<p>If you switch to polar coordinates and use the classical Newtonian gravitation of <span class="math-container">$GM/r^2$</span> (inwards) you get a centrifugal term of <span class="math-container">$v^2/r$</span> (outwards) Since these two terms must cancel out for a circular orbit you end up with <span class="math-container">$v=\sqrt(GM/r)$</span>. This is not the case using your first equation called <span class="math-container">$a_{circ}$</span> above. Instead you will get:</p>
<p><span class="math-container">${GM\over{r^2}}(1-{4GM\over{rc^2}}+v^2/c^2)=v^2/r$</span></p>
<p>This will get you:
<span class="math-container">$v=\sqrt{{GM\over{r}}({1-4GM/rc^2)\over{1-GM/rc^2}}}$</span></p>
<p>for a circular orbit. I do not know if this helps you? I think you are making an &quot;illegal&quot; substitution.</p>
<p>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.)</p>

CC BY-SA 4.0
2023-01-18T12:49:11.910Z
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)
steveOw
CC BY-SA 4.0
2023-01-18T12:53:06.880Z
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.
steveOw
CC BY-SA 4.0
2023-01-19T04:01:13.377Z
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.
steveOw
CC BY-SA 4.0
2023-01-22T01:16:38.890Z
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.
steveOw
CC BY-SA 4.0
2023-01-18T04:00:14.780Z
2023-01-18T04:00:14.780Z
Agerhell
<p><strong>NOTE</strong></p>
<p>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.</p>
<hr />
<p>In the question I overlooked the fact that
<span class="math-container">$\frac{GM}{r^2} = \frac{v_{circ}^2}{r}$</span> only applies strictly in the case of circular orbits.  And therefore <span class="math-container">$v^2_{circ}$</span> should not be used to replace <span class="math-container">$GM/r$</span> when using Equation 1 in the case of elliptical orbits.</p>
<p>The effect of <span class="math-container">$GM/r$</span> can be ascertained elsewhere.  It is known since Newton (see <a href="https://en.wikipedia.org/wiki/Newton%27s_theorem_of_revolving_orbits" rel="nofollow noreferrer">wikipedia:Newton's theorem of revolving orbits</a> ) that the addition of an inverse-distance-cubed (ID3), inward-directed radial acceleration can produce apsidal rotation of an orbit.</p>
<p><a href="https://arxiv.org/abs/1106.1568" rel="nofollow noreferrer">Wells 2011</a> 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:-</p>
<p><span class="math-container">$$
\textbf{a}_r = \frac{-GM}{r^2} \left(1 + \frac{B.GM}{r.c^2}\right)\hat{\textbf{r}}. 
$$</span></p>
<p>My numerical simulator and Wells 2011 equation 22 indicate that a value of <span class="math-container">$B=6$</span> is required in order to produce <span class="math-container">$EAR=\Delta$</span>. (<span class="math-container">$\Delta$</span> is the value of EAR observed for Mercury, Mars and other Solar System bodies).</p>
<p>Now let us look at Equation 1 of the question:-</p>
<p><span class="math-container">$$\textbf{a} = \frac{-GM.\textbf{r}}{r^3}
\left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2}{c^2} 
\right) 
+ \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v}
$$</span></p>
<p>rationalising <span class="math-container">$r$</span> and expanding into four terms:-
<span class="math-container">$$\textbf{a} = 
\left(
\frac{-GM.\hat{\textbf{r}}}{r^2}
\right) 
+\left(
\frac{-GM.\hat{\textbf{r}}}{r^2}
\right).\left( 
 \frac{-4.GM}{c^2.r} 
\right)
+\left( 
\frac{-GM.\hat{\textbf{r}}}{r^2}
\right).
\left( 
 \frac{v^2}{c^2} 
\right)
+\left(
 \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v}
\right)
$$</span></p>
<p>The first term is the standard Newtonian radial acceleration and produces no EAR;</p>
<p>the second term, with <span class="math-container">$\frac{4.GM}{r.c^2}$</span>, will produce <span class="math-container">$EAR=(-4/6)\Delta$</span>.</p>
