I have previously encountered a problem with the equation used by Yeomans and others for determining the gravitational acceleration of comets (see update in my answer to the previous question Can General Relativity indicate phase-dependent variations in planetary orbital acceleration?).

Basically I find that, for hypothetical circular orbits, the predicted direction of radial acceleration is opposite to that predicted by other formulae such as those of Goldstein and Walter. And this would mean that the "Yeomans" equation does not explain the observed Non-Newtonian apsidal rotation (orbital perihelion precession) of the Solar System planets.

Anderson et al 1975 presented two Equations 11 and 12 for the Newtonian and Relativistic accelerations affecting a satellite.

The following equation 3.11 is from Shahid-Saless and Yeomans 1994. It gives the gravitational acceleration of a single body under the influence of another single "source" body. Here $\mu = GM_{Source}/c^2.$ This equation is a selective composite of relevant parts of Equations 11 and 12 from Anderson et al 1975.

Equation 3.11 of Shahid_Saless & Yeomans 1994

Equation 3.11 can be re-written as (and here denoted by Equation 1):- $$\textbf{a} = \frac{-GM.\textbf{r}}{r^3} \left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2}{c^2} \right) + \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v} $$

Note I use different Mathjax fonts to distinguish between scalars $r,v$ and vectors $\textbf{r},\textbf{v}$.

More recently the following Equation 27 is reported in Park et al 2021, The JPL Planetary and Lunar Ephemerides DE440 and DE441. It is similar in parts to the Yeomans Equation but it accounts for the gravitational effects of multiple sources. Note that in Equation 2 of Parks et al, the term $r_{ij}=|\textbf{r}_i-\textbf{r}_j|$ is defined as an unsigned scalar value representing the distance between bodies $i$ and $j$.

D440-D441 Ephemerides Eqtn 27

A very similar equation is presented in Descanso2 Section 4-26. It is based on work by Moyer (1971) and Will(1981). Here $\mu = GM_{body}.$ Note that the typescript in this image makes it easier to tell the difference between vectors and scalars than in the image of the Park et al 2021 Equation 27.

enter image description here

Note: I am aware of important, fundamental differences between Newtonian and General Relativity models of nature (see contribution by /u/Stan Liou/ in this question ). In the following I am neglecting a discussion of these and thinking rather in terms of a Newtonian Model with additional small "exotic" accelerations whose Newtonian pattern is suggested by General Relativity (Slow + Weak regime). I don't think this neglect affects significantly the core problem here, which is the magnitude and direction of the Non-Newtonian acceleration vector and whether or not it explains the observed Non-Newtonian precessions of the planetary orbits.

The Park et al eqtn 27 can be applied to the imaginary case of a test particle (i) in an orbit (with orbital radius and speed similar to a terrestrial planet) around the Sun where the Sun (j) is the only gravitational source in the system. In General Relativity the terms $\beta$ and $\gamma$ are equal to 1. By taking the inertial frame and coordinate system where the position ($r_j$) velocity ($v_j$) and acceleration ($a_j$) of the Sun are zero we find that many of terms become zero.

$$\textbf{a} = \frac{-GM.\textbf{r}}{r^3} \left( 1 - \frac{4.GM}{c^2.r} - 0 + \frac{v^2}{c^2} + 0 -0 -0+0 \right) + \frac{GM}{c^2.r^3}\left( [\textbf{r}] \cdot [4.\textbf{v}-0]\right)\textbf{v} +0$$

And so...

$$\textbf{a} = \frac{-GM.\textbf{r}}{r^3} \left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2}{c^2} \right) + \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v}. $$

... which is identical to my Equation 1 derived from Equation 3.11 of Shaid-Saless & Yeomans 1994.

(Note that the third term inside the curly braces of Park et al's Equation 27, the sum of objects $k$ does not exclude object $i$. But I have ignored this term as the mass of object $i$ relatively very small in this case. The omission appears valid when the Non-Newtonian part of the above equation is compared with Equation 20 from Moyer 1971, Mathematical Formulation of the Double Precision Orbit Determination Program DPODP for the motion of a 'massless' test particle around a single massive body.


