Is the zero gravity experienced in ISS the "artificial" kind?

Question

I always wondered about the following:

An astronaut floating inside a spaceship that is far from Earth or any other other planet will experience true zero gravity because there is negligible gravitational pull coming from any planet nearby. Is this correct?
But an astronaut floating inside the ISS is experiencing artificial zero gravity (I use the term “artificial” because it is not the consequence of the lack of a gravitational pull — the Earth is still there), because the ISS is constantly free-falling towards the earth while at the same time speeding around it. Is this correct?

What I mean is, isn’t the zero gravity experienced by an astronaut inside the ISS similar to what a person would feel if they were standing inside an elevator that was free-falling for a long period of time? Or more precisely, isn’t it similar to the artificial zero-gravity created by those big zero-g airplanes? And we somehow seem to forget that when we look at those ISS videos.

Or am I completely wrong?

IDK if it's a duplicate, but it's answered here: https://space.stackexchange.com/q/54006/6944 tl;dr it's free fall. The ISS is really close to the Earth and strongly affected by its gravity. — Organic Marble, Jul 18 '21 at 21:05
As I said in the linked article above, there is no sane distinction between the two concepts and you should stop thinking of them as different. This is the fundamental principal of General Relativity, which says that there is no experiment you can devise in a closed system that will allow you to determine the difference between being in free fall and being in the absence of gravitational acceleration. — throx, Jul 19 '21 at 05:46
Your scenario (1) ********never******** occurs, anywhere n the universe. So by your definition, all zero gravity is of the "artificial" type. — CuteKItty_pleaseStopBArking, Jul 19 '21 at 06:11
I would have sworn I read the exact same question not a month ago. Cannot find it, though... — I'm with Monica, Jul 19 '21 at 07:45
There's one important difference between "artificial zero gravity" and "no gravity whatsoever". In the "no gravity whatsoever" Newton's Bucket won't work. Since all frames of reference are equivalent, in such conditions there's no difference between a 'static' bucket and 'spinning'. It takes gravity of other bodies to establish the non-rotating pseudo-special frame of reference. Or at least that's what I understood. — SF., Jul 19 '21 at 09:16
Too short for an answer, but congratulations, you discovered the Equivalence Principle (and none of the answers bothered to spell out that name...). :-) Check Wikipedia, they got a pretty extensive writeup. — AnoE, Jul 19 '21 at 09:16
@SF Whether the gravity of other bodies is responsible for Newton's Bucket is an argument among physics theorists that will probably never be resolved. Experiment shows that distant bodies mark the local non-rotating frame of reference, but it is impossible to demonstrate cause and effect. There is no empty universe available in which to perform the experiment. — John Doty, Jul 19 '21 at 13:36
It's all about frames of reference. Einstein's man in the falling elevator doesn't know or care why he's weightless. — RonJohn, Jul 19 '21 at 15:57
Your scenario (1) does occur though, in the vast majority of places. Gravitational pull diminishes rapidly as you move away from masses (inverse square law). In a homogeneuous neighbourhood nearby masses balance out. Interstellar space: galactic centripetal acceleration is in the order of 2e-10 m/s2, less than imperceptible (Milky Way, at our distance from centre). Intergalactic space: many orders of magnitude less. — g.kertesz, Jul 19 '21 at 17:02
@PcMan re "********never******** occurs, anywhere in the universe" except that it does in billions of billions of places in the universe. — uhoh, Jul 20 '21 at 04:55
@uhoh You are using nonsense semantics to promulgate a false statement. This is very unkind to the readers of this forum, and I respectfully request that you do not do so. — CuteKItty_pleaseStopBArking, Jul 20 '21 at 06:26
@throx There is a way to determine. In a freely falling frame particle will move away from each other. In an accelerated frame in outer space they wont. — Deschele Schilder, Jul 20 '21 at 14:33
@DescheleSchilder Fair point, but not really the thrust of the question and unhelpful for a deeper understanding of the physics. The GR equivalence principle assumes a constant gravitational field across the system (or an infinitesimal system), otherwise it is saying it's equivalent to non-uniform acceleration across the system - hence tidal forces. — throx, Jul 20 '21 at 22:18
@throx Fair point but really not höw it is for real physical systems. There is no such thing as a uniform gravity field. — Deschele Schilder, Jul 21 '21 at 00:34
Gravitational time dilation produces a measurable difference between "true" zero gravity and freefall to an outside observer — thegreatemu, Jul 21 '21 at 19:49

