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I always wondered about the following:

  1. 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?

  2. 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?

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    $\begingroup$ IDK if it's a duplicate, but it's answered here: 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. $\endgroup$ Jul 18 at 21:05
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    $\begingroup$ 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. $\endgroup$
    – throx
    Jul 19 at 5:46
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    $\begingroup$ Your scenario (1) **********never********** occurs, anywhere n the universe. So by your definition, all zero gravity is of the "artificial" type. $\endgroup$
    – PcMan
    Jul 19 at 6:11
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    $\begingroup$ I would have sworn I read the exact same question not a month ago. Cannot find it, though... $\endgroup$ Jul 19 at 7:45
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    $\begingroup$ 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. $\endgroup$
    – AnoE
    Jul 19 at 9:16
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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.

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    $\begingroup$ There is one very small difference, since you're so close to a massive object you will experience tidal forces. $\endgroup$ Jul 19 at 21:01
  • $\begingroup$ 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 $\endgroup$ Jul 20 at 2:40
  • $\begingroup$ "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). $\endgroup$ Jul 20 at 2:42
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    $\begingroup$ There is no "zero gravity"; there is only "everything in the vicinity is accelerating in precisely the same way". $\endgroup$ Jul 20 at 19:47
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    $\begingroup$ 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) $\endgroup$
    – Stilez
    Jul 20 at 21:31
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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.

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    $\begingroup$ Between Earth and Mars it's still free-fall."The answer is always free-fall" (except when it's "thermal control") $\endgroup$ Jul 19 at 0:31
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    $\begingroup$ 'Weightlessness experienced' is always caused by the same thing: not undergoing acceleration or being supported by the planet's surface. $\endgroup$
    – Ingolifs
    Jul 19 at 0:32
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    $\begingroup$ 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. $\endgroup$ Jul 19 at 2:13
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    $\begingroup$ @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". $\endgroup$
    – Tim
    Jul 19 at 6:03
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    $\begingroup$ 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. $\endgroup$ Jul 19 at 9:08
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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.

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  • $\begingroup$ 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. $\endgroup$ Jul 19 at 19:23
  • $\begingroup$ 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? $\endgroup$ Jul 20 at 20:18
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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).

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  • $\begingroup$ 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. $\endgroup$ Jul 19 at 13:08
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    $\begingroup$ @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. $\endgroup$ Jul 19 at 13:18
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    $\begingroup$ 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. $\endgroup$ Jul 19 at 14:26
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    $\begingroup$ @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. $\endgroup$ Jul 19 at 23:41
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    $\begingroup$ @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.) $\endgroup$ Jul 20 at 3:47
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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!

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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.

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Supplemental answer:

Being the sticlker that I am I can not remain silent any longer.

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:

  1. Gravity moves at the speed of light To my understanding, things outside the observable universe don't pull on us.
  2. Gravitational acceleration is just the divergence gradient of a scalar potential field and not some amazingly complicated vector field (except near singularities), and (for) 𝑛 masses, then in general, there will be at least 𝑛−1 zeroes (in 1d precisely 𝑛−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 :-)

(for) 𝑛 masses, then in general, there will be at least 𝑛−1 zeroes (in 1d precisely 𝑛−1 zeroes) of the field, all of them isolated, within the convex hull of these masses and none of them stable.

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  • $\begingroup$ 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. $\endgroup$
    – PcMan
    Jul 20 at 6:24
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    $\begingroup$ @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. $\endgroup$
    – uhoh
    Jul 20 at 6:54
  • $\begingroup$ 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. $\endgroup$
    – PcMan
    Jul 20 at 7:51
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    $\begingroup$ 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. $\endgroup$
    – Stilez
    Jul 20 at 8:26
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    $\begingroup$ 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? $\endgroup$
    – Kostas
    Jul 28 at 8:42

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