A thought experiment: Imagine the Sun is suddenly removed. We wouldn't notice a difference for 8 minutes, because that's how long light takes to get from the Sun's surface to Earth.

However, what about the Sun's gravitational effect? If gravity propagates at the speed of light, for 8 minutes the Earth will continue to follow an orbit around nothing. If however, gravity is due to a distortion of spacetime, this distortion will cease to exist as soon as the mass is removed, thus the Earth will leave through the orbit tangent, so we could observe the Sun's disappearance more quickly.

What is the state of the research around such a thought experiment? Can this be inferred from observation?

  • 21
    $\begingroup$ "If however, gravity is due to a distortion of spacetime, this distortion will cease to exist as soon as the mass is removed". Using "spacetime" and "as soon as" in the same sentence is calling for trouble, because there is no "as soon as" in spacetime. $\endgroup$ – Gyro Gearloose Feb 12 '16 at 16:26
  • $\begingroup$ I imagine huge gravitational waves in the result of sudden sun removal. If sun is to be removed in an infinitely small amount the peak amplitude will be infinitely strong and the earth will be ripped in a dust by gravitational wave traversing it. But actually, I wonder that because mass will not have any acceleration, there will be no waves. So it seems to be 8 minutes when the Earth starts going in a straight line. $\endgroup$ – Emil Jul 29 '16 at 16:19
  • $\begingroup$ The earth would rebound at once! $\endgroup$ – J. Chomel Sep 7 '16 at 15:01
  • $\begingroup$ Related MO question: mathoverflow.net/q/112029/13917 $\endgroup$ – Qmechanic Oct 1 '16 at 13:00

10 Answers 10


Since general relativity is a local theory just like any good classical field theory, the Earth will respond to the local curvature which can change only once the information about the disappearance of the Sun has been communicated to the Earth's position (through the propagation of gravitational waves).

So yes, the Earth would continue to orbit what should've been the position of the Sun for 8 minutes before flying off tangentially. But I should add that such a disappearance of mass is unphysical anyway since you can't have mass-energy just poofing away or even disappearing and instantaneously appearing somewhere else. (In the second case, mass-energy would be conserved only in the frame of reference in which the disappearance and appearance are simultaneous - this is all a consequence of GR being a classical field theory).

A more realistic situation would be some mass configuration shifting its shape non-spherically in which case the orbits of satellites would be perturbed but only once there has been enough time for gravitational waves to reach the satellite.

  • $\begingroup$ "Since general relativity is a local theory just like any good classical field theory, the Earth will respond" - well, it's only the theory. I guessed you meant that "...according to theory, the Earth will respond..." $\endgroup$ – Emil Jul 29 '16 at 16:21
  • $\begingroup$ Also, the propagation of changing gravitational forces described in this answer is essentially what Gravitational waves are. It's the same mechanic behind Electromagnetic waves (light) when you factor in effects of special relativity. $\endgroup$ – Trevor Kafka Sep 24 '18 at 0:48

Gravitational influences do propagate at the speed of light, not instantaneously.

The question of what would happen if the Sun instantly disappeared is actually a funny one in general relativity. The equations of general relativity imply as a mathematical consequence that energy must be locally conserved. Therefore, there is no valid solution to the equations that describes the Sun suddenly disappearing (since that scenario violates local energy conservation).

(A similar statement holds in electromagnetism, by the way: charge conservation is a logical consequence of Maxwell's equations, so if someone asks you what the electric field does when a charge suddenly disappears, there is no correct answer.)

But you can sensibly ask what would happen if the Sun suddenly changed its mass distribution -- if it exploded, say, sending its mass in different directions at high speeds. The answer is that the Earth's orbit wouldn't change for 8 minutes.

  • 9
    $\begingroup$ What is the basis for the claim that gravity propagates at the speed of light? $\endgroup$ – Ambrose Swasey Nov 3 '16 at 17:49
  • $\begingroup$ @AmbroseSwasey Quantization of gravity has been used as an effective theory in cosmology for the last decades, and it is the holy grail of theory at present that a definitive quantization model will be demonstrated to fit data. In this model the graviton transfers gravitational effects, it is a zero mass particle and as a zero mass particle has to propagate at velocity c because of special relativity , which is intrinsic in general relativity for flat spaces, at the present time of the universe en.wikipedia.org/wiki/File:History_of_the_Universe.svg . $\endgroup$ – anna v Feb 16 '18 at 5:48

All observations are consistent with standard GR so far, but I don't think the speed of gravity, in particular, has ever been measured.