score 76 · Accepted Answer · edited Jul 21 '21 at 08:11

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Gravity is everywhere. There is never any actual true "zero gravity" in the universe.

But if you're in freefall - meaning following gravity's pull rather than resisting it, or being blocked from following it (by the floor, your nearby planet, spaceship walls as it thrusts, or whatever) - you don't feel it, and that's the thing we call "weightlessness" or (wrongly) "zero gravity".

The weightlessness you feel in a spaceship far from any object, is exactly the same weightlessness you feel on board the ISS orbiting earth. There aren't 2 kinds ("artificial" vs "real"). You can see that because if you zoom out your focus a bit, the spaceship "far from any object" is in fact still falling towards some object, perhaps at very high speed. Its nearest galactic cluster, or supercluster, a few dozen megaparsecs away, perhaps, but it's still falling fast towards it. If it doesn't hit anything, it will follow a path that forms a (probably highly) elliptical orbit over hundreds of millions of years, since it won't lose energy and collide. And the ISS is still following Earth's gravitational pull, it's going to remain in an elliptical orbit too, if you ignore energy loss from the trace atmosphere at that altitude. Identical behaviour, just on different scales.

So there isn't any such thing as "artificial zero gravity", or a distinction between some kind of zero G that's "real" vs. "artificial", apart from simulations like floating in a water tank or other simulators.

If you are freely moving as gravity dictates, you will experience the sensation we call "weightlessness" or "zero gravity". If something stops you doing so, you won't (or will feel it much less). Its that simple. *

* For completeness, if something only slightly stops you from freely following gravity, or the local gravitational field is weak anyway but some object you're pressing against stops you from following it (eg on the moon), you'll feel this as low gravity or microgravity.

edited Jul 21 '21 at 08:11

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answered Jul 19 '21 at 07:48

