<p>Gravity extends to infinity, so no, strictly theoretically speaking there is always some gravity present. In theory, even in this case we could have points in space where gravitational forces cancel out, but given the complexity of our universe, this just won't happen in practice.</p>
<p>As a more relaxed viewpoint - there are special points around orbiting objects called <a href="https://en.m.wikipedia.org/wiki/Lagrange_point" rel="noreferrer">Lagrange points</a>, where gravitational forces of the two objects (eg a star and a planet) sum up to zero. Those points do receive gravitational forces from other objects however (eg from Jupiter in case of the Sun and Earth). Also, as the two objects in question are also moving, these points are moving with them (otoh, there's also simply no absolute positions in space anyways, but that's a title for another story)</p>

CC BY-SA 4.0
2021-06-04T21:44:13.323Z
Even if you just consider the two objects, the gravity does not sum to zero at any of the five Lagrange points.
notovny
CC BY-SA 4.0
2021-06-04T21:47:21.397Z
@notovny yeah, you're right, I worded this inaccurately - I wanted to convey the idea that gravities of those two objects balance out
tuomas
CC BY-SA 4.0
2021-06-04T21:48:33.800Z
Also, you’re still $GM(<R)/R$ deep in your galaxy’s potential.
pela
CC BY-SA 4.0
2021-06-05T01:08:08.167Z
No, Lagrange points are vector zeros in the sum of the real gravitational forces plus centrifugal forces (gradient of the gravitational potential plus a *pseudopotential*). I think that if one considers only Newtonian gravity and establishes some frame of reference, one can easily have zeros in the net gravitational force field.
uhoh
CC BY-SA 4.0
2021-06-05T01:28:37.497Z
cf. https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec18.pdf
uhoh
CC BY-SA 4.0
2021-06-05T06:30:35.073Z
"Gravity extends to infinity". That's something we couldn't possibly know. The theories we currently use to describe gravity assume so, and they work fairly well, but they are theories and not fact, and we know they are not perfect.
Martin Argerami
CC BY-SA 4.0
2021-06-05T08:12:09.773Z
@MartinArgerami yes, I think that makes sense, but I'd prefer to stick with mainstream physics here and not speculate too much. Admittedly in case forces are quantized this will at least practically affect how gravity affects over large distances
tuomas
CC BY-SA 4.0
2021-06-05T17:20:16.247Z
@MartinArgerami and everything we see could be a simulation, or gravity could be a residual effect of all the faeries around us.
eps
CC BY-SA 4.0
2021-06-05T18:37:58.687Z
@eps: either way, there is no way for us to possibly know what happens arbitrarily far away. Actually, if you think about it, most "observations" already depend on theory and assumptions.
Martin Argerami
CC BY-SA 4.0
2021-06-05T18:38:33.097Z
@tuomas: yes if gravity is quantized it should be possible in principle to have places without gravity.
Martin Argerami
CC BY-SA 4.0
2021-06-05T23:31:51.117Z
@MartinArgerami Is there any reason not to think that the same laws of physics we see here would be different at very large distances.  Observations do not depend on theory and assumptions, interpretation of these observation does.
Hollis Williams
CC BY-SA 4.0
2021-06-06T00:29:34.600Z
@MartinArgerami “They are theories and not facts” is a statement that doesn’t get you anywhere. Do you know anything for a fact?
pela
CC BY-SA 4.0
2021-06-06T00:48:55.713Z
@pela: Not sure where you want to go. If you want to discuss philosophical realism, that's a complicated topic, but one that is worthless regarding physics. But to regard theories as facts is a fairly common mistake that scientists have made again and again. Newton's theory was "fact" for hundreds of years, until it wasn't. So, let me ask you, are Galilean transformations a "fact"? are singularities in GR a "fact"? Are Hilbert spaces in Quantum Mechanics "facts"? Physics is made out of theories; "facts" are few.
Martin Argerami
CC BY-SA 4.0
2021-06-06T07:38:01.450Z
Gravity is non-zero at all Lagrange points (only for two stationary objects there will be one point between the objects where gravity is zero). It's *because* of this that Lagrange points exist. Gravitational force is precisely compensated by centrifugal forces at these points.
Deschele Schilder
CC BY-SA 4.0
2021-06-06T13:11:09.520Z
@MartinArgerami That was sort of my point, actually. We don’t *really* know anything for a fact, but at least within mainstream physics, we know some things well enough that calling it “a fact” is quite acceptable, I think (I’m sorry if I came off a bit hostile).
pela
CC BY-SA 4.0
2021-06-06T13:39:22.930Z
@eps: fair enough. I definitely understand where you come from.
Martin Argerami
CC BY-SA 4.0
2021-06-04T21:26:58.397Z
2021-06-04T21:26:58.397Z
tuomas
<p>I think we must be a bit careful what we are discussing here. General relativity states that when an observer is on a geodesic, i.e. in freefall, they are not experiencing gravity. That's Einstein's famous falling elevator <em>gedankenexperiment.</em>  It's not that gravity is there but cannot be measured, or that it is there but canceled out by the equivalent acceleration; it is <em>not there period</em> in the frame of reference of the observer on the geodesic.</p>
<p>In this sense, for every spot in the universe there is a reference frame in which no gravitation is present, namely the one moving along the geodesic. An example would be a beam of light. In its own frame of reference it's moving perfectly straight through space time. (Or rather, sitting perfectly still in an immutable, flat universe.)</p>
<p>Now admittedly, there are no point-shaped observers; what the man in the elevator as well as any other observer with finite size can observe is a <em>gradient</em> causing inner forces within the observer (more precisely, within the observed finite volume of space). The reason is that the geodesics for the different space time locations within the observed volume are <em>not quite</em> parallel in an inhomogeneous gravitational field. <em>A gravity gradient</em> is perfectly well measurable within an extended object in freefall; <a href="https://space.stackexchange.com/questions/40942/how-big-are-the-tidal-accelerations-within-the-iss">an example is the ISS</a>. Only some points on it are in perfect freefall, namely those on the orbital trajectory of its center of mass; the others are pulled or pushed along.</p>
<p>Now the effect is already fairly small even in the comparatively steep gradient close to Earth; if you choose a location with very little gravity to begin with, and choose an inflection point at the right distance from the closest mass clusters, the effect will be extremely small. A good spot might be in an intergalactic void. My gut feeling is that it will be indistinguishable from background noise like photons from the the microwave background radiation, or general vacuum quantum fluctuations. Oh, and the <em>gedankenexperiment</em> also necessitates a massless observer because any local masses would create much stronger gradients, due to their proximity, than even very large masses at large distances, thus drowning out any subtle global gradients.</p>

