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Relmuis
2005-May-11, 07:06 PM
I wonder which different shapes a planet can have. The problem is relatively easy to describe, but extremely hard to solve.

For a planet, one can use a self-gravitating non-compressible liquid, which would have an equal density everywhere. This is not quite true, of course, but very nearly so.

For a sustainable shape, the surface of this liquid must follow an equipotential surface of the gravity field generated by that particular shape. This does't mean that the gravity vector must have the same strength at all points on this surface. But is does mean that the gravity vector must must be normal to this surface at all points on this surface (i.e. a plumbline will everywhere hang at a right angle to the surface).

We know, of course, that the sphere is a sustainable shape for a notrotating planet. If rotation is added, a "centrifugal" potential can be superposed on the gravitational potential, and an ellipsoid of revolution has now become a sustainable shape. But there might well be other sustainable shapes.

I wonder about the torus. If the torus is sustainable, either as a rotating or even as a nonrotating shape, this would not mean that torus-shaped planets would occur in nature. But it would mean that a civilization might build a torus-shaped planet and expect it to retain its shape afterwards.

My reason to expect this possibility is the following: think of a lot of planets spaced equally around the same circular orbit and having the correct speed to orbit the central star at that distance. Now slowly decrease the mass of the central star to zero, all the while increasing the mass of the several planets, in exactly the right amount to retain the same centripetal acceleration (and therefore the same orbital speed). The planets are now essentially keeping each other in the same orbit. Now subdivide the planets until you end up with a circle of contiguous small planets all around the circle. This is, of course, a self-gravitating torus, albeit a very slender one with a very big central hole.

Essentially, the question would now be how much this torus can be fattened up and by how much the central hole can be narrowed for a certain orbital speed. And also, wether there would be a ratio between the two diameters where the orbital speed is allowed to become equal to zero.

A related problem is whether one can have a toroidal star. This problem would be even harder, as in this case the fluid cannot be assumed to be noncompressible.

antoniseb
2005-May-11, 07:23 PM
Presuming you mean for a planet massive enough that gravity, not structural material strength is the main factor in shape.: You can tell if it would be stable by imagining the perfect torus and testing whether the integral of the gravitational influence of some point allows it to stay in position, or pulls it somewhere. In the case of a non-rotating torus, the inner material on the ring will be drawn away toward the center, unless the torus is so large radius and skinny that you could hardly call it a planet.

The next test is to see what happens when you perturb a small part of your structure. Does gravity pull it back, or does it amplify the perturbation.

Are you asking because you are writing a science ficton story, and you are thinking about a natural "Ring World"? Otherwise, this is a pretty improbable scenario.

aeolus
2005-May-11, 07:54 PM
I've looked in a few books and, of course, Googled. I haven't been able to find any references of any torus shaped object floating around out there.

What do I think? It seems possible to exist in the sense that it would be pretty stable and such if it suddenly popped up as a perfect torus with a nice rotation and angular momentum, and all that cool stuff planets ought to have to fit in with the crowd. I see a huge problem with the formation of it though. It doesn't make sense that it would be able to form naturally, given the violent process that is involved in planetary formation. And if it ever did somehow come to exist, sustaining it seems difficult as well, like anton said (about pertrubing the geometry and analyzing how it will behave in the new, non-perfect shape).

I think the closest you'd get is a ring like around the giants or something like the asteroid belt taking the place of a planet around the sun. But a solid object? I, in my non-expert opinion, do not think a toroidal planet could be stable.

Matthew
2005-May-12, 11:04 AM
While the chances of a torus actually being formed is so remorte its not worth considering, what would happeen if one was formed?

If there is an imputity which causes a semi large pievce in the inner circle to break off what would happen? Its centripetal force would stop it from getting to the centre so it would perhaps collide with another part of the ring? And then cause a chain reaction of the whole structure collapsing?

Relmuis
2005-May-12, 02:42 PM
One reason for my wanting to know whether a toroidal planet could exist is pure curiosity. I am, however, writing a science-fiction story in which I would like to use a toroidal planet. But only if such a planet could exist.

One consideration is that a toroidal planet might be more easiliy moveable than the traditional spherical kind. The planet could be used as a spaceship to move large populations from one locale to another. I am thinking of something like a Bussard Ramjet, where the central fusion core could double as an artificial sun.

Even if toroidal planets would be stable, I doubt whether they would occur in nature. Perhaps there is a chance of one in 1.000.000.000.000 that a toroidal planet might spontaneously form, in which case there might be just one in our Milky Way Galaxy.

Actually, the chances for a toroidal star to occur naturally might be somewhat better than for a toroidal planet. Firstly because some stars have such high rotation rates that they are visibly flattened, secondly because in a star most of the matter would be concentrated in the deepest regions (allowing a fat torus to function as if it were a more slender one) and thirdly because radiation pressure would help to keep the central hole open.

bazbsg
2005-May-12, 03:03 PM
Inside a hollow sphere, there will be no net gravity in any direction. Since the force of gravity is proportional to the inverse square of the distance and since the amount of mass at the further distance is proportional to the square of the distance, then gravity from the shell cancels out in all directions. However, a torus doesn't have the same mass far away as close, so it seems to me that gravity (and help from the rotation) should should allow people and objects to "stick" to even the surface of the torus closest to the sun / center. This makes your story better than it would be if anyone who happened to travel to the inner equator were to drift off into inner-space - or maybe not. Good luck with the story...

Yabble
2005-May-13, 02:03 AM
Er guys; last time I looked; Earth was considered to BE a toroidal planet. OK, it's only in the planet's interior; and all it does is generate our magnetic field, the crust is just the lumpy bit floating about on top, but credit where it's due eh?

In an infinite Universe, there must be billions of toriodal structures; if one stable toroidal planet is possible, many would be found. We've only just started looking.

aeolus
2005-May-13, 04:38 AM
I guess a sphere could be considered as a special type of torus, Yabble. You raise a good point, but I think it's safe to say this isn't what Relmius was asking about. He was talking about the whole planet, not just select parts of it's interior.

bazbsg
2005-May-16, 11:45 AM
I'm pretty sure (without browsing the internet to confirm) that a torus and a sphere are considered topologically different.

In what way, Yabble, do you mean that the Earth is a toroid?

biknewb
2006-May-03, 02:52 PM
and thirdly because radiation pressure would help to keep the central hole open.
This I don't get. If there is no center, what is generating radiation then?

I used to have a boomerang in the shape of a ring with four wings. It spinned in a very stable fashion.

regards

tdvance
2006-May-03, 09:52 PM
Ok--you have a solid torus, which can be thought of as the set of points produced by revolving a disk of radius r (the smaller radius) about a point (C) that is distance R (the larger radius) from the center point of the disk.

Consider any point of the torus (any molecule). It should be in orbit around the whole system or else there will be pressures to tear it apart. So--what is its orbital velocity? There's no mass at C so a first thought is "zero". However, more likely it would act as if the entire mass of the torus were concentrated at the point C (just as Earth-Moon or a binary star, it acts as if both bodies orbit about a common center point of mass equal to the total mass--I THINK that's the case!).

So, if r is very small compared to R, it might work. If r is too large, the inside (distance R-r from C) would want to revolve significantly faster than the outside (distance R+r from C) and that would cause the torus to break up, I'd think. If it were a gas/liquid torus, however, this wouldn't be a problem.

Finally, if everything is so that it doesn't tear apart because of orbital concerns, the question is, would it stick together?

Model the torus as shells (sections of a very short, wide cylinder). The shells have radii going from R-r to R+r and are centered at C. Given any point A in the torus distance x from C, a shell of distance greater than x exerts no net gravity on x (a theorem I read in a Feynman lecture book--actually for spherical shells, but symmetry suggests it works for cylindrical too). A shell of radius less than x pulls A toward C. If A is more than R from C, that is toward the center of a disk generating the torus (in favor of stability). If A is less than R from C, it is away from that center and thus will tend to "fall" away from the torus.

Conclusion--it is not stable.


Still, I suspect something amiss with my reasoning, because Kempler rosettes (planets, all the same mass, sharing an orbit about a common center and of fixed angular distance apart) are stable.

Tim Thompson
2006-May-03, 11:26 PM
Gravitational field of a torus (http://www.mathpages.com/home/kmath402/kmath402.htm). It should be mechanically stable in principle, but like anything else, mechanical strength matters. If it's not strong enough to resist its own self-gravity, it will be squished into a sphere, or something close to it. In the far field it looks like a sphere, as one would expect. But in the near field it has quite a complicated gravity, so there aren't going to be any low torus orbiting satellites, I think. And a binary system with a torus would have interesting tidal forces. Maybe the torus could be a multi-axis rotator? Of course, a toroidal planet would never form naturally, but really spiffy alien engineers should be able to build one!

Sock puppet
2006-May-04, 12:09 AM
As long as R>>r. If we nick tdvances reasoning, and set the entire torus spinning as a rigid loop:
Because we have taken r<<R, the rotational velocities of the entire ring can be taken to be approximately the same.
For the same reason, we may approximate it as being flat.
The innermost layers will feel no net gravitational attraction. Centrifugal force will push them outward.
The outermost layers will feel the graviational attraction from the entire torus. This will be greater than the centrifugal force, and will push them in.
Everything in between will be an intermediate case.
All of this has assumed the rotational speed was chosen to be less than the orbital velocity of an object around the mass of the torus at a radius R. If a greater speed was chosen, the torus will fly apart.
Considerable pressure will be exerted upon the central parts (enough to damage it/ create a molten core?)

neilzero
2006-May-04, 03:18 AM
I have little math or physics I can apply to this question, but I will respond anyway. I think we are restricted to sphere: Moderately spheroid, if we rotate the planet. Assuming modest adhesion and cohesion minor features such as mountans and valleys are possible with the help of internal heat and plate tetonics. Earth has such features up to about 0.2% of Earth's radius. 1% would require stonger and/or lighter material than granite, I think, unless we consider a "planet" with much less mass than Mercury.
With materal a million times stronger than the strongest granite, a wide variaty of shapes are possible up to about Mercury mass.
It is generally thought that the asteroid belt has about the mass of Mercury and in theory the belt could be rearranged into a circle with a circumfrence of about 2 billion kilometers. Average spacing (excluding asteroids with a large dimension less than ten meter) would be about 2 kilometers, assuming almost a billion asteroids with a large dimension of more than ten meters. The gaps would be larger than the average diameter, so the gravity would be too low to retain the perfect circle without frequent corrections. More matter could be placed in the string which would improve the stabillity, I think, until the gaps approached zero, after which I think the stability would decrease, making a solid string very difficult without the million times stronger than granite material, or getting rid of the gravitational attraction of Jupiter. Neil

eburacum45
2006-May-04, 06:11 AM
The question is; if you were standing on the torus, would gravity always be normal (perpendicular) to the surface? If the answer is yes, then the torus is stable. If the answer is no, then the substance of the torus (defined by Relmius in the first post as a 'self gravitating non-compressible fluid') will flow, and the torus will change into another shape.

