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Consider Earth moving around the Sun. Is the force of gravity exerted by Earth onto the Sun directed towards the point where Earth is "right now", or towards the point where Earth was 8 minutes ago (to account for the speed of light)? If it's the former, how does the Sun "know" the current orbital position of Earth? Wouldn't this information have to travel at the speed of light first?

If it's the latter, it would force a significant slowdown of Earth's orbital motion, because the force of gravity would no longer be directed perpendicular to Earth's motion, but would lag behind. Obviously, this isn't happening.

So it appears that the force of gravity is indeed directed towards the current orbital position of Earth, without accounting for the delay caused by the speed of light. How is this possible? Isn't this a violation of the principle that no information can travel above the speed of light?

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    $\begingroup$ This is not limited to gravity. The same applies to electromagnetism. $\endgroup$ – safesphere Feb 1 at 1:28
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    $\begingroup$ Related: physics.stackexchange.com/q/101919 $\endgroup$ – Kyle Oman Feb 1 at 2:23
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    $\begingroup$ Related: physics.stackexchange.com/q/5456 $\endgroup$ – Dan Yand Feb 1 at 3:10
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    $\begingroup$ A slowdown of the Earth's orbital motion wouldn't take place (unless you mean a negative slowdown, which I don't believe is what you mean). Instead, the motion would speed up because there would be a tangent force component in the direction of the Earth's motion. Luckily, special relativity comes to the rescue and makes this tangent force disappear as is clearly explained by tparker. So one can say that the fact that the Earth hasn't been pulled out of its orbit is a confirmation of special relativity (the sr effect corresponds to a $\frac v c$ factor, the gr effect to $\frac{v^2}{c^2}$) $\endgroup$ – descheleschilder Feb 2 at 19:31
  • $\begingroup$ Ohlala, so many answers to choose from by people who have never taken relativity in a classroom as opposed to reading pop-science books.... $\endgroup$ – Kostas Feb 2 at 22:41

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Cuckoo asked: So it appears that the force of gravity is indeed directed towards the current orbital position of Earth, without accounting for the delay caused by the speed of light. How is this possible?

If the motion is straight or circular the aberration cancels out, see Steve Carlip: Aberration and the Speed of Gravity:

Steven Carlip wrote: The observed absence of gravitational aberration requires that "Newtonian'' gravity propagates at a speed ς>2×10¹⁰c. By evaluating the gravitational effect of an accelerating mass, I show that aberration in general relativity is almost exactly canceled by velocity-dependent interactions, permitting ς=c. This cancellation is dictated by conservation laws and the quadrupole nature of gravitational radiation.

or to quote the Wikipedia article on the subject:

Wikipedia wrote: Two gravitoelectrically interacting particle ensembles, e.g., two planets or stars moving at constant velocity with respect to each other, each feel a force toward the instantaneous position of the other body without a speed-of-light delay because Lorentz invariance demands that what a moving body in a static field sees and what a moving body that emits that field sees be symmetrical. In other words, since the gravitoelectric field is, by definition, static and continuous, it does not propagate.

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    $\begingroup$ I would be thankful for any comments on my answer below. I kind of made an interpretation of your answer, but I'm not sure I said everything correctly. $\endgroup$ – Elias Riedel Gårding Feb 1 at 2:16
  • $\begingroup$ I too was going to distinguish between Newton's and Einstein's gravitational theories. If assuming Newton's theory, the speed of gravity is usually assumed instantaneous, but Laplace considered a finite speed and showed from astronomical observations it must be millions of times faster than light. If assuming general relativity, then weak gravitational waves propagate at the speed of light, but I'm not sure that strong gravitational waves have a meaningful speed in general (?) $\endgroup$ – Colin MacLaurin Feb 5 at 23:16
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No, gravitational influences never travel faster than the speed of light. However, a naive incorporation of a speed-of-gravity delay would actually lead to the Earth's orbital motion speeding up, not slowing down. (Think about the geometry carefully.) I explained here why that doesn't actually happen in general relativity.