CC BY-SA 4.0
2021-06-06T15:59:01.237Z
A gravity gradient (tidal force) is also present for particles moving on geodesics. If a particle falls freely to Earth then gravity will be zero for the particle, but the gravity gradient not. Only a free particle (somewhere in outer space) that is seen in an accelerated frame will be seen to experience no tidal force.
Deschele Schilder
CC BY-SA 4.0
2021-06-06T17:06:02.160Z
@Barbierium Yes, there is a gradient but it is meaningless (and undetectable) for a point-shaped particle. Even if we can assign a value to the gradient for each point in spacetime it is still a difference between two points.
Peter - Reinstate Monica
CC BY-SA 4.0
2021-06-06T17:13:43.097Z
Yes, that's true. A particle that falls freely to Earth is seen as stationary in a co-falling frame. There is no tidal force on the particle (if a point). Only two (or more) particles will move away from each other slowly (which is the way you can see if you are in a falling frame or an accelerated frame).
Deschele Schilder
CC BY-SA 4.0
2021-06-07T03:37:56.010Z
@tfb You are making the same "mistake" as many make. It's *not* a physically meaningful question to ask if the space is flat somewhere. That's what relativity is all about. Locally, you can't tell if space is flat or curved. We are used to viewing curved space as a 2d sphere with bumps (indentations) in space. But this assumes an absolute space, which isn't there. You could establish an absolute speed of light in it, which not exists. This is even at the base of relativity. An astronaut can always say that it is the Earth accelerating to *him*, while he hangs still.
Deschele Schilder
CC BY-SA 4.0
2021-06-07T07:52:47.713Z
@tfq But measuring curvature requires non-local measurements. How can you measure if space is curved *locally*? With *one* measurement? You will measure zero in the ISS, but non-zero on Earth. If you say the locally every manifold is R, then performing non-local measurements will be like R too. So locally, not all manifolds are like Rflat R. Only when gravity is zero. The *tangents* are like flat R, not space itself.
Deschele Schilder
CC BY-SA 4.0
2021-06-07T12:54:38.737Z
@Barbierium But even the ISS is larger than a point ...
Hagen von Eitzen
CC BY-SA 4.0
2021-06-07T12:56:32.267Z
@HagenvonEitzen But there are infinitely many of them (points). And in every point you can see that gravity is zero in the ISS. Only for *different* points, a tidal force is present (though in the ISS this is practically zer0).
Deschele Schilder
CC BY-SA 4.0
2021-06-07T13:01:22.917Z
@Barbierium The tidal forces on the ISS are certainly not zero (if we can believe [this](https://space.stackexchange.com/a/40943/8922) estimate it should be ~0.1 N for a one-ton module on the periphery).
Peter - Reinstate Monica
CC BY-SA 4.0
2021-06-07T13:04:36.673Z
@HagenvonEitzen For the vast majority of points within the volume of the ISS, gravity is clearly not zero. The ISS' trajectory is a geodesic of its center of mass. By association all points on that trajectory (a one-dimensional, curved line) are on a geodesic and hence don't experience gravity. All other points would rather, however minutely, go elsewhere and are strong-armed along. Note that this assumes a rotation that aligns wit its orbital period; otherwise you have tidal forces everywhere except in the rotational center due to rotation in an inhomogeneous gravitational field.
Peter - Reinstate Monica
CC BY-SA 4.0
2021-06-07T13:16:35.460Z
@HagenvonEitzen The tidal force being 0,1(N)? Then this means that the ISS is not in free fall. If it is in free fall then the tidal force can't be measured because of its smalness.
Deschele Schilder
CC BY-SA 4.0
2021-06-08T04:22:48.670Z
@tfb I agree this is the only meaningful interpretation of the question within GR. If dark energy is indeed well-described by the cosmological constant, then the Einstein equation indicates that there cannot be a region (or even a point) of spacetime on which the curvature tensor vanishes exactly, barring the existence of negative mass. Otherwise, if the dark energy SE tensor fluctuates locally, I don't think there's any reason in principle that there can't be some open subset of spacetime on which the curvature tensor is identically zero, though it seems rather unlikely.
jawheele
CC BY-SA 4.0
2021-06-06T13:59:30.907Z
2021-06-06T14:19:30.760Z
2021-06-06T14:19:30.760Z
Peter - Reinstate Monica
<p>It depends on what your definition of the Universe. As long as there's mass, there's gravity. Gravity acts infinitely at causality speed, also known as speed of light.
Some data suggests that there's a tiny, but everywhere present force that pulls very far away objects in opposite directions, essentially adding more space in between matter, but only where there is very little mass. Since the universe has a lot of low mass bits of space, very far objects would need to travel faster than light to reach us, but we know that's not going to happen. Each moment, some bits of matter far away fall beyond this limit after which their influence cannot influence us.</p>
<p>Now imagine a photon, which is of course massless, traveling into a direction that is empty, apart from objects that are so far away that space is added in between the photon and those objects, at a rate faster than the photon's speed. Once the photon reaches a distance from the source, so that space behind it is added at a higher rate than the speed of the photon, that bit of space where the photon exists would have no theoretical gravitation.</p>
<p>This is more of a thought experiment than anything else. Not only that a massless observer of a gravity free space cannot be created, but by definition, that bit of space would be causally disconnected from everything else. In other words, you asked whether there can be a place that is not related to a force that relates everything, and I'm telling you that yes, but only if we define that space as a space that doesn't relate to anything. You might as well consider that a no.</p>
<p>On top of it, if we have a single traveling photon and nothing else around it, not only that time loses any meaning, but also space. Is it even correct to say that space expands if there is no space? And if a space has no spacetime, is it a universe. Also would you consider that no-space-space, containing a single photon, originating from our universe, as being part of our universe, since they cannot ever interact with each other? They are linked to each other through their past, but surely not through their future.</p>
<p>I'm not an astrophysicist, but I believe that the essence is that the maths and physics that we generally use to describe reality don't forbid the existence of a gravitation-less place, but given the same maths and physics that we use, it's a very boring place void of any meaningful property. A gravitation-less place seems like a fun place to explore, even if only in your mind, but it's about as fun as being forever stranded on a deserted island, where you are the island.</p>

CC BY-SA 4.0
2021-06-05T14:34:32.970Z
2021-06-05T14:34:32.970Z
Andrei
<p>If I understand Einstein's theory (and if it actually is correct) then gravity is the shape of space.  That shape is imposed on it by mass.  Thus if there is mass in the universe, there must be gravity.</p>
<p>Since (per the Big Bang theory) all of space &amp; mass was once a single point (or something really small) which expanded at less than the speed of light, all of space must be affected by that mass, hence gravity must exist everywhere.</p>

