<blockquote>
<p>Would this not mean that information travels infinitely fast and thus contradict Relativity?</p>
</blockquote>
<p>Look more closely at equation 27 in the referenced document. I'll simplify this as
<span class="math-container">$$\boldsymbol{\mathrm a} = \left(\sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}} \Bigl(1 + \text{other terms}\Bigr)\right) + \text{another term} + \text{yet another term}$$</span></p>
<p>Assuming Newtonian gravity, the &quot;other terms&quot;, &quot;another term&quot;, and &quot;yet another term&quot; all vanish, simplifying the equation to
<span class="math-container">$$\boldsymbol{\mathrm a} = \left(\sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}} \Bigl(1 + 0\Bigr)\right) + \boldsymbol{\mathrm 0} + \boldsymbol{\mathrm 0}$$</span>
or just
<span class="math-container">$$\boldsymbol{\mathrm a} = \sum_{B \ne A}\frac{GM_B\,(\boldsymbol{\mathrm r}_B - \boldsymbol{\mathrm r}_A)}{r_{AB}^{\,3}}$$</span></p>
<p>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.</p>

CC BY-SA 4.0
2020-09-13T07:53:52.773Z
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.
Thomas
CC BY-SA 4.0
2020-09-13T08:12:52.697Z
@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.
Harry Johnston
CC BY-SA 4.0
2020-09-13T08:29:16.910Z
@Harry Johnston,  But the equation we are discussing here would incorrectly predict that other planets would be affected instantly.
Thomas
CC BY-SA 4.0
2020-09-13T08:30:19.457Z
@Thomas, yes, in any such scenario the equation gives the wrong answer.
Harry Johnston
CC BY-SA 4.0
2020-09-13T08:47:47.073Z
@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.
David Hammen
CC BY-SA 4.0
2020-09-13T08:48:18.927Z
@Thomas - Think of it this way: for small values of $x$, the sine of x is approximately 0. A better approximation is that $\sin(x) \approx x$. An even better approximation is that $\sin(x) \approx x - x^3/6$. And so on. These truncations of an infinite series is similar to what is done in parameterized post-Newtonian expansions of general relativity.
David Hammen
CC BY-SA 4.0
2020-09-13T16:36:36.537Z
@David Hammen , I am aware that there are different stages of approximation to the full relativiistic equation. I am also aware that the equation (27) in that DE430 publication (which is what they use for their ephemerides calculations) only holds in the weak field limit, appropriate for the solar system. I just would like to know whether this equation takes, explicitly or implicitly, any retardation of the gravitational interaction into account (analogously to the Lienard-Wiechert potential/fields in electrodynamics)
Thomas
CC BY-SA 4.0
2020-09-14T11:39:50.513Z
@Thomas - Please read [Aberration and the Speed of Gravity](https://arxiv.org/abs/gr-qc/9909087). 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."
David Hammen
CC BY-SA 4.0
2020-09-12T23:21:48.577Z
2020-09-12T23:21:48.577Z
David Hammen
<p>Short answer is &quot;no&quot;.</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>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 &quot;Light time delayed Newtonian gravity&quot;.  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.</p>

CC BY-SA 4.0
2020-09-12T21:16:13.207Z
2020-09-13T09:40:32.357Z
2020-09-13T09:40:32.357Z
James K
<p><a href="https://space.stackexchange.com/a/21472/12102">This excellent answer</a> to <em>Besides retarded gravitation, anything else to worry about when calculating MU69's orbit from scratch?</em> 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.</p>
<p>Many answers to <a href="https://space.stackexchange.com/q/23408/12102">How to calculate the planets and moons beyond Newtons's gravitational force?</a> include a way to treat this problem using a well-accepted <em>approximate treatment</em> of general relativity.</p>
<p>From <a href="https://space.stackexchange.com/a/23409/12102">this answer</a>:</p>
<blockquote>
<p>The acceleration of a body in the gravitation field of another body of <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="nofollow noreferrer">standard gravitational parameter</a> <span class="math-container">$GM$</span> can be written:</p>
<p><span class="math-container">$$\mathbf{a_{Newton}} = -GM \frac{\mathbf{r}}{|r|^3},$$</span></p>
<p>where <span class="math-container">$r$</span> is the vector from the body <span class="math-container">$M$</span> 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 <em>the other body</em>, even though the force depends on both masses, because the first mass cancels out by <span class="math-container">$a=F/m$</span>.</p>
</blockquote>
<p>and later:</p>
<blockquote>
<p>The following approximation should be <em>added to</em> the Newtonian term:</p>
<p><span class="math-container">$$\mathbf{a_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right),$$</span></p>
</blockquote>