Sorry, but I don't know whether the gravity produced byy the torus is normal to the surface or not, but I suspect not. For a start, on this page found by Tim Thompson
http://www.mathpages.com/home/kmath402/kmath402.htm
can be found the statement
...This plot shows that there is always a net axial force toward the plane of the ring...

This suggests to me that gravity on a torus will not be normal to the surface, but will tend to pull the matter into a flatter shape- eventually, perhaps, a disk. Disks and spheres are the two common shapes in self-gravitating objects; I think our ring would end up as one or the other.

Ken G
2006-May-04, 09:46 AM
I think the torus is stable. Of course it must be rotating fast enough for the centrifugal force to balance the gravity at the "spine" of the torus, but everywhere else the gravity points in toward the spine. That's all you probably need to keep it contained, pressure forces will balance the crushing gravity. Note the huge angular momentum is the key thing that will keep it from compressing into a sphere or oblate shape. It's a pretty interesting question though.

agingjb
2006-May-04, 10:18 AM
Surely if a torus is spinning fast enough to prevent immediate collapse into an oblate spheroid, then the outer equator will be spinning faster than the (local) orbital velocity and objects on the surface will fly off (not to mention the material of the surface itself).

grant hutchison
2006-May-04, 10:50 AM
Tim's link assumes that the material of the torus is concentrated in an infinitely condensed ring, though the author makes what seems to me like a reasonable argument to extend this to the situation of a real torus so long as we stay outside its surface.
The equation the link provides for gravitational potential could certainly be used to generate a potential map for a cross-section of the torus by anyone with some maths software (or without maths software, but with more time), and the effects of rotation could easily be added to that.
So we should be able to explore eburacum's point about the potential surface for the toroid. If the potential surfaces close to the infinite-density ring are approximately circular in cross section (orthogonal to the ring plane), we might at least be able to guess that the toroid will have some equilibrium surface of constant potential it can relax towards without closing the central hole. But as soon as the torus stops being perfectly circular in cross section orthogonal to the ring plane, the original reasoning allowing us to treat it as an infinite density ring of infinitesimal thickness goes out the window, I think. The mathematics of simulating the exact shape such an object would relax into at gravitational equilibrium seems potentially fraught. (Pun intended ;))

Grant Hutchison

Ken G
2006-May-04, 11:23 AM
Surely if a torus is spinning fast enough to prevent immediate collapse into an oblate spheroid, then the outer equator will be spinning faster than the (local) orbital velocity and objects on the surface will fly off (not to mention the material of the surface itself).
The same argument could be made for people on the night side of the Earth, in regard to Earth's orbit around the Sun. What breaks the reasoning is the gravity of the Earth, or in this case, of the torus (and Tim's link gives a lot of useful info there). I agree with Grant that a full calculation is quite hard, but I suspect that all it would do is alter the cross section of the torus-- it should be able to find some stable configuration, since the torus gravity will hold it together and its angular momentum will keep it from collapsing toward the center. A greater problem might be azimuthal instabilities that could cause it to break up into blobs like beads on a wire. That would probably limit the torus to being pretty thin so that its own tensile strength could help. Then there are wiggles in the thread that might not be stable in the gravity environment of the torus. Most likely there will be some instability that will bite you-- and it only takes one. But you'd still end up with something like the asteroid belt, i.e., somewhat toroidal even if broken up.

mugaliens
2006-May-04, 12:11 PM
The torus planet is not stable and is impossible in nature given the current laws of physics, given your initial condition: "self-gravitating non-compressible liquid, which would have an equal density everywhere."

First, any rapidly spinning mass of matter will assume a flattened disk shape, not that of a planet. Second, even if you had globules orbiting a star in a ring, the best you'd get if they aggregated would be a ring, but only if the star's mass was >> the total mass of the globules. This is because there's always a net force pulling the liquid towards the plane of the disk.

Third, if you were to actually construct a perfect, solid toroid around a star, it would no longer center on the star! The star's gravity would pull on whatever portion is even the slightest bit closer than the rest, and the ring would crash into the star.

Fourth, if the center of the torus weren't around the star and it was liquid, it would flatten to a disk anyway. Even then, whatever portion of that disk was closer to the center of mass would tend move there before moving away, but now that abberation would increase next orbit, and the pertubation rapidly increases until its more kidney-shaped, then it eventually collapses.

Fifth, if you had a solid toroidal planet not centered around a star, even then it would, over time, move towards being a disk, and you'd still have the same problem as with four.

The only way a torroidal planet could be considered to be stable would be if it's shape were defined by it's rigid structure, but even then the gravitational forces would be trying to flatten it and pull it out of round.

Ancillary to this is the Legrange point theory (large and small gravity object like Earth and Moon, and six equidistant points around the Earth, the Moon being at one of them). The points immediately ahead and behind the Moon are the most stable, but this only holds if there are no large gravitational objects at the other points. You'll never see a planet, for example, with six equidistant moons, for there is then no longer an stable points and the moons do the same thing to one another as what would happen in the torus.

Ken G
2006-May-04, 05:37 PM
First, any rapidly spinning mass of matter will assume a flattened disk shape, not that of a planet.

That remains to be seen. I see no reason that can rule out a torus dense and massive enough to self-gravitate.

Second, even if you had globules orbiting a star in a ring, the best you'd get if they aggregated would be a ring, but only if the star's mass was >> the total mass of the globules. This is because there's always a net force pulling the liquid towards the plane of the disk.

That is far from obvious. The net force pulling towards the plane of the disk can be balanced by outward pressure forces. Liquids do have pressure.


Third, if you were to actually construct a perfect, solid toroid around a star, it would no longer center on the star! The star's gravity would pull on whatever portion is even the slightest bit closer than the rest, and the ring would crash into the star.

That argument would suggest that planets could not exist. What saves it is the self-gravity, as for tidal bulges.

The bottom line is that the stability of the torus is a difficult problem, I wonder if it has ever been done. My personal guess is that, since it only takes one instability to disrupt it, there's probably one that will. But then what happens to it is also not obvious. What is clear is that it will not fall in to the center, it has too much angular momentum for that. I continue to think it will eventually fragment into blobs like asteroids, possibly spreading out over time as the asteroids interact with each other. But I can't rule out that the torus will be fully stable.

grant hutchison
2006-May-04, 06:23 PM
Third, if you were to actually construct a perfect, solid toroid around a star, it would no longer center on the star! The star's gravity would pull on whatever portion is even the slightest bit closer than the rest, and the ring would crash into the star.That argument would suggest that planets could not exist. What saves it is the self-gravity, as for tidal bulges.I think mugaliens is referring to the notorious "Ringworld instability", the corollary to the opening discussion in Tim's link, in which it is demonstrated why a test object within the torus' centre hole will be pulled away from the centre to impact the torus.
The same process means that a huge rigid ring centred on a star, like Niven's ringworld, will be pulled progressively more off-centre if it is even slightly perturbed from perfect equilibrium.

Grant Hutchison

mugaliens
2006-May-04, 06:43 PM
Precisely, Grant.

publius
2006-May-05, 03:26 AM
There is something wrong with my brain here. :confused: According to Tim's link, there is net force toward the nearest side inside a circular ring. However, it seems to me I can generate a spherical shell of revolution by simply rotating that ring 180 degrees about the axis formed by the field point and the center of the ring. If a net force exists with the ring, then there should be a net force inside the spherical shell. Where am I going wrong?

Going the other way consider any point on the inside of sphere. We can draw a radius through that point and the center. The intersection of the sphere and any plane containing that radius will be a circular ring, and rotating that plane about that radius will sweep out the sphere. If there was a net force for one of the rings, there would be a net force in all of them, and so there would be a net force for the sphere, which we know isn't the case.

This is assuming the ring is a line-mass. If the ring has some extent in the z direction, making a small cylindrical piece, then the above would break down, as there would not be spherical symmetry of revolution, and I can see a net force toward the closest side there.

So where am I going wrong?

-Richard

Ken G
2006-May-05, 06:24 AM
However, it seems to me I can generate a spherical shell of revolution by simply rotating that ring 180 degrees about the axis formed by the field point and the center of the ring. If a net force exists with the ring, then there should be a net force inside the spherical shell. Where am I going wrong?

I made the exact same mistake in one of my earliest posts on this forum. The problem is, you can't get an equal density shell by rotating a ring-- the parts of the ring that are moving more slowly will "deposit" a higher density of matter into the shell. This is another way of seeing why a ring does have a force toward the ring-- it acts like a spherical shell with an overdensity at the nearby (and opposite) points.



Going the other way consider any point on the inside of sphere. We can draw a radius through that point and the center. The intersection of the sphere and any plane containing that radius will be a circular ring, and rotating that plane about that radius will sweep out the sphere.

But you are multiply counting some of the mass in the spherical shell as being in different rings, when in fact the mass should only be counted once in this construction.


This is assuming the ring is a line-mass. If the ring has some extent in the z direction, making a small cylindrical piece, then the above would break down, as there would not be spherical symmetry of revolution, and I can see a net force toward the closest side there.

I think you have recognized the flaw right there. Even a true ring can be approximated in the limit of smaller and smaller strips-- the symmetry you seek is never recovered.

Ken G
2006-May-05, 06:31 AM
I think mugaliens is referring to the notorious "Ringworld instability", the corollary to the opening discussion in Tim's link, in which it is demonstrated why a test object within the torus' centre hole will be pulled away from the centre to impact the torus.
The same process means that a huge rigid ring centred on a star, like Niven's ringworld, will be pulled progressively more off-centre if it is even slightly perturbed from perfect equilibrium.