<p>As described in the Question:</p>
<p>the third term <span class="math-container">$\frac{v^2}{c^2}$</span> will produce <span class="math-container">$EAR=(+1/3)\Delta$</span></p>
<p>and the 4th term will produce net <span class="math-container">$EAR=(+4/3)\Delta$</span>.</p>
<p>Adding these together gives:- (-4/6) + (1/3) +(4/3) = (3/3).</p>
<p>Therefore, for Elliptical Orbits, Equation 1, predicts 100% of the Exotic Apsidal Rotation.</p>
<hr />
<h3>Still unresolved</h3>
<p>What remains unresolved though is that for a Circular orbit:-</p>
<p>(1) Term 2 in Equation 1 will produce different values of EAR according to whether it is interpreted as:-</p>
<p>(1a) <span class="math-container">$(4.v^2/c^2)$</span>  giving <span class="math-container">$EAR=(-4/3)\Delta$</span>, or</p>
<p>(1b) <span class="math-container">$(4.GM/r.c^2)$</span>  giving, <span class="math-container">$EAR=(-4/6)\Delta$</span> (based on <span class="math-container">$B=6$</span>).</p>
<p>(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 <strong>opposite direction</strong> to that observed in (and predicted by Equation 1 for) elliptical orbits.</p>
<p>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.</p>
<p>However it might cause a theoretician raise an eyebrow or two.</p>
<hr />
<p>An (approximate) equation for the instantaneous acceleration obtained from analysis of Einstein's GR was presented in this earlier question <a href="https://astronomy.stackexchange.com/questions/7900/can-general-relativity-indicate-phase-dependent-variations-in-planetary-orbital?noredirect=1&amp;lq=1">Can General Relativity indicate phase-dependent variations in planetary orbital acceleration?</a>:-</p>
<p><span class="math-container">$$ \mathrm{F/m} \,=  \, \frac {-GM}{r_\theta^2} - \frac {3 GM.GM.P}{c^2 \, r_\theta^4}
$$</span></p>
<p>where <span class="math-container">$P$</span> is the length of the orbit's semi-latus rectum given by <span class="math-container">$P=A(1-e^2)$</span> where <span class="math-container">$a$</span> is the length of the semi-major axis.  For a circular orbit the eccentricity <span class="math-container">$e=0$</span> and the constant radius is <span class="math-container">$R$</span> and <span class="math-container">$R=a$</span>. So we obtain for a circular orbit:-</p>
<p><span class="math-container">$$ \mathrm{F/m} \,=  \, \frac {-GM}{R^2}\hat{\textbf{r}} \left(1 + \frac {3 GM}{c^2 \, R} \right) \,=  \, \frac {-GM}{R^2}\hat{\textbf{r}} \left(1 + \frac {3 v_t^2}{c^2  } \right) 
$$</span></p>
<p>which is the same as the elliptical orbit equation derived from Goldstein and Walter.</p>
<hr />
<p>Wells, 2011 ( page 28, equation 69 ) reports that for the ID3 equation <span class="math-container">$\delta$</span>, the angular rotation of the Line of Apsides per orbit is given by:-
<span class="math-container">$$ 
\delta = \frac{6.\pi.GM}{c^2.a(1-e^2)} =\frac{24\pi^2.a^3}{c^2.a(1-e^2).T^2} 
$$</span>
(where <span class="math-container">$a$</span> is semi-major axis length and <span class="math-container">$e$</span> is eccentricity)
Wells notes that this is the same <span class="math-container">$\delta$</span> as that given by Einstein's (1915) formula.  This would indicate that both ID3 and Einstein(1915) predict the same value of <span class="math-container">$\delta$</span> - which does indeed appear to be the case for elliptical orbits.</p>
<p>For circular orbits, radius <span class="math-container">$r=a=R$</span> is constant and eccentricity <span class="math-container">$e$</span> is zero and so we obtain:-
<span class="math-container">$$ 
\delta_{circ} =\frac{24\pi^2.R^2}{c^2.T^2} 
$$</span></p>
<p>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.</p>
<hr />
<p>But here we face another issue. Consider two similar circular orbits, with the same values of <span class="math-container">$GM$</span> and <span class="math-container">$R$</span> and for which we have an equation for the radial acceleration in the form:-</p>