CC BY-SA 4.0
2020-09-13T08:23:34.690Z
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.
Thomas
CC BY-SA 4.0
2020-09-13T08:43:19.793Z
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.
David Hammen
CC BY-SA 4.0
2020-09-13T08:58:11.407Z
@DavidHammen yes indeed! If you check the [answer](https://space.stackexchange.com/a/23409) that is my link here, it says "You will see various forms in the following links, *most with additional terms not shown here*: "Eq. 27 in https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf" Over the years I've learned to be more cautious when writing about physics and math thanks mostly to your tuition. In these I think I've gotten the wording right.
uhoh
CC BY-SA 4.0
2020-09-13T16:20:57.480Z
@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.
Thomas
CC BY-SA 4.0
2020-09-13T16:25:08.003Z
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 https://physics.stackexchange.com/questions/377707/electric-field-of-uniformly-moving-charge-misprint-in-feynman-textbook-28-3 )
Thomas
CC BY-SA 4.0
2020-09-13T17:14:47.223Z
@Thomas you're now way over my head!
uhoh
CC BY-SA 4.0
2020-09-13T17:15:16.087Z
@DavidHammen I thought your comment was meant for me. I got a notification and there's no `@replies` on it
uhoh
CC BY-SA 4.0
2020-09-14T11:42:23.497Z
@uhoh - My most recent comment (and only comment until now) **was** meant for you. My next comment is meant for Thomas (the OP).
David Hammen
CC BY-SA 4.0
2020-09-14T11:45:57.377Z
@Thomas - That is not how gravitation works. Gravitation changes the shape of space-time while electromagnetism (by itself) does not. Incorporating a retarded position into gravitation does not work.
David Hammen
CC BY-SA 4.0
2020-09-14T20:50:11.083Z
@David Hammen If one additionally ignores the mass term in the bracket for the correction terms above (i.e. for a Eucldean/Minkowski metric) , one should get a form equivalent to the Lienard-Wiechert potential in electrodynamics, but the latter always points to the instantaneous position for a constant velocity (see the diagram in physics.stackexchange.com/questions/377707/… ) whereas the equation above has a term pointing in the direction of the velocity vector (albeit only of second order in v/c)
Thomas
CC BY-SA 4.0
2020-09-14T21:42:09.793Z
@David Hammen -//"  Incorporating a retarded position into gravitation does not work "// The very paper by Carlip you referenced in your comment to your own answer above does in fact explicitly speak about retarded positions for gravity as well (Section 2) " In other words, the gravitational acceleration is directed toward the retarded position of the source quadratically extrapolated toward its ‘instantaneous’ position, up to small nonlinear terms and corrections of higher order in velocities"
Thomas
CC BY-SA 4.0
2020-09-13T00:25:46.337Z
2020-09-13T00:30:51.007Z
2020-09-13T00:30:51.007Z
uhoh
<p>The corrections to the gravitational force due to finite propagation speed are much smaller than you'd expect.</p>
<p>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.</p>
<p>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 &quot;told&quot; to do so by the central mass.</p>
<p>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.</p>

CC BY-SA 4.0
2020-09-14T02:20:53.013Z
2020-09-14T02:20:53.013Z
benrg
<p>The equation of motion used for the calculation of orbits of solar system objects (Eq(27) in <a href="https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf" rel="noreferrer">https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf</a> ) is formulated in terms of instantaneous variables, that is the acceleration of an object is assumed to depend on the <strong>instantaneous</strong> 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?</p>

CC BY-SA 4.0
2020-09-12T21:12:57.890Z
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?
Deschele Schilder
CC BY-SA 4.0
2020-09-13T05:18:12.840Z
@DescheleSchilder I assume so and have suggested that [here](https://astronomy.stackexchange.com/a/38851/7982).
uhoh
CC BY-SA 4.0
2020-09-13T10:08:20.967Z
@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.
Deschele Schilder
CC BY-SA 4.0
2020-09-13T19:55:03.483Z
@DescheleSchilder, you mean, apart from the fact that Newtonian gravity with time delay gives a very wrong answer? :-)
Harry Johnston
CC BY-SA 4.0
2020-09-13T20:50:53.523Z
@HarryJohnston Apart from that indeed... ;-)
Deschele Schilder
CC BY-SA 4.0
2020-09-13T22:48:23.350Z
@DescheleSchilder, I have the uninformed impression that the most important difference is that in GR the gravitational field of a moving object sort of acquires that object's velocity.  The simplest Newtonian-like rule that behaves similarly would be for the gravitational force to point towards where the object would be now if it had travelled at constant velocity during the time it took light from the object to reach you.
Harry Johnston
CC BY-SA 4.0
2020-09-14T12:57:46.073Z
@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.
David Hammen
<gravity><ephemeris>
Instantaneous gravity in equation of motion for ephemerides calculation
Thomas
# Instantaneous gravity in equation of motion for ephemerides calculation

(Asked by: username_0 on 2020-09-12)

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?