I see, but then mugaliens is talking about a different problem-- there is no central star here, the gravity comes from the torus itself.

publius
2006-May-05, 06:23 PM
Ken,

I see, said the blind man. If I had two rings, then I would be adding mass twice where the rings overlapped at the poles.

If O (theta) is the angle is the ring plane, that is the same azimuthal O in a spherical coordinate system. Integrating a line mass density around the ring, the length element would be r*dO. However, for a surface density, the area element is r^2*sin O*dO*d(phi) -- the sin O term is how the pole is only counted once, so to speak.

By trying to simply rotate the ring around, I'm dropping the sin O term, making the mass element be rho*r^2*dO*d(phi), which is equivalent to having a variable surface density inversely proportional to sin O, ie rho = constant/sin O, which would mean infinite density at the poles, which comes from counting the overlapping poles for each of the infinite number of differential rings.

I see it, now. Total mass would remain finite, rho*2pi^2*r^2, as opposed to rho*4pi*r^2, but the infinite mass density at the poles would still give a g singularity at the poles.

-Richard

publiusr
2006-May-05, 08:03 PM
Interesting. I wonder if I could get a big rotating nickel iron asteroid melted--explode a bomb in it--and while the molton, revolving blob was expanding into a hollow shell--hit it dead center with a very fact moving object.

Would it form a metal smoke ring of sorts?

Ken G
2006-May-07, 07:46 AM
I see it, now. Total mass would remain finite, rho*2pi^2*r^2, as opposed to rho*4pi*r^2, but the infinite mass density at the poles would still give a g singularity at the poles.


Yes, that's exactly it. And that surface-density spike is why we get outward forces near the inside of the ring.

grant hutchison
2006-May-10, 11:07 PM
Ah-ha. Here's what you need: OSTRIKER, J. The equilibrium of self-gravitating rings (http://adsabs.harvard.edu/abs/1964ApJ...140.1067O). Astrophysical Journal 1964; 140: 1067-87.
It provides the equilibrium calculations for slender (minor axis << major axis) self-gravitating homogeneous tori which rotate like a solid body. As a bonus it also generalizes to tori composed of compressible substances with a polytropic pressure-density relationship, and tori with a gravitating mass in the central hole.
It seems that there is an equilibrium solution for such an object. It must rotate around its axis at a specific angular velocity to counteract the gravitational tendency to collapse into the central hole. And when it rotates at that angular velocity (mirabile dictu!) the equipotential surfaces are circular in cross section: it settles into a true torus with the acceleration vector everywhere normal to its surface, and everywhere equal in magnitude.

I've just set it up in MathCAD, and a torus as broad across as Jupiter, as thick through the ring as Earth, and with Earth's density, has to rotate with a period of 9.46 hours to be in equilibrium. It has a surface gravity of around 1.3g.

Some problems with that, though: it doesn't seem to meet the criteria for "slender" (since the ring in this case is a tenth as thick as the torus is broad). The surface gravity varies slightly between the inside and outside of the ring, and the equipotential contours are a little egg-shaped. Reducing the ring thickness to a hundredth of the torus width gets a good match, and to a thousandth is even better. It's evident that this is one of those "as the radius tends to zero" solutions.

I have no idea how stable such an object would be to perturbation, though: like Ken, I suspect it might hive off into spherical lumps if you so much as sneezed on it.

Grant Hutchison

biknewb
2006-May-11, 06:27 AM
I have no idea how stable such an object would be to perturbation, though: like Ken, I suspect it might hive off into spherical lumps if you so much as sneezed on it.

Well that is a great plot for a short sf story!
As long as it is sf, the inhabitants might construct a stabilizing centre with negative gravity, or an electrostatic repulsive force.

regards

Ken G
2006-May-11, 03:25 PM
Ah-ha. Here's what you need: OSTRIKER, J. The equilibrium of self-gravitating rings (http://adsabs.harvard.edu/abs/1964ApJ...140.1067O). Astrophysical Journal 1964; 140: 1067-87.

Nice find. And I'm not worried about the egg-shaped nature for finite width tori, it could still be perfectly stable. Might be an interesting wrinkle for a sci fi story if the g varied over the surface much more than the Earth does. (new meaning to "jet lag"!)

grant hutchison
2006-May-11, 04:39 PM
The deviation from a perfect torus may actually not be as large as I found last night. I was using the ApJ paper to calculate the equilibrium angular velocity, and the equations from Tim's link (with additional terms to allow for the rotating reference frame) to look at the equipotential contours and the surface acceleration. (Deliberately mixing the two references together to cross-check my understanding of both.)
But Tim's link treats the torus as an infinitely condensed ring, rather than an extended object, so there will be some inaccuracies in the picture I reported.

I don't pretend to understand the "perturbation analysis" sections in the ApJ paper. Perhaps, to a more mathematically literate brain than mine, they shed some light on the torus's stability in the face of small perturbations?

Grant Hutchison

Sticks
2006-May-11, 05:34 PM
I seem to remember that there was a sequal to Ringworld (http://en.wikipedia.org/wiki/Ringworld), Ringworld Engineers, where the giant torroidial world had gone off centre and the heros had to use solar flares to correct it, but wiping out life on one part of the ring in the process.

grant hutchison
2006-May-11, 05:43 PM
I seem to remember that there was a sequal to Ringworld (http://en.wikipedia.org/wiki/Ringworld), Ringworld Engineers, where the giant torroidial world had gone off centre and the heros had to use solar flares to correct it, but wiping out life on one part of the ring in the process.Yes, that's what I was referring to earlier in this thread (http://www.bautforum.com/showpost.php?p=737842&postcount=22).
Niven didn't think about stability considerations when he wrote the original novel, but shortly after it was published he turned up at an SF conference to find a bunch of delegates wearing T-shirts sporting the phrase "The Ringworld is Unstable!" They were physics students who'd done the maths.

So he had to write a sequel. Or so the story goes ... :)

Grant Hutchison

grant hutchison
2006-May-13, 04:40 PM
I'm enjoying thinking about this one. :)
If we imagine an "Earth-like" toroidal planet, it would presumably have two ring-shaped polar caps, north and south, which would make surface travel between the inner and outer surfaces difficult.

A population on the outer surface would have a pretty Earth-like view of the Universe, except the horizon would be farther away to east and west than to north and south. And there would be a lot more world to explore in an easterly and westerly direction: the first circumnavigation of the torus would be an epic trip. The first explorers to reach and cross the poles would have the experience of seeing the far side of the torus rise into view at the same time as they developed an infinite east-west horizon.

For a population on the inside of the torus, the far side would be visible in the sky at night, and would go through phases as the angle of illumination changed. During the equinoxes, there would be a period of eclipse as the far side of the torus blotted out the sun.

Grant Hutchison

Ken G
2006-May-13, 05:52 PM
For a population on the inside of the torus, the far side would be visible in the sky at night, and would go through phases as the angle of illumination changed. During the equinoxes, there would be a period of eclipse as the far side of the torus blotted out the sun.

It's not immediately obvious to me what orientation of the torus you are imagining-- if it was tilted up on its side, then it would be kind of like Uranus, with no real day or night coming from the rotation, the daylight would be all about the time of year. Or it could be laid flat, and then the outside would have normal day and night, but the inside might be in perpetual night all year long. For intermediate tilts, the inside would be like being above the arctic circle, where the length of the day would vary dramatically over the year. A big issue would be if the opposite side looked larger or smaller than the solar disk, that would change things a lot. It would have some pretty cool phases, like sunsets behind the far horizon. If the tilt was low enough, the central inner ring could be a great place for doing astronomy-- it would be in perpetual night! (It would also be very snow-covered, so you could actually have three "polar" rings!)

grant hutchison
2006-May-13, 06:35 PM
It's not immediately obvious to me what orientation of the torus you are imagining--The hint was in the "Earth-like", but I didn't make it at all clear: I'm imagining a 23.5-degree tilt of the spin axis relative to the orbital plane, and a 365-day orbit around a sunlike star.
The equinoctial eclipses would, as you say, depend on the ratio of major axis to minor axis, which would determine the span of the ecliptic blocked by the far side of the torus. "My" Jupiter-wide, Earth-thick torus would certainly block out the solar disc as it passed through its equinoxes. (Which would be true equinoxes for denizens of the outer surface, but a period of days-long eclipse for inner-surface dwellers.) Let's see: 2.7 times closer than the Moon is to Earth; 3.7 times wider: if it were a disc it would be 5 degrees wide. But it's a band, and the sun is passing diagonally behind it on a track inclined at 23.5 degrees. So the far side of the torus blocks about 12.5 degrees of sky along the ecliptic. Knock off half a degree to get the angular distance over which the sun's disc is completely hidden, figure the sun moves along the ecliptic at a degree per day: it's a 12-(Earth)day eclipse, twice a year.

Grant Hutchison

PS: I love the idea of the third polar ring!

Ken G
2006-May-14, 01:03 PM
PS: I love the idea of the third polar ring!
Yeah, that would be cute, though you're right that for a torus like you are picturing, it would require the tilt be very low, like Venus perhaps. It would certainly be a picturesue world to live on!

blueshift
2006-May-14, 01:10 PM
Why not add a spinning black hole to the situation in the story like Kip Thorne did in his prologue of "Black Holes and Time Warps"? There he imagined a work crew building a girder-work ring around a spinning 45 solar mass black hole with an horizon spin rate of 270 rps and an equatorial circumference of 533 km.

By poking a magnetic field line through the event horizon, the spinning horizon drags local space that interacts with the field for generating electric power so a civilization can live there for quite a deal longer than the meager 8 billion year life cycle of the sun.

You could have civilizations competing over spinning black holes or manipulating the grid to steal power away for some excitement..You best modify it enough to keep Thorne from taking you to court...

Ken G
2006-May-14, 01:21 PM
Can one plagiarize a physical effect? My guess is, once an idea is out, and it just solves the equations, all copyrights are lost. The drawback of using science fact in your science fiction!

Relmuis
2006-May-14, 03:46 PM
Wonderful! So a toroid planet is possible after all.