<p><span class="math-container">$$ \textbf{a}_r = \frac{-GM}{R^2}\left( 1 + K(R)\right) $$</span></p>
<p>and there is no transverse acceleration. Here <span class="math-container">$K(R)$</span> is the Non-Newtonian acceleration factor). Orbit_1 is a purely Newtonian Orbit and so <span class="math-container">$K(R)_1=0$</span> and the orbital period is <span class="math-container">$T_1$</span>.  Orbit_2 has non-zero <span class="math-container">$K(R)_2$</span> whose value is constant because <span class="math-container">$R$</span> 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 <span class="math-container">$K(R)_2$</span> factor, simply by measuring the difference in velocities <span class="math-container">$v_1$</span> and <span class="math-container">$v_2$</span> (all velocities are transverse) and/or the corresponding difference in orbital periods <span class="math-container">$T_1$</span> and <span class="math-container">$T_2$</span>.</p>
<p>We have for Orbit_1:
<span class="math-container">$\frac{v_1^2}{R}=\frac{GM}{R^2}$</span> --&gt; <span class="math-container">$v_1=\sqrt{\frac{GM}{R}}$</span></p>
<p>And for Orbit_2 <span class="math-container">$\frac{v_2^2}{R}=\frac{GM}{R^2}\left( 1+K(R)\right)$</span> --&gt; <span class="math-container">$v_2=\sqrt{\frac{GM}{R}}.\sqrt{1+K(R)}$</span></p>
<p>Therefore the ratio of velocities <span class="math-container">$\frac{v_2}{v_1}$</span> is given by:-
<span class="math-container">$$
\frac{v_2}{v_1} = \sqrt{1+K(R)}
$$</span></p>
<p>In planetary orbits <span class="math-container">$K(R) &lt;&lt; 1$</span> and so we can apply the binomial approximation to first order:-</p>
<p><span class="math-container">$$
\frac{v_2}{v_1} = 1+ \frac{K(R)}{2}.
$$</span></p>
<p>For the ID3 equation <span class="math-container">$B=6$</span> :-
<span class="math-container">$$K(R) = \frac{B.GM}{c^2.R} = \frac{6.GM}{c^2.R} $$</span> and so:-</p>
<p><span class="math-container">$$
\frac{v_2}{v_1}(ID3) = 1 +  \frac{3.GM}{c^2.R} .
$$</span>
and <span class="math-container">$GM = 4\pi^2.R^3/T_1^2$</span> so:-</p>
<p><span class="math-container">$$\frac{v_2}{v_1}(ID3) = 1 +  \frac{12\pi^2.R^2}{c^2.T_1^2} $$</span></p>
<p>The second term on the RHS represents the additional fraction of Orbit_1 travelled by satellite_2 in the period <span class="math-container">$T_1$</span>.  We can convert this into the additional angle <span class="math-container">$\epsilon$</span> through which satellite_2 moves along the circular orbit during period <span class="math-container">$T_1$</span>, by multiplying by <span class="math-container">$2\pi$</span> radians.</p>
<p>So:-</p>
<p><span class="math-container">$$\epsilon_{ID3} =  \frac{24\pi^3.R^2}{c^2.T_1^2}$$</span></p>
<p>Since the difference between <span class="math-container">$T_1^2$</span> and <span class="math-container">$T_2^2$</span> is very small we can approximate <span class="math-container">$T_1$</span> by <span class="math-container">$T_2$</span>. Given that <span class="math-container">$e=0$</span> this expression for the exotic additional rotation of a circular orbit is equivalent to the Einstein (1915) equation for <span class="math-container">$\epsilon$</span> previously given <span class="math-container">$\epsilon_{Einstein} =  \frac{24\pi^3.A^2}{c^2.T^2(1-e^2)}$</span>.</p>
<p>The problem now is that the acceleration equations from Goldstein, Walter and derived from Einstein all have the value <span class="math-container">$B=3$</span>.  Which means that, for circular orbits, they predict half the <span class="math-container">$\epsilon$</span> value predicted by the ID3 equation and by the Einstein 1915 <span class="math-container">$\epsilon$</span> equation.</p>