Experimental measurements of the speed of gravity was quite a controversy a few years ago when a paper came out claiming that the speed of gravity was very close to $c$ as measured by the Shapiro delay. To see papers on the subject google shapiro+speed+gravity:

Clifford Will is an expert in the area and says that there was no measurement. He has a website on the subject that gives links to the various papers:

My guess is that the Will side won. But academia means "never having to admit you were wrong". Here's a pair of dueling papers on the subject, published in the same journal at the same time (that date to after Clifford Will last updated his page above):

Class.Quant.Grav. 22 (2005) 5181-5186, Sergei M. Kopeikin, Comment on 'Model-dependence of Shapiro time delay and the "speed of gravity/speed of light" controversy'

Class.Quant.Grav.22 (2005) 5187-5190, S. Carlip, Reply to "Comment on Model-dependence of Shapiro time delay and the 'speed of gravity/speed of light' controversy"

  • $\begingroup$ a direct evidence Radio Tests of GR with Jupiter and a quasar by Fomalont and Kopeikin. Measures made in 2002 , published in 2009 $\endgroup$ – user46925 Dec 26 '15 at 6:21

From Hacker News https://news.ycombinator.com/item?id=6253263

This is a far more interesting question than it might seem at first glance, and it deserves some attention because it tells us something fundamental and wonderful and just bloody awesome about the universe.

But I don't know how to tell the story succinctly. So I'm going to do that thing I do. I am very, very sorry. Please feel free to move on if this strikes you as tiresome.

Consider the Earth, and you on it. You're not floating freely, so clearly something's going on. We call that "gravity." We can call it, in the most generic sense, an interaction: you and the Earth are interacting somehow, and that's what's keeping you from floating freely. We can then ask what the speed of that interaction is by putting it in these specific terms: How much time will elapse between your changing your position relative to the ground and your beginning to fall?

Yes, it's the Wile E. Coyote problem. Wile E. Coyote runs off a cliff, floats in mid-air long enough to hold up a sign that says "Help," then begins to fall.

Clearly that's an exaggeration. But just how much time does elapse, in real life, between stepping off a cliff and beginning to fall?

We can approach the problem naively by remembering that all propagating phenomena in the universe are limited by the speed of light. Given that fact, it makes sense to hypothesize that the time between the moment when Wile E. steps off the cliff and when he begins to fall will be equal to or more than the distance between him and the ground divided by the speed of light. It certainly can't be less, right?

We can then construct a set of very, very precise experiments with very fine tolerances — probably involving electromagnets and lasers or something — to test this hypothesis. And then we can find that we're totally goddamn wrong.

To the absolute limit of our ability to measure it — and our ability to measure it is really good, since we used electromagnets and lasers and other expensive science things — when an object is dropped, it begins falling instantaneously. Not after a very small interval of time, but absolutely instantaneously. As in zero time elapses between dropping and falling. This is fairly earthshaking, really. Because it implies that somehow a "signal" of some kind is getting from the ground to Wile E. faster than the speed of light. Which is supposed to be impossible.

I'm going to skip ahead a bit here, because I don't feel like explaining the entire theory of general relativity, and it won't be that useful in answering the question anyway. Suffice to say that no, no time elapses between dropping and falling, but at the same time no, no signal or interaction has to propagate upward from the ground to Wile E. in order to make him start falling. In fact, what's going on is that Wile E. Coyote is always falling, due to the curvature of spacetime created by the Earth. Whenever he's standing at the edge of the cliff, on the ground, the ground beneath his feet — paws? — is arresting his fall by, effectively, pushing up against him. The very instant that's removed, he starts falling. So in that sense, gravity has no speed. Because it doesn't actually propagate through space. One way to look at it is to say the gravitational field fills space, so wherever you are, you're already being affected by it all the time. Another way is to say that gravitation essentially is space, so it affects you simply by virtue of existing. The two are essentially equivalent English translations of the equations that actually describe the phenomenon.

But okay, that's half the problem. The gravity of a static body fills space, or is space, and as such can't be meaningfully said to have a speed. But what about the gravity of a changing body? Like you said, what if "suddenly a black hole appeared?" Well, the answer of course is that that never happens, ever. Gravitation doesn't suddenly anything; macroscopic things don't just appear out of nowhere, and teleportation is impossible. So we don't have to think about that … and in fact we couldn't get meaningful answers if we tried.

