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The equation of motion used for the calculation of orbits of solar system objects (Eq(27) in https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf ) is formulated in terms of instantaneous variables, that is the acceleration of an object is assumed to depend on the instantaneous positions, velocities and accelerations of all other objects in the solar system (this has been confirmed to me by the authors of this publication). Would this not mean that information travels infinitely fast and thus contradict Relativity?

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  • $\begingroup$ Hi! One thing. In calculating the orbit of the Earth around the Sun, is an instantaneous influence implied in Newton's equation for gravitational attraction? $\endgroup$ – Deschele Schilder Sep 12 '20 at 21:12
  • $\begingroup$ @DescheleSchilder I assume so and have suggested that here. $\endgroup$ – uhoh Sep 13 '20 at 5:18
  • $\begingroup$ @uhoh I can't see the difference between Newtonian gravity with time delay and GR. Of course, space and time are absolute in Newtonian mechanics, while in GR it's spacetime that's absolute. But nevertheless. $\endgroup$ – Deschele Schilder Sep 13 '20 at 10:08
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    $\begingroup$ @DescheleSchilder, you mean, apart from the fact that Newtonian gravity with time delay gives a very wrong answer? :-) $\endgroup$ – Harry Johnston Sep 13 '20 at 19:55
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    $\begingroup$ @DescheleSchilder - Newton's law of gravitation does indeed assume instantaneous action at a distance. Laplace proved in 1805 that if all one did to Newton's law of gravitation was to add a delay, the speed of gravity would have to exceed the speed of light by a factor of about 10<sup>7</sup> lest the solar system quickly become unstable. General relativity does a whole lot more than say that gravitational waves travel at the speed of light. The delay you are seeking essentially does not exist. $\endgroup$ – David Hammen Sep 14 '20 at 12:57
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Would this not mean that information travels infinitely fast and thus contradict Relativity?

Look more closely at equation 27 in the referenced document. I'll simplify this as $$\boldsymbol{\mathrm a} = \left(\sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}} \Bigl(1 + \text{other terms}\Bigr)\right) + \text{another term} + \text{yet another term}$$

Assuming Newtonian gravity, the "other terms", "another term", and "yet another term" all vanish, simplifying the equation to $$\boldsymbol{\mathrm a} = \left(\sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}} \Bigl(1 + 0\Bigr)\right) + \boldsymbol{\mathrm 0} + \boldsymbol{\mathrm 0}$$ or just $$\boldsymbol{\mathrm a} = \sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}}$$

All of those extra terms in equation 27 in the referenced document are due to a linearization of general relativity assuming that distances are very large compared to the Sun's Schwarzschild radius and that velocities are very small compared to the speed of light.

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  • $\begingroup$ Yes, but all the 'extra' terns on the right hand side of the equation are evaluated at the same time for which you want to know the acceleration of your test mass. So if any of these terms suddenly changes for some reason, the acceleration of your test mass changes instantly as well. $\endgroup$ – Thomas Sep 13 '20 at 7:53
  • $\begingroup$ @Thomas, the derivation of those equations assumes that nothing unexpected happens. If the Death Star flew by and blew up Jupiter, the gravitational effect of Jupiter on the other planets wouldn't actually change instantaneously, there would be a speed-of-light delay. $\endgroup$ – Harry Johnston Sep 13 '20 at 8:12
  • $\begingroup$ @Harry Johnston, But the equation we are discussing here would incorrectly predict that other planets would be affected instantly. $\endgroup$ – Thomas Sep 13 '20 at 8:29
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    $\begingroup$ @Thomas - Re But the equation we are discussing here would incorrectly predict that other planets would be affected instantly. No, they don't. They represent an extremely good approximation of the hairy mess of general relativity. Even Newton's law of gravitation is a fairly good approximation in many circumstances. $\endgroup$ – David Hammen Sep 13 '20 at 8:47
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    $\begingroup$ @Thomas - Please read Aberration and the Speed of Gravity. The abstract reads "The observed absence of gravitational aberration requires that “Newtonian” gravity propagate at a speed $c_g > 2 \times 10^{10}c$. 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 $c_g = c$. This cancellation is dictated by conservation laws and the quadrupole nature of gravitational radiation." $\endgroup$ – David Hammen Sep 14 '20 at 11:39
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Short answer is "no".

