1
$\begingroup$

We know the theory of general relativity (GR) accounts for the illusive $43$ arc seconds of precession of the perihelion of Mercury. The calculation is well known and well studied. How does one account for the remaining planetary precession? So, more specifically, the effect that the other planets have on the precession of Mercury. I know there are methods listed in the paper by Price and Rush, 1978, which states there is more methods presented in undergraduate mechanics books. However, I have never came across such things.

Is there a method in which one can derive both the planetary and GR precession effects or do we need to separate the precession into a Newtonian and relativistic form and then add them together?

EDIT: Furthermore, what about an arbitrarily chosen planet? The ring method mentioned in the answer by @diracology and the paper that I have linked assumes assumes a uniformly distributed ring of mass surrounding Mercury. However, what if we were concerned with planetary and GR precession effects of Jupiter (just as an example) - Do we then proceed to consider two rings around Jupiter?

$\endgroup$
  • 1
    $\begingroup$ I don't understand why this was down voted. I think the question is posed well and is neither too broad nor has the potential to start an unnecessary discussion. $\endgroup$ – Rumplestillskin Feb 27 '17 at 9:38
  • 1
    $\begingroup$ The new title question (v4) [which does not include GR] seems to be a duplicate of physics.stackexchange.com/q/261118/2451 $\endgroup$ – Qmechanic Feb 27 '17 at 15:05
  • $\begingroup$ I've changed it to be more specific and not duplicate previous posts. $\endgroup$ – Rumplestillskin Feb 28 '17 at 0:36
2
$\begingroup$

It is possible to derive the precession of Mercury perihelion in some sort of unified way. However we still need to take into account the Newtonian and Relativistic contributions. They cannot be put in equal foot from the physical point of view.

The idea is to write the effective potential with the centrifugal term, the non-perturbative term that represents the interaction between the Mercury and the Sun, the Newtonian gravitational interaction with the other planets and the Relativistic correction due to spacetime deformations. The last two contributions being small perturbative terms.

The idea is that since this effective potential has a stable minimum, it can be Taylor expanded. The second order term actually gives the period of the radial oscillations from which we can obtain the angular displacement (for small perturbations). If the angular displacement corresponding to one period is different to $2\pi$ it means that the planet's orbit is precessioning.

The complete derivation can be seen in this post Details of Newtonian Prediction for Mercury's Precession. You only need to take into account the relativistic contribution $$U_{GR}=-\frac{GML^2}{mc^2r^3},$$ where $G$, $M$, $c$, $r$, $m$ and $L$ are the Universal constant, Sun's mass, the speed of light, the distance from the Sun to the planet, the mass and the angular momentum of the planet, respectively. The two terms together give a total of $575$ arcsecond per century.

$\endgroup$
  • $\begingroup$ I don't understand what you mean when you say (at the end of your first paragraph): "They cannot be put in equal foot from the physical point of view". Can you explain? $\endgroup$ – NickD Feb 27 '17 at 20:45
  • $\begingroup$ @Nick It just mean that the two terms have different origins, one is Newtonian and the other is Relativistic. $\endgroup$ – Diracology Feb 27 '17 at 20:53
  • $\begingroup$ Great answer! I had come across this "ring" method that is mentioned in the post you have linked too. But what about if it were Jupiter? So the planet feels perturbing forces from both planets closer to the Sun and also further away. Does the ring method still work? $\endgroup$ – Rumplestillskin Feb 28 '17 at 0:45
  • $\begingroup$ @Diracology I've edited the question to incorporate my comment. $\endgroup$ – Rumplestillskin Feb 28 '17 at 0:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.