because they are further away than the distance the signal could travel since the start of the universe. But I'm not a cosmologist, so there may be a flaw in that. – JMac Jan 28 '20 at 17:22
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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.

mmesser314
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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.

Yukterez
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  • You can have cosmological constant in the Newtonian gravity. – A.V.S. Jan 28 '20 at 16:22
  • In that case it should also work then – Yukterez Jan 28 '20 at 16:23
  • What "gravitational force" is there in GR? – ProfRob Feb 06 '20 at 07:15
  • 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. – Yukterez Feb 06 '20 at 07:19