CC BY-SA 4.0
2021-06-05T17:40:19.250Z
2021-06-05T17:40:19.250Z
jamesqf
<p>Suppose the universe consists of only a hollow sphere. The diameter of the sphere doesn't matter, but we require that the shell of the sphere be uniform thickness.</p>
<p>The sphere will have finite mass, therefore, if an observer at any distance from the sphere were to measure the resulting gravitational field, he would find a non-zero field, and hence, a force in the direction of the sphere. That being said, there is no place outside the sphere that doesn't have gravitational forces.</p>
<p>However, inside the sphere is another matter. There will be no force on a object within the shell of the sphere. All of the force contributions from the mass of the sphere will cancel out. In this make-believe universe, there is no gravitational force within the hollow part of the sphere.</p>
<p>Someone mentioned that the Space Station, being in orbit, caused the interior to have no gravity. This isn't quite correct: there are measurable microgravity forces. These are due to the fact that the real world doesn't consist of only the earth and the orbiting vehicle. Gravitational forces from tides, mass concentrations in the earth, drag on the Station from residual atmosphere at it's orbital height: all contribute to a very small gravitational forces inside.</p>

CC BY-SA 4.0
2021-06-05T23:08:17.560Z
There is of course gravity inside an orbiting space station, otherwise it would not orbit. There is almost no *acceleration* of an astronaut relative to the space station because they both orbit together.
uhoh
CC BY-SA 4.0
2021-06-05T21:12:21.003Z
2021-06-05T21:12:21.003Z
Bob
<p>I can think of one place. Inside the ISS gravity is zero! If you go up and enter the station you will experience this. No gravitational force will pull you down.</p>
<p>So, inside the ISS, there are certainly places (points) where there is no gravity. There exists a very very tiny tidal force at these points, but for all practical purposes, this can be completely ignored. Theoretically, there <em>are</em> tidal forces, so there <em>is</em> gravity, but from an experimentalist point of view, these can be ignored.</p>
<p>Only if the ISS were orbiting a black hole, the astronauts would be able to sit on the floor of the ISS. They would probably be stretched beyond comfort...</p>
<p>So two particles in the ISS, initially at rest wrt each other, will always stay at rest wrt each other. In the very far future will they have a small observable velocity. It would be a nice problem to find out how long it takes...</p>
<p>While gravity in the universe can be zero, the potential due to gravity is non-zero everywhere. Even between the most distant galaxies, the potential is non-zero, while the force of gravity between the galaxies can be positive, zero, or negative.</p>

CC BY-SA 4.0
2021-06-06T17:27:24.323Z
Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/126144/discussion-on-answer-by-barbierium-are-there-places-in-the-universe-without-grav).
Connor Garcia
CC BY-SA 4.0
2021-06-06T20:12:16.413Z
Inside the ISS, gravity is about 95% of Earth-surface normal.
Mark
CC BY-SA 4.0
2021-06-06T23:43:38.100Z
@Mark Not if you fall along. For an astronaut, there is 0% of the Earth's gravity.
Deschele Schilder
CC BY-SA 4.0
2021-06-07T02:12:39.040Z
@Barbierium, if you fall along with the ISS then that is *because* of gravity.  That's what keeps the ISS (and everything and everyone in it) in orbit.  That gravity does not press you against the interior of the station in no way means that there is no gravity.
John Bollinger
CC BY-SA 4.0
2021-06-07T02:36:36.007Z
@JohnBollinger But how will you know if the windows are closed? This is exactly what Einstein meant by his equivalence principle. There is no way of knowing. And even if you looked outside, you can always say that it's the Earth accelerating towards you. Envisioning the Earth as being present in an absolutely curved spacetime is wrong. That's what relativity is all about.  It's the equivalence principle, on which general relativity is built. You never can tell who of two observers is in absolute motion. Only the relative motion counts.
Deschele Schilder
CC BY-SA 4.0
2021-06-07T02:48:13.177Z
@JohnBollinger So, if you fall, it's *because* the Earth accelerates to you. Even though the surface of the Earth remains stationary. It can stay stationary because it accelerates outward at the same rate as space moves inward. In this space, the astronauts are stationary (no gravity) and the Earth accelerates (gravity).
Deschele Schilder
CC BY-SA 4.0
2021-06-07T12:50:22.130Z
@Barbierium, if you want to make a an argument based on a non-accelerating frame of reference then you don't need to put a person in earth orbit, much less inside the ISS, to do so.  You can choose a non-accelerating frame anywhere in the universe you like.  However, the resulting absence of net force at that point is a characteristic of the chosen frame of reference, not an inherent property of the location, therefore this approach does not produce a satisfying answer to the question posed.
John Bollinger
CC BY-SA 4.0
2021-06-07T12:55:20.777Z
@JohnBollinger Now you're getting it. You can indeed find these points everywhere in the universe. So the vision that spacetime is absolutely curved is just a wrong assumption. You could just as well envision the whole of spacetime as non-curved. From the start, you learn that the universe is a 3d manifold with local indentations. But this is a relative point of view, not an absolute one. This absoluteness doesn't exist in GR. That's what relativity is all about.
Deschele Schilder
CC BY-SA 4.0
2021-06-07T13:02:47.980Z
If that's what you mean to say, then this answer does a poor job of conveying it. Moreover, I think it misses the OP's point, except to the extent that the passing comment that "the potential due to gravity is non-zero everywhere" addresses what the OP really wants to know.
John Bollinger
CC BY-SA 4.0
2021-06-07T22:46:26.097Z
Everything *inside* the ISS experiences **local** gravity, because those things have mass, and the ISS has mass. If the only “floating” macroscopic thing inside the ISS were a non-charged speck of dust, it would inevitably accelerate towards the shell of the ISS. The only place where gravity doesn’t exist is a place where no mass  can be observed. Even in free fall, it makes no sense to posit the observer as having only one frame of reference.
Chappo Hasn't Forgotten
CC BY-SA 4.0
2021-06-05T09:09:06.803Z
2021-06-07T02:52:32.830Z
2021-06-07T02:52:32.830Z
Deschele Schilder
<p>In two dimensions I think I can infer in a lame, unconvincing and <em>rigorless</em> numerical way that there are likely to be zeros in gravity from a random distribution of objects there can be points of zero gravity.</p>
<p>I create a space with 20 randomly distributed point sources, calculate and plot the force field on a 2000 x 2000 grid then choose the smallest grid point and through a minimization routine find a point with arbitrarily small scalar force.</p>
<p>I've done everything on a log10 scale, the max, min values are of order +8 and -1 but I can easily find <code>log10(force_magnitude)</code> around -14 by specifying that tolerance in the minimization routine.</p>
<p>I can't prove this extends to 3 dimensions nor arbitrarily large space and numbers, but I have a hunch this can be addressed mathematically, so I have just asked in Math SE: <a href="https://math.stackexchange.com/q/4163635/284619">What is the relative density and dimensionality of zeros in inverse square force fields to density of sources in (at least) 1, 2 and 3 dimensions?</a></p>
<p>Here are six cases for flavor:</p>
<h3><em>There be zeros here!</em></h3>
<p><a href="https://i.sstatic.net/JQtRG.png" rel="noreferrer"><img src="https://i.sstatic.net/JQtRG.png" alt="There be zeros here!" /></a></p>
<pre><code>import numpy as np
from scipy.optimize import minimize
import matplotlib.pyplot as plt