But things do move. The moon's moving relative to the surface of the Earth; we can tell, even apart from the fact that we can see it up there, because the moon is the major contributor to the tides, and the tides rise and fall. But what's the relationship between the moon's position in space and the tidal acceleration on the Earth? Are the two somehow always in perfect sync, or is there some lag? If so, how much, and in what direction? That's actually a much harder question to answer than you might think. There was a now-infamous paper some years ago by a fellow named Tom Van Flandern (recently passed, God rest his soul) that asserted that the change in gravitational acceleration in a dynamical system actually propagates many times faster than the speed of light — at least twenty billion times faster than the speed of light — but not instantaneously. This got a lot of attention at the time. If the propagation speed of changes in spacetime geometry were equal to the speed of light, that'd be fine. If it were literally instantaneous, that'd also be fine, more or less, though our theory would need some tweaking. But faster than c but still finite? That was really hard to explain.

It turned out not to be a problem though. Because Van Flandern just made a mistake in his paper. See, the relationship between motion and gravitation is not as straightforward as it might seem. In fact — and I'm glossing over this now, because the maths are damn complicated — whenever a gravitating object moves inertially, the gravitational acceleration vector at a point removed actually points at where the object actually is at a given instant, as opposed to where the object's light is seen to be coming from at that instant. So in that sense, we're back to gravitation being instantaneous again!

But is it really? No. Because you see, if the inertially moving object were to come to a stop instantaneously, the acceleration vector would continue to point toward its future position for a time, as if it were still moving inertially, even though the object is actually somewhere else. The sum of effects that serve to cancel out aberration when everything moves inertially would break down, and the acceleration field would point toward empty space for however long it takes for the change in geometry to propagate through space at the speed of light from the gravitating object to the point in question.

Except things don't stop moving instantaneously. Things accelerate, and acceleration requires energy, and when you factor that in, the equations balance out again. (If you feel up to the challenging of following a lot of advanced mathematics, here's the best paper I know on the subject.) - http://arxiv.org/abs/gr-qc/9909087v2

So what does that mean? It means that the "speed of gravity" is the speed of light … technically. Changes in the geometry of spacetime actually propagate at the speed of light, but the apparent effects of gravitation end up being instantaneous in all real-world dynamical systems, because things don't start or stop moving or gain or lose mass instantaneously for no reason. Once you factor in everything you need to in order to model a real system behaving in a realistic manner, you find that all the aberrations you might expect because of a finite speed of light end up canceling out, so gravity acts like it's instantaneous, even though the underlying phenomenon is most definitely not. The universe is pretty damn cool, if you ask me.

  • $\begingroup$ Alright big issue here: you were saying that when Wile Coyote fell he fell instantaneously and so gravity has ridiculous speed. Who's to say he didn't just walk into a stream of particles with any speed? If I walk into a shower and the water hits me fast it doesn't mean the stream of water has a light speed, it means it was already partways through its pilgrimage from top to bottom. And boy did I not read the whole shebang, but +1 for thoroughness eh. Feel free to show me what I should read in your post good work $\endgroup$ – Andres Salas Jul 24 '14 at 21:03
  • $\begingroup$ This also solves your issue with gravity affecting objects where they stand, since moving through a stream of water does not have to be viewed as an interaction with the source, but with particles already sent. That's why objects will still experience gravity when a distortion occurs until the distortion ripple reaches them at a speed, seemingly c. I'm not super moved as to these observations being issues, you know? But I do wonder why a planet reacts to someone else absorbing their emitted particles, where does teh pushback come from? $\endgroup$ – Andres Salas Jul 24 '14 at 21:08
  • 3
    $\begingroup$ @AndresSalas One way to look at it is to say the gravitational field fills space, so wherever you are, you're already being affected by it all the time. ... technically. Changes in the geometry of spacetime actually propagate at the speed of light, but the apparent effects of gravitation end up being instantaneous in all real-world dynamical systems $\endgroup$ – Basic Oct 2 '14 at 22:20

Your question was first asked by Laplace. The following is from the Wikipedia article on "The speed of gravity"


The first attempt to combine a finite gravitational speed with Newton's theory was made by Laplace in 1805. Based on Newton's force law he considered a model in which the gravitational field is defined as a radiation field or fluid. Changes in the motion of the attracting body are transmitted by some sort of waves.[4] Therefore, the movements of the celestial bodies should be modified in the order v/c, where v is the relative speed between the bodies and c is the speed of gravity. The effect of a finite speed of gravity goes to zero as c goes to infinity, but not as 1/c2 as it does in modern theories. This led Laplace to conclude that the speed of gravitational interactions is at least 7×106 times the speed of light. This velocity was used by many in the 19th century to criticize any model based on a finite speed of gravity, like electrical or mechanical explanations of gravitation.

