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#1




What's gravity like on a planet shaped like a donut?
Assuming the donut is perfectly round and has uniform density, I'm pretty sure the same case applies as for the hollow spherical planet, so the person on the inner rim of the donut would feel no gravitational force at all. You would simply float off with a slight push. You can use calculus to do what Chronos referred to as breaking up the planet into pieces small enough that shape doesn't matter. Gauss's law, which is used to find electric field based on electric charge on surfaces or points, can be changed a bit to work with gravity and mass. Gauss's law says (integral) E*dA = Qin/Eo, where E is the field, A is the area of a closed surface, Qin is the charge inside the surface, and Eo is a constant. For gravity, it changes to (integral) g*dA = Min*4*pi*G. g is gravitational acceleration (=force/mass), A is the area, Min the mass inside the surface, G is the gravitational constant. Chronos assumed that for an object at the inner rim, the only force would be directly towards or away from the center, which I think is right. Given that, if you use a surface as a very thin disk (width w) centered inside the donut with the edge right at the inside rim, all of the gravity will pull in a direction through the thin edge, and almost none up or down through the larger surface. Also, since the outer edge is symmetrical, the force has to be the same everywhere. So the integral becomes simply multiplication g*A = Min*4*pi*G. There's no mass inside the inner rim, so g*A = 0*4*pi*G = 0. There is a small but finite area on the rim of the disk we're looking at (A = w*2*pi*(radius of inner rim)). Since g*A = 0 and A isn't 0, g must be 0 for all points on the inner rim, and in fact anywhere inside the donut that is centered along the axis of the donut. Looking back at that...it's not a very clear explanation. I'm working off of physics notes from 2 years back. It might make a bit more sense using the idea of looking at the mass above versus below you. Chronos said that in the donut, the mass below you would take up all the space below you, while the mass above would be a small area of the "sky". But in this case, the mass above you would look small only because it is far away. It is in fact much more mass though. And the amount of mass above versus below also depends on how high above the surface you're looking from. As your height above the inner rim decreased towards 0 (a tiny dot on the surface), the mass above you would approach infinity, and the mass below you would approach 0, but the distance to it would approach 0 also, and by the 1/r^2 law, the force from it would go to infinity. The only way to resolve the multiplying of 0's and infinities is with limits and calculus...thus the ugliness above. If anyone actually made it through all this and perhaps understood it, maybe you can explain a bit better. By the way, using this reasoning, if you were on the outer rim, you would feel the same pull as if you were on a spherical planet of the same radius and mass as the donut. If you were somewhere else on the surface, you would be pulled towards the center of the big donut ring, and also toward the center of the "tube" which forms the donut. ****************************** I have edited in a link to the Staff Report being discussed  Dex [Edited by C K Dexter Haven on 06082001 at 07:36 AM] 
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#2




[Homer Simpson]
Mmmm........... Planet Donut........ [/Homer Simpson] Anyway...... here's another way to think of it: Think of a normal, round planet, with a moon in orbit (kind of like ours!). When you're on the surface of the planet directly "under" the moon, do you suddenly float off the surface? Nope. I guess if the Donut were small enough, you'd get enough of a dual gravity pull to be able to jump clear off the surface, but with a larger donutplanet, the gravity would be too weak to do anything except create a permanently high tide. I think Chronos was working with the larger donutplanet idea (remember that gravity gets weaker exponentially the farther away from th source that you get). 
#3




Tora, tora, torus!
I'm glad someone asked this question, because I'd often wondered about it myself.
Chronos says: "Your weight will still be somewhat less than if you were on the outer edge of the donut (since there is some gravity from the far side), but you won't float away." Let's assume that the planet also rotates, in such a way that the axis passes through the centre, equidistant from all points along the inner rim. What if any, effect would centrifugal force have on gravity on the inner rim? Would the planet have to spin like a top in order to give, say, Earthlike gravity on the inner rim? Would this tend to throw people on the outer rim off? (I read that, in order to make Earthlike gravity in the space station depicted in 2001: A Space Odyssey, the station would have to spin like a top, not at the leisurely pace depicted.) And how would the inner rim get sunlight? My proposal is for the planet's axis to be at an angle relative to is rotational plane, so that the inner rim would get sunlight at least during certain parts of the year. BTW, you probably all know that only on a torus can it take five, not four, colours to colour a map if you don't want two adjacent countries to be the same colour.
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#4