Stilez

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There is one very small difference, since you're so close to a massive object you will experience tidal forces. – Jason Goemaat Jul 19 '21 at 21:01
Gravitational force may not be zero far out in space, but that's technically why the term "micro gravity" exists which is more appropriate than "zero gravity". It is very, very very low, and yet the result is the same as in a very high gravity situation (orbiting Earth, or the Sun for that matter). And I don't think your fast motion example is particularly convincing as a teaching point because you can never feel motion, whether "falling" is happening or not. That you can't feel falling requires something extra, and that – The_Sympathizer Jul 20 '21 at 02:40
"something extra" has to do with the fact the acceleration given to all parts of your body is essentially uniform (except when it isn't). – The_Sympathizer Jul 20 '21 at 02:42
@J... Exactly what I was thinking. Even in the space between the planets, you're in a free-fall orbit around the sun, with it's bigger gravity. Or in between the stars, you're in free-fall orbit around the centre of the galaxy, with it's humungous gravity, exactly the same in principle as the ISS is around the earth (with it's cute but comfortable gravity). – Reversed Engineer Jul 20 '21 at 18:12
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There is no "zero gravity"; there is only "everything in the vicinity is accelerating in precisely the same way". – Darth Pseudonym Jul 20 '21 at 19:47
@Darth (1) yes, but being understanding of popular terminology, and what OP means by "zero G" or "weightlessness" is easy, and explained in answers. (2) disagree, that begs the question "what counts as the vicinity". I can be at an outside edge of ISS and drifting in space, and ISS will move away (tiny difference of gravity), but I'll still feel "weightless". Or I can be in a void with only my spaceship as company and if its engine is on, we accelerate in "precisely the same way" but I don't feel weightless. So its poorly defined and worse, can fail both ways. So I prefer my given definition. – Stilez Jul 20 '21 at 20:22
@Stilez I'm sorry if that came off as disagreement; I was actually just meaning to restate what you said in your answer about 'freely moving as gravity dictates'. But as to your point 2, you can be in freefall and not feel like you're in zero gravity. If you're falling past a building, you're going to feel like you're falling, not floating, even if the physics are identical to what you'd have if you were in orbit. It's all just following gravity's curve. "Vicinity", like "zero gravity", is a human interpretation that doesn't exist in physics, but it's critical to how we interpret things. – Darth Pseudonym Jul 20 '21 at 20:42
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In that case (falling past a building) what happens is you feel in freefall - if you close your eyes its the same (barring noise, wind pressure, and the thud at the bottom!). You get input from your eyes that you interpret as saying you are falling. But idealised falling is indistinguishable from weightlessness and popular "zero G* (in deep space falling towards the nearest supercluster). (CONT) – Stilez Jul 20 '21 at 21:31
.......You can even flip it round, if I magically put you into deep space, falling towards some supercluster far far away, and then fed a VR video of a passing building and some wind noise to your sensory nerves, that would feel like idealised falling, showing the 2 are the same from a weightlessness perspective, and its only other data that we correlate and compare as context, that tells us which is the "correct" of these words for the identical falling/weightless/"null-G" experience. If you freely follow gravity, that's the defining criterion for the experience the OP means, by any name. – Stilez Jul 20 '21 at 21:32
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Excellent answer! Tiny nit: "or being blocked from following it" I think you mean "or not being blocked from following it". – Nate Barbettini Jul 21 '21 at 15:34
You can't tell the difference in source of weightlessness, but gravitational time dilation depends on the strength of the local gravitational potential, so there is indeed a fundamental difference between the two scenarios...just not observable from within the freefall reference frame – thegreatemu Jul 21 '21 at 19:48

score 26 · Answer 2 · answered Jul 18 '21 at 21:25

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Gravity has infinite range, so there is nowhere in the universe where you can be free from its influence. Sure, there are places such as supervoids where the influence of gravity will be very little, but there's nowhere where it is absent entirely.

The day to day experience of 'gravity' - the feeling of standing on the surface of a planet - isn't really the work of gravity itself. Unless you're very close to a black hole, the pull of gravity is equal on all parts of your body, and so there's no differential force that allows you to feel anything.

What you do feel, while on the surface of a planet, is the ground pushing against your feet, which push against your legs, which push against your torso, which pushes against your head and vestibular system and allows you to tell which way is up. This is the same differential force you feel when accelerating or turning in a car, and it's because this force isn't instantaneously applied evenly through the body that we can feel it.

So yeah. Being in zero-g feels like falling. Since the 'non-artificial' kind of zero-gravity as you state in your question doesn't exist anywhere, the concept of 'artificial' zero gravity isn't an especially useful one.

