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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?

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    $\begingroup$ "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. $\endgroup$ – mins Jun 5 at 11:59
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    $\begingroup$ 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)? $\endgroup$ – Ng Ph Jun 5 at 15:04
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    $\begingroup$ 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. $\endgroup$ – Phil Freedenberg Jun 5 at 16:31
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    $\begingroup$ @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. $\endgroup$ – mins Jun 5 at 21:13
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    $\begingroup$ @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!). $\endgroup$ – Ng Ph Jun 5 at 21:52
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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?

Here are six cases for flavor:

There be zeros here!

There be zeros here!

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()
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    $\begingroup$ I wonder what it takes to extend to 3D. $\endgroup$ – Connor Garcia Jun 5 at 5:57
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    $\begingroup$ @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?)? $\endgroup$ – Ng Ph Jun 5 at 17:54
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    $\begingroup$ I probably missed something: but isn't the center of mass guaranteed to have 0 net gravity from all the points, in any dimension? $\endgroup$ – Eric Duminil Jun 5 at 18:19
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    $\begingroup$ @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. $\endgroup$ – uhoh Jun 5 at 22:09
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    $\begingroup$ @uhoh: Thanks for the answer, it makes sense. Good thing I wrote "I probably missed something". :D $\endgroup$ – Eric Duminil Jun 6 at 11:23
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This answer just amplifies on the correct answer by Barbierium.

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

The answer to this question is that there is no answer to the question. To define whether or not a gravitational force would act on a test particle at a certain point in space, we have to define some frame of reference. But in fact there is no globally preferred frame of reference. The only preferred frame of reference is a local one, which is a free-falling frame of reference, and in such a frame the gravitational force on a test particle is zero.

This is known as the equivalence principle.

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    $\begingroup$ True for a small test particle. But a large object in a free-falling frame of reference still experiences tidal forces. Perhaps the question could be reinterpreted as physics.stackexchange.com/questions/17459/… $\endgroup$ – jpa Jun 6 at 6:25
  • $\begingroup$ @jpa Do you think that tidal forces will affect the astronauts? If their bodies are. say 2 meters long, how big will this tidal force be? I would say that practically the astronauts can be assumed to be small test particles (though they will protest against this assumption!). $\endgroup$ – Methadont Jun 6 at 7:29
  • $\begingroup$ @Barbierium I agree that for most practical purposes the distinction does not matter, but the question is not limited to small objects. $\endgroup$ – jpa Jun 6 at 9:10
  • $\begingroup$ @jpa The question is about places. Places are points in space, so there would surely be places with zero gravity. Inside the ISS there are many of them (considered independently from the other points). $\endgroup$ – Methadont Jun 6 at 9:18
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    $\begingroup$ @TonyK probably the useful definition of 'place' is '(smooth) timelike curve'. If you restrict yourself to timelike geodesic then 'gravity is zero' at any such place. Every point of spacetime has timelike geodesics passing through it. $\endgroup$ – user38308 Jun 6 at 19:09
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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, 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)

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    $\begingroup$ Even if you just consider the two objects, the gravity does not sum to zero at any of the five Lagrange points. $\endgroup$ – notovny Jun 4 at 21:44
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    $\begingroup$ Also, you’re still $GM(<R)/R$ deep in your galaxy’s potential. $\endgroup$ – pela Jun 4 at 21:48
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    $\begingroup$ 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. $\endgroup$ – uhoh Jun 5 at 1:08
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    $\begingroup$ "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. $\endgroup$ – Martin Argerami Jun 5 at 6:30
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    $\begingroup$ @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 $\endgroup$ – tuomas Jun 5 at 8:12
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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. 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.

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  • $\begingroup$ 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. $\endgroup$ – Methadont Jun 6 at 15:59
  • $\begingroup$ @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. $\endgroup$ – Peter - Reinstate Monica Jun 6 at 17:06
  • $\begingroup$ 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). $\endgroup$ – Methadont Jun 6 at 17:13
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    $\begingroup$ So I think this makes the physically meaningful version of this question clear: is it possible that there are there regions of spacetime which are flat? Not approximately or asymptotically flat, but flat. I think the answer is 'no' (in our universe) but I don't have any kind of proof of that. $\endgroup$ – user38308 Jun 6 at 19:14
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    $\begingroup$ @Barbierium: no, I am not making that mistake, trust me. As I have said in another comment: yes, you can't locally measure curvature (locally any manifold is like $\mathbb{R}^n$). And since there are geodesics through every point of spacetime you end up with the answer that gravity is zero at every point of spacetime (aka 'you can always be in free-fall'). That's true but it's also not useful. So instead, to find a useful meaning the obvious one is to measure curvature. Which is perfectly possible, just not at a single point or along a single geodesic. $\endgroup$ – user38308 Jun 7 at 7:02
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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.

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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.

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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.

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    $\begingroup$ 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. $\endgroup$ – uhoh Jun 5 at 23:08
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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.

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    $\begingroup$ Inside the ISS, gravity is about 95% of Earth-surface normal. $\endgroup$ – Mark Jun 6 at 20:12
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    $\begingroup$ @Mark Not if you fall along. For an astronaut, there is 0% of the Earth's gravity. $\endgroup$ – Methadont Jun 6 at 23:43
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    $\begingroup$ @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. $\endgroup$ – John Bollinger Jun 7 at 2:12
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    $\begingroup$ @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. $\endgroup$ – Methadont Jun 7 at 2:36
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    $\begingroup$ 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. $\endgroup$ – John Bollinger Jun 7 at 13:02

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