Actually, it takes seven colors on a torus. What's maddening is that that was proved long, long before the fourcolor theorem for planes and spheres was proved.
By the way, there are plenty of surfaces more complicated than a torus. I don't know about 2001, but the speed of rotation shown for Babylon 5 is about right, and Babylon 5 is about a mile in diameter. A planet is a lot bigger, and would consequently need even less rotation.
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#5




If the gravitational attraction is neutral or toward the "center," wouldn't the planet collapse (since there would be no gravitational force holding it in place)? In short, for the thing to exist at all, seems to me that the gravitational pull has to be as Chronos described.
However, I'm a mathematician, Jim, not a physicist. << By the way, there are plenty of surfaces more complicated than a torus. >> Actually, from a certain mathematical (algebraic topology) point of view, there aren't any two dimensional surfaces other than a plane, a sphere, a torus, and a Klein bottle. Any homological or homotopic differences simply arise from adding handles. Thus, a sugarbowl (two handles) is just a torus (one handle) with another handle added.[/evil grin] 
#6




Dex:
Your logic also applies to hollow spheres. Please note that to the best of my knowledge, these things do not exist. The question posited a hypothetical structure, so for purposes of the question, we assume it exists. On the other hand  depending on the forces, the internal strength (tensile?) of the torus might hold it together. Also note the other point about spinning said object to increase it's stability. 
#7




Good analysis, Dave1701e, but you left out one detail: You're entirely neglecting the flux through the flat faces of your cylinder. Yes, the flux is small, but unless your cylinder has zero length, it'll be nonzero, and you're integrating it over a large area. Of course, if your cylinder does have zero length, then your rim will have zero area. So, we've got net gravitational flux zero through this cylinder, since there's no mass inside it. Through the ends of the cylinder, there's a nonzero flux inwards, so the flux through the rim must be outwards, to balance it.
Dex, even with the force as I describe it, the thing would still collapse, since the inner portion of the "planet" has to support the outer portion, which is being pulled inwards with a greater force. We just have to assume that this planet is made out of some miracle material which is strong enough to support its own weight. If you neglect structural strength and outside forces, then the only stable shape is a sphere, or, if it's rotating, an ellipsoid. This is why all known planets are spheres, rather than tori. By the way, isn't a torus just a sphere with a single handle?
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#8




Thanks for the clarification, Chronos. Yes, you can argue that a torus is just a "single handle" but it's usually considered as a separate 2d object.
You can make the basic 2d objects by starting with a rectangular sheet of paper, think of the corners as A, B, C, and D: A______________________B  ______________________ C D You have to imagine the line BD filled in, I can't draw it here. But we're talking a rectangular piece of paper, OK? IF you glue points A, B, C, and D together, you have a sphere. IF you glue AB to CD, you have a cigarette; now wind around and glue AC to BD and you have a torus (donut). IF you glue AC to BC (so you're putting a half twist in the paper, and then glue AB to ACD, you've got a Klein Bottle. A "handle" is obtained by glueing AB to CD (the cigarette) and then glueing the AD and BD sides to another surface. And that's all there is. ( ta da) 
#9




Quote:
Let's agree that, anywhere inside a hollow sphere, the gravitational force sums to zero; more particularly, anyone "standing" on the inside surface will feel no gravitational force. Now then, picture yourself standing on the inside of a hollow planet, someplace on the equator. You feel no force. I come along with a planetsized machete and lop off some of the shell at the north and south poles. What happens? Well, since the mass I removed is all "above" you, and it can no longer exert gravitional force on you, the gravitational force you feel is no longer balanced. The net force you feel must be in the opposite direction to the removed mass, or outward, toward the shell you are standing on. The more mass (up to a point) I remove with my machete, the more firmly your feet are planted. You can imagine that, at some point, what's left of the sphere looks approximately like a torus. Thus, Chronos is correct.
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#10