CC BY-SA 4.0
2023-01-22T00:54:34.760Z
2023-01-24T20:53:43.477Z
2023-01-24T20:53:43.477Z
steveOw
<p>It helps to be clear at the outset what the effect of different exotic accelerations terms is upon orbit rotation.</p>
<p>Assuming <span class="math-container">$m&lt;&lt;M$</span> and Newtonian Elliptical Orbits we have  Specific Angular Momentum <span class="math-container">$h$</span> given by:-
<span class="math-container">$h=\, v_t.r = \sqrt{GM.a(1-e^2)} = \sqrt{GM.P}.$</span></p>
<p>From which <span class="math-container">$v_t^2 = \frac{GM.P}{r^2}$</span> and <span class="math-container">$\frac{GM}{r}=v_t^2\frac{r}{P}.$</span></p>
<p>Consider the different exotic acceleration components <span class="math-container">$Q$</span> occurring in the following two equation patterns:-</p>
<p>(i) radial acceleration (<span class="math-container">$- =$</span> inwards towards source):
<span class="math-container">$$\textbf{a}_r = \frac{F_r}{m} = \frac{-GM}{r^2} \left( 1 + Q_R\right);$$</span>
and</p>
<p>(ii) transverse acceleration (<span class="math-container">$+ =$</span> in the direction of the transverse velocity <span class="math-container">$\textbf{v}_t$</span>):</p>
<p><span class="math-container">$$\textbf{a}_t = \frac{F_t}{m} = \frac{+GM}{r^2} \left(  Q_T\right)$$</span></p>
<p>and consider their effects upon <span class="math-container">$\epsilon$</span> (the exotic angular rotation per orbit) where <span class="math-container">$\Delta$</span> is the observed value of <span class="math-container">$\epsilon$</span> for a given satellite-orbit of the Sun:-</p>
<p>Radial acceleration terms:-
<span class="math-container">$$\textbf{Q}_{R1} = \frac{GM}{r.c^2}\hat{\textbf{r}} = \frac{v_t^2.r}{c^2.P}\hat{\textbf{r}}  \rightarrow \epsilon_1 =\frac{\Delta}{6}$$</span>
<span class="math-container">$$
\textbf{Q}_{R2} = \frac{GM.P}{r^2.c^2}\hat{\textbf{r}}  = \frac{v_t^2}{c^2}\hat{\textbf{r}}  \rightarrow \epsilon_2 =\frac{\Delta}{3} \text{(elliptical orbits) or} = \frac{\Delta}{6} \text{(circular orbits).}$$</span></p>
<p>Transverse acceleration term:-</p>
<p><span class="math-container">$$\textbf{Q}_{T1} = \frac{|v_r||v_t|(\hat{\textbf{r}}\cdot\hat{\textbf{v}_r})}{c^2}\hat{\textbf{v}_t}
\rightarrow \epsilon_3 =\frac{\Delta}{3} \text{(elliptical orbits) or} \, = 0\, \text{(circular orbits).}$$</span></p>
<hr />
<p>We can apply these values to calculate the effects predicted by Equation 1:-</p>
<p><span class="math-container">$$\textbf{a} = \frac{-GM.\textbf{r}}{r^3}
\left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2}{c^2} 
\right) 
+ \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v}.
$$</span></p>
<p>Rationalising <span class="math-container">$r$</span> 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:-
<span class="math-container">$$\textbf{a} = 
\left[
\frac{-GM}{r^2} \hat{\textbf{r}}
\right] 
+\left[\left(
\frac{-GM}{r^2}
\right)\left( 
 \frac{-4.GM}{c^2.r} 
\right) \hat{\textbf{r}} \right] 
+\left[\left( 
\frac{-GM}{r^2}
\right)
\left( 
 \frac{v_r^2 + v_t^2}{c^2} 
\right)\hat{\textbf{r}}\right] 
+\left[
\left( \frac{GM}{r^2}       \right)
\left( \frac{4|v_r^2|}{c^2} \right)
(\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r)\, \hat{\textbf{v}}_r
\right]
+\left[
\left(\frac{GM}{r^2}\right)
\left(\frac{4|v_r.v_t|}{c^2}\right) 
(\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r)\, \hat{\textbf{v}}_t
\right] 
$$</span></p>
<p>Dropping the standard Newtonian term and dropping terms in <span class="math-container">$v_r^2$</span> (which do not produce significant exotic rotation) and applying <span class="math-container">$Q$</span> terms from above we obtain:-
<span class="math-container">$$\textbf{a} = 
\left[
\left(\frac{-GM}{r^2}\right)
\left( -4\right) \textbf{Q}_{R1} 
\right] 
+\left[
\left( \frac{-GM}{r^2}\right)
\textbf{Q}_{R2} 
\right] 
+\left[
\left(\frac{GM}{r^2}\right)
4.\textbf{Q}_{T1}
\right] 
$$</span></p>
<p>For elliptical orbits we obtain the rotation due to the exotic terms:-
<span class="math-container">$$\epsilon = \Delta*\left(\frac{-4}{6}+\frac{1}{3}+\frac{4}{3}\right) = \Delta $$</span></p>
<p>For circular orbits we obtain:-
<span class="math-container">$$\epsilon = \Delta*\left(\frac{-4}{6}+\frac{1}{6}+0\right) =   -\frac{\Delta}{2}$$</span></p>
<p><strong>CONCLUSION</strong></p>
<ol>
<li>For Solar satellites in elliptic orbits ( with <span class="math-container">$e&gt;0$</span> ) the value of <span class="math-container">$\epsilon$</span> from Equation 1 accounts exactly for the exotic orbital rotations.</li>
<li>For Solar satellites in hypothetical circular orbits ( where <span class="math-container">$e=0$</span> ), the value of <span class="math-container">$\epsilon$</span> from Equation 1 is -50% of that required to account for the exotic orbital rotations.</li>
</ol>
<p><strong>NOTE ON TERM <span class="math-container">$Q_{R2}$</span></strong></p>
<p>The fact that term <span class="math-container">$Q_{R2}$</span> gives different values of the ratio <span class="math-container">$\frac{\epsilon}{\Delta}$</span> 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.</p>