Something strange has been going on: I seem to have lost the subscription to this thread. Usually, when I visit the BAUT forum, I first check User CP, to see whether any threads which I follow got new posts. But this one never showed up in the list, and I discovered all your new posts just today. I am especially pleased with the two new links.

As an aside: Biknewb asked in post number 10 what I meant by the central hole of a toroid star being held open by radiation pressure. The radiation does not, of course, emanate from the center of the torus, but from its surface. Radiation emanating from the surface facing the central hole and striking the surface opposite would exert a pressure helping to keep the hole open. Which is why I suspected that toroid stars would be "more possible" (face less constraints) than toroid planets.

As for planets, if they are considered to consist of an incompressible liquid, and if the cross-section of the torus may be assumed to be circular (consisting of two circles rather than egg-shapes or something else), there is a four-dimensional parameter space: the density D of the liquid, the inner radius (R-r), the outer radius (R+r), and the rotation period T. The set of all stable torus-planets would correspond to a shape in the (D,R,r,T)-space. If I understand Grant Hutchison rightly, this shape would be a piece of hypersurface (i.e. a set of measure zero) as there would be at most one allowable value of T for every choice of D, R and r. And this suggests that the planet would be unstable; even a tiny change in one parameter would move its representing point out of the hypersurface.

Yet, if the liquid were to be covered with a solid crust, this crust might, by its mechanical strength, stabilize what would otherwise be an unstable equilibrium.

I wonder if there is a simple recipe which gives T as a function of D, R and r.

grant hutchison
2006-May-14, 04:18 PM
I wonder if there is a simple recipe which gives T as a function of D, R and r.I set up the necessary equations to do this task in MathCAD, derived from the paper I linked to, and then verified the result by looking at the potentials and accelerations near the torus surface using the simpler assumptions from Tim's link.

The mass (M) of the torus is given by:

M = 2.pi&#178;.R.r&#178;.D

The angular velocity (omega) necessary to equilibrate such a torus is given by:

omega&#178; = [ln(8.R/r) - 4/5].G.M/(2.pi.R&#179;)

The rotation period is of course just 2.pi / omega

Grant Hutchison

Relmuis
2006-May-14, 04:27 PM
Thank you. This is just what I was looking for in starting this thread, and will be most helpful in writing the next part of my story.

TheBlackCat
2006-May-14, 05:59 PM
Everyone is looking at this from a gravitational perspective, but there is another issue to take into account that may also have a large impact: surface tension. Due to the surface tension of fluids (which the interior of this planet would be), a fluid naturally assumes a structure that minimizes its surface area. A torus has an extremely large surface area for a given volume, other shapes like a flattened sphere very well might be just as stable gravitationally but have far small surface areas for a given volume of material. If this is the case a toroidal planet would naturally reform itself until it reached a form with the minimum surface area that is gravitationally stable. The gravitational forces may be enough to keep it stable, I dont know, but it is something to keep in mind.

grant hutchison
2006-May-14, 06:47 PM
The forces arising from surface tension are going to be very small in comparison to gravity at this scale.
For a rough estimate, the surface tension of water under air is about 0.07N/m, which is relatively high as these things go. Laplace's law says that the pressure exerted by a membrane under tension is equal to the tension divided by the radius of curvature (with a small constant of multiplication depending on the geometry). Plugging in Earth's radius to give a planetary scale, that's 0.07/6378000 = ~10-8Pa = ~10-13 atmospheres. Local differences in pressure arising from local differences in curvature (which tend to drive the surface towards spherical) are going to be an order of magnitude smaller again.
So it seems likely that other forces will swamp surface tension: a breeze across a lake is sufficient to bend the surface of the water into waves, for instance, despite its surface tension.

Grant Hutchison

TheBlackCat
2006-May-14, 06:58 PM
Yes, but the forces don't have to be large. Even small forces acting over huge areas and over billions of years could very easily be enough to cause the planet to significantly change its form. We are talking about a mature planet, apparenlty with life. That means it is probably at least a billion years old if it is stable enought to stand on, almost certainly at least several billion if there is complex life. The planet may not instantly collapse but a small, contiuous force can have a large impact over sufficiently long periods of time.

grant hutchison
2006-May-14, 08:04 PM
The planet may not instantly collapse but a small, contiuous force can have a large impact over sufficiently long periods of time.Only if unopposed.
A metre of water under 1g exerts a pressure of 104Pa. So surface tension pressures of 10-8Pa are equivalent to picometre depths of water under 1g.
If gravitational and centrifugal forces create a potential contour that defines a planet-scale torus with a surface gravity of ~1g, then the surface tension forces we've discussed can drive the torus away from that equilibrium shape on the picometre scale, but the shape should then have a new equilibrium at that tiny level of distortion.

Grant Hutchison

Edit 1: Further thoughts: Since surface tension on the inner and outer surfaces of the torus exerts a net tiny force towards the centre of the torus (initiating a collapse towards sphericity), the equilibrium state for the torus would require an equally tiny increment in angular velocity to compensate.

Edit 2: Some tidying up of the language, for clarity and accuracy.

Relmuis
2006-May-15, 10:52 AM
What the surface tension would do, however, is spoil our chance to do small scale experiments (with drops of mercury for example) in weightless conditions.

In the formulae provided by Grant Hutchison in post 43, if the first formula (for the mass M) is substituted into the second one (for omega squared) it becomes visible that the rotation period T depends on r and R solely by their ratio (r/R). Which means that a small "planet" would be a good model for a large one with the same density, rotation period and shape.

This rather surprised me, though with hindsight I could see that it had to be so; no formula could yield an answer in pure units of time, unless all units of length were divided away. So there is a scaling law; a small torus having to rotate at the same rate as a large one. But this scaling law breaks down for very small shapes due to surface tension (and it would break down for very large ones due to the liquid being no longer incompressible under extreme pressure; the kind of pressure where matter becomes degenerate).

Ken G
2006-May-15, 01:12 PM
I think the issue of stability of the torus in regard to surface tension relates to the concept of a local minimum in the energy, as compared to a global minimum. Given the necessary amount of angular momentum to make the torus stable, there might possibly be a lower energy configuration, but I don't think it would look like an oblate spheroid-- I suspect there's too much angular momentum for that. But even if the torus is only a local minimum, it might still be rather difficult to get it to leave the torus configuration, and Grant's point about the weakness of surface tension suggests that alone couldn't do it. Still, TheBlackCat makes a good point that there might indeed be lower energy configurations that conserve angular momentum, though I would bet it would probably be the gravitational potential energy itself that determines the lowest energy configuration that conserves solid-body angular momentum. (In particular, I worry about azimuthal blobs-- maybe surface tension, or even gravity, might favor blobs, and it's not obvious that this was included in Ostriker's stability analysis). Since it only takes one instability to destroy it, I probably wouldn't sign up to live on that planet until it had proven itself reliable!

Relmuis
2006-May-15, 01:34 PM
The planet would be in an unstable equilibrium, akin to a soccer ball lying on top of a car. If the car moves just a little, or if the wind blows, the soccer ball rolls away from its position, and because it then encounters a slope, it starts rolling faster and faster. A similar unstable equilibrium would be that of a rugby ball sitting on one of its points. If it starts tilting over, it will feel a force causing it to keel over all the way.

However, both balls could be stabilized by putting some glue underneath. A small force could not cause them to become unstuck, though a large force could.

For a toroidal planet, the analogon of the glue would be a solid crust, which forces the planet to retain its shape in the face of small perturbations. Large perturbations would cause the crust to shatter or deform, but small ones might have little effect.

The question would be: just how unstable is the equilibrium? And how big are the perturbations which are to be expected?

It would help if the orbit of the planet around its star were to be perfectly circular, if its equatorial plane were to perfectly coincide with its orbital plane, and if moons or other planets were to be absent. But solar tides would be inevitable, even if constant in strength, and uniformly changing in direction.

grant hutchison
2006-May-15, 02:43 PM
Given the necessary amount of angular momentum to make the torus stable, there might possibly be a lower energy configuration, but I don't think it would look like an oblate spheroid-- I suspect there's too much angular momentum for that.Indeed: the moment of inertia of the sort of "slender" tori discussed is very much higher than that of a sphere of the same volume.
My Jupiter x Earth torus has the same volume as a sphere 23915 km in radius. I've just run the relevant details through the equations for a Maclaurin spheroid, and I'm still a factor of 3 short of matching the equilibrium torus's angular momentum even when I push the rotation period of my sphere down to 2.4 hours, at which point it has flattened into a spheroid with an eccentricity of 0.92, and its equatorial gravity is only barely providing the necessary centripetal force to hold it together.
There's a maximum angular velocity for the Maclaurin spheroid, beyond which angular momentum can be added only by increasing the equatorial radius and decreasing the angular velocity. I guess that our torus, collapsing towards this configuration from an initial high-radius low-velocity state would enter that "arm" of the Maclaurin equilibrium, and end up as something resembling a disc. (If it didn't fragment.)

Grant Hutchison

Ken G
2006-May-15, 02:45 PM
It would help if the orbit of the planet around its star were to be perfectly circular, if its equatorial plane were to perfectly coincide with its orbital plane, and if moons or other planets were to be absent. But solar tides would be inevitable, even if constant in strength, and uniformly changing in direction.
But tides don't break stability, they just alter the stable shape. The only instabilities I can see would be ones that also apply to a long thin self-gravitating rod.

Ken G
2006-May-15, 03:29 PM
Indeed: the moment of inertia of the sort of "slender" tori discussed is very much higher than that of a sphere of the same volume.