Dust
I've enjoyed reading the posts on this thread. Thank you all.
I have a rather silly question somewhat related to the topic, and don't really want to start a new thread because it's really too silly to have its own thread. My globe (and yours too, I suspect) has dust all over it, even on those surfaces facing down. What exactly keeps this dust from floating off? Is the dust inherently sticky, or (please refrain from laughing) does the globe actually have enough gravity to attract the dust? Or both? (See, told you it was silly!) Actually, it seems to me that the globe doesn't have enough gravity to attract the dust, and that the dust clings because the globe's surface is microscopically rough. (When I blow the dust off, for example, most of it blows away, leaving only those particles which are either very well attached or so tiny as to be affected by the globe's weak gravity.) But maybe a physicist among you would be kind enough to answer a question that keeps me awake at night (an exaggeration, but it does give me unpleasant dreams).
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#11




Well, I imagine that scant (and I mean VERY scant) amounts of moisture in the air help keep the dust attached. Another possibility is that dust may become "charged" with all this freefloating, and wants to stick to something in order to become grounded (but that's just a wild assumption).

#12




Well? What's the answer...
Well, it's been a while since I did any calc' work like this, but lets
have a whirl at actually SOLVING this problem. Let's start by defining the problem shall we? We have a torus shaped object with a little dude sitting on the inner wall at some point. Fair enough. The torus is made of a material K with some mysterious density P that we don't actually know. The distance from the center of the ring (ie. empty space) to the inner edge of the torus is b meters, and the distance from the center of the ring to the center of the cyclinder making the torus is c meters. This means the cross section of the torus has a radius of (cb) meters. OK! Time to get cracking. Let's look at a cross sectional volume of the torus. It has volume V defined by Pi*R^2 * dx, where dx is the width (heh, you can see where I'm going...) of the little disk we've cut out. It has mass P * V, and a gravitational force Fx associated with it, attracting the little guy sitting on the torus with Fx equal to G m1 m2 / r^2, where m1 and m2 are the mass of the person and disk respectively, and r is the distance from the center of the disk to the person. Fair enough? So we have a bunch of little disks around the torus, each attracting the little dude with some force Fx, depending on the distance of the disk from the person. Now how to calculate that? Well, to start with lets imageine this little torus in a nice 2D plane: //=\\ O \\=// Not much of torus, but never mind. The person is sitting at the bottom. Now, since gravity is vector, we can cancel the components of gravity with go left and right on this little diagram; it's only the up and down forces that will be left. It's a bit hard to see from this diagram, but I'm sure you get the idea. The disk on the right has an equivalent disk on the left with equal magintude gravitational force, but exact oposite components left and right. Now, some maths to play with, we're gonna have some fun with the cosine and sine rules. Remember at the top our b and c? Let's make a nice triangle. This triangle (you'll have imagine it, since I can't draw) has three faces of length b, c and dxy. Dxy is the distance between the dude standing on the inner side of the torus and the center of our little disk. The angles opposite each face are B, C and DXY respectively. To make this a little clearer: At the bottom of the triangle is the dude. This angle is C. Directly above that is side b, which has length b, going to the center of the ring. At the center of the ring DXY is made from that side, and the other side, from the center of the ring to the center of our disk, at distance c. At the center of the disk angle B is made from that side and the final side, of length Dxy, which goes back to our dude sitting on the torus. Draw a picture. It'll help. By the cosine rule, Dxy^2 = b^2 + c^2  2bc cos (DXY). So Dxy = sqrt (b^2 + c^2  2bc cos (DXY)) Additionally by the Sine rule: a / sin A = b / sin B = Dxy / sin Dxy Hence: sin C = c sin DXY / Dxy So what? Well, let's have a look. The force upwards by both the disks on the left and right is equal, and from our triangle there, it turns out the magnitude of that force is given by Dxy Fx cos C. Think of the right angle triangle with hypotonuse (sp?) Dxy, and lower angle C. Hence, our force for one disk pair is given by: Fnet = 2 Dxy cos (C) Fx Now let the width of those disks drop to 0. By