I am having trouble understanding something about the precession of Mercury. The precession takes place in the same direction of motion as the orbit of the planet, so that the ellipse of the orbit becomes angled slightly more forward over time, right? In a gravity simulator, this amounts to an increase of gravity of (G M / r^2) * (1 + 3 *(v/c)^2) to second order in terms of the instantaneous speed of the planet, or likewise with (G M / r^2) (1 + 6 G M / (r c^2)) in terms of the instantaneous distance from the center of the sun. A decrease of the Newtonian gravity gives a precession in the opposite direction. Okay, so the local gravity at r matches the Newtonian gravity G M / r^2, correct? A particle travelling radially at r will be gravitationally time dilated and length contracted according to a distant observer, each by sqrt(1 - 2 G M / (r c^2)), so that the distant observer will say that the particle travels (1 - 2 G M / (r c^2)) slower than the local observer measures, right? So here's my question. If the local gravity is G M / r^2 and a distant observer uses a time dilation and length contraction of the local measurements for what he measures, then just as the distant observer measures a lesser speed of an infalling particle, so too should he measure a lesser acceleration of the particle, and similarly for orbit, although without the length contraction in the tangent direction, shouldn't he? If so, then since the distant observer measures a lesser acceleration of Mercury, the precession should be in the direction opposite the direction of motion, while precession in the forward direction requires an acceleration measured by the distant observer that is greater, not less, than the Newtonian gravity. So what is going on here? How do we get a precession in the positive direction with a lesser acceleration of gravity?