Comment by srl | Hacker News Reader

Comment by srl | Hacker News Reader

(Lest I come across as a complete ass: this is a cool piece, of course.)

At least the discussion of gravitation is wrong. Somewhat unintuitively, the gravitational force on the inner part of the torus (the surface closest to the center, on the plane of the axes) is 0.

The pathological example is a hollow sphere of dense material. Outside the sphere, it "looks like" (if you just measure g) a solid planet. Inside the sphere, there is no gravitational field whatsoever, no matter how close you come to the surface.

Surprisingly, this holds no matter how large the sphere is. Suppose you're sitting on the inside of the surface of the sphere, and you decrease the radius a bit. Now the gravitational pull /away/ from this surface decreases like r^-2, so you would expect the gravity towards the surface (which is essentially unchanged) to increase. The issue is that the amount of material opposite you - the surface area of a sphere, really - also increases like r^2. (This is informal, but the best I can do for an intuitive explanation.)

Tough to find a good explanation of this online. http://physics.stackexchange.com/questions/364/gravity-on-a-...

I haven't done this calculation, so I'm not going to immediately say you're wrong, but the example you use (a hollow sphere) doesn't generalize to other shapes.

Gravity is zero inside a hollow sphere or inside a hollow infinite cylinder. Here's the conceptual reason why. Imagine that you're near the left side of the sphere or cylinder. Then the mass to your left is closer to you, and therefore every kg of it exerts a strong gravitational pull (as force is proportional to 1/r^2). But although the mass to your right is farther away and thus exerts less gravitational force per kg, there's a lot more of it: roughly speaking, the amount of mass in a given direction is proportional to r^2 (times the solid angle it subtends). This factor of r^2 precisely cancels out the effects of the 1/r^2 weaker force per kg. (And yes, a careful derivation using calculus or Gauss's law would be more rigorous.)

None of that applies in the case of a torus. If I'm in the hole of the torus close to the left side, there's still a 1/r^2 difference in gravitational force per kg favoring the pull toward the left. But now, roughly speaking, the extra mass to my right is only greater by a factor of r (coming from the circumference of a circle rather than the surface area of a sphere). So the far side of the torus becomes less and less significant the father away it gets.

On top of all that, my impression from the article was that this sort of planet would require a really fast rate of rotation. I get the sense that the virtual "centrifugal" force plays a major role in the physics here. (Or if you prefer, that it's not reliable to ignore the effects of being in an accelerated reference frame.)

Dammit! You're right, I'm (very) wrong. (The stackexchange answer I used as a "check" was wrong too. That'll show me.) I was lazily using Gauss's law and ignoring the fact that a torus isn't spherically symmetric.

Could you please point out where a stackexchange answer was wrong on that question? I'm a heavy user on this site and have written (and continue to write) a lot on this subject there. Thanks.

Item 3 on the answer from the user "Sklivvz♦". It was corrected in a comment.

I guess an intuitive arguement for why the gravity should be nonzero is to consider the limit as the major radius gets bigger and bigger. When it is very big, for a person on the surface it seems as if you are standing on the surface of a cylinder (the rest of the torus is far away), so you feel some gravity. So it can not be identically zero.

Note, though, that the analogous argument for gravity inside a sphere doesn't work: make the sphere very big and stand at some point on the inside. You'd think that you'd feel the same gravity as you would due to an infinite plane (constant g toward the ground), but in fact it would come out to zero because there's so much distant mass in other directions (the r^2/r^2 argument I made earlier). So it pays to be careful!

Yes. But standing inside the sphere it is kindof intuitive that the far-away parts of the sphere are not negligible---they fill up the entire night sky!

Are you taking into account the correction from centripetal force, though? Even in the case of the hollow sphere, if we spin it fast enough it'll work just like your standard spinning space station, on the inside surface (or at least on a band-like area of the inside).

I find it difficult to believe that the author made such a glaring mistake, even after doing the simulations and all. Also, he says " the surface is an equipotential surface, so gravity (plus the centrifugal correction) is always perpendicular to it. " The spinning is the reason this ridiculous configuration is at least theoretically stable.

I'm not taking into account centripetal acceleration, but it is small in ordinary planets. In order for the planet to actually be a torus, it needs to be spinning extraordinarily quickly, precisely because gravity does not help maintain the torus shape.

From my reading, the author's simulations presupposed what s/he believed the gravitational field is. (This is actually a reasonable thing to do, and I don't fault him for it.)

Well, the planets are tori here because of the extraordinary spinning; they have 3-4 hour long days.

I'm pretty sure the whole point is that the planet has to be spinning fast enough that the apparent gravity on the inside equator is roughly equal to the actual gravity on the outside of the equator.

That's not correct, but it's a deceptive trick. The governing condition is that the surface is equipotential after you include both the gravitational and centripetal fields. It's tempting to then think that gravity will be constant over the surface, but that's not what equipotential means. Gravity is always normal to the surface (straight "up"), but it is allowed to vary in strength. This actually happens for Saturn, where the gravity on poles versus equator felt by someone standing there (not that you can stand there) differs by around 10%.

Reply on news.ycombinator.com