doing this, we can take the sum of an infinite number of disk pairs, each of infantismal width, resulting with the net force total. Ie. Basically it means we can integrate it to get the answer, which is good, 'cause it's less work (than adding all the disks togethor). The trick is, what to integrate over. The definite integral integral (A>B) f(x) dx is defined as lim (n>infinity) * sum (i=1>n) f(xi*) deltaX. Basically a sample of width deltaX, where f(xi*) is the sampling of the actual function. So the infinite sum of samples of the function taken when the sample width approaches 0; exactly what we're doing here. However, in this case we're using DXY to sample the disks, not x, so we need to find integral (0>Pi) f(DXY) dDXY. (=P I didn't explain that very well, but it is true). Now from all that stuff at the top, Fx = G m1 m2 / (Dxy)^2, but m2 = V P, which is given by m2 = P (Pi * (cb)^2) Let's just bunch that whole big term togethor and let K = 2 * P * Pi * (cb)^2 * G * m1. Then we get something reasonable to evaluate: Fnet = K * cos (C) / Dxy Unfortunately we only have this in the form of cos(C), not cos (DXY), but that's easily fixed; recall: sin C = c sin DXY / Dxy. So: Fnet^2 = K^2 * cos(C)^2 / Dxy^2 Fnet^2 = K^2 * (1sin(C)^2) / Dxy^2 Fnet^2 = K^2 * (1  (c sin DXY / Dxy)^2) / Dxy^2 Fnet^2 = K^2 * (1/Dxy^2  c^2 sin (DXY)^2 / Dxy^4) Fnet^2 = K^2 * (Dxy^2  c^2 sin (DXY)^2 / Dxy^4) Fnet^2 = (K^2 / Dxy^4) * (Dxy^2  c^2 sin (DXY)^2) And remember that Dxy = sqrt (b^2 + c^2  2bc cos (DXY)), so that Dxy^2 is b^2 + c^2  2bc cos (DXY). Fnet^2 = (K^2 / Dxy^4) * (b^2 + c^2  2bc cos (DXY)  c^2 sin (DXY)^2) Fnet^2 = (K^2 / Dxy^4) * (b^2 + c^2  2bc cos (DXY)  c^2 (1  cos (DXY)^2)) Fnet^2 = (K^2 / Dxy^4) * (b^2  2bc cos (DXY) + c^2 cos (DXY)^2) Fnet^2 = (K^2 / Dxy^4) * (c cos (DXY)  b)^2 finally: Fnet = K * (c cos(DXY)  b) / Dxy^2 Fnet = K * (c cos(DXY)  b) / (b^2 + c^2  2bc cos (DXY)) Which we can integrate to find the total net force on the little dude! How cool. ...unfortunately, integrating this is too much for my little brain. I can't seem to do it. It does seem to be a reasonable solution however; consider if the radius of the ring (c) where equal to the distance from the center to the edge of the torus (b). Hence, the torus is actually just a piece of string. Since c = b; Fnet = K * (c cos (DXY)  c) / (2c^2  2c^2 cos (DXY)) Fnet = K * (cos (DXY)  1) / (2c  2c cos (DXY)) Fnet = K/2c * (cos (DXY) + 1) / (1  cos (DXY)) Fnet = K/2c !! Which cannot be right! The attraction from all disks is the same! But remember K has a factor (cb)^2 in it; so the attraction from all the disks is actually 0, since they have zero radius, and thus no mass, and, incidentally, the same. But back to the problem at hand. When we graph this function we get something strange, increasing in amplitude as time increases, this is because a greater component of the force is directed towards the up, recall we are looking at the net force up here, not the net force in total. However, I wish I could actually show you these graphs. You don't need to go and solve the integral exactly! It turns out the graph will be of the form of a sine/cosine wave, negative from 0 to N, and positive from N to Pi. When N is > Pi / 2, the net force total is toward the torus. When N is < Pi < 2, the net force total is towards the center of the ring. When N exactly equal to Pi / 2, the net force total will be zero. These forces are the forces the little dude experiences I mean. What is N though? Simple, N is when Fnet is 0. That is, when: Fnet = K * (c cos(DXY)  b) / (b^2 + c^2  2bc cos (DXY)) = 0 Hence, when: (c cos(DXY)  b) = 0 OR: b = c cos (DXY) b = c cos(DXY) b/c = cos (DXY) And we all know, in the range 0>Pi, cos (DXY) = 0 only when DXY = Pi/2 *=======================================================================* SO!! Finally, in summery, let me conclude: 1) Thank you for your patience 2) Thank you for this problem, it was facinating 3) Sorry if I made any mistakes! 4) Thanks to JJJ for pointing the website out. 5) THE ANSWER: The answer is, whether or not you will be attacked to the surface of said torus depends on the ratio of the distance to the edge of the torus surface from the center of the ring to the radius of the actual torus ring. If the ratio said is > Pi/2, you would be attracted to surface under you, if ratio is > Pi/2, you would be attracted to the surface way above you. If it was exactly Pi/2, you would feel nothing. NOTE: This weird manner of evaluting the intergral with out actually evaluating it can ONLY be done since it turns out the integral at 0 is 0. Otherwise there would be some constant shift of the ratio left or right. Once again; sorry for any mistakes/spelling errors. It's like 2.03 am in the morning. =P Enjoy! Hope this helps answer your question! 
#13