answered Jul 18 '21 at 21:25

Ingolifs

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So the weightlessness experienced in the ISS is due to free fall (and not due to minimal gravitational pull), whereas the weightlessness experienced somewhere between e.g. Earth and Mars is due to the lack of a strong gravitational pull coming from a nearby planet correct ? So the cause is indeed different in each case right ? – Sprout Coder Jul 18 '21 at 23:54
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Between Earth and Mars it's still free-fall."The answer is always free-fall" (except when it's "thermal control") – Organic Marble Jul 19 '21 at 00:31
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'Weightlessness experienced' is always caused by the same thing: not undergoing acceleration or being supported by the planet's surface. – Ingolifs Jul 19 '21 at 00:32
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The second to last paragraph is key. You don't feel gravity, your body has nothing that's not affected by gravity which it could compare itself against. You only feel the planet (or a spacecraft) pushing you away from the path through spacetime you would otherwise take. If nothing's doing that, you're in freefall. Freefall's no more or less "real" if you're on a suborbital trajectory that will hit Earth's surface before it completes an orbit, in Earth orbit, in solar or galactic orbit, or in some intergalactic void, unbound to any galaxy cluster, supercluster, etc. – Christopher James Huff Jul 19 '21 at 02:13
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@SproutCoder the weightlessness experienced between Earth and Mars would be different - instead of free-falling around Earth, you're free-falling around the sun. If you did the same between Sol and Proxima Centari, you would be free-falling around the galactic core. Between the Milky Way and Andromeda, you'd be free-falling around the center of the galactic cluster. "The answer is always free-fall". – Tim Jul 19 '21 at 06:03
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Ok then you all for clarifying. I was obsessing over the planetary gravitational pull vs. free-fall but I understand there's not such distinction. – Sprout Coder Jul 19 '21 at 09:08
actually: weigthlessness does not feel like falling, unless you fall in very good vacuum or in an accelerated elevator (which I don't hope). When falling on Earth, you usually move through air which you feel. In weightlessness you just float in the air without any relative velocities - it's a distinct perceived difference to anything usually described as falling on Earth. – planetmaker Jul 19 '21 at 16:17
In Lagrangepoints gravity is zero. All origins in falling frames have zero gravity. – Deschele Schilder Jul 19 '21 at 19:19
Zero gravity does exist. All freely falling disconnected point masses experience zero gravity. They dont experience any basic force pushing or pulling them. Gravity doesnt push or pull. The three basis forces do (inertial forces). – Deschele Schilder Jul 19 '21 at 19:29
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@DescheleSchilder https://en.wikipedia.org/wiki/Lagrange_point#/media/File:Lagrangianpointsanimated.gif If gravity were zero, objects in the Legrange points would not be going in circles (accelerating). – user3067860 Jul 19 '21 at 20:05
I meanzero gravity in the sense that no gravity is felt at that point. An accelerometer would point to zero. The ssme holds for the origins of all freely falling frames (the Lagrange point is such an origin). – Deschele Schilder Jul 19 '21 at 20:41
+1 but a tiny nit about "... there is nowhere in the universe where you can be free from its influence." – uhoh Jul 20 '21 at 00:35
Excellent answer! Tiny nit: "or being blocked from following it" I think you mean "or not being blocked from following it". – Nate Barbettini Jul 20 '21 at 18:55
@NateBarbettini I think you mean Stilez's answer, not mine. – Ingolifs Jul 21 '21 at 01:01
@Ingolifs Doh, clicked the wrong comment button. Sorry! – Nate Barbettini Jul 21 '21 at 15:33

score 8 · Answer 3 · answered Jul 19 '21 at 01:23

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The right way to think about it is that, always and everywhere, weightlessness is the "artificial" kind. It is certainly true that the gravitational field is very weak far from any masses, but on the way to the moon the astronauts were coasting in free fall so it made no difference to their experience what the gravitational field strength was. Even some comet halfway between the sun and Alpha Centauri is moving quite fast around the center of the galaxy.

answered Jul 19 '21 at 01:23

Mark Foskey

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In every freely falling there is a point with zero gravity. In fact, you can choose all points to have zero gravity. Nothing artifical about it. – Deschele Schilder Jul 19 '21 at 19:23
Am I right in thinking that even in a hypothetical thought-experiment universe with no matter in it whatsover, if I were to be magic'd into existence there, I would still feel some gravity, if only e.g. due to the mass in my legs being attracted toward the mass of my arms, and vice versa? – Jeremy Friesner Jul 20 '21 at 20:18

