4 ms·
> create gravity on the outer surface That's a tricky one. Do you mean make the ring thick enough that it has enough mass to have its own gravity?
by powmonk 10y ago
> create gravity on the outer surface
That's a tricky one. Do you mean make the ring thick enough that it has enough mass to have its own gravity?
- rbanffy 10y agoNo. When the ring rotates at orbital speed, there is no perceived gravity on either side. If the ring rotates faster than that, one would perceive a force pulling them away from the central star. If, however, the ring rotates below orbital speed its mass would be pulled toward the central star (along with anyone standing on its outer surface). The forces pulling it towards the star would compress its structure (the opposite of the traction forces a faster ring with positive gravity on the inner surface would experience). I think (I haven't checked the numbers) the ring's mass can't be neglected, but, because of the shape, it'd act as an extra mass within the central star. The ring is, in fact, in its own orbit.
- rbanffy 10y agoGood and bad news. A stationary ring around a star with one solar mass would be very close, about a tenth of the distance Mercury orbits the sun, to make its inhabitants experience one Earth gravity on its outer surface. The good news is the ring would be much smaller than 1AU and it'd be easier to do it around a large white dwarf and still get a lot of energy from it. Not sure how much pressure solar wind would be, but, if the star is active enough, we could use it to partly sustain the ring structure.