# https://astronomy.stackexchange.com/questions/44184/are-there-places-in-the-universe-without-gravity

def mini_me(xy, positions):
    return np.log10(get_force_magnitudes(positions, xy))

def get_force_magnitudes(positions, XY):
    r = positions[:, None, None, :] - XY
    forces = r * (((r**2).sum(axis=-1))[..., None]**-1.5) # all vectors
    force_field = forces.sum(axis=0)  # vector field
    return np.sqrt(force_field**2).sum(axis=-1)

N = 20
positions = np.random.random((N, 2))

x = np.linspace(0, 1, 2000)
X, Y = np.meshgrid(x, x, indexing='xy')
XY = np.stack((X, Y), axis=2)

force_magnitude = get_force_magnitudes(positions, XY)
indices = np.unravel_index(np.argmax(-force_magnitude), force_magnitude.shape) # find the smallest one on the grid
xy0 = XY[indices] # starting point for minimization

result = minimize(mini_me, xy0, args=(positions, ),
                  method='Nelder-Mead', tol=1E-08)

if True:
    fix, ax = plt.subplots(1, 1)
    extent = 2*[0, 1]
    thing = ax.imshow(np.log10(force_magnitude), origin='lower',
               extent=extent, vmax=2)
    x, y = positions.T
    ax.plot(x, y, '.r')
    x, y = result.x
    label = str(round(result.fun, 2))
    ax.plot([x], [y], 'o', color='none', markeredgecolor='red',
            markersize=14, markeredgewidth=2)
    ax.text(x+0.02, y, label, color='red', fontsize=14)
    plt.colorbar(thing, ax=ax)
    plt.title('log10 normalized scalar force')
    plt.show()
</code></pre>

CC BY-SA 4.0
2021-06-05T05:57:18.573Z
I wonder what it takes to extend to 3D.
Connor Garcia
CC BY-SA 4.0
2021-06-05T06:13:16.760Z
@ConnorGarcia first one should probably check if the dimensionality of the minimas in 2D and 3D are always 0D (i.e. points) or if they can be 1D curves or 2D surfaces, because *I think that's more interesting*, but to answer your question one can simply add an extra axis to this calculation and run it for minutes or hours. BUT the better way would be to apply some smart techniques like a [Monte Carlo technique](https://en.wikipedia.org/wiki/Monte_Carlo_method) and perhaps somehow a [1/r^2 weighted voronoi diagram](https://en.wikipedia.org/wiki/Voronoi_diagram#Generalizations_and_variations)...
uhoh
CC BY-SA 4.0
2021-06-05T06:13:59.407Z
to find *all minima* within a given region, thereby dramatically speeding up the technique and making it smarter and more exhaustive before pushing it to n-dimensions. Let's keep an eye on the question I mentioned that I've just posted in Math SE.
uhoh
CC BY-SA 4.0
2021-06-05T14:54:07.037Z
@uhoh, which force is keeping your random point masses in equilibrium in your 2D universe?
Ng Ph
CC BY-SA 4.0
2021-06-05T15:19:59.843Z
@NgPh If you read through my post carefully you'll note it's just an exploratory calculation to see what happens with a finite space and 20 objects. It's not a universe. Objects can certainly be moving, this is a snapshot in time assuming instantaneous Newtonian gravity. Of course one can assume it's much larger or infinite, but that won't affect these results too much, as things farther away will naturally begin to average to zero. This really needs to be addressed *mathematically* which is why, if you read through my post carefully you'll note that I've posted a question in Math SE.
uhoh
CC BY-SA 4.0
2021-06-05T17:54:08.087Z
@uhoh, Basically, you are saying that a mathematical model can be built so that there are always areas of zero potential when a number of random point mass is given as input. I can accept that, but does this answer in any way the original question (or show a possible path towards a rigorous answer?)?
Ng Ph
CC BY-SA 4.0
2021-06-05T18:19:51.383Z
I probably missed something: but isn't the center of mass guaranteed to have 0 net gravity from all the points, in any dimension?
Eric Duminil
CC BY-SA 4.0
2021-06-05T21:59:49.443Z
@NgPh yes, in that it *begins to* answer it. Sometimes the generation of an answer to a question take time to develop, sometimes it a collaborative effort among several users working together. While SE does have some superficial gamification elements, the primary goal is of course the generation of good answers to on-topic questions. Sometimes it takes five minutes, but once in a while it can take weeks or months or even longer. I find this question quite a challenging and interesting one so I've made a contribution towards answering it and I'll keep working on it.
uhoh
CC BY-SA 4.0
2021-06-05T22:09:36.140Z
@EricDuminil center of mass is $$\frac{\sum_i m_i \mathbf{r_i}}{\sum_i m_i}$$ and the gravitational acceleration is $$G\sum_i \frac{m_i }{r_i^2}\mathbf{\hat{r}_i}.$$ In other words CM is the masses weighted by distance and acceleration by inverse square distance, so I don't think that that can be true.
uhoh
CC BY-SA 4.0
2021-06-06T09:57:14.903Z
@ConnorGarcia Wouldn't there need to be a move to 4D? 3D is all fine and dandy but things change with time so 3D might tell you where a spot with no gravity WAS...
WernerCD
CC BY-SA 4.0
2021-06-06T11:23:20.733Z
@WernerCD: Step by step. One step might be to prove that there always exists a zero somewhere, in 3D space. And then to check if this zero moves continuously through time, and doesn't simply jump from one position to another.
Eric Duminil
CC BY-SA 4.0
2021-06-06T11:23:58.333Z
@uhoh: Thanks for the answer, it makes sense. Good thing I wrote "I probably missed something". :D
Eric Duminil
CC BY-SA 4.0
2022-04-27T21:39:49.557Z
Wouldn't the edges of the universe have zero gravity? Your test does not seem to show such, though.
Alexis Wilke
CC BY-SA 4.0
2022-04-27T22:40:21.223Z
@AlexisWilke I don't know if the universe has an edge, but $1/r^2$ never goes to zero.
uhoh
CC BY-SA 4.0
2021-06-05T03:34:30.397Z
2021-06-05T04:18:40.330Z
2021-06-05T04:18:40.330Z
uhoh
<p>Not sure if that is possible as I couldn't find an answer about it.</p>
<p>Are there places in the Universe where there are no gravitational forces?</p>