!! Oops. Forgot.
Roughly!! I forgot to add: ROUGHLY. Remember, we're approximating the gravitational attraction of a series of particles by those disks because it makes life easy. The solution is ROUGHLY correct, but because of the method not completely correct. Doing it properly would be a nightmare. =P
caio! 
#14




The first error I noticed in that derivation is that you can't treat a disk as a point mass. To really do the problem correctly, you'd need to do three integals, to cover the threedimensional body. You'd need to pick an appropriate coordinate system (cylindrical is the best I can come up with), and make sure that your limits of integration are welldefined (which would mean some ugliness in r and z).
However, I'm not going to do that unless they start paying me an awful lot more than a coffee mug. The Gauss's Law argument is enough to show that the force is always towards the planet, and we know qualitatively that the force will be greater on the outside than on the inside. That's enough for me. 


#15




My biggest concern about the explanation is that the reasoning of "what you see is what is gravity" is simply wrong. It would be correct if everything you saw were very thin shellsthen your vision would give a good idea of what the local gravitation would be. However that's not the case here.

#16




Point mass
You're right. I was just mucking around with number really.
Do you spose you could actually solve it by using the disk approach, but saying every disk is a sum of particles in 2D forming the surface (since the disk width approaches 0), and using a different integral to sum the forces from each of these particles. You'd only need two integrals then; dA for Area and d0 for the angle. Just a thought. I think if you did that you could get a reasonable answer to the question. 
#17




Sorry, Shadow, I'm no good at anything other than the basic maths... I diddle mostly with thought experiments. And what I'm picturing is, basically, the simple fact that gravity gets REALLY weak at longer distances, and that any ring large enough to hold any significant amount of matter onto the surface would also have to be wide enough to prevent it from collapsing in on itself (if the gravitational pull on the inside is strong enough to cause any significant pull on you, it'd probably also cause a much more significant pull on the opposite side of the planet... if you get my meaning).
Not that I consider this any sort of authoritative conclusion, that's just my two cents... Impressive calculations, though, even if you were just "mucking around with numbers". 
#18




Re: Well? What's the answer...
Quote:
(Sorry, couldn't resist...) ;)
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#19




What about a tunnel?
OK, so on a similar note...
Take for granted that the planet we're talking about is solid. If I was to drill a tunnel from one side of the planet to the next (UK to New Zealand style) and jump down, what would happen? Would the force be enough to propel me out the other side? Secondly, if I was to do this and not exit on the other side, would I keep falling to the same 'hight' at the other side, or would the height diminish? (accepting there is atmosphere in the tunnel). If It was to diminiush, would I be kept in zerog at the centre? Thanks everyone.... 


#20




PS...
I should mention that I'm thinking both about donut planets and spherical ones... This is a question I've been pondering for a while, but bringing donuts into the picture makes it unbearable...

#21




We actually received the initial question well over a year ago, and the reason it has taken so long to publish is that no one wanted to do the math (British: maths). Chronos managed to sidestep the math with his very neat argument, and we all applauded in awe.

#22




Re: What about a tunnel?
Quote:
For a toroidal planet (frictionless, airless, etc), I suspect that the point where you turned around would depend on the aspect ratio of the torus (the "fatness" of the donut), and whether you're talking about standing on the inside or the outside of the donut. For example, if you were on the outside of a skinny donut, you'd jump in the hole, fall through the rock, exit out the other side (The "inside" of the torus), turn around, fall back, and return to the point that you started from (conservation of energy). If the torus was substantially fatter, you'd fall through the rock, exit to the inside of the torus, continue through the center, enter the rock on the other side, keep going until you got to the outside surface (symmetry), turn around, and return. 
#23




Yeah, you could use the disk approach as per your modification there, Shadow Mint, but that still leaves three integrals, really, because integrating over dA for each disk is really two integrals. It's often represented as one, as a shorthand, but that doesn't really decrease the amount of work involved. You might also try dividing it up into a bundle of thin rings, and doing the other integrals afterwards... I'm not sure which would be easier.