**Tags:** gravity, ephemeris

> **Comments:**
> **username_1**: 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?  
> **username_2**: @username_1 I assume so and have suggested that [here](https://astronomy.stackexchange.com/a/38851/7982).  
> **username_1**: @username_2 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.  
> **username_3**: @username_1, you mean, apart from the fact that Newtonian gravity with time delay gives a very wrong answer? :-)  
> **username_1**: @username_3 Apart from that indeed... ;-)  
> **username_3**: @username_1, I have the uninformed impression that the most important difference is that in GR the gravitational field of a moving object sort of acquires that object's velocity.  The simplest Newtonian-like rule that behaves similarly would be for the gravitational force to point towards where the object would be now if it had travelled at constant velocity during the time it took light from the object to reach you.  
> **username_4**: @username_1 - 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.  

---

## 4 answers

### Answer by username_4 on 2020-09-12. Score: 7

> 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.

> **Comments:**
> **username_0**: 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.  
> **username_3**: @username_0, 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.  
> **username_0**: @username_3,  But the equation we are discussing here would incorrectly predict that other planets would be affected instantly.  
> **username_3**: @username_0, yes, in any such scenario the equation gives the wrong answer.  
> **username_4**: @username_0 - 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.  
> **username_4**: @username_0 - Think of it this way: for small values of $x$, the sine of x is approximately 0. A better approximation is that $\sin(x) \approx x$. An even better approximation is that $\sin(x) \approx x - x^3/6$. And so on. These truncations of an infinite series is similar to what is done in parameterized post-Newtonian expansions of general relativity.  
> **username_0**: @username_4 , I am aware that there are different stages of approximation to the full relativiistic equation. I am also aware that the equation (27) in that DE430 publication (which is what they use for their ephemerides calculations) only holds in the weak field limit, appropriate for the solar system. I just would like to know whether this equation takes, explicitly or implicitly, any retardation of the gravitational interaction into account (analogously to the Lienard-Wiechert potential/fields in electrodynamics)  
> **username_4**: @username_0 - Please read [Aberration and the Speed of Gravity](https://arxiv.org/abs/gr-qc/9909087). 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."  

---

### Answer by username_5 on 2020-09-12. Score: 5

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.


---

### Answer by username_2 on 2020-09-13. Score: 3

[This excellent answer](https://space.stackexchange.com/a/21472/12102) 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?](https://space.stackexchange.com/q/23408/12102) include a way to treat this problem using a well-accepted *approximate treatment* of general relativity.

From [this answer](https://space.stackexchange.com/a/23409/12102):

> The acceleration of a body in the gravitation field of another body of [standard gravitational parameter](https://en.wikipedia.org/wiki/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:
>
> $$\mathbf{a\_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right),$$

> **Comments:**
> **username_0**: 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.  
> **username_4**: 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.  
> **username_2**: @username_4 yes indeed! If you check the [answer](https://space.stackexchange.com/a/23409) that is my link here, it says "You will see various forms in the following links, *most with additional terms not shown here*: "Eq. 27 in https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf" Over the years I've learned to be more cautious when writing about physics and math thanks mostly to your tuition. In these I think I've gotten the wording right.  
> **username_0**: @username_2 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.  
> **username_0**: 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 https://physics.stackexchange.com/questions/377707/electric-field-of-uniformly-moving-charge-misprint-in-feynman-textbook-28-3 )  
> **username_2**: @username_0 you're now way over my head!  
> **username_2**: @username_4 I thought your comment was meant for me. I got a notification and there's no `@replies` on it  
> **username_4**: @username_2 - My most recent comment (and only comment until now) **was** meant for you. My next comment is meant for username_0 (the OP).  
> **username_4**: @username_0 - That is not how gravitation works. Gravitation changes the shape of space-time while electromagnetism (by itself) does not. Incorporating a retarded position into gravitation does not work.  
> **username_0**: @username_4 If one additionally ignores the mass term in the bracket for the correction terms above (i.e. for a Eucldean/Minkowski metric) , one should get a form equivalent to the Lienard-Wiechert potential in electrodynamics, but the latter always points to the instantaneous position for a constant velocity (see the diagram in physics.stackexchange.com/questions/377707/… ) whereas the equation above has a term pointing in the direction of the velocity vector (albeit only of second order in v/c)  
> **username_0**: @username_4 -//"  Incorporating a retarded position into gravitation does not work "// The very paper by Carlip you referenced in your comment to your own answer above does in fact explicitly speak about retarded positions for gravity as well (Section 2) " In other words, the gravitational acceleration is directed toward the retarded position of the source quadratically extrapolated toward its ‘instantaneous’ position, up to small nonlinear terms and corrections of higher order in velocities"  

---

### Answer by username_6 on 2020-09-14. Score: 2

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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