CC BY-SA 4.0
2021-06-05T11:59:32.240Z
"*Are there places in the Universe where there are no gravitational forces*": I'm not sure  saying there is no force is equivalent to saying there are forces but they result in no effect at some point. This should be clarified first in answers.
mins
CC BY-SA 4.0
2021-06-05T15:04:23.110Z
Agree with user mins, the question is ambiguous. All contributors down here seem to take the second interpretation: Gravity law applies universally, but there are mass distrbutions in such a way that gravity forces (and only these) cancel out at particular locations. The other interpretation of the question: Can there be mass w/o gravity somewhere in this Universe (and why not)?
Ng Ph
CC BY-SA 4.0
2021-06-05T16:31:33.640Z
Even  if such places (with zero gravity) exist, a mass located there would not be in stable equilibrium. In other words, the slightest perturbation in its position would disturb the equilibrium.
Phil Freedenberg
CC BY-SA 4.0
2021-06-05T17:25:17.013Z
@mins didn't Einstein believe that you would never be able to detect gravitational waves from black holes because the effect was too insignificant?  Where do you actually draw a line where something has "no effect"?
eps
CC BY-SA 4.0
2021-06-05T21:13:06.353Z
@eps. I meant no effect in theory. But my point was different: One can create motion making a body weightless (orbital flight / parabolic trajectory), meaning microgravity. Microgravity isn't synonymous of reduced gravity field (in spite of the word), but of reduced *apparent* weight. Do I cancel gravity when I free fall back on Earth after a 30 cm jump? All the contrary, feeling weightless means gravity is the sole force acting on me, I'm in free fall, meaning *my fall is due to gravity* and isn't perturbed by any other force.
mins
CC BY-SA 4.0
2021-06-05T21:52:33.020Z
@aybe, It would help if you could clarify. The bottom-line is: do you mean zero-gravity or do you mean "no gravity", in the sense that Newton's theory doesn't apply? In the first meaning, if one theoretically removes one of the objects in presence, the zero-gravity state vanishes. In the second, the presence or absence of gravity-abiding objects around the special "no-gravity" area/point doesn't affect its special status of gravity-exempt (sounds like a good subject for sci-fi novel!).
Ng Ph
CC BY-SA 4.0
2021-06-07T17:31:19.177Z
At an infinite distance away from the boundary where stuff and light from the big bang has reached there is an infinitely low gravity force.
HarryH
<gravity>
Are there places in the Universe without gravity?
aybe
# Are there places in the Universe without gravity?

(Asked by: username_0 on 2021-06-04)

Not sure if that is possible as I couldn't find an answer about it.

Are there places in the Universe where there are no gravitational forces?

**Tags:** gravity

> **Comments:**
> **username_1**: "*Are there places in the Universe where there are no gravitational forces*": I'm not sure  saying there is no force is equivalent to saying there are forces but they result in no effect at some point. This should be clarified first in answers.  
> **username_2**: Agree with user username_1, the question is ambiguous. All contributors down here seem to take the second interpretation: Gravity law applies universally, but there are mass distrbutions in such a way that gravity forces (and only these) cancel out at particular locations. The other interpretation of the question: Can there be mass w/o gravity somewhere in this Universe (and why not)?  
> **username_3**: Even  if such places (with zero gravity) exist, a mass located there would not be in stable equilibrium. In other words, the slightest perturbation in its position would disturb the equilibrium.  
> **username_4**: @username_1 didn't Einstein believe that you would never be able to detect gravitational waves from black holes because the effect was too insignificant?  Where do you actually draw a line where something has "no effect"?  
> **username_1**: @username_4. I meant no effect in theory. But my point was different: One can create motion making a body weightless (orbital flight / parabolic trajectory), meaning microgravity. Microgravity isn't synonymous of reduced gravity field (in spite of the word), but of reduced *apparent* weight. Do I cancel gravity when I free fall back on Earth after a 30 cm jump? All the contrary, feeling weightless means gravity is the sole force acting on me, I'm in free fall, meaning *my fall is due to gravity* and isn't perturbed by any other force.  
> **username_2**: @username_0, It would help if you could clarify. The bottom-line is: do you mean zero-gravity or do you mean "no gravity", in the sense that Newton's theory doesn't apply? In the first meaning, if one theoretically removes one of the objects in presence, the zero-gravity state vanishes. In the second, the presence or absence of gravity-abiding objects around the special "no-gravity" area/point doesn't affect its special status of gravity-exempt (sounds like a good subject for sci-fi novel!).  
> **username_5**: At an infinite distance away from the boundary where stuff and light from the big bang has reached there is an infinitely low gravity force.  

---

## 7 answers

### Answer by username_9 on 2021-06-05. Score: 27 | Accepted answer

In two dimensions I think I can infer in a lame, unconvincing and *rigorless* numerical way that there are likely to be zeros in gravity from a random distribution of objects there can be points of zero gravity.

I create a space with 20 randomly distributed point sources, calculate and plot the force field on a 2000 x 2000 grid then choose the smallest grid point and through a minimization routine find a point with arbitrarily small scalar force.

I've done everything on a log10 scale, the max, min values are of order +8 and -1 but I can easily find `log10(force_magnitude)` around -14 by specifying that tolerance in the minimization routine.

I can't prove this extends to 3 dimensions nor arbitrarily large space and numbers, but I have a hunch this can be addressed mathematically, so I have just asked in Math SE: [What is the relative density and dimensionality of zeros in inverse square force fields to density of sources in (at least) 1, 2 and 3 dimensions?](https://math.stackexchange.com/q/4163635/284619)

Here are six cases for flavor:

