If gravity disappeared, would Newton's third law make everything that was pressed to the ground by gravity get pushed upwards?


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    $\begingroup$ Seems more of a Worldbuilding question than physics, to me. $\endgroup$ – David Richerby Nov 3 '18 at 17:58
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    $\begingroup$ It's relevant whether gravity vanishes suddenly, or over a time like an hour or day. $\endgroup$ – Volker Siegel Nov 3 '18 at 20:48
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    $\begingroup$ @DavidRicherby - The question would be great in Wordbuilding, but the goal and answer is different here. Here the goal is to understand Newton's third law. In wordbuilding we would account for a lot of interesting phenomena, like those outlined in the comments of my answer. $\endgroup$ – Pere Nov 3 '18 at 22:10
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    $\begingroup$ In fact, related questions have already been asked in Wordbuilding. worldbuilding.stackexchange.com/… $\endgroup$ – Pere Nov 3 '18 at 22:13
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    $\begingroup$ @CasimirRönnlöf Honestly, many questions that would be asked by a person who hasn't actively studied physics (usually at the college level, sometimes high school) aren't a good fit for this site. Don't worry about it, though; there's nothing wrong with you asking the question, seeing how it's received, and then learning from the experience. I'd say it's good to keep in mind that even though you got your answer this time, similar questions might not be so well received in the future. $\endgroup$ – David Z Nov 5 '18 at 18:54

12 Answers 12


As other answers explain, Newton's third law wouldn't push you upwards, because reaction disappears as soon of action (gravity) vanishes.

However, we need to keep in mind that we are siting on several thousand kilometres of rock heavily compressed by its own weight. If weight suddenly disappears, that rock will react like a spring and project itself and anything in the surface at very high speed to space. In fact, even the most conservative ballpark estimates of the elastic deformation of Earth in its present state are in the order of several kilometres, so that's the quite instant rebound we can expect.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Nov 7 '18 at 17:19

Yes, but in almost all cases the push would be imperceptible.

Reaction forces from surfaces occur when the molecules in the wall are displaced from their equilibrium position. The harder they are pushed the more they are displaced, and the more they are displaced the harder they pushed back. When you stand on a surface without falling it is because you have displaced the surface enough for the reaction force to match your weight.

As an extreme example of this imagine standing on a trampoline. Heavier people make the trampoline surface sink lower than lighter people. The same is true on hard surfaces, but the displacement is basically imperceptible.

If you suddenly removed gravity, the reaction force from the displaced surface would still be there, and it would push you away until the equilibrium state of the surface is restored.

Again, imagine standing on a trampoline holding heavy weights. When you throw the weights away the trampoline will begin to push you up until you reach a new equilibrium. If the weights were heavy enough it could even launch you into the air.

The same thing would happen harder surfaces, but the amount of time the remaining reaction force would act on you would be tiny, and you would notices hardly any effect.

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    $\begingroup$ I'm trying to imagine holding weights on a trampoline, and throwing them off, to be launched into the air and I can't help but think it doesn't add up? $\endgroup$ – djsmiley2k - CoW Nov 3 '18 at 19:59
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    $\begingroup$ It would take very heavy weights to actually launch you off the trampoline, but even with light weights you would be moved upwards when you toss them away. Find a trampoline and do the experiment. $\endgroup$ – Luke Pritchett Nov 3 '18 at 22:51
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    $\begingroup$ I created an account on this stack just to up-vote this answer. It’s an excellect description of the forces at work, with a perfect illustrative example that makes it intuitively understandable. Moreover, it exactly addresses the point that the asker was actually gerting at. $\endgroup$ – Chris Melville Nov 4 '18 at 21:34
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    $\begingroup$ On a harder surface, one basic-physics way to look at it is that the stored energy (force * distance) is much smaller, because the force is the same (your weight) but the distance is much smaller. (Barely perceptible, by definition for a hard surface.) Of course the force isn't constant over the distance, but Hookean spring stored energy $kx^2/2$ goes down as k grows and x shrinks, holding $F = kx$ constant. $\endgroup$ – Peter Cordes Nov 5 '18 at 1:53
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    $\begingroup$ @djsmiley Had the same problem with that analogy. I think an easier one to imagine is that you're on a giant slingshot (like the carnival ride) attached to a giant heavy mass as an anchor. As soon as you detatch from the mass that's anchoring you - the elastic now returns to it's equilibrium and in doing so - catapults you into the air. It's the same effect, but at an easier to imagine extreme. $\endgroup$ – Bilkokuya Nov 6 '18 at 10:44

You can simulate this experiment in real life with an electromagnet.