From a modern point of view, Laplace's analysis is incorrect. Not knowing about Lorentz invariance of static fields, Laplace assumed that when an object like the Earth is moving around the Sun, the attraction of the Earth would not be toward the instantaneous position of the Sun, but toward where the Sun had been if its position was retarded using the relative velocity (this retardation actually does happen with the optical position of the Sun, and is called annual solar aberration). Putting the Sun immobile at the origin, when the Earth is moving in an orbit of radius R with velocity v presuming that the gravitational influence moves with velocity c, moves the Sun's true position ahead of its optical position, by an amount equal to vR/c, which is the travel time of gravity from the sun to the Earth times the relative velocity of the sun and the Earth. The pull of gravity (if it behaved like a wave, such as light) would then be always displaced in the direction of the Earth's velocity, so that the Earth would always be pulled toward the optical position of the Sun, rather than its actual position. This would cause a pull ahead of the Earth, which would cause the orbit of the Earth to spiral outward. Such an outspiral would be suppressed by an amount v/c compared to the force which keeps the Earth in orbit; and since the Earth's orbit is observed to be stable, Laplace's c must be very large. In fact, as is now known, it may be considered to be infinite, since as a static influence, it is instantaneous at distance, when seen by observers at constant transverse velocity.

In a field equation consistent with special relativity (i.e., a Lortentz invariant equation), the attraction between static charges is always toward the instantaneous position of the charge (in this case, the "gravitational charge" of the Sun), not the time-retarded position of the Sun. When an object is moving at a steady speed, the effect on the orbit is order v2/c2, and the effect preserves energy and angular momentum, so that, and orbits do not decay. The attraction toward an object moving with a steady velocity is towards its instantaneous position with no delay, for both gravity and electric charge.

  • 1
    $\begingroup$ And they used to think that gravity was the longitudinal wave to go with the transverse wave of electromagnetism (i.e. as in the P and S waves of seismology). $\endgroup$ – Carl Brannen Feb 22 '11 at 3:53
  • $\begingroup$ From a GR perspective, does this imply that the space curvature moves at the same velocity as the large gravitational object? If the object were accelerating, would the space curvature lag behind? $\endgroup$ – Timothy Miller Jan 15 '16 at 18:30
  • $\begingroup$ @CarlBrannen That's an interesting piece of history I didn't know, Can you point me to a reference or an article. We're presumably talking about the time period between 1865 (Maxwell's equations) and 1915 (GTR), right? $\endgroup$ – WetSavannaAnimal Jul 11 '16 at 10:46
  • $\begingroup$ Nice to know!! I too would like some references. This was possibly one of the inspirations that led Einstein to predict gravitational waves. $\endgroup$ – ivbc Jun 6 '17 at 16:22

the fact that distortion travels 'as soon' as a mass is removed or not is not implied in any way by gravity being due to a distortion of spacetime. In fact distortions of spacetime are as limited to travel to the speed of light as any other physical influence.


Nothing in the universe can travel faster than light. Because of this only light is the cosmic speed limit as per STR. Even gravitational waves cannot travel faster than light. If sun is removed we would see its effect after 8 minutes. And earth would be free to move , it will then start revolving after it has found a celestial body which is greater in mass than earth and it will start revolving around it as the celestial body has bent more space time according to the Einstien's most profound and greatest "THE GENERAL THEORY OF RELATIVITY "

  • 1
    $\begingroup$ Welcome to stackexchange. Your answer is essentially ok, but what does it add to the other answers? $\endgroup$ – Martin Jan 30 '15 at 13:28
  • 2
    $\begingroup$ I agree it doesn't really add anything. But also it isn't necessarily the case that Earth would orbit around any new body that is larger in mass. $\endgroup$ – Jim Jan 30 '15 at 18:46
  • $\begingroup$ then that means it would move tangentially from its orbit $\endgroup$ – Tushar Bhalla Feb 3 '15 at 14:37

Several correct answers stating that gravity propagates at the speed of light have already been given but there is a connected problem which is far more difficult. In your scenario, you unrealistically completely removed the Sun but, think about it, the Sun constantly slips away, at a speed of 230 km/s with respect to the center of the galaxy. Nevertheless the gravitational force felt by Earth, which takes its source at the Sun, always points toward the center of the Sun. How come? If gravity propagates at the speed of light $c$, that force at a time $t$ should be directed toward the so-called retarted position of the Sun, i.e. the position at time $t - d/c$ where $d$ is the distance from the Sun to the Earth, shouldn’t it? How could the force field in the vicinity of Earth “know” the position of the Sun instantly? The “information” about that position can only propagate at the speed of light, or so we said.

