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Is there an analytical technique for determining the effect of a small variable transverse acceleration upon the rate of aspides precession (strictly not a precession but rotation of the line of aspides) of a planet orbitting around the Sun in a 2D plane according to Newtonian gravity law?

I have modelled such effects in a reiterative computer model and would like to verify those measurements.

The transverse acceleration formula is

$$At = (K/c^2)*Vr*Vt * Ar.$$

Where:-

c is speed of light,

K is a constant of magnitude between 0 and +/-3, such that $K/(c^2) << 1$.

Ar is the acceleration of the planet towards the Sun due to Newtonian gravitational influence of the Sun, ($Ar = GM/r^2$).

Vr is radial component of planet velocity relative to the Sun (+ = motion away from the Sun)

Vt is transverse component of planet velocity relative to the Sun (+ = direction of planet forward motion along its orbital path)

Assume planet mass is small relative to the Sun

No other bodies are in the system

All motions and accelerations are confined to the two-dimensional plane of the orbit.

The reason why this is interesting to me is that a value of K = -3 in my computer model produces anomalous (Non-Newtonian) precession equal to that observed by Le Verrier and predicted by General Relativity.

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Are you still seeking an answer? –  Walter Aug 27 at 15:52
    
@Walter. Yes I am. I have asked similar question at physics.stackexchange.com/questions/123685/… but no solid answer received yet. –  steveOw 2 days ago
    
@Walter. I also asked at math.stackexchange.com/questions/866836/…. –  steveOw 2 days ago
    
Yes, there are approximate analytical methods (perturbation theory), valid in the limit of $K\ll1$. Perhaps you can clarify your question a bit. What's the direction of the transverse acceleration (I understand 'transverse' to mean perpendicular to the instantaneous velocity, but it's not clear whether the acceleration is in the plane of the orbit or perpendicular or a mixture). –  Walter yesterday
    
There is a difference between your question here and that on mathematics (and physics): here the transverse acceleration is proportional to the radial acceleration and $K$ is a dimensionless number, there the radial acceleration has no effect on the transverse acceleration and $K$ must be an acceleration (though you talk about a 'number'). –  Walter yesterday

1 Answer 1

You may want to use perturbation theory. This only gives you an approximate answer, but allows for analytic treatment. Your force is considered a small perturbation to the Keplerian elliptic orbit and the resulting equations of motion are expanded in powers of $K$. For linear perturbation theory, only terms linear in $K$ are retained. This simply leads to integrating the perturbation along the unperturbed original orbit. Writing your force as a vector, the perturbing acceleration is $$ \ddot{\boldsymbol{r}} = \boldsymbol{a}_p \equiv K \frac{GM}{r^2c^2}v_r\boldsymbol{v}_t $$ with $v_r=\dot{\boldsymbol{r}}{\cdot}\hat{\boldsymbol{r}}$ the radial velocity and $\boldsymbol{v}_t=(\dot{\boldsymbol{r}}-\hat{\boldsymbol{r}}\dot{\boldsymbol{r}}{\cdot}\hat{\boldsymbol{r}})$ is the rotational component of velocity (the full velocity minus the radial velocity). Here, the dot above denotes a time derivative and a hat the unit vector.

Now, it depends what you mean with 'effect'. Let's work out the changes of the orbital semimajor axis $a$ and eccentricity $e$.

1 change in semimajor axis

From the relation $a=-GM/2E$ (with $E=\frac{1}{2}\dot{\boldsymbol{r}}^2-GMr^{-1}$ the orbital energy) we have for the change of $a$ due to an external (non-Keplerian) acceleration $\ddot{\boldsymbol{r}}$ $$ \dot{a}=\frac{2a^2}{GM}\dot{\boldsymbol{r}}{\cdot}\ddot{\boldsymbol{r}}. $$ Inserting $\ddot{\boldsymbol{r}}=\boldsymbol{a}_p$ (note that $\dot{\boldsymbol{r}}{\cdot}\boldsymbol{v}_t=h^2/r^2$ with angular momentum vector $\boldsymbol{h}\equiv\boldsymbol{r}\wedge\dot{\boldsymbol{r}}$), we get $$ \dot{a}=\frac{2a^2Kh^2}{c^2}\frac{v_r}{r^4}. $$ Since the orbit average $\langle v_r f(r)\rangle=0$ for any function $f$ (see below), $\langle\dot{a}\rangle=0$: the semimajor axis is unaffected in the orbit-averaged sense.