### *There be zeros here!*

[![There be zeros here!](https://i.sstatic.net/JQtRG.png)](https://i.sstatic.net/JQtRG.png)

```
import numpy as np
from scipy.optimize import minimize
import matplotlib.pyplot as plt

# https://astronomy.stackexchange.com/questions/44184/are-there-places-in-the-universe-without-gravity

def mini_me(xy, positions):
    return np.log10(get_force_magnitudes(positions, xy))

def get_force_magnitudes(positions, XY):
    r = positions[:, None, None, :] - XY
    forces = r * (((r**2).sum(axis=-1))[..., None]**-1.5) # all vectors
    force_field = forces.sum(axis=0)  # vector field
    return np.sqrt(force_field**2).sum(axis=-1)

N = 20
positions = np.random.random((N, 2))

x = np.linspace(0, 1, 2000)
X, Y = np.meshgrid(x, x, indexing='xy')
XY = np.stack((X, Y), axis=2)

force_magnitude = get_force_magnitudes(positions, XY)
indices = np.unravel_index(np.argmax(-force_magnitude), force_magnitude.shape) # find the smallest one on the grid
xy0 = XY[indices] # starting point for minimization

result = minimize(mini_me, xy0, args=(positions, ),
                  method='Nelder-Mead', tol=1E-08)

if True:
    fix, ax = plt.subplots(1, 1)
    extent = 2*[0, 1]
    thing = ax.imshow(np.log10(force_magnitude), origin='lower',
               extent=extent, vmax=2)
    x, y = positions.T
    ax.plot(x, y, '.r')
    x, y = result.x
    label = str(round(result.fun, 2))
    ax.plot([x], [y], 'o', color='none', markeredgecolor='red',
            markersize=14, markeredgewidth=2)
    ax.text(x+0.02, y, label, color='red', fontsize=14)
    plt.colorbar(thing, ax=ax)
    plt.title('log10 normalized scalar force')
    plt.show()

```

> **Comments:**
> **username_19**: I wonder what it takes to extend to 3D.  
> **username_9**: @username_19 first one should probably check if the dimensionality of the minimas in 2D and 3D are always 0D (i.e. points) or if they can be 1D curves or 2D surfaces, because *I think that's more interesting*, but to answer your question one can simply add an extra axis to this calculation and run it for minutes or hours. BUT the better way would be to apply some smart techniques like a [Monte Carlo technique](https://en.wikipedia.org/wiki/Monte_Carlo_method) and perhaps somehow a [1/r^2 weighted voronoi diagram](https://en.wikipedia.org/wiki/Voronoi_diagram#Generalizations_and_variations)...  
> **username_9**: to find *all minima* within a given region, thereby dramatically speeding up the technique and making it smarter and more exhaustive before pushing it to n-dimensions. Let's keep an eye on the question I mentioned that I've just posted in Math SE.  
> **username_2**: @username_9, which force is keeping your random point masses in equilibrium in your 2D universe?  
> **username_9**: @username_2 If you read through my post carefully you'll note it's just an exploratory calculation to see what happens with a finite space and 20 objects. It's not a universe. Objects can certainly be moving, this is a snapshot in time assuming instantaneous Newtonian gravity. Of course one can assume it's much larger or infinite, but that won't affect these results too much, as things farther away will naturally begin to average to zero. This really needs to be addressed *mathematically* which is why, if you read through my post carefully you'll note that I've posted a question in Math SE.  
> **username_2**: @username_9, Basically, you are saying that a mathematical model can be built so that there are always areas of zero potential when a number of random point mass is given as input. I can accept that, but does this answer in any way the original question (or show a possible path towards a rigorous answer?)?  
> **username_23**: I probably missed something: but isn't the center of mass guaranteed to have 0 net gravity from all the points, in any dimension?  
> **username_9**: @username_2 yes, in that it *begins to* answer it. Sometimes the generation of an answer to a question take time to develop, sometimes it a collaborative effort among several users working together. While SE does have some superficial gamification elements, the primary goal is of course the generation of good answers to on-topic questions. Sometimes it takes five minutes, but once in a while it can take weeks or months or even longer. I find this question quite a challenging and interesting one so I've made a contribution towards answering it and I'll keep working on it.  
> **username_9**: @username_23 center of mass is $$\frac{\sum_i m_i \mathbf{r_i}}{\sum_i m_i}$$ and the gravitational acceleration is $$G\sum_i \frac{m_i }{r_i^2}\mathbf{\hat{r}_i}.$$ In other words CM is the masses weighted by distance and acceleration by inverse square distance, so I don't think that that can be true.  
> **username_24**: @username_19 Wouldn't there need to be a move to 4D? 3D is all fine and dandy but things change with time so 3D might tell you where a spot with no gravity WAS...  
> **username_23**: @username_24: Step by step. One step might be to prove that there always exists a zero somewhere, in 3D space. And then to check if this zero moves continuously through time, and doesn't simply jump from one position to another.  
> **username_23**: @username_9: Thanks for the answer, it makes sense. Good thing I wrote "I probably missed something". :D  
> **username_25**: Wouldn't the edges of the universe have zero gravity? Your test does not seem to show such, though.  
> **username_9**: @username_25 I don't know if the universe has an edge, but $1/r^2$ never goes to zero.  

---

### Answer by username_6 on 2021-06-04. Score: 17

Gravity extends to infinity, so no, strictly theoretically speaking there is always some gravity present. In theory, even in this case we could have points in space where gravitational forces cancel out, but given the complexity of our universe, this just won't happen in practice.

As a more relaxed viewpoint - there are special points around orbiting objects called [Lagrange points](https://en.m.wikipedia.org/wiki/Lagrange_point), where gravitational forces of the two objects (eg a star and a planet) sum up to zero. Those points do receive gravitational forces from other objects however (eg from Jupiter in case of the Sun and Earth). Also, as the two objects in question are also moving, these points are moving with them (otoh, there's also simply no absolute positions in space anyways, but that's a title for another story)

> **Comments:**
> **username_7**: Even if you just consider the two objects, the gravity does not sum to zero at any of the five Lagrange points.  
> **username_6**: @username_7 yeah, you're right, I worded this inaccurately - I wanted to convey the idea that gravities of those two objects balance out  
> **username_8**: Also, you’re still $GM(<R)/R$ deep in your galaxy’s potential.  
> **username_9**: No, Lagrange points are vector zeros in the sum of the real gravitational forces plus centrifugal forces (gradient of the gravitational potential plus a *pseudopotential*). I think that if one considers only Newtonian gravity and establishes some frame of reference, one can easily have zeros in the net gravitational force field.  
> **username_9**: cf. https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec18.pdf  
> **username_10**: "Gravity extends to infinity". That's something we couldn't possibly know. The theories we currently use to describe gravity assume so, and they work fairly well, but they are theories and not fact, and we know they are not perfect.  
> **username_6**: @username_10 yes, I think that makes sense, but I'd prefer to stick with mainstream physics here and not speculate too much. Admittedly in case forces are quantized this will at least practically affect how gravity affects over large distances  
> **username_4**: @username_10 and everything we see could be a simulation, or gravity could be a residual effect of all the faeries around us.  
> **username_10**: @username_4: either way, there is no way for us to possibly know what happens arbitrarily far away. Actually, if you think about it, most "observations" already depend on theory and assumptions.  
> **username_10**: @username_6: yes if gravity is quantized it should be possible in principle to have places without gravity.  
> **username_11**: @username_10 Is there any reason not to think that the same laws of physics we see here would be different at very large distances.  Observations do not depend on theory and assumptions, interpretation of these observation does.  
> **username_8**: @username_10 “They are theories and not facts” is a statement that doesn’t get you anywhere. Do you know anything for a fact?  
> **username_10**: @username_8: Not sure where you want to go. If you want to discuss philosophical realism, that's a complicated topic, but one that is worthless regarding physics. But to regard theories as facts is a fairly common mistake that scientists have made again and again. Newton's theory was "fact" for hundreds of years, until it wasn't. So, let me ask you, are Galilean transformations a "fact"? are singularities in GR a "fact"? Are Hilbert spaces in Quantum Mechanics "facts"? Physics is made out of theories; "facts" are few.  
> **username_12**: Gravity is non-zero at all Lagrange points (only for two stationary objects there will be one point between the objects where gravity is zero). It's *because* of this that Lagrange points exist. Gravitational force is precisely compensated by centrifugal forces at these points.  
> **username_8**: @username_10 That was sort of my point, actually. We don’t *really* know anything for a fact, but at least within mainstream physics, we know some things well enough that calling it “a fact” is quite acceptable, I think (I’m sorry if I came off a bit hostile).  
> **username_10**: @username_4: fair enough. I definitely understand where you come from.  