For instance, you can hold a vertically oriented steel plate by a horizontally oriented electromagnet, so that the plate is free to fall down, when the electromagnet is deenergized.

If the falling plate has a horizontal velocity component and describes a parabola, you can conclude that the normal force has pushed it. If the plate falls straight down, you can conclude that the there was no push.

Even without performing such experiment, you, probably, can predict that the plate will fall straight down. This is because the normal force is a reaction force and it never exceeds the applied force causing it, be that electromagnet attraction or gravity. So, as the applied force disappears, gradually or suddenly, the normal force will disappear with it and, therefore, there won't be any push.

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    $\begingroup$ As another answer has alluded to, the compressibility of the material has an effect - if compressible, the system under gravity will be storing energy. imagine a spring between your magnet and plate. With the magnet on, it is held in equilibrium, with the spring compressed. Once the magnet is turned off, the spring will expand, pushing the plate away. $\endgroup$ – Baldrickk Nov 5 '18 at 10:39
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    $\begingroup$ @Baldrickk Interesting point. Did not cross my mind that the earth would store so much potential energy due to its weight. If you take a gigantic solid object and make a cavity inside it, the cavity would not collapse under the weight of the object, since the stress due to the weight is distributed all around. Of course, the earth is not quite solid. Obviously, need to improve my simulation model:) $\endgroup$ – V.F. Nov 5 '18 at 11:34
  • $\begingroup$ It didn't occur to me either, until I read one of the above answers $\endgroup$ – Baldrickk Nov 5 '18 at 13:36

No. The only reason a reaction force exists is because you are pushing down on the floor as a result of gravity pulling you down.

Perhaps one way to visualise this is to imagine a block on a slope at an angle $\theta$ to the horizontal.

When $\theta=0$ (i.e. the slope is flat), the block has force $mg$ down and so the reaction upwards is $R=mg$.

As $\theta$ grows steadily, the downwards force is still $mg$, but now the reaction force (which is the force at a right angle to the slope) becomes $R=mg\cos\theta$.

Imagine this surface has very large friction, so that you can get quite a large $\theta$ without the block slipping down. When you finally do reach a large enough $\theta$, the block will slide down parallel to the slope. Note that if $R$ had retained its value of $mg$, the block would have by now accelerated away from the slope, which doesn't happen.

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    $\begingroup$ Sorry, but this answer is just wrong because 1. this is the description of forces at equilibrium but suddenly eliminating gravity will not be an equilibrium state and 2. it completely ignores the underlying phenomenon as @Luke Pritchett points out. "Reaction" force is not one of the 4 fundamental forces, it's just a convenient rule of thumb for when the block has sunk enough into the material for the electric forces to counteract gravity. And it will bounce away a little when gravity disappears. $\endgroup$ – csiz Nov 3 '18 at 15:39
  • $\begingroup$ Ok, you are right, the "reaction force" is indeed a result of electrostatic interactions between the block and the floor. However, as Luke Pritchett also says, these forces are tiny, and are only significant because we're pushing down on them. If there were no gravity, there would be no pushing down on these molecules, and hence no reaction force. I believe my argument (at least for a simple classical point of view) is still valid. $\endgroup$ – Garf Nov 3 '18 at 15:47
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    $\begingroup$ The OP needs to qualify his question a bit. How fast does gravity disappear? If very slowly, the answer is probably no. If instantly, the answer is yes. $\endgroup$ – David White Nov 3 '18 at 16:29

$\def\vB{\vec B} \def\vR{\vec R} \def\vW{\vec W}$ I wish to object to the very concept of "action and reaction". It's true it goes back to Newton, who says (Principia) Actioni contrariam semper et aequalem esse reactionem [To every action there is always opposed an equal reaction].

Yet he also adds sive corporum duorum actiones in se mutuo semper esse aequales et in partes contrarias dirigi [or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.]