Now let us consider the case where the orbit is circular, then the test particle velocity ($\textbf{v}$) is perpendicular to its position vector ($\textbf{r}$) and so the dot product is zero and so the final term in Equation 1 goes to zero, leaving:-

$$\textbf{a}_{circ} = \frac{-GM.\textbf{r}}{r^3} \left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2_{circ}}{c^2} \right) $$

In a (Newtonian) circular orbit we have $\frac{GM}{r}=v_{circ}^2$ and so we can write: $$\textbf{a}_{circ} = \frac{-GM.\textbf{r}}{r^3} \left( 1 - \frac{4.v^2_{circ}}{c^2} + \frac{v^2_{circ}}{c^2} \right) $$

Then... $$\textbf{a}_{circ} = \frac{-GM}{r^2} \left( 1 - \frac{3v^2_{circ}}{c^2} \right) \hat{\textbf{r}} $$

Finally $$\textbf{a}_{circ} = -\frac{GM}{r^2} \hat{\textbf{r}} + \frac{GM}{r^2} \frac{3v^2_{circ}}{c^2}\hat{\textbf{r}} $$

On the RHS the first term is the usual Newtonian acceleration (towards the Sun) and the second term is the Non-Newtonian acceleration (away from the Sun). And so we have again the same problem as I had with the Yeoman's equation - the Equation 1 predicts that the Non-Newtonian acceleration for a circular orbit is outward (away from the Sun) which would produce negative precession (retardation) of the orbit rather than the positive precession observed for the terrestrial planets and certain asteroids and predicted by the equations of Goldstein and Walter (referenced in this earlier SE Astronomy question of mine ), effectively:-

$$\textbf{a}_r = -\frac{GM}{r^2} \hat{\textbf{r}} - \frac{GM}{r^2} \frac{3v^2_{circ}}{c^2}\hat{\textbf{r}} $$

However, perfectly circular orbits are a bit special. Just because the Equation 1 and the Goldstein/Walter equation give radically different results for circular orbits does not necessarilly mean that they give radically different results for orbits which are significantly elliptical such as Mercury, Mars and certain asteroids. So we should look at Equation 1 in the case of Elliptical orbits.


Equation 1 derived from the equations presented by both Shahid-Saless and Yeomans 1994 and Park 2021 gives the acceleration of an object in an Elliptical Orbit. In previously restricting it to circular orbits I dropped the term, which I denote as $\textbf{a}_E$:- $$ \textbf{a}_E = \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v} $$

The velocity vector $\textbf{v}$ can be decomposed into the two mutually-perpendicular component velocity vectors $\textbf{v}_r$ and $\textbf{v}_t$ directed radially and transversely respectively. And then we can apply the distributive rule for dot products...

$$ \textbf{a}_E = \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot (\textbf{v}_r + \textbf{v}_t) \right)\textbf{v} = \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}_r + \textbf{r} \cdot \textbf{v}_t) \right)\textbf{v} $$

Now $\textbf{r}$ and $\textbf{v}_r$ are parallel so their dot product has magnitude $|r.v_r|$ and sign $(\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r)=+/-1$ according to whether they are or aren't pointing in the same direction. Whereas $\textbf{r}$ and $\textbf{v}_t$ are perpendicular so their dot product is zero.

$$ \textbf{a}_E = \frac{4GM}{c^2.r^3}\left( |r.v_r|(\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) + 0 \right)\textbf{v} $$

Cancelling $r$ then decomposing $\textbf{v}$... $$ = \frac{4GM}{c^2.r^2}\left( |v_r|(\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) \right)\textbf{v} = \frac{4GM}{c^2.r^2} (|v_r| (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) \textbf{v}_r + |v_r| (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) \textbf{v}_t) $$

We are left with two perpendicular components of acceleration...

$$ \textbf{a}_{Er} = \frac{4GM}{c^2.r^2} |v_r^2| (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) \hat{\textbf{v}}_r $$ $$ \textbf{a}_{Et} = + \frac{4GM}{c^2.r^2} |v_r.v_t| (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) \hat{\textbf{v}}_t $$ Now it can be shown that the radial component $\textbf{a}_{Er}$ (in $v_r^2$) is always directed radially away from the source and does not have a relatively large affect on apsidal rotation for Solar planets.

Whereas the transverse acceleration component $\textbf{a}_{Et}$ (in $v_r.v_t$) points in the same direction as $\hat{\textbf{v}}_t$ during the half-orbit between perihelion and aphelion but points in the direction opposite to $\hat{\textbf{v}}_t$ during the other half-orbit. It has the effect of producing apsidal rotation equal to $(+4/3)\Delta$ where $\Delta$ is the observed Non-Newtonian apsidal rotation for Mercury and Mars.

