Dale.Amon@CMU-RI-FAS@sri-unix.UUCP (09/17/84)
Over a pizza last night, I realized that my previous analysis could be stated better. Assume a diffuse, evenly distributed mass M contained with radius R. At any radius r within R, the portion of mass interior to r can be modeled as a point mass and a square law applied. All mass contained in volume infinitisemals exterior to the radius cancel out. If you move outwards by dr, the volume (and thus enclosed mass) increases by a cube, while the attraction from the original radius decreases by a square, resulting in a linear INCREASE in force. This holds for each r+dr =< R. For r+dr>R, a straight square law DECREASE in force occurs. Consider moving from the center of the Earth outwards. At the exact center all mass is exterior to you and you are at 0g. As you move outward, the gravity increase linearly to 1G at the surface, and then decrease via a square law as you move away from the surface. I'd try to do better, but 1) I don't have my copy of Thomas here. 2) Ascii doesn't handle integral equations, and a triple integral in spherical co-ordinates is the real way to express all this garbage.