There is an important difference between the two wordings. In the first there is an asymmetry, which can be seen as temporal (action before, reaction after) or even causal (action = cause, reaction = effect). The second instead is completely symmetrical: there are mutual actions of two bodies, always equal but opposite.

Unfortunately in common parlance (not only in english) the first phrasing has prevailed and is used also in contexts very far from physics. But the symmetric form is much better and nearer to our present way of seeing phenomena, where we are always in presence of an interaction, with no causal or temporal asymmetry.

Let me explain all this with a familiar example. There is a table, or a marble block with a horizontal surface. I have a brick in my hand, and gently put it on the surface. It stays there - what's happened? Which forces, and how were they born?

Analysis repeats what was already said in other answers. Two opposite forces are acting on the brick: one $\vW$ downwards (gravity) the other $\vR$ upwards (the plane's reaction). I know they are opposite thanks to the second law: if the brick stands still, the net force acting on it must be zero: $$\vR=-\vW.\tag1$$

Then third law says that there must be a force $\vB$ which the brick applies to the surface: $$\vB = -\vR.\tag2$$ From (1) and (2) we deduce $$\vB = \vW.$$ In words: the brick applies to the surface beneath a force equal to its weight. It is very important to note that this is true because the brick is still. If I had left it to fall from a distance, during collision we would have measured $|\vB|\gg|\vW|$, with (2) remaining valid.

Up to now I only answered a partial question: which forces are there? And I remarked that my analysis refers to the equilibrium state. It says nothing as to what happens before, and why these forces develop.

A deeper answer requires a closer look at surfaces of table and brick. These are made of atoms, but we must note at once that at atomic scale (and at a much higher scale as well) the surfaces are far from being smooth. They are full of pits, picks, scratches, crests ... of all sorts of irregularities, much bigger than individual atoms. The result is that the encounter between brick and table initially interests a minimal fraction of the atoms present at the surfaces.

In order to see clearly what happens it is useful to put ourselves in a reference frame moving at a speed half the brick's $v$. In this frame the brick is moving downwards with speed $v/2$, the table upwards with the same speed. Note that this isn't c.o.m. frame, but gives us a symmetric view of the interaction which is taking place.

During the approach (initially guided by my hand, don't forget) distance between table and brick decreases, until some atoms come into contact. More precisely: it is known that interatomic forces are strongly dependent on distance. They are negligible if atoms' centers are more than a nanometer's fraction apart, and grow very strong and repulsive at a lesser distance. Of course forces act on both atoms approaching each other, and are equal in magnitude (Newton's third law).

The immediate effect of these forces is to displace the interacting atoms from their equilibrium positions in the solids they belong. As approaching goes on the number of interacting atoms grows bigger and their displacements grow as well. When the number of contacting atoms is sufficiently great, resultant forces on table and brick become appreciable at a macroscopic scale. The one acting on table has no effect, as it is fixed to ground. On the contrary, the force on the brick is contrasting gravity - then in an automatic way my hand responds by reducing its own force, which was necessary to sustain the brick in its slow downwards motion. Finally a point is reached when my hand gets unnecessary: the force of table over brick equals gravity and even overcomes it, slowing brick's motion and reducing it to a halt. Now final equilibrium is established.

Forces intensities are those previously described. The only additional feature of final equilibrium - although a scarcely appreciable one - is a static deformation of both bodies. Not table alone, brick too. How much each body gets warped depends on its rigidity: a wooden table will yield more than a brick, a marble one less (I believe).


There is something similar going on right now ...

Post-glacial rebound names the effect of the land moving upwards, after the glaciers of the last ice age, which pressed down on the land, melted away.

This is an indirect incarnation of the phenomenon you described in your question. Take away the thing, that is compressing stuff, and the stuff will relax.


I agree with David's comment that it is world building question. But treating it as a real answer I would say, take science and bin it. Newton's third law is not an independent thing, it is basically conservation of momentum that will be broken. It implies breaking of translational invariance. This probably means breaking of Poincaré invariance at the microscopic level. I.e. Particles as we know them don't exist anymore and we are left with one big mess. Does physics rearrange itself in some other set of representations that is still self consistent? Or are we left with a universe that simply isn't consistent at all (hard to understand what this would mean)?