This is getting more and more interesting. I think the point is, for any amount of mass and geometry, there's a rotation rate that will allow gravitational support at the equator, and a moment of inertia that will give the total angular momentum. So you might think that configuration could support that angular momentum, but you still don't know if the shape is gravitationally stable. I wouldn't have thought that a disk would be stable, even at critically high rotation rates, because if you rotate a sphere it starts throwing off its equator before it ever gets to a disk. The ring solution has even higher angular momentum than the critically rotating spheroid, and can perhaps be thought of as the limit where all of the sphere has been thrown off into a ring. The ring should be the shape that supports the maximum amount of angular momentum in a stable geometry, so if you get to that shape, there should be no other solidly rotating shape that will work, other than if it breaks into beads along the ring (since that's pretty much the same). If you lose the solid rotation requirement, all bets are off. I think Ostriker was seriously asking if we should expect to find torii in situations with extremely high angular momentum, but so far that doesn't seem to happen.

grant hutchison
2006-May-15, 05:28 PM
Ah-ha.
Although it's possible to satisfy the Maclaurin equations with that second limb of extremely flattened objects I mentioned, it turns out they're unstable. Lang's Astrophysical Formula describes Maclaurin spheroids as "secularly unstable to viscous processes" when the eccentricity exceeds 0.81267, and "dynamically unstable" beyond 0.95289.
Beyond an eccentricity of 0.81267, the Maclaurin spheroids convert to Jacobi ellipsoids. When the Jacobis in turn become unstable, something pear-shaped described by Poincaré (which I've never encountered before) takes over, according to Lang.

Grant Hutchison

eburacum45
2006-May-16, 03:30 PM
But toroids can be stable, more or less, with the right rotation?
If so we might as well have one in OA.

I have made a start on making a Celestia model of one of these; the eventual model will be about 70,000 km in radius with a (hopefully) Earthlike texture.
http://img66.imageshack.us/img66/9606/hoopworld1xf.jpg

What would be the minor radius of a 70,000 km radius toroid with Earthlike density and gravity, I wonder?

grant hutchison
2006-May-16, 05:22 PM
What would be the minor radius of a 70,000 km radius toroid with Earthlike density and gravity, I wonder? You knew I wouldn't be able to resist that, didn't you? :)
2000km gives a surface acceleration 0f 9.8 m/s&#178;, a mass of 3x1025kg and a rotation period for stability of 25.9 hours.
Since you're building one in Celestia, it occurs to me that cloud patterns might well be rather different on such a world. The difference in radius of gyration with latitude will be less than on a conventional spherical world, so Coriolis effect will be less of a player in atmospheric circulation, I think. Maybe we'd get a single Hadley cell spanning from equator to "pole". Spiral storms might not form: but if they did, they'd not only be mirror-reversed between the northern and southern "hemispheres", but between the inner and outer surfaces, too.

Grant Hutchison

eburacum45
2006-May-16, 05:34 PM
Excellent; thank you.
25.9 hours is remarkably Earthlike; I was expecting about ten hours.

eburacum45
2006-May-17, 12:47 PM
Work in progress- one long ice cap on the top and bottom of the world, reduced coriolis effect leading to fewer cyclones...
http://img66.imageshack.us/img66/2898/hoopworld33tj.th.jpg (http://img66.imageshack.us/my.php?image=hoopworld33tj.jpg)

Edit; now the whole ring is finished. I cheated a little and used a modified Earth texture.
http://img66.imageshack.us/img66/3199/hoopworld66er.th.jpg (http://img66.imageshack.us/my.php?image=hoopworld66er.jpg)

Ken G
2006-May-17, 02:49 PM
That's a cool pic. I'm not sure I understand the basis for reducing the coriolis effect, however, as it seems to me it would be virtually identical to that of the Earth's. The acceleration is basically just the speed of the air, divided by the rotational period, so assuming that air currents attain a similar speed, one might expect a similar effect. I think Grant's reasoning is that the gradient in the centripetal acceleration is less, but I don't think that will be the case-- the gradient is locally induced by the pressure gradient, not the Earth's radius. If you have a low pressure center, the air moving into it will still be deflected similarly as in an Earth cyclone. It is sounding like your model planet can end up being quite Earthlike in many ways!

grant hutchison
2006-May-17, 03:17 PM
I'm not sure I understand the basis for reducing the coriolis effect, however ...
Me neither ... :)
I think I just made the classic mistake of ascribing the magnitude of the Coriolis force to the difference in rotational velocity between various parts of the rotating reference frame. :o
I reasoned that a movement from equator to pole on the torus world would be analogous to a much smaller latitudinal movement on Earth (because the surface of the torus is everywhere more or less the same distance from its centre, as are the near-equatorial regions of Earth).
But it's a bad analogy. An air mass making a small latitudinal movement on Earth travels a short distance, and covers it (as you say) at the same velocity as we might imagine prevailing on our torus world. So Coriolis force has less time in which to work, and that's why equatorial air masses on Earth suffer only slight deflection if they make small latitudinal movements.

Sorry: scratch the weak Coriolis ...

Grant Hutchison

Edit: Reconstruction of original post after deleting in error.

Ken G
2006-May-17, 03:24 PM
I actually was convinced initially by your argument, it only occured to me later the problem. A subtle issue indeed-- but your analysis of this whole situation has been outstanding and fascinating.

eburacum45
2006-May-17, 03:38 PM
The surface area is about eleven Earths; the mass is five Earths. This is a way of getting twice as much surface area for your mass -
but compared to ringworlds, which give lots of surface area for a given amount of mass, there is only a little advantage in having a torus rather than a sphere.

neilzero
2006-May-17, 03:56 PM
Togs analogy is about right if you count the trillion times a trillion = 10^24 asteroids between one nanometer and one centimeter. Neil

mugaliens
2006-May-17, 05:55 PM
I just spent the last hour reviewing the posts since I last posted here, and it appears there are four phenomena which are routinely being overlooked as people figure out how this thing might look.

Before I start, let me say I'm knocking the ideas - I think a torroidal planet would be cool! But the following four things would somehow have to be overcome for it to actually work.

First, some assumptions:

1. The toroid would have to be thick enough such that local gravity would be roughly equivalent to that of Earth.

2. The toroid would be spinning at an angular velocity such that the inertia to fly apart balances the pull of gravity towards the center of the toroid.

3. The material would be similar to that of Earth.

Given these three considerations, if such a planet were to exist, it would be much like Earth.

Here's why it'll never work given the conditions above:

1. Planet Earth is a fluid, even the crust, but more so in the molten core. As such, over time, it behaves like a fluid, warping itself into an oblate spheroid. Unless it was perfectly solid, the toroid would not remain intact, and would be subject to flattening and warping.

2. A fluid toroid would encounter tidal effects that would throw it out of round, and the gravitational mechanics of a non-perfect toroid are such that it would continue to become less and less round, until it looked more like the outline of a kidney pool, and so on, until two parts of the once-toroid would touch and the planet would rapidly shred itself.

2. The gravitational mechanics of a fluid toroid with even the slightest imperfection would cause it to aggregate in balls rather than remain a toroid, similar in appearance to (although not by the same means) a stream of water breaks up into water droplets.

3. Any velocity differences (such as currents) would cause those with a lower velocity about the toroidal axis to move towards the axis, and those with a higher velocity to move away, resulting in both a tendancy towards more flattening. In time, the toroid would continue to flatten, and as tidal effects continued, would slowly aggregate towards the axis and wind up in a ball.

4. Any tidal effects would cause the toroid to slow its rotation about it's axis, with the tidal effects being transformed into heat, as they are in the Earth and Moon. The Moon's rotation has stopped because tidal effects overpowered it's rotation long ago. As the rotation slows, the toroid would collapse in towards it's axis, creating widespread and quite massive earthquakes, as well as unbalancing assumption 1, above, which would also tend to move the toroid out of round.

Number 3 would take the longest to achieve, but in a very short order (days? years?) the effects described in numbers 2 and 4 immediately above would destroy the toroidal shape - long before weather patterns established themselves, and much to short for life to have evolved.

In short, because of destabilizing tidal effects, a toroid is not a stable shape for any fluid bound by it's own gravity. An oblate spheroid, however, is stable, the toroid would quickly assume this stable shape.

grant hutchison
2006-May-17, 05:59 PM
Sorry: scratch the weak Coriolis ...
However, one aspect of my original thinking about Coriolis does have relevance, although not quite in the way I thought.
This (ie eburacum45's) world is small in north-south extent: with a minor radius of 2000km, air masses have only a third the latitudinal room to move in, compared to Earth (but very much more longitudinal room).
So a polar air mass will reach the tropics in the same distance (and with the same Coriolis deflection, and the same time to warm up) that would take it no farther than the temperate zone on Earth: that's what started me off wondering about weather patterns and the possibility of a single Hadley cell between pole and equator.
I also wonder what cyclones might look like on such a world: will they be the same size as Earth's, superimposed on a smaller world? Or smaller and weaker in proportion to the reduced north-south extent of the weather patterns?

Grant Hutchison

mugaliens
2006-May-17, 07:09 PM
Good question, Grant!

I think one of the most interesting comments to date has been that there are no continual wind sources on earth due to the fact that that physics mandates a minimum of at least one no-wind point on a spehere, but that a toroidal planet this doesn't hold true.

I believe there may be severe winds on a toroidal planet due to this fact, and once the wind engine gets going, it might be rather severe, with winds constantly rotating the toroid at up to hundreds of miles per hour, as they do in certain areas of Jupiter, which to even large "local" effects the planet is sufficiently large to allow the effective appearance of being absent a zero-wind point.

Then again, I'm no climatologist, so...

grant hutchison
2006-May-17, 09:31 PM
Unless it was perfectly solid, the toroid would not remain intact, and would be subject to flattening and warping.The paper I linked to shows that this isn't so: a fluid toroid is stable, if it rotates at the correct rate.


... and the gravitational mechanics of a non-perfect toroid are such that it would continue to become less and less round, until it looked more like the outline of a kidney pool ...

... The gravitational mechanics of a fluid toroid with even the slightest imperfection would cause it to aggregate in balls rather than remain a toroid ... Ken worried about these potential modes of failure early in the thread (#19), and I reiterated in #30: so I don't know if that would count as "routinely being overlooked". :) While I wouldn't be surprised if the toroid was unstable to quite minimal disturbance to its shape, I'd also not be surprised if it turned out to have a degree of robustness, tending to move back to the previous equilibrium shape if not too drastically disrupted. Do you have some evidence that it's definitely subject to runaway instability in the way you suggest?


Any velocity differences (such as currents) would cause those with a lower velocity about the toroidal axis to move towards the axis, and those with a higher velocity to move away, resulting in both a tendancy towards more flattening. In time, the toroid would continue to flatten, and as tidal effects continued, would slowly aggregate towards the axis and wind up in a ball.I don't see why such internal currents need necessarily occur. Edit: And, given that the torus is moving at several kilometres per second, such currents would have to be very fast to threaten the cohesion of the structure: fluid moving with the direction of rotation would become a little heavier on the inner surface, a little lighter on the outer; fluid moving in the opposite would have the reverse experience. It doesn't seem like it would be immediately disruptive to the whole structure. End edit.
You'd also need to somehow dump a lot of angular momentum for that ball to form.


