Torus-Earth(aleph.se)

http://www.aleph.se/andart/archives/2014/02/torusearth.html

66 comments

dzuc12년 전

A few years ago I was wondering what this would look like and (less accurately) modeled it in 3d: http://farm3.staticflickr.com/2630/4085959963_8dc8217283_o.j...

roywiggins12년 전

That was fun. If you like this sort of thing, "Heavy Planet"[1] is a good hard scifi book set on Mesklin, a fictional super-Earth which spins fast enough to make gravity significantly less at the equator. It's also a good adventure story, if you like 1950s scifi.

[1] http://en.wikipedia.org/wiki/Mesklin

AlanSE12년 전

How does this compare to Mission of Gravity? I've had that on my shelf and haven't got around to it yet. It's a very cool setting, and the author was writing it before we had all the research on the subject we have today. And he was correct thankfully!

gliese133712년 전

He actually wasn't 100% correct. If you get the omnibus "Heavy Planet" edition (including a sequel novel "Starlight" and some short stories), it is explained in an appendix that Hal Clement's estimation of the gravitational field was somewhat off, resulting in erroneous values for the maximum field strength at the poles.

Still, Hal Clement did a bang-up good job, and the error (or correction there of) doesn't negatively affect the story at all.

cossatot12년 전

This is cool! But the important question of what mantle convection, and therefore tectonics, looks like on a toroidal earth is not addressed... for the 'donut' earth where surface gravity varies by a factor of 3 on the surface, one could envision subductions systems, possibly double (two-slab) types, setting up at the northern and southern polar rings, while spreading systems would set up at the inner and outer equators. I have no idea what would happen to lighter, more differentiated arc rocks... would they gather in the middle of a two-sided subduction system, even though that is a gravitational high? Given the instability of these planets to perturbations, it's possible that sufficient redistribution of mass due to tectonics and crustal differentiation would be enough to rip them apart.

Also, setting up a geodynamo would be tough...

gliese133712년 전

It seems more likely that subduction would happen at the inner equator, and spreading systems at the outer. This is addressed briefly in the article- when plates move around the torus, they must compress to fit into the smaller diameter of the inner equator, and stretch out when moving towards the larger outer equator.

srl12년 전

(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-...

Steuard12년 전

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.)

srl12년 전

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.

AlanSE12년 전

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.

sourkremlin12년 전

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

vilhelm_s12년 전

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.

Steuard12년 전

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!

vilhelm_s12년 전

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!

Kutta12년 전

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.

srl12년 전

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.)

Kutta12년 전

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

wyager12년 전

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.

AlanSE12년 전

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%.

[deleted]12년 전

(1)

aruggirello12년 전

IMHO the surface of such a planet should be nothing more than molten lava, as the outer parts of the torus are probably going to rotate much slower than the inner parts (much like the rings of Saturn do). This would not allow for any solid crust to form.

I agree with srl: gravitation must get gradually closer to zero in the inner part, where the molten lava, not being pulled down by any gravity, continually explodes due to the pressure of the internal gases, forming a fuzzy ring of (cooling?) debris, a gradually spreading chaotic mix of rocks, the greater part going towards the torus' real centre of gravity, leaving underlying molten rock exposed to follow their fate later on - in short a crumbling effect that cannot be arrested. This rocky chaos may later start collapsing in the middle of the hole, forming the seed of the future planet. BTW any large enough asteroid impact would also spell doom to the whole torus planet and either wreak havoc, rip it apart (in interesting ways, depending on location and direction of the impact) or cause the central "hole" to collapse sooner, finally reducing the planet into a more spherical shape - enter the Torusmageddon :)

Just my two cents.

rcthompson12년 전

A torus is not a hollow sphere. The OP's article stated that the computed surface of the torus was an equipotential surface, meaning that the apparent gravitational force (a combination of actual gravity and centrifugal force) at every point on that surface is straight down into the ground. OP even calculates the strength of the apparent gravity at every point on and around the torus. Assuming the planet started out in this toroidal shape, it would be stable since the net force at every location is "straight down", so there would not be anywhere where pieces would just fly off into space.

_quasimodo12년 전

"perhaps due to an advanced civilization with more aesthetics than sanity" :)

vidarh12년 전

Iain M. Banks last Culture book has a "advanced civilization with more aesthetics than sanity" that had polished a moon much like a marble, that'd then been lowered down towards the planet it orbited while adjusting the orbital speed accordingly. They'd cut a trench around the entire planet, so that said moon were eventually orbiting below the surrounding planet surface.... One of the main civilizations in the book had then "inherited it" and used the moon as the headquarters of one of their military branches.

It's fun to speculate just how bizarre results you could get if you had a civilization with enough resources to do planetary scale arts projects just for the heck of it...

nitrogen12년 전

Wouldn't the planet's atmosphere and oceans fall into the trench?

CanSpice12년 전

Yes, which is why you dig the trench deep enough to cover it over again, make it airtight, and pump out all of the atmosphere to make a vacuum torus inside the planet.

drakeandrews12년 전

There were either walls on either side of the trench, or the planet was a barren rock to begin with.

throwaway_yy2Di12년 전

Of course. An advanced civilization with engineering sense wouldn't waste precious building materials on thousands of kilometers of planetary bedrock.