#24




donut gravity
So long as you are talking about being symmetrically located relative to the edges of the strip, you can simply ignore that and only consider a ring with no dimension out of plane of the ring. The nonradial forces all cancel out leaving only radial forces, which you might as well collapse into a ring with no width.
The basic argument should follow along the lines of symmetry. At the center, all forces cancel perfectly; it's a metastable point. At the edge, most of the force is towards the center, since the mass that's close to you on either side cancel each other out. The mass that's directly opposite the center from you gets increasingly smaller for the purposes of determining radial force. T As you get farther and farther away from the center, the mass that pulled you away start pulling you back toward the center, but since it's a metastable point, the force must peter out before you get to the center. This suggests that the behavior is like an inverse Lagrangian point, but ringshaped. The maximum force occurs along a concentric ring. TTFN 


#25




I like Zut's intuative analysis.
Hey cronos  here's something that I 'learned' a long time ago but I'm not so sure is true. In fact, I'm pretty sure it's not true (by doing some Zutstyle intuative analysis), but figured I would ask in this thread. I was told that if you take a collection of discrete masses (stars perhaps in a cluster), find the center of gravity, and center there an imaginary sphere that is large enough to enclose the masses, then for any object on the outside of that sphere the system appears gravitationally the same as a single point source in the center with the same mass (ignoring GR). True? 
#26




Not quite true, sford, but it looks like that might be a reference to the aforementioned Gauss's Law. Gauss's Law, in its gravitational form (it can also be applied to electric forces, or magnetic, or a few other more obscure applications) says that if you have a closed surface, any closed surface, and you add up the gravitational field over the entire surface, then the total you get will be directly proportional to the mass inside the surface. Now, if you've got the right sort of symmetry to your problem (usually spherical), then you know that the gravitational field is the same everywhere on your surface, so it would only depend on the total mass enclosed, not on the distribution.
However, if the mass distribution is not spherically symmetric, then you will be able to tell the difference, from just outside the distribution. For instance, satellites' orbits are actually changed slightly by the fact that the Earth is flattened, rather than a perfect sphere. If you get far enough away, then any distribution of mass looks almost spherical, but right up next to the masses, it does make a difference. 
#27




All of this confusing stuff. Has anyone dealt with the fact that it depends on the donut?
Sorry, calc. makes me eyes glaze over, but... Look, whether the side of the donut "overhead" attracts one toward the middle of the donut hole depends on the ratio of the thickness of the donut to the diameter of the holeor the diameter of the whole, as the case may be. If the hole is a small area within a much larger body, the planet will collapse into it, and then no more hole. If we're talking Larry Niven's Ringworld, then the other side is too far away to have a noticeable gravitational pull. So, the answer is, "It depends!" 
#28




Quote:
Solving the definite integral is difficult because it leads to elliptical integrals. However, it can be solved numerically. 
#29




No matter what the size and proportions of the donut, it'll still collapse on itself, if it's not strong enough. Therefore, we're implicitly assuming that it is strong enough. Given that, the proportions of the donut don't have any effect on the direction of the force on a person on the inner rim.
The Ringworld isn't actually relevant to the gravitational question at all, since it's spinning: The primary force keeping folks' feet on the ground there is centrifugal. (Yes, I know that centrifugal force is fictitious. I also know that it doesn't matter that it is. :P) 


#30




<< Solving the definite integral is difficult because it leads to elliptical integrals. However, it can be solved numerically. >>
... Difficult AND tedious, which is why no one wanted to bother. 
#31




The argument I used in the report is admittedly not ironclad. I used that argument because I figured it would be the easiest to understand. The Gauss's Law argument, however, is ironclad, and it gives the same answer.
Of course, both arguments give just the direction of the force, and not the magnitude. If we want the magnitude, then I don't know how to avoid the integrals. Someone send me a paycheck for it, and I'll tell you the magnitude but not before. 
#32




