[Loser]

The gravity of a shell is not precieved by objects within the shell. For this reason, the gravity you experience will grow less and less as you penetrate deeper and deeper into a body such as the Earth.

This is how black holes work. There are many, many stars with sufficient mass to become a black hole. However, they are too big and continue growing as they get hotter with age. It is the heat that pushed the mass of a star ourward from the nuclear reaction that occurs at or very close to its core. At some point a the fusion reactor at the center of the star runs out of fuel that it can fuse at the pressure exerted by the mass of the star pulled inward by its own gravity. At this point the star collaspes in on itself, as it is not producing enough heat to keep itself puffed up. When a sufficient amount of the stars mass gets compressed into a sufficiently small space then you will have a black hole.

At least, I'm pretty sure that's how that works. I am sure about the shell part.

From the outside, the shell will will be an immense amount of gravity. But if the outside surface is spinning, and it will be spinning much faster than the earth's orbital velocity, then at the equator one would be thrown off at around 1 G. In thruth it would be pretty damn hard to ladn on, you'd have to be on one wicked eliptic... though I could have the scale wrong.

Additionally, no human would be able to run or jump fast enough agaisnt the spin of the world to escape it and start floating. I believe, in order to get the 1 G, you'd have to be cranking damn fast.

You actually have some options, where radius and rpm are concerned, since the equation for relative centrifugal force goes like this.

RCF (in g forces) = 1.119x10^-5 x rpm^2 x radius (in cm)

hmm... that rpm is going to be troublesome... oh, centimeters, too... Yeah, the units in this equation are going to kill me.

In any case the gravity will definately be sufficient to keep the atmosphere pinned into the 'bowl' at the equator.

Kamog, that is an excellent solution for the "gradient gravity of a sphereworld" problem. Note, however, that you would need huge, ringworld-style walls between the rotating sections or the atmosphere would all flow downhill. Note also that when the plane formed by that ring does not bisect the star's center of gravity the ring is not in a 'stable orbit' the way a ringworld or solid sphereworld are Though the outer sphere could just provide a mounting point for these various rings, which I think you suggested. Remember that they would, however, be subjected to a pull away from the sphere and toward the star, and the 'ground' would have to be inclined to compensate.

[ November 26, 2003, 18:27: Message edited by: Loser ]