---

### Answer by username_13 on 2021-06-06. Score: 5

I think we must be a bit careful what we are discussing here. General relativity states that when an observer is on a geodesic, i.e. in freefall, they are not experiencing gravity. That's Einstein's famous falling elevator *gedankenexperiment.* It's not that gravity is there but cannot be measured, or that it is there but canceled out by the equivalent acceleration; it is *not there period* in the frame of reference of the observer on the geodesic.

In this sense, for every spot in the universe there is a reference frame in which no gravitation is present, namely the one moving along the geodesic. An example would be a beam of light. In its own frame of reference it's moving perfectly straight through space time. (Or rather, sitting perfectly still in an immutable, flat universe.)

Now admittedly, there are no point-shaped observers; what the man in the elevator as well as any other observer with finite size can observe is a *gradient* causing inner forces within the observer (more precisely, within the observed finite volume of space). The reason is that the geodesics for the different space time locations within the observed volume are *not quite* parallel in an inhomogeneous gravitational field. *A gravity gradient* is perfectly well measurable within an extended object in freefall; [an example is the ISS](https://space.stackexchange.com/questions/40942/how-big-are-the-tidal-accelerations-within-the-iss). Only some points on it are in perfect freefall, namely those on the orbital trajectory of its center of mass; the others are pulled or pushed along.

Now the effect is already fairly small even in the comparatively steep gradient close to Earth; if you choose a location with very little gravity to begin with, and choose an inflection point at the right distance from the closest mass clusters, the effect will be extremely small. A good spot might be in an intergalactic void. My gut feeling is that it will be indistinguishable from background noise like photons from the the microwave background radiation, or general vacuum quantum fluctuations. Oh, and the *gedankenexperiment* also necessitates a massless observer because any local masses would create much stronger gradients, due to their proximity, than even very large masses at large distances, thus drowning out any subtle global gradients.

> **Comments:**
> **username_12**: A gravity gradient (tidal force) is also present for particles moving on geodesics. If a particle falls freely to Earth then gravity will be zero for the particle, but the gravity gradient not. Only a free particle (somewhere in outer space) that is seen in an accelerated frame will be seen to experience no tidal force.  
> **username_13**: @Barbierium Yes, there is a gradient but it is meaningless (and undetectable) for a point-shaped particle. Even if we can assign a value to the gradient for each point in spacetime it is still a difference between two points.  
> **username_12**: Yes, that's true. A particle that falls freely to Earth is seen as stationary in a co-falling frame. There is no tidal force on the particle (if a point). Only two (or more) particles will move away from each other slowly (which is the way you can see if you are in a falling frame or an accelerated frame).  
> **username_12**: @tfb You are making the same "mistake" as many make. It's *not* a physically meaningful question to ask if the space is flat somewhere. That's what relativity is all about. Locally, you can't tell if space is flat or curved. We are used to viewing curved space as a 2d sphere with bumps (indentations) in space. But this assumes an absolute space, which isn't there. You could establish an absolute speed of light in it, which not exists. This is even at the base of relativity. An astronaut can always say that it is the Earth accelerating to *him*, while he hangs still.  
> **username_12**: @tfq But measuring curvature requires non-local measurements. How can you measure if space is curved *locally*? With *one* measurement? You will measure zero in the ISS, but non-zero on Earth. If you say the locally every manifold is R, then performing non-local measurements will be like R too. So locally, not all manifolds are like Rflat R. Only when gravity is zero. The *tangents* are like flat R, not space itself.  
> **username_14**: @Barbierium But even the ISS is larger than a point ...  
> **username_12**: @username_14 But there are infinitely many of them (points). And in every point you can see that gravity is zero in the ISS. Only for *different* points, a tidal force is present (though in the ISS this is practically zer0).  
> **username_13**: @Barbierium The tidal forces on the ISS are certainly not zero (if we can believe [this](https://space.stackexchange.com/a/40943/8922) estimate it should be ~0.1 N for a one-ton module on the periphery).  
> **username_13**: @username_14 For the vast majority of points within the volume of the ISS, gravity is clearly not zero. The ISS' trajectory is a geodesic of its center of mass. By association all points on that trajectory (a one-dimensional, curved line) are on a geodesic and hence don't experience gravity. All other points would rather, however minutely, go elsewhere and are strong-armed along. Note that this assumes a rotation that aligns wit its orbital period; otherwise you have tidal forces everywhere except in the rotational center due to rotation in an inhomogeneous gravitational field.  
> **username_12**: @username_14 The tidal force being 0,1(N)? Then this means that the ISS is not in free fall. If it is in free fall then the tidal force can't be measured because of its smalness.  
> **username_15**: @tfb I agree this is the only meaningful interpretation of the question within GR. If dark energy is indeed well-described by the cosmological constant, then the Einstein equation indicates that there cannot be a region (or even a point) of spacetime on which the curvature tensor vanishes exactly, barring the existence of negative mass. Otherwise, if the dark energy SE tensor fluctuates locally, I don't think there's any reason in principle that there can't be some open subset of spacetime on which the curvature tensor is identically zero, though it seems rather unlikely.  

---

### Answer by username_16 on 2021-06-05. Score: 0

It depends on what your definition of the Universe. As long as there's mass, there's gravity. Gravity acts infinitely at causality speed, also known as speed of light.
Some data suggests that there's a tiny, but everywhere present force that pulls very far away objects in opposite directions, essentially adding more space in between matter, but only where there is very little mass. Since the universe has a lot of low mass bits of space, very far objects would need to travel faster than light to reach us, but we know that's not going to happen. Each moment, some bits of matter far away fall beyond this limit after which their influence cannot influence us.