From previously, the apsidal rotation produced by the rest of Equation 1 is $(-3/3)\Delta$. So the total rotation produced by Equation 1 is $(+1/3)\Delta$ which is a third of the observed planetary orbit rotations.

This compares with the $(+3/3)\Delta$ produced by the equation from Goldstein/Walter which predicts 100% of the observed apsidal rotations.


So my question is: What is the explanation for the "wrong" direction of the relativistic acceleration predicted by Equation 1 which was derived from equation 27 in the Park et al 2021 JPL D440-D441 Ephemerides. Is my analysis wrong or is the equation wrong?

  • 1
    $\begingroup$ I was about to add a bounty to your question but then realized you might be thinking of migrating it to Physics SE or Physics Overflow. If you'd like to keep it here for the next 7+1 days let me know and I'll add a bounty to it. $\endgroup$
    – uhoh
    Jan 21, 2023 at 1:21
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    $\begingroup$ @uhoh Very kind thanks but I think I have resolved the primary issue and have posted an answer to that effect, which I might then accept. So I'm not sure about the etiquette of accepting a bounty. I'm not presently thinking of moving the question to another forum. But if you would like to attract better answers than my own by starting a bounty then dont let me stop you :) . $\endgroup$
    – steveOw
    Jan 22, 2023 at 1:01
  • $\begingroup$ Okay thanks for the update and congratulations on the progress! $\endgroup$
    – uhoh
    Jan 22, 2023 at 1:18

3 Answers 3


If you switch to polar coordinates and use the classical Newtonian gravitation of $GM/r^2$ (inwards) you get a centrifugal term of $v^2/r$ (outwards) Since these two terms must cancel out for a circular orbit you end up with $v=\sqrt(GM/r)$. This is not the case using your first equation called $a_{circ}$ above. Instead you will get:


This will get you: $v=\sqrt{{GM\over{r}}({1-4GM/rc^2)\over{1-GM/rc^2}}}$

for a circular orbit. I do not know if this helps you? I think you are making an "illegal" substitution.

If you throw (3.11) into a numerical integrator you will get the right value quite precisely for the weak field precession we have in our solar system. The time it takes to complete a year (revolution around the Sun) will be longer using PPN than classical Newton (somewhat less than half a second for the Earth if I remember correctly.)

  • 1
    $\begingroup$ The (purely radial) acceleration formula obtained from Goldstein and Walter is $a_r = (GM/r^2)(1+3v^2/c^2)\hat{r}$. The NN (Non-Newtonian) factor is $(+3v^2/c^2)$. Equation 3.11 (my analysis for circular orbits) gives an NN factor of $(-3v^2/c^2)$ which will produce "precession" of the same magnitude but in the opposite direction (i.e. negative precession or retardation). They cannot both be correct (for circular orbits) . (...continued on next comment) $\endgroup$
    – steveOw
    Jan 18, 2023 at 12:49
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    $\begingroup$ For elliptical orbits:- the "Goldstein-Walter" equation gives the correct NN precession for the terrestrial planets (both by numerical integration and analytically); whereas my analysis indicates that Eqtn 3.11 gives the wrong NN precession. I have not reported that analysis so far in order to try and keep things simple. $\endgroup$
    – steveOw
    Jan 18, 2023 at 12:53
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    $\begingroup$ In a Newtonian model, if the planet's orbital apsides are rotating due to an additional (supra-Newonian) centrally-directed (i.e. attractive) force (whose magnitude falls off with distance $r$ as $K/r^{2+q}$ where $K$ is a constant and $q>0$, then the time taken to complete a revolution around the Sun will be less than it would be without that force. This is consistent with what we observe in the Solar System. $\endgroup$
    – steveOw
    Jan 19, 2023 at 4:01
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    $\begingroup$ Your phrase "illegal substitution" got me thinking and I realized I should not be using $(v^2/r)=(GM/r^2)$ for elliptical orbits. Subsequently I have added an answer which (maybe) resolves the primary question. $\endgroup$
    – steveOw
    Jan 22, 2023 at 1:16


There are some errors in this answer but I will leave it posted for now as it contains definitions and sources used in my other (shorter) answer.

In the question I overlooked the fact that $\frac{GM}{r^2} = \frac{v_{circ}^2}{r}$ only applies strictly in the case of circular orbits. And therefore $v^2_{circ}$ should not be used to replace $GM/r$ when using Equation 1 in the case of elliptical orbits.