Deschele Schilder · Answer 4 · 2021-07-19T14:37:26.560

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It is exactly the same zero gravity as you experience in a plane paraboling to Earth. It 's a bit different from the gravity you experience in far-from-mass gravity in outer space though. In the ISS gravity is nearly zero at every point. But not precisley (though it is not easy toy measure if not impossible). There is always a gradient giving rise to tidal forces. This is a global force. It only exists for two separated locals (ponts). Two point masses in the ISS will eventually separate. There is always a certain point in the ISS though for which the gravity is exactly zero. Somewhere in the middle of the ISS. This is happening in a falling elevator too. Somewhere inside the falling elevator the force of gravity is exactly zero. If you place a pont mass in the middle of the elevator it will stay put. If you place it nearer to the bottom or the ceiling of the elevator the force will be still zero but it will accellerate to the bottom or ceiling because of the tidal nonlocal force. You can also say that the ceiling, the bottom and the entire lift exlerience a tidal force and not the mass, which is experiencing no force at all.

This is how you can discriminate between a freely falling elevator and one in free space (or between a lift stationary on Earth and one accelerated in space). In the falling frame (or your falling body) the is always an experience of tidal forces (which are electromagnetic, strong, or weak in Nature).

edited Jul 19 '21 at 14:37

answered Jul 19 '21 at 12:40

Deschele Schilder

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At the very precise level, there is no such thing as a uniform gravitational field, so tidal forces exist everywhere. There's really no distinction between the near-mass and far-from-mass cases, just varying degrees of strength of the tidal force. It's already utterly insignificant on the ISS, clocking in at about a millionth of a G - the ventilation system is more powerful than the tidal force. Trying to distinguish between a freely falling elevator and a stationary one in deep space won't work, as you'll always find a tidal force with a sensitive enough instrument. – Nuclear Hoagie Jul 19 '21 at 13:08
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@NuclearHoagie If the space is flat there is no tidal force. In a Lagrange poinr the tidal force is opposite in both directions. If the tidal force is very small (much smaller than in a gravity field of a planet) you can be sure you are in deep space. I you can easure it. If you lived long enough to watch the behavior of two small masses you could know where you find yourself. – Deschele Schilder Jul 19 '21 at 13:18
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My point is that there is no cutoff for "very small tidal force" that would allow you to classify scenarios into two different categories. There is fundamentally no difference between near-mass and far-from-mass cases, it's just a continuum of tidal effects. The tidal force on the ISS is far, far smaller than the force of gravity acting on the ISS, but it's not deep space. The hypothetical scenario of an elevator falling through a uniform gravitational field is simply a good approximation when far from mass, but it's never actually the case. – Nuclear Hoagie Jul 19 '21 at 14:26
@NuclearHoagie In an elevator falling in a uniform field there is no tidal force at all. The tidal force in deep space are many orders smaller than those in the ISS. These are very very very small indeed but the relative difference is huge. – Deschele Schilder Jul 19 '21 at 14:32
except: there is nowhere in the universe such thing as a 'uniform field'. – planetmaker Jul 19 '21 at 16:13
@planetmaker You can imagine one. On both sides of an infinite slab of mass. In this case there is no tidal force. So measuring zero tidal effect could mean you find yourself above such a plate. Bu how far from it would be impossible to know. The whole universe could be in such a field right now. – Deschele Schilder Jul 19 '21 at 16:35
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@NuclearHoagie: There’s no hard cutoff, but distinctions like “force X is negligibly small compared to force Y” vs “forces X and Y are both significant” are very meaningful and useful. A large enough quantitative distinction becomes a qualitative distinction. – Peter LeFanu Lumsdaine Jul 19 '21 at 23:41
I don't agree with this answer. Using the example, yes there is a point in ISS where if a mass will "stay put" relative to ISS. But that isnt really relevant to this question. All it means is you're moving due to gravity the exact same as ISS, not that "gravity is exactly zero" at all. To show why this is so, imagine being at another point midair in ISS. Now close your eyes. Both points will feel identical, until and unless you bump into the ISS wall for the 2nd of them. So the feeling of weightlessness on ISS is not related to any relative motion to ISS. Its all about moving with gravity. – Stilez Jul 20 '21 at 00:54
Also unless you're being ripped apart or using incredibly sensitive instruments, you'll never feel a tidal force from gravity, either. To the extent that a mass drifts up or down, it is still experiencing weightlessness in that scenario. Its just that freely following gravity means it's not quite on the same path as the elevator, not that it isnt experiencing what we call "weightlessness". The drift and tidal issues here as well, are all but irrelevant to the issues posed by the question as well, because the OP asks if there are multiple kinds of weightlessness, which this doesn't answer. – Stilez Jul 20 '21 at 00:59
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@NuclearHoagie: In the real universe, the question is how low tidal forces are (and/or how flat the local curvature of space-time is?), not whether they're present or absent. It's a different question from the one the querent is asking, but despite being quantitative instead of qualitative, they're related, and could perhaps help the querent put their finger on a physics-based way to make the distinction they're trying to make / thinking about. I upvoted this answer for that reason. (Although it says some other stuff so probably I should post my own.) – Peter Cordes Jul 20 '21 at 03:47
@Stilez You are considering different falling frames.I consider only the frame with the origin in the middle of the ISS. Seen from there there is a force on both sides of the ISS. These forces are oppositely directed. If you consider different origins the force in that new origin is still zero. But the opposite forces change and if you put your origin on one of both sides (the one closest and the one farthest) then all forces will be in one direction only. Only inward ör only outward. This doesnt make much difference though for the experienced EM forces. EM forces are theonly onces experienced – Deschele Schilder Jul 20 '21 at 05:13
@Stilez You can never experience gravity directly. Only by means of the EM interaction. – Deschele Schilder Jul 20 '21 at 05:15
@PeterCordes I like your profile picture! How many times my girlfriend has called already... With a steadily raising voice... – Deschele Schilder Jul 20 '21 at 05:23
Deschele - I'm purely considering the question. Itis not asking if you can detect what's going on with your senses, or lab devices, or by seeing if you collide with an elevator or ISS wall. Its not asking if you can notice differential movement or forces due to tiny differences in gravity between you and an elevator or ISS wall, or detection of EM interactions occurring. Its purely asking, will you experience "weightlessness" or what is often (though wrongly) called "zero G", does negligible gravity cause that sensation, and are there >=2 kinds of 0G. It just doesnt answer the actual question. – Stilez Jul 20 '21 at 08:12
@Stilez There is indeed no feeling as the feeling of standing on earth. Is this what the OP wants to know? Does he think you still are squeezed up there? Or pulled? One can ask if weighlessness up there is the same weighlessness as in a swimmingpool where the astronauts practice. – Deschele Schilder Jul 20 '21 at 08:34
@Stilez What about the first sentence in the answer? – Deschele Schilder Jul 20 '21 at 08:42