Any tidal effects would cause the toroid to slow its rotation about it's axis ...This is a nice one. :) Although there's no doubt a new equilibrium shape will be traced out by the equipotential contours when the torus is subjected to mild tidal deformation (as Ken mentioned), the torus is constantly having to reconform to that equilibrium shape as it rotates, so it's going to lose energy to internal heating.
One has to wonder what shape it will adopt as it loses energy, because it's still initially going to have too much angular momentum to form an equilibrium spheroid. Slowing down my toroid below its equilibrium rotation rate in MathCAD makes the potential contours move inwards in closed loops, separating inside the torus and bunching outside. That suggests that the toroid might initially develop an inner equatorial bulge, with lighter surface gravity inside than out. Once it got slow enough for the inner surface acceleration to hit zero, it would start to haemorrhage material from its inner surface into the central hole.

Grant Hutchison

grant hutchison
2006-May-17, 11:06 PM
I think one of the most interesting comments to date has been that there are no continual wind sources on earth due to the fact that that physics mandates a minimum of at least one no-wind point on a spehere ...Was that on this thread? I must have missed it. :( Presumably it's the same logic I've seen applied to hair, attempting to explain why our hair radiates from at least one "crown" point.

Sphericity doesn't prevent super-rotation, of course: look at Venus, where the upper winds rotate 60 times faster than the planet.

Grant Hutchison

grant hutchison
2006-May-17, 11:08 PM
... but your analysis of this whole situation has been outstanding and fascinating.Thanks for the kind words. :)

Grant Hutchison

Ken G
2006-May-18, 01:06 AM
In short, because of destabilizing tidal effects, a toroid is not a stable shape for any fluid bound by it's own gravity. An oblate spheroid, however, is stable, the toroid would quickly assume this stable shape.
As Grant pointed out, the stabilizing effect here is the huge angular momentum, which you appear to be overlooking. There is no spheroid shape that can accomodate this much angular momentum, that's kind of the point of the toroidal solution. How you actually get that much angular momentum conveyed equally to all the material at once is a problem we will leave to Relmuis!

Relmuis
2006-May-18, 12:56 PM
In the story I am writing, the toroidal world is an artificial one, so I assume that the makers took smaller worlds and accelerated them to exactly the right speeds to congregate in a toroidal region with sufficient speed to form a toroidal blob of magma with the right rotation rate.

But, as I originally conceived the toroidal world, it didn't need a special rotation rate. I hoped that a torus would turn out to be stable even if non-rotating.

If the universe wasn't so big, I would be confident that toroidal worlds do not occur naturally. But some inflation scenarios suggest a size to the tune of 10120 light years, with room for 10130 solar systems. Perhaps in one of those systems, the billiard game of collisions and gravitational interactions contrived to create a toroidal world.

mugaliens
2006-May-18, 05:09 PM
Was that on this thread? I must have missed it. :( Presumably it's the same logic I've seen applied to hair, attempting to explain why our hair radiates from at least one "crown" point.

Sphericity doesn't prevent super-rotation, of course: look at Venus, where the upper winds rotate 60 times faster than the planet.

Grant Hutchison

Well, I thought it was here. Might have been on one of the linked websites. Nevertheless, Zero-velocity wind point is a constraint of spheres, not toroids. This doesn't imply that you can't have super-velocity winds. What it means is that there must be at least one point (and in reality there are many, even on Venus) where the velocity relative to the surface is zero.

With a torid, this doesn't hold true, and we may very well find the norm is hurricane-force or even tornadic-force winds spiraling around the toroid is the norm.

Ken G
2006-May-18, 05:20 PM
And I'll bet there has to be two such points on a sphere. But I don't think it matters terribly-- wind patterns are generally not global anyway, there are the Hadley cells that Grant referred to. Outside the Hadley cell, I doubt the weather cares a hoot whether there is a zero-wind point somewhere else.

mugaliens
2006-May-18, 05:34 PM
You're correct - a minimum of two points. But there can be more, of course.

But I think you're missing the point. To put not too fine a point on it, this tends to anchor weather patterns. Without such an anchor you might very well find that the wind is fairly constant on Planet Tora, but that it's always spiralling in the same direction. Since any weather system must overcome friction with the surface, it would only be tornadic in strength if the energy input into that system were very strong.

It's similar in concept to the earliest two-stroke diesel engines. They would run in either direction, but once you started them running in one direction, so long as you continued providing them with energy, they continued to run.

mugaliens
2006-May-18, 05:43 PM
As Grant pointed out, the stabilizing effect here is the huge angular momentum, which you appear to be overlooking. There is no spheroid shape that can accomodate this much angular momentum, that's kind of the point of the toroidal solution. How you actually get that much angular momentum conveyed equally to all the material at once is a problem we will leave to Relmuis!

It only appears to be stable in theory. What happens is that once it's slightly out of round, the increased gravitational forces between the two closer sides pulls them in. Naturally, they rebound outward, but then the cycle is repeated as they oscillate back in. This continues, and over many years, continues to get worse. It doesn't just restabilize in a perfect circle.

And local gravitational effects will pull even the least slightly imperfect toroid into globules that orbit one another.

I have a piece of software left over from my DOS days that allow me to set up to 27 "planets" in motion around a "star." I can set initial parameters such as mass, initial positions and velocities. They're just represented by small circles on the screen, but the equations are extremely accurate. If I try to simulate a toroid by setting the star's mass to zero, and giving the six planets the same mass and velocity, just separated by 60 degrees between them, they initially follow a circular orbit. Then, that orbit decays until chaos rules. With six it just doesn't work, folks - nor five nor four nor three. Two works ok, or many provided the star is massive compared to the planets and they have significantly different orbits. If I recreate the Earth, Sun, Moon, and say a small asteroid at one of the Lagrange points, however, it's stable.

But any more than two planets in the same orbit (and a toriod would be many planets in the same orbit) it simply falls apart.

mugaliens
2006-May-18, 05:55 PM
PS: If you want to see chaos in motion, there's another program you can use. Kind of cool! It's here: http://loke.as.arizona.edu/~ckulesa/astrosoft.html

Set it up as follows:

Position: -50, 50, and 0
Velocity: 0, -0.1, and -0.13
Masses: 7 and 5
Size ratio: 0.8
Uncheck fast preview and QuickDraw, but check Reverse. Then, put 99999 iterations.

They shred!

mugaliens
2006-May-18, 06:07 PM
Another bunch of settings that's particularly cataclysmic:

40, 25, 0

.4, .5, 0

5, 5

1, with the rest of the settings as above.

Talk about head-on collisions!

grant hutchison
2006-May-18, 06:09 PM
To put not too fine a point on it, this tends to anchor weather patterns. Without such an anchor you might very well find that the wind is fairly constant on Planet Tora, but that it's always spiralling in the same direction.I'm afraid I'm not getting this. Why should a couple of wind-free points (which are free to move around) serve to anchor weather systems?


With six it just doesn't work, folks - nor five nor four nor three.Klemperer looked at such rosettes in a very general way in the 1960s, and showed that they were all stable if unperturbed, and all unstable to the slightest perturbation. This means they're generally unstable in computer simulation, because the rounding errors of the digital representation soon propagate through the simulation and dominate.

You can't reason from the Klemperer instability to the torus, however. Your individual masses are all surrounded by their own little wells of gravitational potential, and they're therefore quite keen to fall towards each other. The toroid is at the bottom of a toroidal potential well, and is therefore quite keen to stay there.
While I think what you're suggesting (runaway instability to trivial perturbation) is quite possible, I don't think you've presented anything so far to make it more plausible to me than the alternative (stable against minor perturbations).

Grant Hutchison

mugaliens
2006-May-18, 06:12 PM
This is fun - I'm going to start a new thread!

Ken G
2006-May-19, 01:28 PM
I agree with Grant, although we both allow for the possibility of the "globular" instability that mugaliens favors, which we mentioned early on. As for wavelike instabilities along its length, I'm not so sure about those, Ostriker's analysis doesn't seem to include that either, but I would tend to think the angular momentum would be highly stabilizing against decreases in radius or pear-shaped effects. Again, I wouldn't buy real estate on that torus until it proved it can stay around for awhile!

mugaliens
2006-May-19, 04:37 PM
Occam's razor demonstrates that if it were possible, we'd see a lot more evidence of it. The rings of Saturn, Jupiter, and Uranus, not to mention the flatness of galaxies, demonstrate that accretion disks are the norm when angular momentum is too high to form a roundish planet or star, and that roundish planets and stars are the norm when the angular momentum of the system falls below a certain point.

mugaliens
2006-May-19, 05:00 PM
I'm afraid I'm not getting this. Why should a couple of wind-free points (which are free to move around) serve to anchor weather systems?

Klemperer looked at such rosettes in a very general way in the 1960s, and showed that they were all stable if unperturbed, and all unstable to the slightest perturbation. This means they're generally unstable in computer simulation, because the rounding errors of the digital representation soon propagate through the simulation and dominate.

You can't reason from the Klemperer instability to the torus, however. Your individual masses are all surrounded by their own little wells of gravitational potential, and they're therefore quite keen to fall towards each other. The toroid is at the bottom of a toroidal potential well, and is therefore quite keen to stay there.
While I think what you're suggesting (runaway instability to trivial perturbation) is quite possible, I don't think you've presented anything so far to make it more plausible to me than the alternative (stable against minor perturbations).

Grant Hutchison

I think it's time someone modelled this using finite element analysis, but over time (and Mathcad won't cut it).

I'll spend a little extra time at work and see what I can get the boys on the supercomputer to come up with.

We'll assume the following:

1. Water is the liquid (the supercomputer is already tuned to analyze water six ways to Sunday, including it's various effects).

2. It is incompressible (have to do that, as it's a surface analysis model - no compressibility factors to speak of on the surface).

3. Beginning conditions are that the water won't have any currents (can't model that anyway).

4. Toroid's surface gravity will average 1 G. This can be very closely approximated by assuming a very long cylinder and calculating radius that'll produce 1 G.