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    $\begingroup$ Related: space.stackexchange.com/questions/21467/… $\endgroup$ – ahiijny Feb 1 at 15:11
  • $\begingroup$ A strange comment considering that the OP already concluded correctly that the force must be along the present "true" line-of-sight. $\endgroup$ – Kostas Feb 2 at 22:31
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    $\begingroup$ This does not answer the question. $\endgroup$ – Andrew Steane Feb 2 at 23:13
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    $\begingroup$ @tparker a ref to another answer is fine, but is a comment. To answer merely "it is light speed limited" but without setting out the mechanism in physical terms is not, I think, to answer the question. But in saying this, my tone of voice is intended to be calm not strident! $\endgroup$ – Andrew Steane Feb 3 at 13:02
  • $\begingroup$ @tparker you said "gravitational influences never travel faster than the speed of light". This is THE wrong answer to the question. Actually, the static potential is instantaneous and along line of sight, but do you know the reasons why such instantaneous interaction does not violate the postulates of relativity? The easy answer is that a static potential is not a wave, but this is not entirely correct either. The full answer is complicated, graduate course level stuff... $\endgroup$ – Kostas Feb 4 at 11:01
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Let's temporarily pretend that we can speak of gravitation as being released in pulses, which is a weird way to speak, but I think is part of your mental model. Let's also pick a reference frame : suppose the Sun is stationary in our laboratory.

The Earth right now is not reacting to the right now pulse of the Sun's gravity. The Earth right now is intercepting the pulse of gravity released $8$ minutes ago, which happens to point directly back to the Sun (because it is stationary).

For a careful, mathematical treatment of this, see The Feynman Lectures in Physics, vol. II, section 26-2, where an electric field in laboratory coordinates is found to have magnetic components in the moving frame which (to first order) cancel the aberration (angular deflection) caused by retarding waves by their travel times. The same thing happens in gravitation: the off-diagonal elements of the tensor pick up the terms necessary to cancel retarded aberration (to first order).

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  • $\begingroup$ Simplest answer for a layman, by far. Hope it's correct because this is the one I'll remember. $\endgroup$ – Craig Hicks Feb 1 at 18:20
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    $\begingroup$ In fact, that's what magnetism is: it's what we see in a reference frame in which electric charges are moving. It consists of exactly the "aberrations" that are necessary to preserve the Lorentz invariants. $\endgroup$ – Acccumulation Feb 1 at 20:06
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    $\begingroup$ @CraigHicks Yes it's right. The Sun sends out its "marching orders" to the Earth in the form "ok whoever is out there at 8 minutes away: when you receive this instruction from me, adjust your velocity towards where I, the mighty Sun, will be in 8 minutes, assuming I am moving inertially". $\endgroup$ – Andrew Steane Feb 2 at 23:22
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It seems that your intuition is tricking you. If I understood you correctly, you're thinking of the Sun's gravitational pull as directed at the Earth, like it's pointing at the specific "target spot" in space, and when gravity reaches that spot, it pulls there.

But it's better to think of gravity as pulling from all directions at once: so the Sun is simultaneously pulling both the spot where the Earth is now, and the spot where the Earth will be eight minutes from now, and every spot in between, and every other spot in our orbit, and so on.

Think of it the same way you think about sunlight: the sun is shining in all directions, so it never needs to know where we are going to be - some light just goes out in a straight line in all directions, and so we always miss the light that was headed towards us when it left the sun, and hit the light that left the sun pointing at where we would end up eight minutes later.

In this case, you can think of them the same way: The sun isn't sending those things towards us specifically, it's just sending them out everywhere at once.

