Okay. So I've been trying to fit the precession of the planets with the time gravity would take to travel the distance to the planets and the distance the sun might move during that time. Didn't work. So last night, I'm sitting here thinking about how the orbit of Mercury would effect the sun's motion to begin with. Even if it did to any significant degree, one must also figure in how the other planets effect it as well. So Mercury's precession should be effected by the sun's motion due to all planets, probably a little herky-jerky, but not enough to do anything major. Besides, all of the motions of the planets might about even out in some way, so there should be very little motion of the sun anyway.
So I was back at square one. I thought, what is it about the sun that is different than the Newtonian values by considering it a stationary point? Well, it is not a point, of course. There is a gradient across its width. So, what is it about its width that is different from one side to the other? Well, it spins, of course. And then I had it.
Okay. So the sun spins. The Doppler effect directly toward the sun would then be 1+(sin 0)v/c on one side and 1-(sin 0)v/c on the other, where v is the velocity at the equator of the sun. This would average out across its gradient as ([1+(sin 0)v/c]+[1-(sin [i]0[i])v/c])/2=1. Well, no luck there. Gravity simply evens out across the gradient directly toward the sun. But what about tangent to it? In this case, we would take the difference between the gravity acting on one side of an object to the other, averaged over the gradient from the center, and the result is ([1+(cos 0)v/c]-[1-(cos 0)v/c])/2=(cos 0)v/c. The cosine of the gradient is Rsun/(Rsun2+d2)1/2. But even for a planet as close as Mercury, the radius of the sun is not much compared to the distance of Mercury, so the entire formula reduces to simply (v/c)(Rsun/d).
Well, that's it. That's the whole formula. No way it could be that simple, you say? Let's find out. By the way, another way to think about what is going on here is to imagine the velocity of rotation of the sun at the surface, (v/c), and then project it to some other distance, as if the gravitational field rotates with the sun itself. It is the same as a gravitational field projecting itself in a straight line with the original velocity of a body, even after the body has changed directions. And the same thing with EM waves, I suppose.
The rotation of the sun will act as if it would turn an entire elliptical object as large as the solar system with it, so also an orbit little by little. The most common form of units for precession seems to be that of arc-seconds per century. This formula is a ratio for the amount of rotation of orbit per orbit. So we will convert it. To do so, we simply multiply it times the ratio of the time per century to the time per orbit, and then by degrees of orbit. The formula is now (v/c)(Rsun/d)(tcentury/Torbit)(360 degrees * 3600). Now we will apply it to Mercury and see what happens. First, though, we will replace v for the sun with 2piRsun/Tsun, which amounts to the same thing. Our entire formula is now precessionMercury=2piRsuntcentury(360 degrees * 3600)/TsuncdMercuryTMercury. Next we must find the average distance of Mercury from the sun. Its closest distance is Dc=4.6*1010 meters and its farthest is Df=7*1010 meters. We could find the average with (Dc+Df)/2=5.8*1010 m, or (DcDf)1/2=5.675*1010 m, or we could take the distance Mercury would be if it were to orbit in a perfect circle in the same time for orbit, or dMercury=(GMsunTMercury2/4pi2)1/3=5.791*1010 m. I'm actually not sure which one should be used here, but the last one seems most likely, and it doesn't differ much from the others anyway.
So finally, here we go. For Rsun=6.96*108 m, Tsun=2.193*106 sec, dMercury=5.791*1010 m, TMercury=7.6005*106 sec, and tcentury=3.1536*109 sec, we get a precession of Mercury of 42.992 arc-seconds per century. Not bad, huh? Of couse, this can also be applied to the other planets as well. It could also be applied to stars orbitting the galaxy, since the galaxy also rotates. For stars orbitting on the outer edge, Rgalaxy/d=1, so they would have a large precession. But since this precession is within their orbit, it would simply add to their velocity, and they will orbit much faster than it would seem they should. This might explain that whole dark matter thing as well.