(And that's to say nothing of that area/volume ratio, what that implies for radiative heat rejection!)

sharpneli12년 전

And a termite might think that an advanced civilization would never spend tons of valuable vegetable matter for purely aesthetic reasons.

It's all about the relative abilities. It's not completely unconcievable that some might transform planets the way we build funny houses, even if they are 'waste of precious building materials'.

arethuza12년 전

I seem to recall that the Culture does regard planets as being rather wasteful of mass - much more efficient to build an O.

NB I'm sure some Culture ships have probably built planets in idle moments - just for the hell of it.

josh-wrale12년 전

“Welcome to Magrathea, where we construct planets made to order.” (Source: The Hitchhiker’s Guide to the Galaxy)

debt12년 전

This reminds of something I've seen earlier: what if the Earth had rings like Saturn? I'm more curious as to how this would affect us culturally; the effect a toroidal earth would have on our collective consciousness, how we treat each other, and our evolution.

https://www.youtube.com/watch?v=UT2sQ7KIQ-E

johnwatson1121812년 전

Not sure if it was in the article or not but I think the Hairy Ball theorem from Topology would be important. Since earth is a sphere the wind can't blow all in the same direction. Some currents must oppose others. In a torus the wind could start blowing in one direction and stay that way forever.

rtkwe12년 전

That's not what the Hairy Ball Theorem (HBT) says, the HBT says that there must be a point on a sphere where a continuous vector field is zero. It doesn't say they can't be pointing in the same direction except that point.

johnwatson1121812년 전

this link http://uncyclopedia.wikia.com/wiki/Hairy_ball_theorem talks about the application to the torus. It states that it is possible to completely comb the hair on a doughnut. That is the point I was trying to make earlier. I think it does imply that the wind could start blowing in the same direction if we lived on a torus. I think it would be the same if we lived inside like in science fiction movies or if we lived on the surface with a normal atmosphere.

valtron12년 전

I can't tell if you're being sarcastic... but if not, this is the first legitimate use of uncyclopedia as a source I've seen.

pndmnm12년 전

In fact, it does -- to demonstrate, assume otherwise and use the definition of continuity.

rtkwe12년 전

sigh To quote because topology is a pain:

> "there is no nonvanishing continuous tangent vector field on even dimensional n-spheres. For the ordinary sphere, or 2‑sphere, if f is a continuous function that assigns a vector in R3 to every point p on a sphere such that f(p) is always tangent to the sphere at p, then there is at least one p such that f(p) = 0"

Which means they can all blow in a continuous path except for at least one point, so the wind at every point but 2 could be blowing east on a sphere.

http://en.wikipedia.org/wiki/Hairy_ball_theorem

lutorm12년 전

Yes, but the velocity must go to zero as you approach the poles. I took the OP to mean "with the same velocity", not "in closed loops". Obviously wind must blow in closed loops, since there are no sources or sinks of air.

dogecoinbase12년 전

So to you, "pointing in the same direction" includes "pointing in exact opposite directions within an epsilon-ball for any epsilon"?

xyzzyz12년 전

There's a whole continuum between "pointing in the same direction" and "pointing in exactly opposite directions".

rcthompson12년 전

Wind currents would be determined by convection and Coriolis effects, so it would always be going in different directions at different locations.

GhotiFish12년 전

if planets are effectively amorphous liquids with no surface tensions at planetary scales, then why would you see areas on the surface that experience different amounts of gravitation. The rock of the planet experiences that gravitation all the same as the surface of the rock, mass experiencing low gravitation must equilibrate with mass experiencing high gravitation. So what stops mass experiencing 0.65 units of gravity from displacing mass experiencing 0.3 units? You would think this would result in somewhat consistent gravity around the planet. This result is unexpected to me.

gliese133712년 전

Because "magnitude of the gravitational force" =/= "gravitational potential" and potential is what matters- the shape is stable if the potential is equal across the entire surface. This corresponds to the gravitational force vector always being perpendicular to the surface (i.e., gravity always points 'down'). As long as the gravitational force vector is perpendicular to the surface, nothing will get pulled sideways, therefore nothing will move, therefore the shape is statically stable, and the magnitude of the perpendicular force can vary as much as you like.

Now, you may be thinking "but shouldn't the bits of planet-fluid under stronger gravity sink and push on the fluid around them and thus indirectly cause stuff to move around to the sides?" This is where equal potential really becomes important. For one bit of the planet to sink (losing potential energy), another bit of the planet would have to rise (gaining potential energy). If the shape of the planet currently conforms to an equipotential surface, any marginal redistribution of which bits are up and which bits are down will end up requiring at least as much total energy as the original configuration. Thus, the shape is a local optimum in the energy configuration space, and nothing will move.

GhotiFish12년 전

> Now, you may be thinking "but shouldn't the bits of planet-fluid under stronger gravity sink and push on the fluid around them and thus indirectly cause stuff to move around to the sides?"

that was what I was thinking.