score 1 · Answer 5 · answered Jul 19 '21 at 17:16

Or more precisely isn't it similar to the zero-gravity created by those big zero-g airplanes

Yes, it's identical

Note that you use the phrase:

artificial zero-gravity

There is no such thing as "artificial zero-gravity". It's a meaningless phrase.

Note that you use the phrase:

zero-gravity

There is no such things as zero-gravity.

Pilots etc. use the phrase "zero-gravity" or "zero-g", just roughly, to mean "the feeling when you're in one of those planes or the ISS".

Essentially, everything in your question is correct, and more so!

score 1 · Answer 6 · answered Jul 20 '21 at 04:08

There isn't a qualitative difference between the two situations you've presented, only quantitative.

(And neither one can meaningfully be called "artificial" - both situations are real microgravity, as explained in other answers. You might apply "artificial" to astronaut training neutral-buoyancy tanks (although it seems that the usual term is "microgravity simulator"), or a hypothetical science-fiction gravity manipulating device.)

The actual quantitative differences between local (theoretically) observable gravity conditions are:

Tidal forces: far from anything, the gravity field is fairly close to uniform. The forces squishing or pulling on things are significantly smaller than in low orbit of a planet such as Earth in the vicinity of the Sun. See my answer on Could a space colony 1g from the sun work? for some examples of tidal forces on an extended object, and https://en.wikipedia.org/wiki/Tidal_force
Curvature of space-time: far from anything, space-time is very close to flat (or to the natural curvature of the whole universe?). I'm very much not an expert on this. Close to a planet, spacetime is curved so a freefall path curves towards or around it. (Again, I'm probably butchering some terminology or worse.)