First we should note that the issue is totally universal and potentially disastrous. Let us consider the Earth-Moon system for example. Let us analyse it in it center of gravity frame to show the choice of frame is not the issue (to be equivalent with the Sun example, I should have used a frame centered on the Sun this time for example). The force $F_E$ exerted by the Earth on the Moon points toward the center of the Earth, even though the Earth moves about this center of gravity. Conversely, the force $F_M$ exerted by the Moon on the Earth points toward the center of the Moon even though the Moon moves about the center of gravity. A naive application of the principle that gravity shall propagate at the speed of light would have $F_E$ points toward the retarded position of the Earth and $F_M$ points toward the retarded position of the Moon. As a result, those two forces would not be aligned anymore and they would create a torque which would change the angular momentum of the Earth-Moon system. This is squarely ruled out by observation. The same would happen for any two celestial bodies.

Some will immediately objects that using the language of forces and more generally of Newtonian mechanics is totally inapropriate and that this is the source of the issue. It is not so. Gravity is weak enough in the Solar system and speeds are small enough compared to the speed of light that we do not need to use the fully developed General Relativity. To a very good approximation, we can use Newtonian mechanics with some corrections. The question is then: how does it come this approximation eventually makes gravitational forces point toward instantaneous positions and not retarded positions? It is not that easy to wave it away: after all it takes 1 sec for a signal propagating at light speed to travel from the Earth to the Moon and 8 min from the Sun to the Earth. Those times are clearly not negligible and at first sight it is hard to understand why it is as if they get discarded in this approximation. Some subtle effect seems to be at play which somehow moves the force direction from the retarted position toward the instantaneous one as this approximation is developed.

This is in fact exactly what happens. The full mathematical treatment is far too complex to be given here but the result can be loosely stated as follow. As one approximate General Relativity for low speed and weak gravity, we end up with a gravitational force pointing toward the retarted position of the source, extrapolated quadratically toward its instantaneous position. Mathematically, if $n(t)$ is the unit vector pointing from the centre of the Moon toward the retarded position of the center of the Earth, the force felt by the Moon points in the direction

$$n(t) + \tau \frac{dn}{dt} + \tau^2 \frac{dn}{dt}$$

where $\tau = \frac{d}{c}$ is the retard. This equation is only illustrative: its correct mathematical form would bring complexities I do not want to dvelve into. I only wrote it to show that quadratically extrapolated was meant in the sense of a second-order Taylor expansion in $\tau$.

This cancellation of aberation is therefore only approximate. As a result, there is a residual change of angular momentum but it is made too small to matter in the Solar system. But for two neutron stars orbiting close enough to each other, angular momentum decays at a non-negligible rate and it has been measured (c.f. the ultra-famous Hulse–Taylor system) in extremely good agreement with the theory. Nevertheless this cancellation is very much "good enough" but it is by no means a miracle. However this answer is already too long, especially considering it is a tangent on the question of the OP and I will not elaborate.

Steve Carlip wrote a very good article [1] addressing this whole issue, including an enlightening comparison with electromagnetism where aberration is also partially cancelled, as well as explaining the fundamental reasons for this cancellation.

[1] S. Carlip, Aberration and the speed of gravity, Physics Letters A 267 (2000), 81 - 87 https://arxiv.org/abs/gr-qc/9909087

  • $\begingroup$ Reading this and Carlips paper it is interesting to note that the math does not say that the attractive force vector is directed 'exactly' towards the instantaneous (however defined) position of the source and not towards it's retarded position, so the speed of gravity cancels out almost 'exactly'. To get to exactly you seem to have to use laws of conservation of energy and momentum, but why can you not show directly (charged particle) from GR and Maxwell that this is true 'exactly' and not just approximately to some higher order of v/c or the like? $\endgroup$ – Jens Jul 14 '17 at 13:51
  • $\begingroup$ I am not sure I understand your question. What do you mean by "exactly" between quotes? $\endgroup$ – user154997 Jul 14 '17 at 15:51
  • $\begingroup$ Like x"= -k^2 x gives x=A sine(B+kt) as an 'exact' solution. Thus postulating an 'exact' non-relativistic gravitational equation of motion, which includes delays of propagation (light and gravity), one should be able to give an 'exact' solution (closed form or series expression) that demonstrates whether or not the attractive force is seen as 'exactly' directed towards the instantaneous position. If it is just a second order approximation it really does'nt say much about the basic question. What we need is proof that c-gravity is exactly cancelled by c-light in the solution! $\endgroup$ – Jens Jul 19 '17 at 18:11

