Mercury's Orbital Precession in Special Relativity I am researching Mercury's orbital precession. I have considered most perturbations and general relativity. I am still not satisfied. I need your help.
I need a solution to Exercise 13, Chapter 6, in Ref. 1 (which is Exercise 26, Chapter 7, in both Ref. 2 and Ref. 3).
The exercise is copied below:

Show that the relativistic motion of a particle in an attractive inverse square law of force is a precessing motion. Compute the precession of the perihelion of Mercury resulting from this effect. (The answer, about 7" per century, is much smaller than the actual precession of 43" per century which can be accounted for correctly only by general relativity.)

I have the solution to Exercise 7, Chapter 3.
References:


*

*H. Goldstein, Classical Mechanics, 1st edition, 1959. 

*H. Goldstein, Classical Mechanics, 2nd edition, 1980. 

*H. Goldstein, Classical Mechanics, 3rd edition, 2000. 
 A: I think this is the Thomas precession, which is a kinematical effect that depends on the shape of the worldline and is independent of the nature of the force.
Wikipedia gives a low-speed approximation for the Thomas precession of $ω_T=av/2c^2$. For a circular orbit with radius $r$ and speed $v$, the precession per orbit is $$Δθ = (2πr/v)ω_T = πra/c^2 = πv^2/c^2$$
which agrees with the low-speed, low-eccentricity formula in Fausto Vezzaro's answer (using $v^2/r=GM/r^2$).
This preprint gets a Thomas precession of 7.163″/hyr from a more careful calculation that takes the eccentricity into account. It also says that this is a problem for general relativity, which isn't true (calculations in GR automatically include SR "effects"), but I suppose the special-relativistic calculation is correct in spite of that.
This preprint, which was mentioned in a comment by Pulsar, derives a similar result with no mention of Thomas precession, and then a result twice as large (14.3″/hyr, one third of the GR prediction) from what is allegedly a more careful treatment.