So there's nothing fundamentally different from falling around a planet continually vs. falling through intergalactic space (in orbit or not around a nearby galaxy or supercluster). It's just a matter of degree.

And yes, you could in theory make measurements of those factors (at least tides) inside a sealed elevator with no windows. Especially if you could isolate your experiment from any inconvenient humans moving around, and wait months or years to see how quickly some objects initially at rest relative to the ship accelerate (very tiny acceleration integrated over long timespans).

Hopefully identifying what those quantitative differences are can help you put a finger on what you were wondering about / thinking about when you came up with the natural / artificial distinction you were trying to make.

score -1 · Answer 7 · edited May 30 '22 at 08:46

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Supplemental answer:

tl;dr: There are billions of billions of points in the universe with zero gravitational acceleration!

While each answer includes some form of the "gravity is everywhere because it never ends" (i.e. $1/r^2$ never goes to zero so everything pulls on everything) I have to inject two caveats:

~~Gravity moves at the speed of light~~ To my understanding, things outside the observable universe don't pull on us.
Gravitational acceleration is just the ~~divergence~~ gradient of a scalar potential field and not some amazingly complicated vector field (except near singularities), and (for) masses, then in general, there will be at least −1 zeroes (in 1d precisely −1 zeroes) of the (acceleration) field, all of them isolated, within the convex hull of these masses and none of them stable.

Those are mathematical points of course, but some argue that math is more real than anything else :-)

edited May 30 '22 at 08:46

Rory Alsop

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answered Jul 20 '21 at 00:29