CC BY-SA 4.0
2023-01-24T20:49:53.887Z
2023-01-25T22:16:59.987Z
2023-01-25T22:16:59.987Z
steveOw
<p>I have previously encountered a problem with the equation used by Yeomans and others for determining the gravitational acceleration of comets (see update in <a href="https://astronomy.stackexchange.com/a/25429/7982">my answer to the previous question <em>Can General Relativity indicate phase-dependent variations in planetary orbital acceleration?</em></a>).</p>
<p>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 &quot;Yeomans&quot; equation does not explain the observed Non-Newtonian apsidal rotation (orbital perihelion precession) of the Solar System planets.</p>
<p><a href="https://adsabs.harvard.edu/full/1975ApJ...200..221A" rel="nofollow noreferrer">Anderson et al 1975</a>
presented two Equations 11 and 12 for the Newtonian and Relativistic accelerations affecting a satellite.</p>
<p>The following equation 3.11 is from <a href="https://adsabs.harvard.edu/full/1994AJ....107.1885S" rel="nofollow noreferrer">Shahid-Saless and Yeomans 1994</a>.   It gives the gravitational acceleration of a single body under the influence of another single &quot;source&quot; body. Here <span class="math-container">$\mu = GM_{Source}/c^2.$</span>   This equation is a selective composite of relevant parts of Equations 11 and 12 from Anderson et al 1975.</p>
<p><a href="https://i.sstatic.net/OoPWu.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OoPWu.png" alt="Equation 3.11 of Shahid_Saless &amp; Yeomans 1994" /></a></p>
<p>Equation 3.11 can be re-written as (and here denoted by Equation 1):-
<span class="math-container">$$\textbf{a} = \frac{-GM.\textbf{r}}{r^3}
\left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2}{c^2} 
\right) 
+ \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v}
$$</span></p>
<p>Note I use different Mathjax fonts to distinguish between scalars <span class="math-container">$r,v$</span> and vectors <span class="math-container">$\textbf{r},\textbf{v}$</span>.</p>
<p>More recently the following Equation 27 is reported in <a href="https://iopscience.iop.org/article/10.3847/1538-3881/abd414" rel="nofollow noreferrer">Park et al 2021, The JPL Planetary and Lunar Ephemerides DE440 and DE441</a>. 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 <span class="math-container">$r_{ij}=|\textbf{r}_i-\textbf{r}_j|$</span> is defined as an unsigned scalar value representing the distance between bodies <span class="math-container">$i$</span> and <span class="math-container">$j$</span>.</p>
<p><a href="https://i.sstatic.net/ip3yI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ip3yI.png" alt="D440-D441 Ephemerides Eqtn 27" /></a></p>
<p>A very similar equation is presented in <a href="https://descanso.jpl.nasa.gov/monograph/series2_section.html" rel="nofollow noreferrer">Descanso2</a> Section 4-26. It is based on work by Moyer (1971) and Will(1981). Here <span class="math-container">$\mu = GM_{body}.$</span> 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.</p>
<p><a href="https://i.sstatic.net/T4Lpr.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/T4Lpr.png" alt="enter image description here" /></a></p>
<p>Note: I am aware of important, fundamental differences between Newtonian and General Relativity models of nature (see contribution by /u/Stan Liou/ in <a href="https://astronomy.stackexchange.com/questions/7700/what-is-the-correct-ratio-of-newtonian-to-general-relativistic-gravitational-eff?rq=1">this question</a> ). In the following I am neglecting a discussion of these and thinking rather in terms of a Newtonian Model with additional small &quot;exotic&quot; 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.</p>