I can model gravity by Newton's law of gravitation, and it gives extremely accurate results in most situations. Even though it is an approximation to the more accurate model of gravity in General Relativity, it is accurate enough to predict the location of the planets well enough for most purposes.

The actual models used are based on Kepler's laws (which can be derived from Newton's law with two point sources of gravity) + perturbations (interactions with other bodies, effects of non-spherical bodies, and relativistic effects). However rather than attempt to solve the full General Relativity for the solar system, relativistic effects are treated as a perturbation of the Newtonian orbits.

General relativity is not used naively, as it compuatationally hard. Direct numerical solution of the equations of spacetime in the vicinity of the sun takes too long, and doesn't give substantially better answers.

We choose a model because it usefully predicts some aspect of reality. But I hope it is obvious that the choice of model doesn't change reality. If I choose a model that neglects friction, or bending, or some other aspect, that doesn't mean that friction doesn't exist, only that my model may be somewhat less accurate than a more complex model that does incorporate these aspects.

The fact that a model of the solar system is based on the instantaneous positions of the planets doesn't mean relativity is wrong, merely that relativity (in its full details) is not needed to predict the postions of the planets over the next few thousand years, to more accuracy than you will ever need.

It might be worth noting that in gravitational systems such as the solar system (with fairly weak gravity compared to black holes) Instantaneous Newtonian Gravity is actually a much better model of General Relativity than "Light time delayed Newtonian gravity". It is a better model in the sense that its predictions are much closer to reality. When you introduce General Relativity and gravity acts at the speed of light, there are other terms that almost (but not quite) cancel the effect of a finite speed of light, so it almost (but not quite) appears as if gravity acts instantly.

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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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  • $\begingroup$ Where does your last expression exactly com from? In Eq.(27) of the DE430 publication I linked to in my opening post, the acceleration vector has only components along the (instantaneous) radius vector, not the velocity vector. $\endgroup$ – Thomas Sep 13 '20 at 8:23
  • $\begingroup$ The approximation in this answer is much simpler than that of equation 27 in the referenced publication. However, both the approximation in this answer and the approximation used in the referenced document are just that: approximations. $\endgroup$ – David Hammen Sep 13 '20 at 8:43
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    $\begingroup$ @uhoh Sorry, I noticed now that I made a mistake when evaluating Eq.(27) in that reference (I got confused by the many brackets there). If you set there position, velocity and acceleration of mass B =0, you get in fact the expression quoted by you above. However, since the resultant acceleration is not exactly along the radius vector, ir means there is an aberration (albeit only second order in v/c), but the authors of the DE430 ephemerides publication assured me that they ignore any aberration in their calculations. $\endgroup$ – Thomas Sep 13 '20 at 16:20
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    $\begingroup$ Also, if one additionally ignores the mass term in the bracket above, one should get a form equivalent to the Lienard-Wiechert potential in electrodynamics, but this always points to the instantaneous position for a constant velocity (see the diagram in physics.stackexchange.com/questions/377707/… ) $\endgroup$ – Thomas Sep 13 '20 at 16:25
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    $\begingroup$ @uhoh - My most recent comment (and only comment until now) was meant for you. My next comment is meant for Thomas (the OP). $\endgroup$ – David Hammen Sep 14 '20 at 11:42
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The corrections to the gravitational force due to finite propagation speed are much smaller than you'd expect.

This is for a couple of reasons. First, the gravitational field of a uniformly moving body doesn't lag behind its motion—that would violate the principle of relativity. If you push a semi-rigid object through the air at a constant speed, the parts you aren't pushing will bend and lag behind, but that's only because of the force of the wind. If you try the same thing in vacuum where there's no wind, you won't have to exert any force with your finger to keep the object moving, and it will return to its preferred shape.

Second, everything is affected equally by gravity, including the gravitational field itself. So when a bunch of bodies accelerate under their mutual gravitational influence, their gravitational fields accelerate along with them, without being "told" to do so by the central mass.

The finite speed of gravity can often be neglected because the field is very good at guessing where it ought to be, especially when the sources of the field are moving only under the influence of gravity.

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