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How would a metal bar built around the world react to gravity?
November 1, 2007 12:38 PM   RSS feed for this thread Subscribe

How would a metal bar built around the world react to gravity?

I’m not sure how to properly phrase this question. But, how would a metal rod with a slightly larger diameter built around the earth react to gravity? Say it’s built as to avoid extremes in altitude, mountains and the sort. Would it hover in place, spin at a set point, spin randomly, or buckle? Is there anyway to know the answer other than building one?

I created a simple visual to help explain what I mean.
posted by studentbaker to science & nature (36 comments total) 16 users marked this as a favorite
Probably like a planetary ring.

I love your visual.
posted by stereo at 12:43 PM on November 1, 2007


I don't know the complete answer, but I suspect that the whole model is complicated by the fact that the Earth is not a sphere, but is an oblate spheroid.
posted by The World Famous at 12:44 PM on November 1, 2007


I think it would just rest up against the side of the earth and not do anything. Boring, I know.
posted by lohmannn at 12:49 PM on November 1, 2007


The earth isn't a perfect sphere with uniform gravity. There's flattened poles and a bulging equator and topography variations that cause fluctuations in the earth's gravitational field.

I don't think it'd spin or stay still. I think it'd buckle and crash into the earth. Where would depend on how you situated the ring and what it overlapped.
posted by empyrean at 12:49 PM on November 1, 2007


If this band is made of ferrous metal, you'll get interactions with the magnetic field in addition to gravity.

Let's make a couple assumptions:
-Get rid of the Earth and replace it with a gravity source of equivalent magnitude that is infinitely small.
-The band is a perfect circle with its center exactly coinciding with the point gravity source
-The gravity of the band is negligible compared to the gravity source.
-The entire universe is comprised of the gravity source, the band, and nothing else.

If these conditions are met, I would argue that the band stays perfectly still, centered on the gravity source.

In the real world, I think one of several things could happen:
-The band will be large enough that its own gravity will cause it to collapse.
-It's really next to impossible to perfectly center the band around the gravitational center of the Earth, so one side will get a stronger pull than the other and it will crash in to the Earth. Any other imperfections will cause the same.

Assuming you can build something like this, it will probably rotate like an unbalanced wheel for awhile. Its center of rotation and the gravitational center of the Earth will drift, so you'll get an eccentric rotation that will increase in amplitude until it crashes in to the planet.
posted by backseatpilot at 12:51 PM on November 1, 2007


Apparently, a solid ring would be unstable. Click here for James Clerk Maxwell's 1859 paper On the Stability of the Motion of Saturn's Rings. He cites Laplace, who demonstrated "that a uniform solid ring cannot revolve permanently about a planet," but I haven't been able to find a link.
posted by stereo at 12:52 PM on November 1, 2007 [1 favorite]


In general, rings or spheres are not affected by any force, like gravity or magnetism, inside them. In the case of a ring, it's unaffected by the part of the earth in the same plane as the ring, but might be pulled on by the parts off the earth that go beyond the ring i.e. the poles if the ring was at the equator.

So, it would just sit there. You'd be able to pick it up but it would just move randomly until it bumped into the earth and would probably just get stuck. It would not necessarily float above the earth's surface. Nor would it necessarily move. But chaces are that it would appear more-or-less weightless relative to the earth.
posted by GuyZero at 12:55 PM on November 1, 2007


All of these behaviors are plausible at some level of approximation, depending on the initial position, velocity and angular momentum of the ring. For example, if it's on the equatorial plane of a prefectly spherical planet, rotating with an angular velocity appropriate for its diameter, it will just stay there. For maximum hand-waving, I'm assuming it will be large enough to be outside the atmosphere.


Also, if you are not a designer you should be.
posted by the number 17 at 12:55 PM on November 1, 2007


The stresses would probably tear it apart.
posted by vacapinta at 1:13 PM on November 1, 2007


Also the moon would exert gravitational pull on it. That would a) move it off center, and b) exert tidal stress on the structure.
posted by frieze at 1:17 PM on November 1, 2007


When I was 10 or 11 years old I asked this exact question in the form of a post on halfbakery.

While I was quickly disabused of the notion of weightlessness being transmitted by touch, I occasionally though about this problem in the intervening years. Thanks for asking this though, because now you've reminded me and I want to know the answer!

The center of mass of the ring is in it's center. So gravitational force on the entire thing (assuming it's rigid) could be modeled as acting on that single point. In this case, that point would be attracted to the center of mass of the earth (which we'll assume is a uniformly massive perfect sphere). So it's rest configuration would center it's center of mass and the earth's, and it would resist perturbation and try to remain there.