<p>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 <span class="math-container">$\beta$</span> and <span class="math-container">$\gamma$</span> are equal to 1.  By taking the inertial frame and coordinate system where the position (<span class="math-container">$r_j$</span>) velocity (<span class="math-container">$v_j$</span>) and acceleration (<span class="math-container">$a_j$</span>) of the Sun are zero we find that many of terms become zero.</p>
<p><span class="math-container">$$\textbf{a} = \frac{-GM.\textbf{r}}{r^3}
\left( 1 - \frac{4.GM}{c^2.r} - 0 + \frac{v^2}{c^2} + 0 -0 -0+0
\right) 
+ \frac{GM}{c^2.r^3}\left( [\textbf{r}] \cdot [4.\textbf{v}-0]\right)\textbf{v}
+0$$</span></p>
<p>And so...</p>
<p><span class="math-container">$$\textbf{a} = \frac{-GM.\textbf{r}}{r^3}
\left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2}{c^2} 
\right) 
+ \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v}.
$$</span></p>
<p>... which is identical to my Equation 1 derived from Equation 3.11 of Shaid-Saless &amp; Yeomans 1994.</p>
<p>(Note that the third term inside the curly braces of Park et al's Equation 27, the sum of objects <span class="math-container">$k$</span> does not exclude object <span class="math-container">$i$</span>. But I have ignored this term as the mass of object <span class="math-container">$i$</span> 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 <a href="https://ntrs.nasa.gov/citations/19710017134" rel="nofollow noreferrer">Moyer 1971, Mathematical Formulation of the Double Precision Orbit Determination Program DPODP</a> for the motion of a 'massless' test particle around a single massive body.</p>
<h3>CIRCULAR ORBITS</h3>
<p>Now let us consider the case where the orbit is circular, then the test particle velocity (<span class="math-container">$\textbf{v}$</span>) is perpendicular to its position vector (<span class="math-container">$\textbf{r}$</span>) and so the dot product is zero and so the final term in Equation 1 goes to zero, leaving:-</p>
<p><span class="math-container">$$\textbf{a}_{circ} = \frac{-GM.\textbf{r}}{r^3}
\left( 1 - \frac{4.GM}{c^2.r} + \frac{v^2_{circ}}{c^2} 
\right) 
$$</span></p>
<p>In a (Newtonian) circular orbit we have <span class="math-container">$\frac{GM}{r}=v_{circ}^2$</span> and so we can write:
<span class="math-container">$$\textbf{a}_{circ} = \frac{-GM.\textbf{r}}{r^3}
\left( 1 - \frac{4.v^2_{circ}}{c^2} + \frac{v^2_{circ}}{c^2} 
\right) 
$$</span></p>
<p>Then...
<span class="math-container">$$\textbf{a}_{circ} = \frac{-GM}{r^2}
\left( 1 - \frac{3v^2_{circ}}{c^2}  
\right) \hat{\textbf{r}}
$$</span></p>
<p>Finally
<span class="math-container">$$\textbf{a}_{circ} = -\frac{GM}{r^2} \hat{\textbf{r}}
   + \frac{GM}{r^2}  \frac{3v^2_{circ}}{c^2}\hat{\textbf{r}}
$$</span></p>
<p>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 <a href="https://astronomy.stackexchange.com/questions/7900/can-general-relativity-indicate-phase-dependent-variations-in-planetary-orbital/25429#25429">this earlier SE Astronomy question of mine</a>  ), effectively:-</p>
<p><span class="math-container">$$\textbf{a}_r = -\frac{GM}{r^2} \hat{\textbf{r}}
   - \frac{GM}{r^2}  \frac{3v^2_{circ}}{c^2}\hat{\textbf{r}}
$$</span></p>
<p>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.</p>
<h3>ELLIPTICAL ORBITS</h3>
<p>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 <span class="math-container">$\textbf{a}_E$</span>:-
<span class="math-container">$$
\textbf{a}_E = \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}\right)\textbf{v}
$$</span></p>
