I thought there was no gravity on the inside of a ring so the exact middle of the inside face would not have any gravity at all.
Very true. However, if my memory’s not deceiving me, he did have a toroidal … um, well, planet is an exaggeration. I dunno where my copy of Protector is, but there should be a made toroid (with non-touching sphere in center) somewhere around 2/3 or the way through the book.
Larry Niven did built a toroidal world in the 1960s – it’s in Protector, and the first paperback edition features the toroid world on the cover.
IIRC, the human Pak Brennan who built it used super-science to put it and keep it together. I can’t imagine a toroidal planet, even of the size shown on the cover of Protector, being at all stable.
This is what I learned in physics class as well, but I can’t find my textbook for a cite. The basic rule I remember is that for a radially symettric object, if you draw a circle around its center you only need to consider the mass inside the circle when calculating gravitational force on that circle. If we draw a circle along the inside of our toroid, there is nothing inside our circle and therefore no gravity all along the inside edge.
In this case, I would be rather worried about the air and water all leaving after several thousand years because of this. While the atmosphere wouldn’t be instantly pulled off the surface, the sun-warmed atoms would start to migrate to the center and form a disk shaped collection inside the ring. If this happened enough, the atmosphere and water could start to have a weak gravitational pull of its own, accelerating the process. Hmm, now I need to download a physics simulation and actually prove this.
The problem is that your torus isn’t “a radially symmetrical object”. In order for that law about ignoring things outside the circle to work, it must be spherically symmetric. In a torus, you’ll have conmtributions inside the ring due to gravitational influences outside it.
Heck, the calculation isn’t that hard for a simple case – imagine a point on a circle, and calculate the gravitational forces due to the rest of the circle. You’ll find there’s a small force pulling the object toward the center.
The problem is that your torus isn’t “a radially symmetrical object”. In order for that law about ignoring things outside the circle to work, it must be spherically symmetric. In a torus, you’ll have conmtributions inside the ring due to gravitational influences outside it.
Heck, the calculation isn’t that hard for a simple case – imagine a point on a circle, and calculate the gravitational forces due to the rest of the circle. You’ll find there’s a small force pulling the object toward the center.
That works out to a 26-36% grade. That’s a pretty steep slope for loose soil. It seems to me that whenever the wind picks up some dust it would fall toward the inner part of the torus so the world would wind up lopsided over time.
If the planet was spinning around an imaginary axis at the center of the hole, would not the centrifugal forces on the inner edge of the torus counteract any gravitational effects?
That, I think, was largely the point of Niven’s Ringworld. But the gravitational force is still there, even if centrifugal force gives a fictitious force in the other direction (It’s there in Ringworld, too). But the original premise said nothing about spinning. In fact, why assume spinning about that axis , or only that axis. Spinning about an axis that passed through the torus itself would give more interesting, if less useful, results.
Blah, I’ve been out of physics class too long to add anything useful. I was thinking of a spherically symmetric object, and that doesn’t apply in this situation.
However, This link, suggests that the torus would hold together, and the inside of the ring would feel attracted out toward the ring. Ask a scientist agrees, points inside the torus would be pulled outward. I guess this could work after all, even without having to spin it.
I’m sorry, maybe I’m being dense (not to mention 3 days late), but I don’t think that awldune’s question has been answered yet. I still don’t understand what is meant by “Our spherical Earth only wraps East-West.” On a sphere, what difference does it make what direction you travel? If you travel long enough, you eventually come back to where you started (or “‘appear’ on the the other” side).
If the Earth were a cylinder I could understand this, but it isn’t, ergo, I can’t.
If you walk off the west edge of a map of the earth, you reappear on the eastern edge of that map. But, if you walk off the north edge, you don’t reappear on the south.
That’s entirely an artifact of how we draw maps, not a property of a sphere.
If we use essentially any other rectangular coordinate system, then walking due North would eventually have you reappearing on the other side of the map. We just happen to use unique distorted mapping, where the entire northern edge is really a single point, which we deem “as far north as you can get”.
It’s not an entirely arbitrary choice --it has much to recommend it-- but still, it’s a cartographic property, not a geometric one. Cartographic properties don’t affect physics in the slightest; they’re social choices.
In fact, the whole issue of “the opposite edge” presumes a rectangular map (which has little merit for drawing the surface of a globe). As individuals, we may carry this prejudice over to, say a Mercator projection, because we’re accustomed to seeing them drawn on rectangular sheets, but all we should really say about a Mercator or any other projection is "if you keep walking in any direction, you will hit some edge, and reappear on some other edge. THAT would be the only appropriate and generally meaningful geometric interpretation of a “wrap”.
Unless you live in a Square RPG.
But the point is, if you had that “wrap” property on both dimensions of the map, the map would be describing a torus.
Here’s my quick and dirty solution for the case where you are standing on top of the donut.
It looks like you would feel a force of 1.02 times your normal weight inclined at 11.3 deg. when the opposite side’s center is 2r away. In other words the case where the donut has no hole. That’s worst case.
Correction, correction. I went to all the trouble of figuring the angel of the gravitation from the opposite side of the torus and then forgot to use it (hardly anyone is perfect any more).
The corrected worst case is that the total force would be 1.1 inclined at an angle of just over 9.3[sup]o[/sup].
I know it’s just a typo, but I love the idea of “the Angel of Gravitation”, and I’d like to see a picture.
I’m searching for a picture of something like a giant suction cup with slimy tentacles pulling us down into Great Dismal Swamp.
And how many times to you preview to make sure that Gaudere’s Law didn’t bite you?