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  • $\begingroup$ Another, closely related (equivalent?) way to put it: think of it as the Earth being pulled, rather than the Sun reaching out to the Earth and pulling. $\endgroup$ – outis Feb 2 at 2:02
  • $\begingroup$ The analogy with sunlight is probably the most intuitive explanation I've seen here. $\endgroup$ – Lightness Races in Orbit Feb 3 at 12:58
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Gravity is the bending of the space and the sun's mass causes bending of the space around it and in a circular path around the sun it is uniform. The earth travels in elliptical orbit so the force of gravity is more at some places i. e. the bending of space is more at a place closer to the sun and less at a place far from it. The force of gravity is not exerted by the sun on the earth but the bending caused keeps the earth near the sun and keeps it revolving and the velocity ensures that the earth does not just fly away in the direction of the tangent drawn to its orbit. So there is nothing like the sun "knows" where to exert the gravity but as I have said that it is uniform in a circular orbit so the bending is the force of gravity acting on it. The angle between the earth's motion and gravity(acting as the centripetal force here) will be 90°.The earth's motion will be in the direction of the tangent drawn at the orbit. As you might know that the centripetal force acts towards the center of the circular path and the angle between a tangent and the line passing through the center is always 90° so that law is never violated.

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  • $\begingroup$ What is gravity? - Some non-sense about space-time, which is where I get off, +1 $\endgroup$ – Mazura Feb 1 at 17:25
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    $\begingroup$ This is broadly correct statement of some facts about gravity but it does not really address the question asked by the OP, except perhaps for circular orbits where a symmetry argument is possible. $\endgroup$ – Andrew Steane Feb 2 at 23:18
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I was going to write an incorrect answer at first, but having researched a bit what Yukterez wrote, I think I may be able to offer some intuition.

First, let's look at the solar system in the centre-of-mass frame, where the Sun is essentially stationary in the middle. Here, the force on the Earth is directed towards the Sun's position 8 minutes ago, which conveniently is the same as its current position. So this doesn't affect the Earth's position in the way you say.

If the Sun and Earth were moving at a constant velocity to each other, they would, like Yukterez wrote, still be attracted to the instantaneous position of the other body. This is not because of a faster-than-light influence, but because gravity is more than a simple attractive force between objects (analogous to the electric field in electromagnetism). For moving sources, there are other components of the gravitational field (roughly analogous to the magnetic field). You can view this as the force travelling at the speed of light, but being directed towards the position of the source "predicted" based on its past linear motion.

However, if the source is accelerating, like the Earth around the Sun, the effects are less obvious. Then, you really can't view it only as an attractive force towards any position (I think, correct me if I'm wrong), but you get more complex things like gravitational waves. But still, nothing ever travels faster than light.


As I am more familiar with electromagnetism, let's take an example from there. If two oppositely charged bodies ($A$ and $B$) are moving with constant velocity towards each other, we can view the system in the rest frame of $A$. Here, it emits only a static electric field, given by the Coulomb formula, and $B$ therefore feels an attraction directly towards it. In the rest frame of $B$, this means that the attraction is towards the instantaneous position. But how can this be since the electric field is propagating at a finite speed? The answer is that since now $A$ is moving, it also emits a magnetic field. While this doesn't affect $B$ directly (since $B$ is stationary here), it does affect the electric field and makes it point in a different direction (towards the instantaneous position).

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  • $\begingroup$ According to relativity, all coordinate-independent properties of gravitational attraction would have to be the same in a reference frame in which the bodies are moving together, as opposed to one in which they are stationary. $\endgroup$ – Acccumulation Feb 1 at 20:01
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Think of the gravitational field as curved space. Like a bump in space time. This bump will stay constant over time more or less. The planet then just looks at the bump and moves accordingly. However if the sun were to explode it might cause a small ripple in spacetime that would reach the earth in 8 minutes

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The earth moves in the static, unchanging gravitational field of the sun. Its potential has the form $V(r) = -G_N M/r$. This field acts on all objects exactly the same way, and the force is equal to the gradient of this potential times mass of the object: $F=mg$ and $ g= -{\partial V \over \partial r}$. The gravity, g, has magnitude and a direction and it points toward the sun! Why not accept this simple, correct, standard answer?