The effect of $GM/r$ can be ascertained elsewhere. It is known since Newton (see wikipedia:Newton's theorem of revolving orbits ) that the addition of an inverse-distance-cubed (ID3), inward-directed radial acceleration can produce apsidal rotation of an orbit.

Wells 2011 describes how Exotic Apsidal Rotation (EAR, my shorthand term) of an orbit can be produced by adding an exotic radial ID3 acceleration to the standard Newtonian radial ID2 acceleration in the form:-

$$ \textbf{a}_r = \frac{-GM}{r^2} \left(1 + \frac{B.GM}{r.c^2}\right)\hat{\textbf{r}}. $$

My numerical simulator and Wells 2011 equation 22 indicate that a value of $B=6$ is required in order to produce $EAR=\Delta$. ($\Delta$ is the value of EAR observed for Mercury, Mars and other Solar System bodies).

Now let us look at Equation 1 of the question:-

$$\textbf{a} = \frac{-GM.\textbf{r}}{r^3} \left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2}{c^2} \right) + \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v} $$

rationalising $r$ and expanding into four terms:- $$\textbf{a} = \left( \frac{-GM.\hat{\textbf{r}}}{r^2} \right) +\left( \frac{-GM.\hat{\textbf{r}}}{r^2} \right).\left( \frac{-4.GM}{c^2.r} \right) +\left( \frac{-GM.\hat{\textbf{r}}}{r^2} \right). \left( \frac{v^2}{c^2} \right) +\left( \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v} \right) $$

The first term is the standard Newtonian radial acceleration and produces no EAR;

the second term, with $\frac{4.GM}{r.c^2}$, will produce $EAR=(-4/6)\Delta$.

As described in the Question:

the third term $\frac{v^2}{c^2}$ will produce $EAR=(+1/3)\Delta$

and the 4th term will produce net $EAR=(+4/3)\Delta$.

Adding these together gives:- (-4/6) + (1/3) +(4/3) = (3/3).

Therefore, for Elliptical Orbits, Equation 1, predicts 100% of the Exotic Apsidal Rotation.

Still unresolved

What remains unresolved though is that for a Circular orbit:-

(1) Term 2 in Equation 1 will produce different values of EAR according to whether it is interpreted as:-

(1a) $(4.v^2/c^2)$ giving $EAR=(-4/3)\Delta$, or

(1b) $(4.GM/r.c^2)$ giving, $EAR=(-4/6)\Delta$ (based on $B=6$).

(2) The two alternative values of EAR from Equation 1 for circular orbits are both negative, which means that they both predict rotation of circular orbits in the opposite direction to that observed in (and predicted by Equation 1 for) elliptical orbits.

It might be argued that perfectly circular orbits (for planets, moons and spacecraft, etc.) are effectively impossible in nature (due to things like unsymmetrical mass distributions, thermal variations and external perturbations). And therefore the seeming deficiencies of Equation 1 (and similar equations used by Anderson, Yeomans, Parks and others) in the circular case should not prevent it from being used for modelling real orbits.

However it might cause a theoretician raise an eyebrow or two.

An (approximate) equation for the instantaneous acceleration obtained from analysis of Einstein's GR was presented in this earlier question Can General Relativity indicate phase-dependent variations in planetary orbital acceleration?:-

$$ \mathrm{F/m} \,= \, \frac {-GM}{r_\theta^2} - \frac {3 GM.GM.P}{c^2 \, r_\theta^4} $$

where $P$ is the length of the orbit's semi-latus rectum given by $P=A(1-e^2)$ where $a$ is the length of the semi-major axis. For a circular orbit the eccentricity $e=0$ and the constant radius is $R$ and $R=a$. So we obtain for a circular orbit:-

$$ \mathrm{F/m} \,= \, \frac {-GM}{R^2}\hat{\textbf{r}} \left(1 + \frac {3 GM}{c^2 \, R} \right) \,= \, \frac {-GM}{R^2}\hat{\textbf{r}} \left(1 + \frac {3 v_t^2}{c^2 } \right) $$

which is the same as the elliptical orbit equation derived from Goldstein and Walter.