Although the Gauss' Law argument is correct, doing the integration numerically is simple and that easily answers some of the other questions. I have tried this, and the force at the inner rim, of course, does pull away from the hole. On the surface, the force has a tangential component which tends to pull toward the inner rim, so this is where oceans would collect on a nonspinning donut planet. For the question about a hole drilled through and whether an object dropped through would ever reach the center of the main hole, zut is correct. If b/a>0.2736 the object will fall all the way through to the other side (b=radius of the circle which is revolved about an axis to define the torus and a=distance from the axis to the center of the revolved circle).
Since Chronos wouldn't give the analytical solution unless paid, I felt obliged. There was a recent GQ asking if partial differential equations could be used for anything, and this problem is a good example. Instead of doing the integration, the partial differential equation for potential field could be solved. The expression I found for the potential is an infinte series, but just a couple of terms give a value within 1% of the numerical results, unless b/a is very small (a skinny torus). When b/a is small, the solution is well approximated by an elliptic integral expression. Here is the solution for potential field (normalized by G, density of the torus, and mass of an object in this field) around a torus: Let c=sqrt(a^{2}b^{2}) and H be defined from tanh(H)=sqrt(1b^{2}/a^{2}). The normalized potential field is given by a sum over n=0 to infinity of terms with the form: A_{n}*cos(n*t)*P(n0.5,0,cosh(h))*sqrt(cosh(h)cos(t)) Where h and t are functions of spatial location (the point where the field is being evaluated), and A_{n} is a coefficient dependent only on the geometry of the torus. The field is axisymmetric so, using cylindrical coordinates, depends on r and z. The variables h and t are defined by: tanh(h)=2*c*r/(r^{2}+z^{2}+c^{2}) and cos(t)=cosh(h)c/r*sinh(h) P(n,m,x) is the associated Legendre Function of the first kind. The first two terms of the series are usually sufficient for accuracy, and the first two A_{n} are closely approximated by: A_{0}=c^{2}*pi^{2}*sqrt(2)*(11/8/(sinh(H))^{2}+3*log(tanh(H))/4) A_{1}=5*c^{2}*pi^{2}*sqrt(2)*(1/8/(sinh(H))^{2}+log(tanh(H))/4) 
#33




I bow down before you, Manlob. Admins, we've got to send this guy a coffee mug, at least.

#34




SPOOFE
Quote:
Quote:



#35




Exponent = 2 I believe.

#36




"Exponential" means a constant raised to a variable, not a variable raised to a constant.

#37




Well, you've just eradicated a tiny corner of ignorance Ryan. Correction noted!

#38




ring gravity
I made a plot of the gfield of a ring (actually 200 point masses arranged in a ring). No surprises.
http://ogre.nu/images/gringi.jpg The plane of the image is perpendicular to that of the ring. The force is stronger where the equipotential curves are closer together. 
#39




spin gravity; topology
The centripetal acceleration of an object moving in a circle is 4 pi^2 R / T^2, where R is the radius of the circle and T is the period.
I find it convenient to remember that one rpm gives one gee at 900 meters (oh all right, 894) and work up or down from that:  if you have the radius, divide it by 894m and take the square root to get the period in minutes;  if you have the period in minutes, square it and multiply by 894m to get the radius. From this we learn incidentally that a ringworld that spins in one day (1440 min), for one gee, has a radius of 1.854E9 meters, or 6.185 lightseconds, or 4.823 lunar orbits.  Dexter wrote: > Actually, from a certain mathematical (algebraic topology) > point of view, there aren't any two dimensional surfaces > other than a plane, a sphere, a torus, and a Klein bottle. And projective plane: what you get if you glue opposite edges of a rectangle, with a half twist in each pair. > Any homological or homotopic differences simply arise > from adding handles. Well then, you could leave the torus out of the list  as well as the plane, which is equivalent to a sphere with one point removed.  Four colors on a sphere, six on a Klein bottle or projective plane, seven on a torus. See also http://www.math.niu.edu/~rusin/paper...hromatic.genus (I dunno what happens to the coloringnumber if you add handles to a nonorientable surface.) 


#40




This is great stuff, bronto. I hope you're planning on sticking around and posting more? We can always use more indepth, wellthought posts.

#41




How flattering. But I fear this forum is too vast for me!

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