2 change of eccentricity

From $\boldsymbol{h}^2=(1-e^2)GMa$, we find $$ e\dot{e}=-\frac{\boldsymbol{h}{\cdot}\dot{\boldsymbol{h}}}{GMa}+\frac{h^2\dot{a}}{2GMa^2}. $$ We already know that $\langle\dot{a}\rangle=0$, so only need to consider the first term. Thus, $$ e\dot{e}=-\frac{(\boldsymbol{r}\wedge\dot{\boldsymbol{r}}){\cdot}(\boldsymbol{r}\wedge\boldsymbol{a}_p)}{GMa} =-\frac{r^2\;\dot{\boldsymbol{r}}{\cdot}\boldsymbol{a}_p}{GMa} =-\frac{Kh^2}{ac^2}\frac{v_r}{r^2}, $$ where I have used the identity $(\boldsymbol{a}\wedge\boldsymbol{b}){\cdot}(\boldsymbol{c}\wedge\boldsymbol{d}) =\boldsymbol{a}{\cdot}\boldsymbol{c}\;\boldsymbol{b}{\cdot}\boldsymbol{d}- \boldsymbol{a}{\cdot}\boldsymbol{d}\;\boldsymbol{b}{\cdot}\boldsymbol{c}$ and the fact $\boldsymbol{r}{\cdot}\boldsymbol{a}_p=0$. Again $\langle v_r/r^2\rangle=0$ and hence $\langle\dot{e}\rangle=0$: the eccentricity is unaffected in the orbit-averaged sense.

3 change in the direction of periapse

The eccentricity vector $$ \boldsymbol{e}\equiv\frac{\dot{\boldsymbol{r}}\wedge\boldsymbol{h}}{GM} - \hat{\boldsymbol{r}} $$ points (from the centre of gravity) in the direction of periapse, has magnitude $e$, and is conserved under the Keplerian motion (validate all that as an exercise!). From this definition we find its instantaneous change due to external acceleration $$ \dot{\boldsymbol{e}}= \frac{\boldsymbol{a}_p\wedge(\boldsymbol{r}\wedge\dot{\boldsymbol{r}}) +\dot{\boldsymbol{r}}\wedge(\boldsymbol{r}\wedge\boldsymbol{a}_p)}{GM} =\frac{2(\dot{\boldsymbol{r}}{\cdot}\boldsymbol{a}_p)\boldsymbol{r} -(\boldsymbol{r}{\cdot}\dot{\boldsymbol{r}})\boldsymbol{a}_p}{GM} =\frac{2K}{c^2}\frac{h^2v_r\boldsymbol{r}}{r^4} -\frac{K}{c^2}\frac{v_r^2\boldsymbol{v}_t}{r} $$ where I have used the identity $\boldsymbol{a}\wedge(\boldsymbol{b}\wedge\boldsymbol{c})=(\boldsymbol{a}{\cdot}\boldsymbol{c})\boldsymbol{b}-(\boldsymbol{a}{\cdot}\boldsymbol{b})\boldsymbol{c}$ and the fact $\boldsymbol{r}{\cdot}\boldsymbol{a}_p=0$. The orbit averages of these expression are considered in the appendix below. If we finally put everything together, we get $$ \dot{\boldsymbol{e}}=\boldsymbol{\omega}\wedge\boldsymbol{e} $$ with $$ \boldsymbol{\omega}=\Omega K \frac{v_c^2}{c^2} \hat{\boldsymbol{h}}\; \frac{\sqrt{1-e^2}}{2\pi}\int_0^{2\pi} \frac{(2+e^2-e\cos\eta)\sin^2\!\eta}{(1-e\cos\eta)^5}\mathrm{d}\eta. $$ Unfortunately, I couldn't find a closed expression for this integral. Note that this is a rotation of periapse in the plane of the orbit with angular frequency $\omega=|\boldsymbol{\omega}|$. In particular $\langle e\dot{e}\rangle=\langle\boldsymbol{e}{\cdot}\dot{\boldsymbol{e}}\rangle=0$ in agreement with our previous finding.

Don't forget that due to our usage of first-order perturbation theory these results are only strictly true in the limit $K(v_c/c)^2\to0$. At second-order perturbation theory, however, both $a$ and/or $e$ may change. In your numerical experiments, you should find that the orbit-averaged changes of $a$ and $e$ are either zero or scale stronger than linear with perturbation amplitude $K$.

disclaimer No guarantee that the algebra is correct. Check it!


Appendix: orbit averages

Orbit averages of $v_rf(r)$ with an abitrary (but integrable) function $f(r)$ can be directly calculated for any type of periodic orbit. Let $F(r)$ be the antiderivative of $f(r)$, i.e. $F'\!=f$, then the orbit average is: $$ \langle v_r f(r)\rangle = \frac{1}{T}\int_0^T v_r(t)\,f\!\left(r(t)\right) \mathrm{d}t = \frac{1}{T} \left[F\left(r(t)\right)\right]_0^T = 0 $$ with $T$ the orbital period.