Wells, 2011 ( page 28, equation 69 ) reports that for the ID3 equation $\delta$, the angular rotation of the Line of Apsides per orbit is given by:- $$ \delta = \frac{6.\pi.GM}{c^2.a(1-e^2)} =\frac{24\pi^2.a^3}{c^2.a(1-e^2).T^2} $$ (where $a$ is semi-major axis length and $e$ is eccentricity) Wells notes that this is the same $\delta$ as that given by Einstein's (1915) formula. This would indicate that both ID3 and Einstein(1915) predict the same value of $\delta$ - which does indeed appear to be the case for elliptical orbits.

For circular orbits, radius $r=a=R$ is constant and eccentricity $e$ is zero and so we obtain:- $$ \delta_{circ} =\frac{24\pi^2.R^2}{c^2.T^2} $$

The sense of rotation is positive (i.e. it moves around the Sun in the same sense as the revolving planet). We would expect, in the case of a circular orbit, this to apply to both the Einstein 1915 equation and the ID3 equation.

But here we face another issue. Consider two similar circular orbits, with the same values of $GM$ and $R$ and for which we have an equation for the radial acceleration in the form:-

$$ \textbf{a}_r = \frac{-GM}{R^2}\left( 1 + K(R)\right) $$

and there is no transverse acceleration. Here $K(R)$ is the Non-Newtonian acceleration factor). Orbit_1 is a purely Newtonian Orbit and so $K(R)_1=0$ and the orbital period is $T_1$. Orbit_2 has non-zero $K(R)_2$ whose value is constant because $R$ is constant. There is no line of Apsides in a circular orbit, but we can measure the extra orbital rotation in Orbit_2 caused by the $K(R)_2$ factor, simply by measuring the difference in velocities $v_1$ and $v_2$ (all velocities are transverse) and/or the corresponding difference in orbital periods $T_1$ and $T_2$.

We have for Orbit_1: $\frac{v_1^2}{R}=\frac{GM}{R^2}$ --> $v_1=\sqrt{\frac{GM}{R}}$

And for Orbit_2 $\frac{v_2^2}{R}=\frac{GM}{R^2}\left( 1+K(R)\right)$ --> $v_2=\sqrt{\frac{GM}{R}}.\sqrt{1+K(R)}$

Therefore the ratio of velocities $\frac{v_2}{v_1}$ is given by:- $$ \frac{v_2}{v_1} = \sqrt{1+K(R)} $$

In planetary orbits $K(R) << 1$ and so we can apply the binomial approximation to first order:-

$$ \frac{v_2}{v_1} = 1+ \frac{K(R)}{2}. $$

For the ID3 equation $B=6$ :- $$K(R) = \frac{B.GM}{c^2.R} = \frac{6.GM}{c^2.R} $$ and so:-

$$ \frac{v_2}{v_1}(ID3) = 1 + \frac{3.GM}{c^2.R} . $$ and $GM = 4\pi^2.R^3/T_1^2$ so:-

$$\frac{v_2}{v_1}(ID3) = 1 + \frac{12\pi^2.R^2}{c^2.T_1^2} $$

The second term on the RHS represents the additional fraction of Orbit_1 travelled by satellite_2 in the period $T_1$. We can convert this into the additional angle $\epsilon$ through which satellite_2 moves along the circular orbit during period $T_1$, by multiplying by $2\pi$ radians.


$$\epsilon_{ID3} = \frac{24\pi^3.R^2}{c^2.T_1^2}$$

Since the difference between $T_1^2$ and $T_2^2$ is very small we can approximate $T_1$ by $T_2$. Given that $e=0$ this expression for the exotic additional rotation of a circular orbit is equivalent to the Einstein (1915) equation for $\epsilon$ previously given $\epsilon_{Einstein} = \frac{24\pi^3.A^2}{c^2.T^2(1-e^2)}$.

The problem now is that the acceleration equations from Goldstein, Walter and derived from Einstein all have the value $B=3$. Which means that, for circular orbits, they predict half the $\epsilon$ value predicted by the ID3 equation and by the Einstein 1915 $\epsilon$ equation.


It helps to be clear at the outset what the effect of different exotic accelerations terms is upon orbit rotation.