<p>The velocity vector <span class="math-container">$\textbf{v}$</span> can be decomposed into the two mutually-perpendicular component velocity vectors <span class="math-container">$\textbf{v}_r$</span> and <span class="math-container">$\textbf{v}_t$</span> directed radially and transversely respectively. And then we can apply the distributive rule for dot products...</p>
<p><span class="math-container">$$
\textbf{a}_E 
= \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot (\textbf{v}_r + \textbf{v}_t) \right)\textbf{v}
= \frac{4GM}{c^2.r^3}\left( \textbf{r} \cdot \textbf{v}_r + \textbf{r} \cdot  \textbf{v}_t) \right)\textbf{v}
$$</span></p>
<p>Now <span class="math-container">$\textbf{r}$</span> and <span class="math-container">$\textbf{v}_r$</span> are parallel so their dot product has magnitude <span class="math-container">$|r.v_r|$</span> and sign <span class="math-container">$(\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r)=+/-1$</span> according to whether they are or aren't pointing in the same direction.  Whereas
<span class="math-container">$\textbf{r}$</span> and <span class="math-container">$\textbf{v}_t$</span> are perpendicular so their dot product is zero.</p>
<p><span class="math-container">$$
\textbf{a}_E 
= \frac{4GM}{c^2.r^3}\left( |r.v_r|(\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) + 0 \right)\textbf{v}
$$</span></p>
<p>Cancelling <span class="math-container">$r$</span> then decomposing <span class="math-container">$\textbf{v}$</span>...
<span class="math-container">$$
= \frac{4GM}{c^2.r^2}\left( |v_r|(\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) \right)\textbf{v}
= \frac{4GM}{c^2.r^2} 
(|v_r| (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) \textbf{v}_r 
+ |v_r| (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r)  \textbf{v}_t)
$$</span></p>
<p>We are left with two perpendicular components of acceleration...</p>
<p><span class="math-container">$$
\textbf{a}_{Er} = 
\frac{4GM}{c^2.r^2} |v_r^2| (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) \hat{\textbf{v}}_r
$$</span>
<span class="math-container">$$
\textbf{a}_{Et} =  
+ \frac{4GM}{c^2.r^2} |v_r.v_t| (\hat{\textbf{r}} \cdot \hat{\textbf{v}}_r) \hat{\textbf{v}}_t
$$</span>
Now it can be shown that the radial component <span class="math-container">$\textbf{a}_{Er}$</span> (in <span class="math-container">$v_r^2$</span>) is always directed radially away from the source and does not have a relatively large affect on apsidal rotation for Solar planets.</p>
<p>Whereas the transverse acceleration component <span class="math-container">$\textbf{a}_{Et}$</span> (in <span class="math-container">$v_r.v_t$</span>) points in the same direction as <span class="math-container">$\hat{\textbf{v}}_t$</span> during the half-orbit between perihelion and aphelion but points in the direction opposite to <span class="math-container">$\hat{\textbf{v}}_t$</span> during the other half-orbit. It has the effect of producing apsidal rotation equal to  <span class="math-container">$(+4/3)\Delta$</span> where <span class="math-container">$\Delta$</span> is the observed Non-Newtonian apsidal rotation for Mercury and Mars.</p>
<p>From previously, the apsidal rotation produced by the rest of Equation 1 is <span class="math-container">$(-3/3)\Delta$</span>.  So the total rotation produced by Equation 1 is <span class="math-container">$(+1/3)\Delta$</span> which is a third of the observed planetary orbit rotations.</p>
<p>This compares with the <span class="math-container">$(+3/3)\Delta$</span> produced by the equation from Goldstein/Walter which predicts 100% of the observed apsidal rotations.</p>
<h3>THE QUESTION</h3>
<p>So my question is: <em>What is the explanation for the &quot;wrong&quot; 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?</em></p>

