## Gravity speed of 10.6c gives Mercury 43 arcs/cent.

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crawler
Posts: 553
Joined: Sun Oct 28, 2018 5:33 pm

### Gravity speed of 10.6c gives Mercury 43 arcs/cent.

Mathpages – 6.2 – says ......... https://www.mathpages.com/rr/s6-02/6-02.htm
......Hall noted that he could account for Mercury's precession if the law of gravity, instead of falling off as 1/r^2, actually falls of as 1/r^n where the exponent n is 2.00000016.......

I crunched some numbers in Excel & i found that Hall's exponent of 2.00000016 was equivalent to gravity having a speed of 10.6c.
Mercury's orbit goes from 46 million km to 70 million km in 44 days, which is an average change of 6.3131 km/s.
The average orbit is 58 million km.
If R^2.00000016 equals r^2, then R equals 57.999 885 million km, which is 115 km less.
Mercury on average takes 18.2154 sec to change its orbit by 115 km (ie at 6.3131 km/s).
A speed of 57 999 885 km in 18.2154 sec is 3.1841 million km/s, which is 10.6c.

I fail to understand how a symmetrical 2-body force can give precession, & all orbital 2-body forces are symmetrical.
However assuming that i am missing something, & assuming that Hall's equations were correct (but the mechanics not), the 43 arcs/cent can be explained by the exponent for r being 2.000000000, but with gravity having a speed of 10.6c.

Me myself i prefer a much faster speed for gravity, affecting precession in a tangential way, rather than the above (symmetrical) longitudinal gravity effect. I will have a think about it. Firstly, a tangential effect is likely to be non-symmetrical, so that would be a good start.

Mathpages -- 8.10 -- has some good stuff re Gerber. And Einstein's earlier 18 arcs/cent.
https://www.mathpages.com/rr/s8-10/8-10.htm
Mathpages – Gerber's Gravity.
https://www.mathpages.com/home/kmath527/kmath527.htm
Pejic mentions Asaph Hall (1894).
https://math.berkeley.edu/~mpejic/pdfdo ... ession.pdf
Page 1275 mentions Hall's theory.

crawler
Posts: 553
Joined: Sun Oct 28, 2018 5:33 pm

### Re: Gravity speed of 10.6c gives Mercury 43 arcs/cent.

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.

crawler
Posts: 553
Joined: Sun Oct 28, 2018 5:33 pm

### Re: Gravity speed of 10.6c gives Mercury 43 arcs/cent.

crawler wrote:
Fri Feb 28, 2020 3:39 am
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..........

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...........

All planetary precessions are due to the finite speed of gravity (ignoring tidal effects).
My red wordage is mistaken. I can see that a (circular) ring-planet will always give a forwards tangential tug to (an elliptical) Mercury, except of course when that tug falls to zero at the two ends of Mercury's ellipse. But the tug is never backwards (in the simple ring-planet model).
In other words the ring-planet's tangential forwards tug is in that sense similar to the never-ending tangential forwards tug arising from the finite speed of gravity (except that this never falls to zero at any point).

This means that the effect of a finite speed of gravity does not have to fill the whole of the 574.10 arcs/cent, it only needs to fill say 45 arcsec.
Otherwise i stand by everything else that i wrote.
And silly spacetime of course adds zero to Mercury's precession.

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