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Imagine two massive objects, with the same mass (M) circling around their center of mass (C.M.). Let's assume that the distance between them is 1 light hour. Don´t the two bodies get accelerated and move away from each other because they feel the gravity of each other as it was 1 hour ago, because of which a force tangent to the direction of the speed develops?
Can it be that gravity acts instantaneously? I've read about experiments proving the speed of gravity to be equal to the speed of light, but these were disputed.
Isn't there both in Newtonian mechanics and GR there a time delay? The difference being though that space and time in Newtonian mechanics are separate and absolute (implying instantaneity?), while in GR the both are a whole and absolute.

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    $\begingroup$ Gravity isn't a force in GR. Gravity is a force in Newtonian mechanics and Newtonian gravity makes incorrect detailed predictions about orbits. I'm unclear what kind of answer you are expecting. In GR the bodies do not move away from each other (ignoring any tidal effects), they move closer together because the system radiates gravitational waves. $\endgroup$
    – ProfRob
    Sep 13, 2020 at 11:38
  • $\begingroup$ @RobJeffries So the answer is no? Isn't the so-called force transmitted by gravitons in string theory, in a flat spacetime? Which is the same mechanism as the transmission of other forces? So, actually, we can speak of a force? Mass-energy (the energy-momentum tensor) being the charge? $\endgroup$
    – Deschele Schilder
    Sep 13, 2020 at 18:42
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    $\begingroup$ I am not sure the position where the other mass was 1 hour ago is even well-defined. It depends on the speed of the frame of reference. $\endgroup$
    – Florian F
    Sep 13, 2020 at 20:35

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Can two heavy objects circling around their C.M. be separated because of the speed of gravity?

No. Newtonian mechanics does quite well assuming instantaneous gravity. You can get relativistically correct orbits either by doing a complete GR calculation, or using a simple approximation as already eluded alluded to in @antispinward's answer.

This excellent answer to Besides retarded gravitation, anything else to worry about when calculating MU69's orbit from scratch? in Space Exploration SE explains that we get the wrong answer if purely Newtonian mechanics is used except for slowing down the speed of gravity, but that's because one is neither treating gravity correctly nor using Newtonian mechanics correctly.

Many answers to How to calculate the planets and moons beyond Newtons's gravitational force? include a way to treat this problem using a well-accepted approximate treatment of general relativity.

From this answer:

The acceleration of a body in the gravitation field of another body of standard gravitational parameter $GM$ can be written:

$$\mathbf{a_{Newton}} = -GM \frac{\mathbf{r}}{|r|^3},$$

where $r$ is the vector from the body $M$ to the body who's acceleration is being calculated. Remember that in Newtonian mechanics the acceleration of each body depends only on the mass of the other body, even though the force depends on both masses, because the first mass cancels out by $a=F/m$.

and later:

The following approximation should be added to the Newtonian term:

$$\mathbf{a_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right),$$

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    $\begingroup$ If I could give +10 (I mean 10 upvotes), I would! $\endgroup$
    – Deschele Schilder
    Sep 13, 2020 at 10:31
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    $\begingroup$ By the way, I like what you said about the White Cube... $\endgroup$
    – Deschele Schilder
    Sep 13, 2020 at 10:34
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    $\begingroup$ Sigh. Foiled again. A certain Vonnegut quote comes to mind: "“Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. For he is full of murderous resentment toward the unlearned who have come by their ignorance the easy way." $\endgroup$
    – Tom Russell
    Sep 14, 2020 at 2:57
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    $\begingroup$ @TomRussell see also a related question at Physics.SE: Virtual Photon transmission speed of a Static Electric Field? $\endgroup$
    – Ruslan
    Sep 14, 2020 at 7:39
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    $\begingroup$ "a simple approximation as already eluded to" alluded $\endgroup$
    – Acccumulation
    Sep 16, 2020 at 3:29
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Patching a finite speed of gravity on top of Newtonian physics, as you have done in your example, doesn't work: it gives predictions that are inconsistent with observation, among which is that a two-body system would become rapidly unstable through the mechanism you depicted.

You need to use general relativity to analyse this situation, and it turns out that when you do that, the physics (at least in the regime where gravity is weak) almost exactly compensates for the effect of a finite speed of gravity.

As noted in this article by Steve Carlip, a similar effect occurs in electromagnetism, where the force is directed towards a charge despite the force being propagated at the speed of light.

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  • $\begingroup$ The compensation is not exact, which can be observed in certain systems, for example the anomalous precession of Mercury's orbit Isn't the spacetime curvature "around" the Sun static? Which means that the speed of gravity is not the cause of Mercury's orbit? Isn't the cause simply the distorted spacetime? $\endgroup$
    – Deschele Schilder
    Sep 12, 2020 at 22:14
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    $\begingroup$ @DescheleSchilder - Distorted spacetime is gravity, and the Sun is also affected by Mercury's gravity. In any case, the point is that general relativity does not yield the exact results of Newtonian physics, and if you want to incorporate the finite speed of gravity, you have to use general relativity not Newton+slow gravity. And to be honest, general relativity makes my head hurt in the case of a single mass, a two-body system with anything other than a test particle is a nightmare. $\endgroup$
    – user24157
    Sep 12, 2020 at 22:23
  • $\begingroup$ But how does the speed of gravity make the orbit of Mercury precess? $\endgroup$
    – Deschele Schilder
    Sep 12, 2020 at 22:51
  • $\begingroup$ @DescheleSchilder - it's a consequence of general relativity, which you need to use in order to incorporate a finite speed of gravity because Newton+slow gravity does not match observations. Nevertheless my reference is clearly causing confusion and leading to this going off-topic from the original question, so I've removed it from my answer. $\endgroup$
    – user24157
    Sep 12, 2020 at 22:53
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It is not only possible, we have almost this situation in our neighborhood. Alpha Centauri only 4 lightyears away with two Stars. Alpha Centauri A and B. A has 1.1 times the mass of the sunn and B 0.9. This is very similar. And the distance is is 11 AU. 1.5 ~ 90 Lightminutes.

Alpha Centauri Wikipedia

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    $\begingroup$ This doesn't really answer the question, which is about the direction of the gravitational force when gravity has a finite speed. The Alpha Centauri binary is not becoming unbound due to the finite speed of gravity. $\endgroup$
    – user24157
    Sep 14, 2020 at 6:08

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