Basically if you change one small detail from existing physics you probably end up with an inconsistent mess.

P.S.: I still like the question and the other answers that focus on just this law and ignore possible inconsistencies with the rest of science. But I thought an answer of this sort was still missing.


If gravity disappeared, the Earth would be torn apart due to it rotation, the pieces flying off in all directions in the rotation plane, just like an object that you're spinning on a string will shoot away if the string breaks.

The planet already isn't a perfect sphere, but somewhat squashed. This is because of the centrifugal force. It's gravity that is keeping it as spherical as it is. Without gravity, it would squash grotesquely and disintegrate.

You might not live long enough to witness much of this. The moment gravity disappears, the atmosphere would rapidly depressurize. Without gravity there is no air pressure. You will likely be severely injured by the depressurization and lose consciousness.

Let's think for a minute, though, about a planet which isn't spinning and forget the atmosphere for a second. Gravity creates immense pressures in the planet's interior. If gravity is suddenly gone, that pressure is released, like releasing a wound spring. That's going to cause tectonic and volcanic activity, at least for the brief moment before everything starts to completely break up. In that instant of gravity loss, the whole planet will suddenly expand like a rubber ball that has been compressed from every direction and suddenly released. This sudden expansion will have the effect of ejecting everything on the surface, like a kick from within.


In the absence of gravity, nothing gets pushed up because there is no force pushing it. Every fixed object will remain fixed and all the other tend to be floating, because nothing pulls it down.

  • $\begingroup$ Something already falling when you pull the gravity switch will continue to fall, though it will stop accelerating... and eventually it will impact.... which might be very slow though if aerodynamics interfere strongly... $\endgroup$ – rackandboneman Nov 6 '18 at 2:48

I think there's some confusion here between force pairs and opposing force. Suppose you have a person, a floor, and the Earth. Let's call $F_1$ the force the Earth exerts on the person. The person will also exert a force on the Earth. Call that $F_2$. The person will exert a force on the floor. Call that $F_3$. The floor exerts a force on the person. Call that $F_4$. $(F_1,F_2)$ and $(F_3,F_4)$ are force pairs. $F_4$ is an opposing force to $F_1$. For both force pairs and opposing forces, there are two forces in opposite directions, but with force pairs the source body and affected body are swapped. For instance, for $F_1$, the source is the Earth and the affected body is the person, for $F_2$ the source is the person and the affected body is the Earth. For opposing forces, the affected body is the same.

Newton's third law says that each force comes in a force pair, but it does not say that every force has an opposing force; if that were true, nothing would ever move. If you were to eliminate $F_1$, then $F_4$ would remain, and thus the person would be pushed upwards. However, this is not due to Newton's Third Law, except in that the reason that $F_4$ exists in the first place is due to physical processes, of which Newton's Third Law is one factor: $F_1$ pushes the person into the floor, this causes $F_3$, and $F_4$ comes along with $F_3$ as part of the force pair. While $F_4$ is ultimately caused by $F_1$, it is not the "equal and opposite force" to $F_1$ that Newton's Third Law speaks of. That force is $F_2$, the force of attraction the Earth feels towards the person (because of the Earth's large mass the acceleration that this force causes is extremely small, and so it is often ignored). If $F_1$ were to disappear, $F_2$ would also disappear. $F_4$ would remain, at least for a moment (once the person is pushed high enough, $F_4$ would disappear), and so $F_3$ would also remain. That is, the person would no longer be pushed into the floor by gravity, but they would still be exerting a downward force on the floor. Imagine a spring pressed against a block: if you let go, the spring will push the block, and the spring will then be accelerated in the other direction.

More relevant than the Third Law is Newton's Second Law: if a person is not experiencing acceleration (in the Earth's frame of reference) with gravity, then the net forces acting on them must be zero, so there must be some force other than gravity. If that force remains after gravity disappears, then the net force is now upwards.


As we are in Physics.SE, this answer focuses on the immediate fundamental interactions in the close vincinity (e.g., OP standing on a slab of hard material) as opposed to anything else (like Earth itself being compressed like a spring and suffering spontaneous disintegration).

You are not standing on Earth because of Newtons Laws. Those laws just inform you about what happens with bodies when forces act on them, it says nothing at all about where those forces are coming from.

