I fail to see how a radial force of any kind can precess an orbiting 2-body system (ignoring changing shape, & tidal effects)(if the radial force is constant or not). If the radial force is symmetrical (ie follows a fixed repeating equation) then there can be no such precession, but there can be precession due to a happy choice of starting point, but this happy precession will be prograde or retrograde, & an average (for the full range of starting points) must i reckon give zero precession.

In other words the ring-planet model precessions are rubbish, & the Einsteinian precessions too. In the long run (if not the short run) they must go to zero. What we need is a non-symmetrical force, ie a force that doesn't negate during a half of each orbit.

One such force is gravity if gravity has a finite speed. Here in a 2-body system each body feels that the gravitational attraction of the other body comes from an old (retarded) position of that other body.

If gravity has a speed of 20 billion c then the Sun feels that Mercury (in its fast big elliptical orbit) is retarded by 0.4638 mm, & Mercury feels that the Sun (in its slow little elliptical orbit around the Sun/Mercury barycentre) is retarded by 76.98 pico metre (a ratio of 1 to 6,025,265).

The Sun's ave orbit around their barycenter = 87.97 days, 9.6261 km radius, 7.958 mm/s.

According to Van Flandern the Laplace 1825 equation is A^2 = a^2 + 4*(G*m)*(T – t) / V.........

https://www.intalek.com/Index/Projects/ ... ntsSay.htm
……… where (A) is the new semi-major axis, (a) is the initial semi-major axis (70,000,000,000 m for Mercury), (G) is Newton's G, (m) is the mass of the Sun (1.989 by 10^30 kg), (T – t) is the interval (1 year)(31,557,600 sec), & (V) is the speed of gravity.

For V = 20 billion c we find that (a) increases by 39.9260 m in one year (4 km in 100 years).

Some might say that 40 m/yr appears too large, ie it would show up in modern measurements. In that case the speed of gravity might be more than 20 billion c, but i reckon that much of that 40 m/yr is negated by friction & drag due to the Sun's atmosphere, reducing the 40 to perhaps say 1 m/yr. Or if u prefer a faster speed of gravity then it might be 10 m/yr reducing to 1 m/yr (or even 2 m/yr reducing to 1 m/yr).

I think that NASA calculate that Earth's orbit increases by 15 mm/yr due to the Sun losing mass in accordance with E = mcc. In which case the number for Mercury is say 6 mm/yr based on an orbit of 58 million km compared to Earth's 149.6 million km. I think that NASA reckon that they can measure Mercury's orbit to about 0.1 m, & that annual variations can total say 40 m due to many factors.

Me myself i dont believe that E = mcc, & i reckon that energy has zero mass. I reckon that photons have mass, & if the Sun emits photons then of course the Sun loses mass. Neutrinos are paired photons, & hencely neutrinos have twice the mass of a photon. I dont know whether a photon's mass depends on the photon's energy, i suppose it does.

Anyhow, if gravity gives Mercury a larger orbit then i suppose that it (at the same time) gives Mercury a precession. I don’t know what the equation looks like, but i suspect that it might yield 45 arcsec/cent.

In addition, i reckon that calculation will yield a total of 574.10 arcsec/cent (ie the aforementioned full observed number) when the real orbit of the Sun around the solar barycentre is taken into account.

The aforementioned 76.98 pm retarded position of the Sun doesn't ever become an advanced position, hencely the 45 arcsec/cent wont depend on starting point. I think that the same can be said re the 574.10 arcsec/cent.

All planetary precessions are due to the finite speed of gravity (ignoring tidal effects).

And i don't believe that the Sun's oblateness produces any precession.

And of course silly faux-spacetime cant produce precession of any spacetime kind.

I fail to see how a radial force of any kind can precess an orbiting 2-body system (ignoring changing shape, & tidal effects)(if the radial force is constant or not). If the radial force is symmetrical (ie follows a fixed repeating equation) then there can be no such precession, but there can be precession due to a happy choice of starting point, but this happy precession will be prograde or retrograde, & an average (for the full range of starting points) must i reckon give zero precession.

In other words the ring-planet model precessions are rubbish, & the Einsteinian precessions too. In the long run (if not the short run) they must go to zero. What we need is a non-symmetrical force, ie a force that doesn't negate during a half of each orbit.

One such force is gravity if gravity has a finite speed. Here in a 2-body system each body feels that the gravitational attraction of the other body comes from an old (retarded) position of that other body.

If gravity has a speed of 20 billion c then the Sun feels that Mercury (in its fast big elliptical orbit) is retarded by 0.4638 mm, & Mercury feels that the Sun (in its slow little elliptical orbit around the Sun/Mercury barycentre) is retarded by 76.98 pico metre (a ratio of 1 to 6,025,265).

The Sun's ave orbit around their barycenter = 87.97 days, 9.6261 km radius, 7.958 mm/s.

According to Van Flandern the Laplace 1825 equation is A^2 = a^2 + 4*(G*m)*(T – t) / V.........

https://www.intalek.com/Index/Projects/Research/TheSpeedofGravity-WhattheExperimentsSay.htm

……… where (A) is the new semi-major axis, (a) is the initial semi-major axis (70,000,000,000 m for Mercury), (G) is Newton's G, (m) is the mass of the Sun (1.989 by 10^30 kg), (T – t) is the interval (1 year)(31,557,600 sec), & (V) is the speed of gravity.

For V = 20 billion c we find that (a) increases by 39.9260 m in one year (4 km in 100 years).

Some might say that 40 m/yr appears too large, ie it would show up in modern measurements. In that case the speed of gravity might be more than 20 billion c, but i reckon that much of that 40 m/yr is negated by friction & drag due to the Sun's atmosphere, reducing the 40 to perhaps say 1 m/yr. Or if u prefer a faster speed of gravity then it might be 10 m/yr reducing to 1 m/yr (or even 2 m/yr reducing to 1 m/yr).

I think that NASA calculate that Earth's orbit increases by 15 mm/yr due to the Sun losing mass in accordance with E = mcc. In which case the number for Mercury is say 6 mm/yr based on an orbit of 58 million km compared to Earth's 149.6 million km. I think that NASA reckon that they can measure Mercury's orbit to about 0.1 m, & that annual variations can total say 40 m due to many factors.

Me myself i dont believe that E = mcc, & i reckon that energy has zero mass. I reckon that photons have mass, & if the Sun emits photons then of course the Sun loses mass. Neutrinos are paired photons, & hencely neutrinos have twice the mass of a photon. I dont know whether a photon's mass depends on the photon's energy, i suppose it does.

Anyhow, if gravity gives Mercury a larger orbit then i suppose that it (at the same time) gives Mercury a precession. I don’t know what the equation looks like, but i suspect that it might yield 45 arcsec/cent.

In addition, i reckon that calculation will yield a total of 574.10 arcsec/cent (ie the aforementioned full observed number) when the real orbit of the Sun around the solar barycentre is taken into account.

The aforementioned 76.98 pm retarded position of the Sun doesn't ever become an advanced position, hencely the 45 arcsec/cent wont depend on starting point. I think that the same can be said re the 574.10 arcsec/cent.

All planetary precessions are due to the finite speed of gravity (ignoring tidal effects).

And i don't believe that the Sun's oblateness produces any precession.

And of course silly faux-spacetime cant produce precession of any spacetime kind.