Now imagine a photon, which is of course massless, traveling into a direction that is empty, apart from objects that are so far away that space is added in between the photon and those objects, at a rate faster than the photon's speed. Once the photon reaches a distance from the source, so that space behind it is added at a higher rate than the speed of the photon, that bit of space where the photon exists would have no theoretical gravitation.

This is more of a thought experiment than anything else. Not only that a massless observer of a gravity free space cannot be created, but by definition, that bit of space would be causally disconnected from everything else. In other words, you asked whether there can be a place that is not related to a force that relates everything, and I'm telling you that yes, but only if we define that space as a space that doesn't relate to anything. You might as well consider that a no.

On top of it, if we have a single traveling photon and nothing else around it, not only that time loses any meaning, but also space. Is it even correct to say that space expands if there is no space? And if a space has no spacetime, is it a universe. Also would you consider that no-space-space, containing a single photon, originating from our universe, as being part of our universe, since they cannot ever interact with each other? They are linked to each other through their past, but surely not through their future.

I'm not an astrophysicist, but I believe that the essence is that the maths and physics that we generally use to describe reality don't forbid the existence of a gravitation-less place, but given the same maths and physics that we use, it's a very boring place void of any meaningful property. A gravitation-less place seems like a fun place to explore, even if only in your mind, but it's about as fun as being forever stranded on a deserted island, where you are the island.


---

### Answer by username_17 on 2021-06-05. Score: -1

If I understand Einstein's theory (and if it actually is correct) then gravity is the shape of space. That shape is imposed on it by mass. Thus if there is mass in the universe, there must be gravity.

Since (per the Big Bang theory) all of space & mass was once a single point (or something really small) which expanded at less than the speed of light, all of space must be affected by that mass, hence gravity must exist everywhere.


---

### Answer by username_18 on 2021-06-05. Score: -2

Suppose the universe consists of only a hollow sphere. The diameter of the sphere doesn't matter, but we require that the shell of the sphere be uniform thickness.

The sphere will have finite mass, therefore, if an observer at any distance from the sphere were to measure the resulting gravitational field, he would find a non-zero field, and hence, a force in the direction of the sphere. That being said, there is no place outside the sphere that doesn't have gravitational forces.

However, inside the sphere is another matter. There will be no force on a object within the shell of the sphere. All of the force contributions from the mass of the sphere will cancel out. In this make-believe universe, there is no gravitational force within the hollow part of the sphere.

Someone mentioned that the Space Station, being in orbit, caused the interior to have no gravity. This isn't quite correct: there are measurable microgravity forces. These are due to the fact that the real world doesn't consist of only the earth and the orbiting vehicle. Gravitational forces from tides, mass concentrations in the earth, drag on the Station from residual atmosphere at it's orbital height: all contribute to a very small gravitational forces inside.

> **Comments:**
> **username_9**: There is of course gravity inside an orbiting space station, otherwise it would not orbit. There is almost no *acceleration* of an astronaut relative to the space station because they both orbit together.  

---

### Answer by username_12 on 2021-06-05. Score: -3

I can think of one place. Inside the ISS gravity is zero! If you go up and enter the station you will experience this. No gravitational force will pull you down.

So, inside the ISS, there are certainly places (points) where there is no gravity. There exists a very very tiny tidal force at these points, but for all practical purposes, this can be completely ignored. Theoretically, there *are* tidal forces, so there *is* gravity, but from an experimentalist point of view, these can be ignored.

Only if the ISS were orbiting a black hole, the astronauts would be able to sit on the floor of the ISS. They would probably be stretched beyond comfort...

So two particles in the ISS, initially at rest wrt each other, will always stay at rest wrt each other. In the very far future will they have a small observable velocity. It would be a nice problem to find out how long it takes...

While gravity in the universe can be zero, the potential due to gravity is non-zero everywhere. Even between the most distant galaxies, the potential is non-zero, while the force of gravity between the galaxies can be positive, zero, or negative.

> **Comments:**
> **username_19**: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/126144/discussion-on-answer-by-barbierium-are-there-places-in-the-universe-without-grav).  
> **username_20**: Inside the ISS, gravity is about 95% of Earth-surface normal.  
> **username_12**: @username_20 Not if you fall along. For an astronaut, there is 0% of the Earth's gravity.  
> **username_21**: @Barbierium, if you fall along with the ISS then that is *because* of gravity.  That's what keeps the ISS (and everything and everyone in it) in orbit.  That gravity does not press you against the interior of the station in no way means that there is no gravity.  
> **username_12**: @username_21 But how will you know if the windows are closed? This is exactly what Einstein meant by his equivalence principle. There is no way of knowing. And even if you looked outside, you can always say that it's the Earth accelerating towards you. Envisioning the Earth as being present in an absolutely curved spacetime is wrong. That's what relativity is all about.  It's the equivalence principle, on which general relativity is built. You never can tell who of two observers is in absolute motion. Only the relative motion counts.  
> **username_12**: @username_21 So, if you fall, it's *because* the Earth accelerates to you. Even though the surface of the Earth remains stationary. It can stay stationary because it accelerates outward at the same rate as space moves inward. In this space, the astronauts are stationary (no gravity) and the Earth accelerates (gravity).  
> **username_21**: @Barbierium, if you want to make a an argument based on a non-accelerating frame of reference then you don't need to put a person in earth orbit, much less inside the ISS, to do so.  You can choose a non-accelerating frame anywhere in the universe you like.  However, the resulting absence of net force at that point is a characteristic of the chosen frame of reference, not an inherent property of the location, therefore this approach does not produce a satisfying answer to the question posed.  
> **username_12**: @username_21 Now you're getting it. You can indeed find these points everywhere in the universe. So the vision that spacetime is absolutely curved is just a wrong assumption. You could just as well envision the whole of spacetime as non-curved. From the start, you learn that the universe is a 3d manifold with local indentations. But this is a relative point of view, not an absolute one. This absoluteness doesn't exist in GR. That's what relativity is all about.  
> **username_21**: If that's what you mean to say, then this answer does a poor job of conveying it. Moreover, I think it misses the OP's point, except to the extent that the passing comment that "the potential due to gravity is non-zero everywhere" addresses what the OP really wants to know.  
> **username_22**: Everything *inside* the ISS experiences **local** gravity, because those things have mass, and the ISS has mass. If the only “floating” macroscopic thing inside the ISS were a non-charged speck of dust, it would inevitably accelerate towards the shell of the ISS. The only place where gravity doesn’t exist is a place where no mass  can be observed. Even in free fall, it makes no sense to posit the observer as having only one frame of reference.  

---

