Stephen A. Lawrence wrote: > > > Harry Veeder wrote: >> Consider the situation far from any planets or stars. >> >> If the ball-bearing is initially at the centre of the shell it will remain >> there. If it is initially off centre, the ball bearing and the shell will >> move so as to minimize the distance between the point on the shell that was >> initially closest to the ball bearing. >> >> >> This is my prediction. It does not violate conservation of momentum, >> but it is not based on Newton's or Einstein's conception of gravity. > > I haven't worked it out, but I think a force going as 1/r^3 would have > that effect. Actually any rate of falloff faster than 1/r^2 should do > that, I think.
I wouldn't try to "work it out". The theory which gives rise to this prediction is in the very early stages of gestation. > But ... the fact that Mercury's orbit precesses as it does is evidence > that "real" gravity around a spherically symmetric object doesn't > actually fall off as 1/r^2 (falls off a bit faster, IIRC), which makes > me wonder whether "real" gravity would also show that effect, albeit > weakly (my "proof" using the Ricci tensor notwithstanding ... among > other things I assumed a massless ball bearing, which is a little wrong). > In Newton's mathematical model of gravity an equivalency exists between power-of-inertia and power-of-attraction. The properties of inertia (a.k.a inertial mass-points) are transferred to points-of-attraction (a.k.a. gravitational mass points) through the equivalency. Also since the power-of-inertia is communicated by contact the transference leads to the paradox of action at a distance. One way to address the paradox is to extend inertia into space and time, which is what Einstein did. In this way we are never out of "touch" with inertia. My way is to begin by giving gravity a distinct power from that of inertia. Harry