You stand on a surface thanks to two of the four fundamental interactions (and sure, this is technically simplified, for gory details check the link and if you are really interested, the primary sources linked there - but the principle applies):

  • Gravity: you know about this one.
  • Electromagnetism: this is not only running your computer, but also keeps the atoms in your boots from sinking into the atoms of the floor. As you know, even neutrally charged atoms consist of the outer shell, which is negatively charged, and the core, which is positively charged. When individual atoms or molecules move closer together, the first interaction is between their outer electron shells - as those get closer and closer together, they repel. If you are not a Large Hadron Collider, this force stops atoms getting real close to each other very effictively; as the electrons (shells) get ever closer together, this force gets ridiculously large compared to gravity.
  • You can forget about the other two, they only appear at much smaller scales, within the nucleus.

These interactions cause all forces we know about. Two of them cause all macroscopic forces (between atoms); the other two cause all forces within atoms. If you so wish, the first two explain how atoms interact; the other two give a reason why the subatomic parts of the atom stick together at all (instead of everything being just a soup of quarks).

Regularly, gravity pulls you and Earth together until the atoms which make up the outermost layer in your boots get close to the atoms of the floor, ever increasing the electromagnetic repelling force acting in the opposite direction - until those cancel out and movement stops. At this point, by Newton's 1st Law, those forces are equal but opposite.

If Gravity suddenly disappears, then at this very moment, only the force caused by Electromagnetism stays around. This will not only push you up, but, by Newton's 3rd Law, also push Earth down. All other effects aside (speaking absolutely theoretically, without air, etc.), the visible effect will be you moving, as by the 2nd Law, it takes somewhat more energy to visibly move Earth, but as we are talking theoretically here, this is what the 3rd Law says - both of you move. The forces act on both parties involved.

How large is this force? Exactly as large to cancel you being pulled down by gravity, before. But as explained before, the electromagnetic interaction acts over incredibly short distances. As soon as the atoms get a bit away from each other (atomic scale), the force will effectively stop (although, like gravity, in principle it has unlimited range, it falls off with the square of the distance). While Alpha Centauri in theory pulls on you while you are on Earth, the effect is not relevant; the same is true when the atoms of the soles of your boots get away the tiniest amount of distance from the floor.

The previous paragraph would be the end of it if your body and boots were ideally incompressible, and absolutely stiff. But obviously this is not the case; all of these parts are very squishy (think about your flesh etc.; and all boot materials flex very much for comfort). All of this behaviour is also, again, caused by the same electromagnetic interaction. So there is a chain of repellation between all the atoms involved, and in our natural state (under the influence of gravity), they are all very much compressed like a spring. This spring expands when gravity goes away, so this will be the visible force acting (caused, again, strictly by the electromagnetic interaction).

So, TL;DR: No, Newton's 3rd Law does not make you fly up. Yes, you and Earth will technically separate, caused by the electromagnetic interaction, which was responsible for you not merging with Earth when gravity was still around. No, the immediate atomic border between you and the Earth plays only a negligible part, since most squishy parts of your body and clothing make an effective spring, so there are humongous amounts of atom "layers" involved. I am in no position to calculate how much force actually acts; obviously it will be small, but as you can see and feel the spring action in work (especially if you are wearing springy runner's boots...), you can assume that it will get you moving at least a little, perceptible bit, before other elements (drafts of air, Earth exploding etc.) take over.

  • $\begingroup$ Wouldn't you immediately separate from the planet simply due to its rapid spinning just like a rock being loosed from a sling? $\endgroup$ – CramerTV Nov 6 '18 at 2:09
  • $\begingroup$ Yes, of course, @CramerTV. Many effects would happen. But I feel like OP's question is really about the immediate "standing, then being 'pushed' apart" issue - I'm specifically ignoring all side effects like spinning of the earth, air, explosions, etc.. $\endgroup$ – AnoE Nov 6 '18 at 6:25

If Gravity disappeared things that had been effected by Gravity would not get pushed upward. However the Earths equator is traveling at about 465m/s relative to the center, and without Gravity things that were not held to the Earth by other forces would tend to maintain the same tangental velocity, and so would ultimately escape the Earth.


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