Assuming $m<<M$ and Newtonian Elliptical Orbits we have Specific Angular Momentum $h$ given by:- $h=\, v_t.r = \sqrt{GM.a(1-e^2)} = \sqrt{GM.P}.$

From which $v_t^2 = \frac{GM.P}{r^2}$ and $\frac{GM}{r}=v_t^2\frac{r}{P}.$

Consider the different exotic acceleration components $Q$ occurring in the following two equation patterns:-

(i) radial acceleration ($- =$ inwards towards source): $$\textbf{a}_r = \frac{F_r}{m} = \frac{-GM}{r^2} \left( 1 + Q_R\right);$$ and

(ii) transverse acceleration ($+ =$ in the direction of the transverse velocity $\textbf{v}_t$):

$$\textbf{a}_t = \frac{F_t}{m} = \frac{+GM}{r^2} \left( Q_T\right)$$

and consider their effects upon $\epsilon$ (the exotic angular rotation per orbit) where $\Delta$ is the observed value of $\epsilon$ for a given satellite-orbit of the Sun:-

Radial acceleration terms:- $$\textbf{Q}_{R1} = \frac{GM}{r.c^2}\hat{\textbf{r}} = \frac{v_t^2.r}{c^2.P}\hat{\textbf{r}} \rightarrow \epsilon_1 =\frac{\Delta}{6}$$ $$ \textbf{Q}_{R2} = \frac{GM.P}{r^2.c^2}\hat{\textbf{r}} = \frac{v_t^2}{c^2}\hat{\textbf{r}} \rightarrow \epsilon_2 =\frac{\Delta}{3} \text{(elliptical orbits) or} = \frac{\Delta}{6} \text{(circular orbits).}$$

Transverse acceleration term:-

$$\textbf{Q}_{T1} = \frac{|v_r||v_t|(\hat{\textbf{r}}\cdot\hat{\textbf{v}_r})}{c^2}\hat{\textbf{v}_t} \rightarrow \epsilon_3 =\frac{\Delta}{3} \text{(elliptical orbits) or} \, = 0\, \text{(circular orbits).}$$

We can apply these values to calculate the effects predicted by Equation 1:-

$$\textbf{a} = \frac{-GM.\textbf{r}}{r^3} \left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2}{c^2} \right) + \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v}. $$

Rationalising $r$ and expanding into five terms, where the first term is the standard Newtonian acceleration, and the remaining four terms are due to the exotic acceleration:- $$\textbf{a} = \left[ \frac{-GM}{r^2} \hat{\textbf{r}} \right] +\left[\left( \frac{-GM}{r^2} \right)\left( \frac{-4.GM}{c^2.r} \right) \hat{\textbf{r}} \right] +\left[\left( \frac{-GM}{r^2} \right) \left( \frac{v_r^2 + v_t^2}{c^2} \right)\hat{\textbf{r}}\right] +\left[ \left( \frac{GM}{r^2} \right) \left( \frac{4|v_r^2|}{c^2} \right) (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r)\, \hat{\textbf{v}}_r \right] +\left[ \left(\frac{GM}{r^2}\right) \left(\frac{4|v_r.v_t|}{c^2}\right) (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r)\, \hat{\textbf{v}}_t \right] $$

Dropping the standard Newtonian term and dropping terms in $v_r^2$ (which do not produce significant exotic rotation) and applying $Q$ terms from above we obtain:- $$\textbf{a} = \left[ \left(\frac{-GM}{r^2}\right) \left( -4\right) \textbf{Q}_{R1} \right] +\left[ \left( \frac{-GM}{r^2}\right) \textbf{Q}_{R2} \right] +\left[ \left(\frac{GM}{r^2}\right) 4.\textbf{Q}_{T1} \right] $$

For elliptical orbits we obtain the rotation due to the exotic terms:- $$\epsilon = \Delta*\left(\frac{-4}{6}+\frac{1}{3}+\frac{4}{3}\right) = \Delta $$

For circular orbits we obtain:- $$\epsilon = \Delta*\left(\frac{-4}{6}+\frac{1}{6}+0\right) = -\frac{\Delta}{2}$$


  1. For Solar satellites in elliptic orbits ( with $e>0$ ) the value of $\epsilon$ from Equation 1 accounts exactly for the exotic orbital rotations.
  2. For Solar satellites in hypothetical circular orbits ( where $e=0$ ), the value of $\epsilon$ from Equation 1 is -50% of that required to account for the exotic orbital rotations.


The fact that term $Q_{R2}$ gives different values of the ratio $\frac{\epsilon}{\Delta}$ for elliptical and circular orbits is interesting. It suggests that this term does not provide a good representation of actual behavior. I will take this up in another question.


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