CC BY-SA 4.0
2023-01-21T01:21:16.540Z
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.
uhoh
CC BY-SA 4.0
2023-01-22T01:01:19.257Z
@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 :) .
steveOw
CC BY-SA 4.0
2023-01-22T01:18:23.340Z
Okay thanks for the update and congratulations on the progress!
uhoh
<orbit><general-relativity><apsidal-precession>
Problem with the Orbital Precession predicted by the Relativistic Acceleration equation used in the JPL D440-D441 Ephemerides - please advise
steveOw
# Problem with the Orbital Precession predicted by the Relativistic Acceleration equation used in the JPL D440-D441 Ephemerides - please advise

(Asked by: username_0 on 2023-01-16)

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?*](https://astronomy.stackexchange.com/a/25429/7982)).

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](https://adsabs.harvard.edu/full/1975ApJ...200..221A)
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](https://adsabs.harvard.edu/full/1994AJ....107.1885S). 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](https://i.sstatic.net/OoPWu.png)](https://i.sstatic.net/OoPWu.png)

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](https://iopscience.iop.org/article/10.3847/1538-3881/abd414). 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](https://i.sstatic.net/ip3yI.png)](https://i.sstatic.net/ip3yI.png)

A very similar equation is presented in [Descanso2](https://descanso.jpl.nasa.gov/monograph/series2_section.html) 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](https://i.sstatic.net/T4Lpr.png)](https://i.sstatic.net/T4Lpr.png)

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](https://astronomy.stackexchange.com/questions/7700/what-is-the-correct-ratio-of-newtonian-to-general-relativistic-gravitational-eff?rq=1) ). 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](https://ntrs.nasa.gov/citations/19710017134) for the motion of a 'massless' test particle around a single massive body.

### CIRCULAR ORBITS

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](https://astronomy.stackexchange.com/questions/7900/can-general-relativity-indicate-phase-dependent-variations-in-planetary-orbital/25429#25429) ), 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.

### 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.

### THE QUESTION

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?*

**Tags:** orbit, general-relativity, apsidal-precession

> **Comments:**
> **username_1**: 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.  
> **username_0**: @username_1 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 :) .  
> **username_1**: Okay thanks for the update and congratulations on the progress!  

---

## 3 answers

### Answer by username_2 on 2023-01-18. Score: 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:

${GM\over{r^2}}(1-{4GM\over{rc^2}}+v^2/c^2)=v^2/r$

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.)

> **Comments:**
> **username_0**: 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)  
> **username_0**: 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.  
> **username_0**: 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.  
> **username_0**: 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.  

---

### Answer by username_0 on 2023-01-22. Score: 2

**NOTE**

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](https://en.wikipedia.org/wiki/Newton%27s_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](https://arxiv.org/abs/1106.1568) 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.

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### 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?](https://astronomy.stackexchange.com/questions/7900/can-general-relativity-indicate-phase-dependent-variations-in-planetary-orbital?noredirect=1&lq=1):-

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

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.

So:-

$$\epsilon\_{ID3} = \frac{24\pi^3.R^2}{c^2.T\_1^2}$$

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.


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### Answer by username_0 on 2023-01-24. Score: 2

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

**CONCLUSION**

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.

**NOTE ON TERM $Q\_{R2}$**

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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