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    $\begingroup$ Wait, you criticize other answers for being based merely on pop science (including the ones that cite non-pop-science works), and this is your answer? There's not even any discussion of time dependence in it. It doesn't talk about propagation of gravity, at any speed, which was the point in question, at all. $\endgroup$ – Obie 2.0 Feb 3 at 12:50
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    $\begingroup$ I mean, it's right, but there's a reason it's right, which is what the question is getting at. And there are limitations on how right it is, too.... $\endgroup$ – Obie 2.0 Feb 3 at 12:58
  • $\begingroup$ One could also ask the same question about the instantaneous Coulomb electrostatic interaction, and the bunch of people who are pretending to answer would not know the right answer in that case either. And that is what, 3rd year of college physics? Gravity is typically a graduate level course. $\endgroup$ – Kostas Feb 4 at 10:40
  • $\begingroup$ One could. And the explanation for both equivalent situations is in the answer linked in my comment. But you don't have any of that here.... $\endgroup$ – Obie 2.0 Feb 4 at 11:01
  • $\begingroup$ Your comment links to a long winded answer by tparker in which he ultimately says: "It turns out that at this order the problem becomes mathematically equivalent to a familiar problem in special relativity: that of the electric field generated by a charge moving at constant relativistic velocity. It's a standard result from classical E&M (see, for example, Purcell) that in this situation, the electric field actually points exactly to the particle's present position, not to its retarded position as one might expect." That is true, but if you count that as an answer, please yourself. $\endgroup$ – Kostas Feb 4 at 11:13
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In Newton's theory of gravity, the gravitational force is defined as being proportional to the product of the two attracting masses and inversely proportional to the square of the distance between the masses. The force is instantaneous. As Newton said himself, this is philosophically unsatisfying - for example, at any instant, how does Earth know how much mass the Sun has, how does it know how far away from the Sun it is, to know what gravitational force is being applied? But Newton's theory works and is very accurate.

It is in the theory of General Relativity that signals can't travel faster than the speed of light. This is carried over from the Special Theory of Relativity along with the difficult to visualise idea of 4 dimensional space-time. In outline, General Relativity defines gravity in a different way, there is no force of gravity as such. Bodies like Earth follow a 'straight' path (or geodesic) through space-time. Far from gravitational sources space-time is flat, at a gravitational source space-time is stretched or squeezed, in between the space-time must smoothly transition from squeezed/stretched to flat. Bodies like the Earth follow the straightest path through space-time at each local point. [Note, General Relativity does not say that mass (or strictly speaking the total mass/energy) causes space-time to squeeze/stretch but defines the correlational between the two).

To good approximation the Sun is symmetrical, and doesn't change over short time spans, and the space-time around it is not squeezed or stretched much (it is a relatively weak gravitational source). This means that the shape of space-time near the Sun is static and there is no issue with changes to the space-time having to be propagated from the Sun to the location of the Earth. In General Relativity the Earth follows the 'straight' line through space-time locally and so avoids the philosophical problems in Newton's theory.

I think this is an interesting question, it draws out that different theories define physical properties in different ways (gravity in this case), that although Newton's theory of gravity seems simple on the surface it can be, at least philosophically, hard to understand.

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The sun exerts gravitational waves in all directions, not just at other bodies of mass like the Earth. When the Earth arrives at some point in space, it is attracted to the Sun by gravitational waves of the Sun that are already there.

For example, say the Earth is at point A and will arrive at point B in 1 minute. When it gets at point B, it will be pulled by gravitational waves that have already been traveling for 7 minutes from the Sun to arrive at point B.

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    $\begingroup$ I don't think that's what gravitational waves are. Aren't waves the propagation of a fluctuation in gravitational field? It's just the field that we're talking about, and a relatively stable one at that. (not a physicist!) $\endgroup$ – Lightness Races in Orbit Feb 3 at 12:59
  • $\begingroup$ What Lightness said. Only quadrupole (and higher) moments can radiate gravitationally. So, for example, a perfect sphere emits no gravitational waves, but it certainly has a gravitational field. $\endgroup$ – PM 2Ring Feb 3 at 13:23
  • $\begingroup$ @Lightness Races in Oribt No you're right, I was confused. Big fan btw! $\endgroup$ – Inertial Ignorance Feb 3 at 16:30
  • $\begingroup$ Well get in where do I collect my PhD :D $\endgroup$ – Lightness Races in Orbit Feb 3 at 17:06

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