uhoh

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All the questions pertaining to this ask whether there is a true zero-gravity situation for a macroscopic object, typically an astronaut or spaceship. Yes, there will be locations of zero size of gravitic equipotential, but you could not even fit a single atom in there, much less an actual object. – CuteKItty_pleaseStopBArking Jul 20 '21 at 06:24
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@PcMan yep, that's the nature of mathematical points, and for solid objects we only need to get their center of mass to that point for the net acceleration to be zero. – uhoh Jul 20 '21 at 06:54
How does one fit an object with nonzero dimensions, to align exactly with a nondimensional point(or line). Even one planck length off, and you are out of the "zero" location. Putting a macroscopic object (or its center of mass) on a nondimensional point is as nonsensical as asking someone to manufacture a measuring stick that is exactly one meter in length, with zero deviation. The universe does not do such things, outside of mathematics books. – CuteKItty_pleaseStopBArking Jul 20 '21 at 07:51
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I find this both fascinating, yet also useless because it's something that does not occur in any practical sense for any real-world micro or macro object. Maybe its biggest relevance (other than theoretical cosmology and model development) is the concept, and also these kinds of scalar force maps that suggest (possibly huge) regions of gravity "as small but nonzero" as you'd like. – Stilez Jul 20 '21 at 08:26
This answer is wrong from beginning to end 1) scalar gravity potential is instantaneous in both Newtonian and Einsteinian gravity. 2) it is not the "divergence" but gradient of the scalar potential. – Kphysics Jul 21 '21 at 07:07
@Kostas I think "beginning to end" exaggerates gratuitously. I mixed "divergence" and "gradient" the same way I mix "ItsI and "it's"; my words don't always come out right. Once solve the Laplace equation and get $\phi$ I can write $\nabla \phi$ and call it "del" without having to say "gradient" out loud or even type it. But I don't yet believe you that gravity is instantaneous and can come from beyond the observable universe, can you cite a source for that or at least link to an authoritative and well-received Stack Exchange answer? Thanks! – uhoh Jul 21 '21 at 11:05
@uhoh There are endless discussions all over the net about it. Usually they ask: if the earth was attracted to the retarded location of the sun it would it not slow down very quickly? And the answer is yet it would, but the scalar potential is instantaneous, see here https://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html About objects outside the observable universe, it is nonsense to even talk about it: thats not the way GR works. – Kphysics Jul 23 '21 at 07:17
@Kostas to provide you more space I've asked Is it "nonsense to even talk about" objects outside the observable universe not having gravitational influence on us? (finite speed of gravity) in Astronomy SE. – uhoh Jul 23 '21 at 23:41
Since you don't like drive-by downvotes, I'll explain my downvote. (1) I agree with @Kostas. (2) Accelerating with respect to what? Your answer implicitly assumes the Newtonian concept of universal inertial frames of reference. (3) We know that that is not the way the universe works. GR appears to be the way the universe works. (GR might not be right; physicists for the most part assume that it is not perfectly correct, for multiple reasons. But that breaking point has not yet been found.) – David Hammen Jul 24 '21 at 09:40
@DavidHammen maybe so, I'll keep thinking about it. I've opened a question in astronomy (linked above) to provide more space for explanation – uhoh Jul 24 '21 at 14:16
Somebody makes three errors in three sentences, acknowledges two of them, but does not edit his answer to correct them and then insists that the third sentence is correct anyway... rewrite your answer and if there is something else to talk about I might come back. – Kphysics Jul 26 '21 at 10:26
@Kostas I've posted the other question to help me get a clearer picture of the subject. I am not yet sure how to adjust/finesse the wording to make it acceptable to everybody at the same time. Not everybody speaks full GR and not all answers about space exploration embrace complete GR solutions to every problem. I thought I did switch it to gradient but it seems I didn't. Perhaps I was waiting for clarity on the other stuff then forgot. That's easy to fix. – uhoh Jul 26 '21 at 10:32
@Kostas But I do not yet acknowledge that stuff outside the visible universe is still pulling on us, and simply "That's not how GR works" is not in and of itself convincing to me nor helpful. Are you saying there are no zeros in acceleration in GR as there are for Newtonian gravity? Are you saying there is no acceleration in GR? Your comments are not illuminating nor actionable (except for the gradient bit. Please elaborate in an illuminating and actionable way. Thanks! – uhoh Jul 26 '21 at 10:35
@uhoh The second mistake was: "Gravity moves at the speed of light" - I provided you with a reputable source that the scalar potential (which is what we are talking about here) does not propagate with a delay. – Kphysics Jul 27 '21 at 12:28
@uhoh And regarding the third thing, I am not saying that the stuff outside the visible universe is still pulling on us, only that it is impossible to measure that and therefore nonsense to talk about it. – Kphysics Jul 27 '21 at 12:30
@Kostas in my limited capacity I'm still struggling with gravitational waves traveling at the speed of light but the scalar potential being instantaneous, but that's my own struggle and I'll have to deal with it somehow. It seems that I've already been told not to use a delay and then embraced no delay here. I've adjusted the wording of point #1; which is meant to address only that there is more universe than observable universe. – uhoh Jul 28 '21 at 00:15
@Kostas SE answers can sometimes be imperfect; has this answer approached a level of tolerability you can accept? If glaring errors remain please let me know. Thanks! fyi I’ve just asked Why do gravitational waves "only" travel at the speed of light but the gravitational scalar potential is instantaneous? and referenced there my scolding here. – uhoh Jul 28 '21 at 01:57
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There is nothing useful that can be said about it here, or in these new Q you asked on astronomy.SE - requires full understanding of GR. When @Ben-Crowell there tells you that saying the scalar potential is instantaneous in GR is wrong, he is right as well. At least please accept that you cannot improve Newton by adding a delay? – Kphysics Jul 28 '21 at 08:42
@Kostas My previous comment "It seems that I've already been told not to use a delay and then embraced no delay here." and its links have already done that. – uhoh Jul 28 '21 at 08:53

Is the zero gravity experienced in ISS the "artificial" kind?

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