5. Toroid's angular velocity will be sufficient to counter the tendancy of the toroid's gravitation pull.

6. It will have any atmosphere, but since the density of Earth's oceans is about 547 times greater than it's atmosphere, we'll leave that out of the calculation for now.

7. I'll try running it with the smallest pertubation I can that generations any result, simulated, say, a small comet passing through the system. Nothing as radical as the tide it would have as it orbited the sun!

Results in a bit.

grant hutchison
2006-May-19, 05:13 PM
I'll spend a little extra time at work and see what I can get the boys on the supercomputer to come up with.Excellent! :)

Grant Hutchison

Ken G
2006-May-19, 06:44 PM
I'll try running it with the smallest pertubation I can that generations any result, simulated, say, a small comet passing through the system. Nothing as radical as the tide it would have as it orbited the sun!
Yes, this should be extremely informative. I agree a two-step approach is nice: first look for inherent instabilities, then see what the solar tides do. I don't think solar tides will make it any more unstable, but who knows? It will certainly induce tidal currents, which will then interact mightily with the coriolis effect to knock the shape out of round, but that might be the end of it. I suppose a real fluid might then induce dissipation, which would rob energy from the planet rotation. This will have to conserve angular momentum, so the lost rotation will have to show up in the orbital angular momentum-- the planet will spin down and move outward, I believe, but it might take a long time. Of course, this could be fixed by placing the planet close to the M-type star, at a distance where it has an orbital period of about a day, so its rotation can be tidally locked with its orbit! In that case, the whole calculation can be done in the static effective corotating potential. Probably too close for comfort, might have to live on the inner ring of a horizontal torus, or the backside of a vertical one...

mugaliens
2006-May-20, 11:49 AM
Ugh - what a night. I gave my buddy a copy of J. Ostriker's "The equilibrium of self-gravitating rings," a couple of beers, and we got to work. The math is way beyond me, but he went through it like a knife through butter and after about twenty minutes of taking notes, he began creating a dataset modelling the torus.

It was a lot more complicated than I thought, and took him about three hours to create the dataset and run the sims.

Here's are the results:

1. It is critically stable with an ovoid cross-section. Any variations from perfect symmetry render it unstable.

2. The first mode of failure is what I dub "globulization," with the thing disintegrating into globules of varying size. These tended to attract one another, so larger and larger globules were formed.

3. The second mode of failure as this happened was rather chaotic. Due to the varying sizes of the globules, local gravity vectors were no longer symmetric and some globules missed one another, hurling eachother into eccentric orbits. Most of the orbits remained in the ecliptic plane, but a few, primarily the smaller ones, canted themselves.

4. Naturally, there were collisions. We couldn't model those, so we assumed that when two globules touched they simple became one with the combined mass and system vector of the two. Had we been able to module collisions, I'm sure a lot more smaller globules would have been formed, significantly adding to the chaos and mass outside the plane of the ecliptic.

5. The end result was a series of wild gyrations of a decreasing number of globules. We watched the thing for half an hour at nine globules, but even though their orbits intersected, they managed to avoid hitting one another and thereby further reducing their numbers. I suspect that if we let it continue, and provided the orbits crossed one another, that the maximum number would be two (two objects in orbit around one another will never cross orbital paths).

6. When we ran it using masses at the center, the center masses, if small compared to the mass of the toroid, tended to become a part of this chaos. If much larger, they did a good job of mopping up some of the more erratic globules, but the end result was the same, with few globules remaining in somewhat erratic orbits, but mostly in the plane of the ecliptic. One interesting note is that this tended towards a more ring-like structure, since the inner erratic orbits were mopped up by the central mass. Without the central mass erratic orbits continued to cross the systems center of mass.

Again, if we'd had the ability to model collisions, I suspect the number of discrete globules would be much higher, and with a sizeable central mass it would trend towards the form that we see today in the rings about planets, but without a sizeable central mass it would trend towards a disk.

Relmuis
2006-May-20, 12:34 PM
Wonderful. Here are the answers to questions which I have pondered for years. And now I can also describe the destruction of the planet with some confidence.

By the way, would it feasible to include a solid crust into the simulation? Something which will elastically (reversably) deform under small tensions and create a force tending to undo the deformation? But which will fall apart when the tensions become too large?

Ken G
2006-May-20, 01:12 PM
Very nice simulation. Yeah, the solid-crust aspect might be important, given how the oceans on Earth react much better to the tidal gravitational potential. Maybe a solid crust could oppose the globulization unless a big perturbation occured. That your code could even handle this instability shows that it must be very well constructed. If I were you, I'd look into the follow-ons to Ostriker's paper and see if you have a publishable result here.

grant hutchison
2006-May-20, 03:06 PM
Yes, this is great stuff. Thanks to you, mugaliens, and your buddy.

Couple of questions, fueled by curiosity:

Did "critically stable" fall out of the equations in some way, or is that a conclusion based on applying perturbations and watching what happened?

Did you happen to see what effect small changes in angular velocity had on the object?
Thinking about it, and then looking at my model of Ostriker's equations, it would appear that as soon as the torus loses angular velocity it should sag inwards, which seems like it should precipitate some sort of runaway situation in which the reduced rotational velocity is even less adequate to support the structure against its now-increased centripetal gravity. But the torus will also be able to support itself in compression, because its self gravity will hold it together along its minor radius: so it can behave a little like a cylindrical balloon standing on end under compression. Some sort of buckling then seems like a likely mode of real-world failure, at which point it would presumably shake apart into globules again. But in a mathematical world of perfectly poised equilibrium, I wonder what happens to it.

Grant Hutchison

Ken G
2006-May-20, 05:58 PM
Thinking about it, and then looking at my model of Ostriker's equations, it would appear that as soon as the torus loses angular velocity it should sag inwards, which seems like it should precipitate some sort of runaway situation in which the reduced rotational velocity is even less adequate to support the structure against its now-increased centripetal gravity.
I'm not sure I understand this argument, normally the gravity scales like 1/r^2 and the ficticious centrfugal force like 1/r^3 at constant angular momentum. Thus if the torus were to spin more slowly, but maintain a new angular momentum, I'd expect r to drop until the rising 1/r^3 overtakes the rising 1/r^2, possibly overshooting and oscillating. That much is like what happens in an elliptical orbit. So I might imagine a pulsational mode that could exist, but would dissipate in time, rather than an r instability. It seems like the r coordinate is pretty stable, but the theta coordinate may have all kinds of problems.

Relmuis
2006-May-20, 06:02 PM
If a torus lost angular momentum, might it not simply change its shape; that is, the ratio r/R? After all, the rotation rate depends on this ratio only, and it could have all kinds of values for the same volume (which is proportional to R * r2).

By the way, I have done some calculations of the rotation period, T.

Using the formularium from post 43:

T = 1/sqrt[{elog8(R/r)-4/5} * pi * G * D * (r/R)2]

e4/5 is smaller than 8, so the expression between the curled brackets { } will always be positive, at least while R is larger than r (which it must be, for the torus to have a central hole). So, there is no allowable value of R/r where the torus doesn't need to rotate.

Calculating the value between the curled brackets yields:
for R = r : 1.28
for R = 10 * r : 3.58
for R = 100 * r : 5.88
for R = 1000 * r : 8.18
etcetera (2.3 more for every extra zero).

pi is of course roughly equal to 3.14, and G to 6.7 * 10-11 m3/kg*s2

Setting the density D to 1,000 kg/m3 (the density of water), and multiplying everything together, we get a certain numerical value after the sqrt sign.

for R = r : 27 * 10-8 1/s2
for R = 10 * r : 75 * 10-10 1/s2
for R = 100 * r : 123 * 10-12 1/s2
for R = 1000 * r : 171 * 10-14 1/s2

This yields a rotational period T which is roughly equal to

for R = r : 2 * 103 s (somewhat over half an hour)
for R = 10 * r : 1.1 * 104 s (somewhat more than three hours)
for R = 100 * r : 0.9 * 105 s (a little above 24 hours)
for R = 1000 * r : 0.75 * 106 s (8.7 days)

These values must be divided by the square root of the relative density (5.5 for Earth; the square root is 2.35).

For slender toruses T will be nearly linearly proportional to R/r.

(Edited to fix a calculation error.)

grant hutchison
2006-May-20, 10:26 PM
I'm not sure I understand this argument ...It wasn't intended as a definitive argument, sorry. I was talking myself through a scenario that didn't make sense to me, in the hope someone would point out the problem, or I'd think of it myself.
I was also trying to think about the tidal evolution of such an object, if angular momentum were being steadily transferred from the rotation of the torus to its orbit.
An infinitesimal slowing of the torus should make its inner surface lighter and its outer surface heavier, producing an isostatic uplift of the inner surface (which fits with the behaviour of the equipotential lines when I play with my model). That, in turn, moves mass inwards. So the mean radius of gyration of the torus mass decreases. It now experiences greater centripetal acceleration, and the angular velocity necessary for equilibrium is correspondingly higher. But the torus is moving at an angular velocity that was too low to sustain it at its previous height. Seems like it should bulge ever more and move ever lower.
So what happens? Does it become a pulsating torus that accelerates inwards and then decelerates outwards bodily, in the analogy to an elliptical orbit you describe? That doesn't feel right to me for something achieved through slow tidal evolution.
What I was thinking about is that for a torus of constant volume to get smaller, it must also become thicker: the minor radius should increase as the square root of the decrease in major radius. So is the potential energy gained by the torus collapsing to a smaller major radius offset by the fact that its mass must also everywhere rise against its own self gravity to accommodate the major-radius shrinkage? It seems like these two interdependent and mutually opposed sources / sinks of potential energy might allow the torus to rotate at non-equilibrium velocities, as if it were capable of supporting compression (when moving too slowly) or resisting tension (when moving too quickly).
But probably all the above demonstrates is that I'm struggling to imagine the evolution of such an object under gradual tidal slowing. :)

Grant Hutchison

grant hutchison
2006-May-20, 10:31 PM
If a torus lost angular momentum, might it not simply change its shape; that is, the ratio r/R? After all, the rotation rate depends on this ratio only, and it could have all kinds of values for the same volume (which is proportional to R * r2).I think this is something similar to what I'm trying to imagine: that tidal evolution might not make the torus turn more and more slowly, but make it settle centre-wards at a constant angular velocity, becoming fatter as it did so.