It would take less than 8 mins. It depends on elasticity of space time fabric. Consider Put a marble on a cloth and then observe how much it descends and curves cloth. Now, suddenly remove marble The time taken by cloth to regain its original position, so that it ends point feel no curvature, clearly depends on elasticity of fabric and amount of depth it had sunk. Our space, it is very much elastic and for curvature of sun, it would take few seconds to regain its original position.

Gravity and gravitational waves are different.

Let me clear it to you. Consider a taut string 100m long. Send an impulse to it. Clearly,it has some speed of propagation, nearly 2m/s(depends on material). Now, cut one end of it, how much time does it take for other side of string to know it.Somewhat 1 second. Isn't it amazing.

Similarly, gravitational waves travel at c but gravity of object is much faster detected.


I feel like this question is being asked wrong and/or it is being interpreted wrong for what you're actually asking. It is understood that the propagation of anything cannot exceed 'c', but I don't think propagation is necessary to answer the question, or to create a valid thought experiment. First off, gravity is not fully understood by any mainstream science and a lot of the paradoxial problems inherent within our current accepted understanding tend to leave many scratching their heads. I'm no physicist, or scientist for that matter, but this has been on my mind for a very long time and I decided to throw it out here and allow you all to tear it to pieces or at least lead me in a better direction lol.

The question, how would the sudden disappearance of the sun affect gravitation, and would it follow 'c' or happen instantaneously?

My answer is Both.

Lets look at gravity in a couple of different ways to explain why I believe this. I see a lot of references to gravity as a wave...I assume this is because of the apparent "propagation" that occurs within a gravitationally active region. I accept that any physical change made by object A that "could" effect object B must travel to object B no faster than 'c'. So yeah, sun goes poof, we wait the 8 minutes before gravity is released. Here's where I go left.....That "wave" isn't necessary to get information from A to B instantly. Look at it backwards, mass is the force (cause), gravity is the result of that force (effect). I don't view gravity as we observe it as a force but the released energy of another force.....displacement. The region that would see a net change if the sun went poof would be space-time. Look at it in a simplified way, I stand at one end of a field and you at the other with 2 cans and a string, pull it taunt and yell into it.....the vibrations travel down the string to my can at the speed of sound and I can hear it. For the sake of this example, lets assume the speed of sound represents 'c', and the sound wave represents gravity....the string would represent space-time. Everything works just as you would expect it. Now, i would ask you to make a constant humming noise into the can. Several milliseconds later, I begin to hear it. Suddenly, you pass out from humming instead of breathing and drop the can. Again, I must wait several milliseconds before I realize something has happened and you've stopped. What I failed to realize was that I already had that information. As the can left your hand (the pull of your gravity), the gravitational constant in local space-time was changed (the tension on the string went slack). Does this not happen instantly? Granted, I know of no device that can measure the gravitational constant in a specific region of space-time but is this not a method of reading the net effect of a massive and sudden gravitational change? What if I lay at the bottom of a pool with an air hose and blow bubbles? The bubbles travel to the surface at (hypothetical) 'c' but the bubbles themselves displace the water causing it to slightly rise in apparent volume. Does this increase in net volume not happen the instant the bubble displaces the water?

Bottom line, I agree that if the sun vanished, it would take 8 minutes for a change in its gravitational influence on the Earth to be observed, but I believe that the net effect on the region of space-time between the earth and sun could be observed instantly using the proper equipment to detect those changes.

  • 2
    $\begingroup$ This looks more like a separate (though related) question than an answer. I suggest you make it into a question. $\endgroup$ – Wouter Mar 3 '13 at 11:07
  • 1
    $\begingroup$ Slack won't be noticed instantaneously. It will take some time for electrostatic forces to transfer this slackness. $\endgroup$ – Anubhav Goel Oct 4 '16 at 1:10

protected by Qmechanic Jan 30 '15 at 13:20

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.