For the orbit averages required in $\langle\dot{\boldsymbol{e}}\rangle$, we must dig a bit deeper. For a Keplerian elliptic orbit $$ \boldsymbol{r}=a\left((\cos\eta-e)\hat{\boldsymbol{e}}+\sqrt{1-e^2}\sin\eta\,\hat{\boldsymbol{k}}\right)\qquad\text{and}\qquad r=a(1-e\cos\eta) $$ with eccentricity vector $\boldsymbol{e}$ and $\hat{\boldsymbol{k}}\equiv\hat{\boldsymbol{h}}\wedge\hat{\boldsymbol{e}}$ a vector perpendicular to $\boldsymbol{e}$ and $\boldsymbol{h}$. Here, $\eta$ is the eccentric anomaly, which is related to the mean anomaly $\ell$ via $ \ell=\eta-e\sin\eta, $ such that $\mathrm{d}\ell=(1-e\cos\eta)\mathrm{d}\eta$ and an orbit average becomes $$ \langle\cdot\rangle = (2\pi)^{-1}\int_0^{2\pi}\cdot\;\mathrm{d}\ell = (2\pi)^{-1}\int_0^{2\pi}\cdot\;(1-e\cos\eta)\mathrm{d}\eta. $$ Taking the time derivative (note that $\dot{\ell}=\Omega=\sqrt{GM/a^3}$ the orbital frequency) of $\boldsymbol{r}$, we find for the instantaneous (unperturbed) orbital velocity $$ \dot{\boldsymbol{r}}=v_c\frac{\sqrt{1-e^2}\cos\eta\,\hat{\boldsymbol{k}}-\sin\eta\,\hat{\boldsymbol{e}}}{1-e\cos\eta} $$ where I have introduced $v_c\equiv\Omega a=\sqrt{GM/a}$, the speed of the circular orbit with semimajor axis $a$. From this, we find the radial velocity $v_r=\hat{\boldsymbol{r}}{\cdot}\dot{\boldsymbol{r}}=v_c e\sin\eta(1-e\cos\eta)^{-1}$ and the rotational velocity $$ \boldsymbol{v}_t = \frac{\sqrt{1-e^2}(\cos\eta-e)\,\hat{\boldsymbol{k}}-(1-e^2)\sin\eta\,\hat{\boldsymbol{e}}}{(1-e\cos\eta)^2}. $$

With these, we have $$ \left\langle \frac{h^2v_r\boldsymbol{r}}{r^4}\right\rangle = \Omega v_c^2\, \hat{\boldsymbol{k}}\, \frac{e(1-e^2)^{3/2}}{2\pi}\int_0^{2\pi}\frac{\sin^2\!\eta}{(1-e\cos\eta)^5}\mathrm{d}\eta \\ \left\langle \frac{v_r^2\boldsymbol{v}_t}{r}\right\rangle = \Omega v_c^2\, \hat{\boldsymbol{k}}\, \frac{e^2(1-e^2)^{1/2}}{2\pi}\int_0^{2\pi}\frac{\sin^2\!\eta(\cos\eta-e)}{(1-e\cos\eta)^5}\mathrm{d}\eta $$ and thus $$\left\langle 2\frac{h^2v_r\boldsymbol{r}}{r^4}-\frac{v_r^2\boldsymbol{v}_t}{r}\right\rangle =\Omega v_c^2\,\hat{\boldsymbol{k}}\, \frac{e\sqrt{1-e^2}}{2\pi}\int_0^{2\pi} \frac{(2+e^2-e\cos\eta)\sin^2\!\eta}{(1-e\cos\eta)^5}\mathrm{d}\eta $$

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Thanks for the detailed analysis. It will take me some time to digest. My numerical experiments indicate that $a$ and $e$ are indeed constant but the direction of $a$ rotates by a fixed angle each orbit. If I use different values of $K$ then the magnitude of the rotation angle is linearly proportional to $K$. It is prediction of the rotation angle which I am most interested in. –  steveOw yesterday
    
You mean the direction of periapse. Yes, that can be looked at using perturbation theory too. –  Walter 20 hours ago
    
To be precise I seek the apsidal precession which according to wikipedia is "the first derivative of the argument of periapsis, " en.wikipedia.org/wiki/Apsidal_precession, ideally in terms of other orbital parameters such as period, eccentricity and semi-latus rectum. –  steveOw 16 hours ago
    
A quick trial of your equation for ω (rotation of periapse with angular frequency) for Mercury, with K=3, gives 47 arc seconds per century (109% of the official value 43"). Would you say that error of 9% is reasonable for 1st-order peturbation analysis? Anyway I am greatly impressed with the result, but will check further tomorrow:-). –  steveOw 5 hours ago

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