Grant Hutchison

Ken G
2006-May-20, 11:31 PM
That, in turn, moves mass inwards. So the mean radius of gyration of the torus mass decreases.

I'm with you there.


It now experiences greater centripetal acceleration, and the angular velocity necessary for equilibrium is correspondingly higher. But the torus is moving at an angular velocity that was too low to sustain it at its previous height.

That's the disconnect-- what is conserved is angular momentum, not velocity, so as it gets smaller it speeds up. Same as for an object in elliptical orbit.


Does it become a pulsating torus that accelerates inwards and then decelerates outwards bodily, in the analogy to an elliptical orbit you describe?

That was in response to idea of reducing the angular velocity a bit. If it is continuous, then the deformation would be continuously shrinking.


What I was thinking about is that for a torus of constant volume to get smaller, it must also become thicker: the minor radius should increase as the square root of the decrease in major radius. So is the potential energy gained by the torus collapsing to a smaller major radius offset by the fact that its mass must also everywhere rise against its own self gravity to accommodate the major-radius shrinkage?
This sounds analogous to what is happening to the Earth. It is slowing, so its equator is not bulging as much, while trying to preserve its volume so its polar radius will increase a little. But it has no trouble navigating those competeing trends, and I suspect the torus will be the same.

grant hutchison
2006-May-20, 11:51 PM
That's the disconnect-- what is conserved is angular momentum, not velocity, so as it gets smaller it speeds up.That's precisely why I was fiddling around with the first instant of tidal evolution. Any degree of slowing results in the toroid shrinking, and therefore speeding up. But if it doesn't slow, it will have no means of speeding up. So how does it ever get slow enough to speed up in the first place?
That's why I liked the idea of tidal losses constantly braking the thing so that it somehow just deflated gently at a constant angular velocity.

But I guess it doesn't pay to reflect too much on the interactions of infinitesimals. :( Phobos seems to be managing quite well to increasing its angular velocity around Mars as a result of being continuously slowed by torque from the Martian tidal bulge, despite all my misgivings ...

Grant Hutchison

mugaliens
2006-May-21, 12:38 AM
I suspect that it's possible, Relmius, but I don't know how accurate it might be, due to the widely varying materials and masses of the Earth's crust, mantle and core. It's pretty difficult modeling simple vehicle collisions.

We might be able to do an abstraction, one with a water surface to a depth of about 8,000', average for the Earth's oceans, with an inner core of a known material approximating the large-scale physical characterics of the Earth.

I can't begin to even guess at what factors to use for it's modulus of elasticity might be.

I'd need the Young's, sheer, and bulk modulii, as well as the ultimate/breaking strength (I'm assuming the material is brittle). In addition to breaking from positive or negative strain, there's also buckling to consider.

We wouldn't begin to be able to examine issues such as slip, though

Is there a geographical physicist in the house who might help with coming up with a well-known material that's structurally similar to the average strength of the Earth?

Meanwhile, I think the biggest problem would be tensile strength. The Earth is full of cracks - only gravity holds the rocks together. Any tension at all in the toroid and it would fall apart. Over time even slight variations in gravity would produce tides and it's rotation would slow. Even a slight slowing would put tremendous compression stresses on the material.

Relmuis
2006-May-21, 04:47 PM
A ballpark estimate for the tensile (breaking) strength of a common material might be 108 N/m2. Deformation would be elastic to roughly half that tension, with a reversible elongation of 0.1 per cent over a tension of 5 * 107 N/m2. The thickness of the crust might be 25 km.

Elastic resistance to shear would be worse by a factor 2, I think, but breaking shear might be equal to breaking tensile strength, and the onset of plastic (irreversible) deformation would still be half the breaking value.

Actually, of course, the stiffness of the material would decrease with depth and reach zero (the liquid state) at perhaps 50 km.

Ken G
2006-May-21, 05:11 PM
That's why I liked the idea of tidal losses constantly braking the thing so that it somehow just deflated gently at a constant angular velocity.

Your equations tell us its size as a function of its angular momentum, so if we know how fast it is losing angular momentum, we'll also know how fast it is shrinking.



Phobos seems to be managing quite well to increasing its angular velocity around Mars as a result of being continuously slowed by torque from the Martian tidal bulge, despite all my misgivings ...

Look no farther than the Earth/Moon system-- the Earth's rotation is slowing down as the Moon picks up orbital angular momentum also.

mugaliens
2006-May-21, 08:16 PM
A ballpark estimate for the tensile (breaking) strength of a common material might be 108 N/m2. Deformation would be elastic to roughly half that tension, with a reversible elongation of 0.1 per cent over a tension of 5 * 107 N/m2. The thickness of the crust might be 25 km.

If it were truly solid, I would agree with you. But the Earth is by no means solid - the crust is full of cracks. It's more like an aggregation of rock, sand, alluvial stuff. There may be large chunks that are truly solid, but I'm not sure they're much bigger than Texas. Throughout the toroid wouldn't there be similar cracks in it's crust? I haven't a clue as to the consistancy of the mantle.


Elastic resistance to shear would be worse by a factor 2, I think, but breaking shear might be equal to breaking tensile strength, and the onset of plastic (irreversible) deformation would still be half the breaking value.

Actually, of course, the stiffness of the material would decrease with depth and reach zero (the liquid state) at perhaps 50 km.

Ok. I'll make some assumptions and see if he wants to spend any more time on it.

Relmuis
2006-May-22, 10:37 AM
A ballpark estimate for the tensile (breaking) strength of a common material might be 108 N/m2.

Actually, I may have been too pessimistic. The Encyclopaedia Brittannica gives the rigidity of Quartz (which is pure SiO2, the main constituent of the Earth's crust), as 4.4 * 1010 N/m2!

I don't expect this to be a valid rigidity for the entire crust, but 1 * 1010 N/m2 seems reasonable in this light. That's one thousand times as rigid as I thought.

By the way, I think that breaking strain is less important than onset of plasticity. As soon as we leave the elastic realm, the crust can no longer rebound to its original shape once the perturbation has stopped. This would herald the disaster. The disaster itself would be upon the planet when breaking strain is achieved; holes and tears would appear, magma would well up, and there would no longer be any (mechanical) constraint left.



Actually, of course, the stiffness of the material would decrease with depth and reach zero (the liquid state) at perhaps 50 km.

I realized that there must be a discontinuity between the solid and the liquid; I think it's the so-called Mohorovicic Discontinuity. The stiffness will decrease with depth, to a certain value, and then suddenly drop to zero.

grant hutchison
2006-May-22, 11:03 AM
Yes, Solar System Dynamics gives 5x1010 N.m-2 as a reasonable rigidity under tidal stress for a rocky world and 4x109 N.m-2 for an icy world.
That's the response to shear stress, yes?

Grant Hutchison

HenrikOlsen
2006-May-22, 11:14 AM
Not to skip too far from the subject, but the real test for planetary engineering would be to make two of these planets and link then together.

grant hutchison
2006-May-22, 11:21 AM
Look no farther than the Earth/Moon system-- the Earth's rotation is slowing down as the Moon picks up orbital angular momentum also.Yes, that one makes intuitive sense to me, if I draw the tidal deformation and the forces at work.
The fact that the toroid is in orbit around itself, rather than supported by compression forces, like the Earth, is what's messing with my head, I believe, though I'm not sure why.
Please don't think I'm arguing against the conservation of angular momentum. I'm just wondering if this thing could support itself by compression at slower-than-equilibrium revolution rates. But I'm sure you're right: if there's a smooth succession of equilibrium conditions (in this case, equilibrium rotation rates), the body will migrate through them.

Grant Hutchison

Ken G
2006-May-22, 02:41 PM
I think the fact that the center of the torus is in orbit around itself is perhaps not so different from the Earth after all-- the center of the Earth is also, in a sense, in orbit around itself too. It's just the trivial case of a zero speed, zero gravity orbit! But what is happening to the rest of the planet, either way, relies primarily on pressure support, not rotation.

grant hutchison
2006-May-22, 03:07 PM
I think the fact that the center of the torus is in orbit around itself is perhaps not so different from the Earth after all-- the center of the Earth is also, in a sense, in orbit around itself too. It's just the trivial case of a zero speed, zero gravity orbit! But what is happening to the rest of the planet, either way, relies primarily on pressure support, not rotation.Nice point. And of course there must be compression forces around the major circumference of the torus to balance the gravitational compression forces along the minor radius, so that everything is isostatic.
But reasoning from the "zero orbit" condition of the Earth to the condition of the torus will probably be just as bad for my head as it was to start fiddling about with those infinitesimals. :)

Grant Hutchison

Relmuis
2006-May-22, 03:15 PM
Every particle inside the torus is describing a circle. (If the torus is in equilibrium, that is.) Therefore, every particle feels a net force pointing to the axis of the torus. This force is the sum of the force of gravity and the forces from neighbouring particles. A tiny cube of liquid feels six forces on its six faces and a seventh force, the force of gravity in its center of mass. The six forces cannot create a torque, because of the liquid state, and therefore the three differences of the three pairs of force must be normal to the three pairs of faces. Therefore the force of gravity minus the required centripetal force can be projected into three components normal to the three pairs of faces, and each of these components will be equal and opposite to the difference of the two forces on the two corresponding faces.

grant hutchison
2006-May-22, 03:24 PM
Yes, Solar System Dynamics gives 5x1010 N.m-2 as a reasonable rigidity under tidal stress for a rocky world and 4x109 N.m-2 for an icy world.
That's the response to shear stress, yes?Oh, and isn't it customary to assume that planet-sized objects have no tensile strength at all? That's the assumption underlying the calculation of the Roche limit, for instance.

Grant Hutchison

Relmuis
2006-May-22, 03:32 PM
Maybe the figure quoted refers to the crust only, not to the liquid interior.

grant hutchison
2006-May-22, 03:40 PM
Maybe the figure quoted refers to the crust only, not to the liquid interior.No, it's for the whole body: it's in the context of tidal despinning, so it would make no sense to deal with anything other than a parameter for the whole object.

Grant Hutchison