So the reason we see gradients is because areas that are experiencing low force have lots of mass, and areas experiencing high force have little mass.

Looking at the diagrams, I guess I can see that is the case.

Thanks for clearing that up!

AlanSE12년 전

I'm not sure if I understand entirely what you're asking, but I think it comes down to "planetary differentiation". It's the same process by which water floats on top of oil because of density. The heavy stuff floats to the bottom and the light stuff floats to the top. That process isn't 100% complete obviously, but that's because near the surface material strength matters a great deal. We still have mountains on Earth, which are clearly not behaving like a liquid. If it was behaving like a liquid, then you're right that the heavier parts would upturn the lighter parts and sink. But Earth still has density variations because the process isn't complete, and tectonic movement keeps rearranging it all.

dredmorbius12년 전

Friction and timescales also matter.

The Earth is highly plastic ... on geological timescales. The continents move, the Himalaya are rising at the rate of about 5 mm per year, and one theory of the post-Messinian flooding of the Mediterranean basin (https://en.wikipedia.org/wiki/Post-Messinian_flood_of_the_Me...) is that a fragment of one of the colliding plates in the region broke off and sank into the lithosphere.

The whole system is dynamic, and driven by both latent gravitational heat of formation and by radioactive decay (including postulated naturally occurring fission reactors within the outer core / inner mantel: http://www.nature.com/news/2008/080515/full/news.2008.822.ht... -- a fission reactor near the surface was found in Gabon, within Africa, active about 1.7 billion years ago: https://en.wikipedia.org/wiki/Natural_nuclear_fission_reacto...).

But it's an interesting rock we crawl about.

moron4hire12년 전

I think it is funny to note that the old style of video game map, a square area where walking over one edge makes you appear at the opposite edge, more closely maps to a toroid than a sphere.

quarterto12년 전

> more closely maps to a toroid than a sphere

It's the very definition of a torus, at least when we're talking about fundamental polygons[0].

[0]: http://en.wikipedia.org/wiki/Fundamental_polygon

moron4hire12년 전

Yes, I just meant that, if you did wrap a square video game map into a torus, it would look kind of funny as the middle would get stretched and the top and bottom edges squashed. But it would certainly all be there, unlike a sphere that would need parts of the map cut out.

dognotdog12년 전

I do wonder if something like this exists, now. I'm familiar with the Dyson ring and sphere, but I never thought of a torus. Even though after the sphere, it seems the most plausible 'mode.'

disputin12년 전

Useful for launches - a space cable car instead of a space elevator.

dnautics12년 전

to be more specific, a funicular. A cable car would take energy to counteract the lift on one side; a funicular would be very nicely counterbalanced.

TehCorwiz12년 전

Larry Niven posited that a ring-shaped world would be roughly equivalent to a partial Dyson sphere. Although I seem to recall that his were only habitable on the inner surface.

Suddenly I'm wondering how much of this applies to toroidal space stations.

AlanSE12년 전

None of this (in the link here) applies to toroidal space stations. The torus shape referred to in the link is maintained by its spin and its self-gravity. People live on the outside of it, and atmosphere is held in the same way that Earth holds atmosphere, by gravity.

A space station that rotates for artificial gravity is completely different. In fact, its gravity is so little that it doesn't matter for any of the structural engineering. It can be completely neglected. The difference in mass between the two proposals is huge, by a factor of like 10^12.

arethuza12년 전

I think Culture Orbitals are more elegant than Niven Rings as you don't need the shadow squares and wouldn't have the stability problems that a Ring has:

http://en.wikipedia.org/wiki/Orbital_%28The_Culture%29

For another take on mega-scale artificial worlds there is Alderson disk:

http://en.wikipedia.org/wiki/Alderson_disk

utoku12년 전

Unfortunately they are not roughly equivalent. A solid shell or a ring around a star is not a Dyson sphere. The correct Dyson sphere is composed of many separate orbiting elements. A solid shell or semi-shell is unstable and is bound to collide with the star or just wander away. Yes, the Ringworld is unstable. This was one of Niven's mistakes and misunderstanding of Dyson spheres.

tutuca12년 전

I wouldn't say a "mistake" he explores that inestability in "Return to Ringworld"...

jessaustin12년 전

http://tvtropes.org/pmwiki/pmwiki.php/Main/Retcon

anon412년 전

Looking at the orbit of the moon which does figure-8s through the hole in the middle, isn't it conceivable that a sun can orbit that same way? Or rather, the two will orbit each other in a strange non-trivial way.

iliis12년 전

It would have to be a very small sun tough in order to fit trough the hole and not burn the planet to a crisp. So we arrived at the Discworld, where the elephants carrying the world sometimes have to lift their legs to let the sun pass.

Such a small star would probably have to be kept burning artificially, as the smallest mass still allowing nuclear fusion is around 80 times the mass of Jupiter.

tgb12년 전

The ringworld has a radius approximately that of the radius of the orbit of earth. A quite large sun could fit easily through such a 'toroidal planet.'

NAFV_P12년 전

Sounds more plausible than this tripe:

http://en.wikipedia.org/wiki/Hollow_Earth

Some idiots still believe it.

[deleted]12년 전