That said, I could be totally wrong in this reasoning - as that laplace paper would seem to indicate.

I'm going to ask my physics prof tommorrow.
posted by phrontist at 1:18 PM on November 1, 2007


It's true that the overall force of gravity on said solid ring would be neutral. But there would be compressive force on the ring. What you have is something akin to the arch of an arch bridge, which is held up by the compressive strength of the material from which it is built. (An ideal arch bridge is a subsection of an ellipse which has one focus at the center of gravity of the planet.)

If the force on the arch (due to weight and loading) exceeds the compressive strength of the material, it collapses. Equally, if the compressive force on your hypothetical ring exceeds the compressive strength of the steel from which it is made, it buckles and collapses.

And it turns out that's what would happen, because the compressive force of a ring that large would exceed the compressive strength of all known materials -- by a very large margin.
posted by Steven C. Den Beste at 1:19 PM on November 1, 2007


Echoing comments on the quality of the example graphic... Consider infographic design, professionally or as a hobby.
posted by cadastral at 1:24 PM on November 1, 2007


There is an escape from what I just said: If the ring is rotating, then centripetal force will cause tension that counteracts the compression caused by weight. When the ring rotates at a rim-speed equal to a neutral orbit at that altitude, the tension and compression will be equal and there won't be any force on the ring at all.

If it spins too rapidly, then it will cause enough tension to exceed the tensile strength of the material of the ring, and then the ring will fly apart.
posted by Steven C. Den Beste at 1:24 PM on November 1, 2007


It might hula-hoop in sync with the moon's revolutions, crushing or displacing whatever's on the opposite side of the planet.
posted by Reggie Digest at 1:25 PM on November 1, 2007


If this band is made of ferrous metal, you'll get interactions with the magnetic field

actually, you could get interactions with any metal.
/Physics rulez
posted by ArgentCorvid at 1:26 PM on November 1, 2007


backseatpilot: one side will get a stronger pull than the other and it will crash in to the Earth.

See, someone brought that up on the halfbakery thread as well, and it seems reasonable, but contradicts the argument I gave above.

Think of it this way. Imagine the off center circle and it's position relative to the earth (I actually hauled out my compass and drew this). While the gravitational force is more on the closer portions of the ring, much more of the mass is on the far side so the ring would be pulled back in the equilibrium.
posted by phrontist at 1:31 PM on November 1, 2007


I would thin that a closer model for comparison might actually be a fictional construct - The Ringworld Is Unstable! In short, solar winds, atmospheric drag & moon aside your ring would have a tendency to (eventually) crash into the earth.

Because the earth's mass is at the center the effect of it's gravity would be constant at all points - a zero sum game with no 'correcting factor'. As soon as the ring came a little off center it'd have a tendency to continue moving in the same direction (relative to the earth) and crash.
posted by mce at 1:33 PM on November 1, 2007


I firmly agree with Steven Den Beste - no material could withstand the force. The math should be simple... I may actually try it out in a bit, but the compressive force would be immense (that's why I've always assumed "perfect rigidity").

So yeah, if you actually tried to build such a thing it would never stay up. But if you had some magic material it would float and be hard to peturb. You could put a monorail or something on it! As long as whatever you hang on it has negligible mass compared to the ring, it wouldn't even wobble.

Problem is, no such material exists.
posted by phrontist at 1:34 PM on November 1, 2007


There are multiple equilibria for this. Among them, staying fixed in one place, or spinning about its axis are unstable equilibria. Any perturbation of such a state would lead the ring to seek a different stable state, which - in as much as anything about this can be called "practical" - would, practically speaking, almost certainly have one section of the ring lodged up against the earth. That would be a stable equilibrium.
posted by Wolfdog at 1:35 PM on November 1, 2007


Oh, and yes, other gravitational fields would pose a problem, but I would think they could be compensated for (with uneven weighting of the ring perhaps?).
posted by phrontist at 1:38 PM on November 1, 2007


phrontist:
While the gravitational force is more on the closer portions of the ring, much more of the mass is on the far side so the ring would be pulled back in the equilibrium.
Gravity follows an inverse square law - the larger mass isn't large enough to account for the larger distance.
posted by mce at 1:39 PM on November 1, 2007


Wolfdog: I don't see how that's possible without the ring disintegrating... where is my reasoning above flawed?
posted by phrontist at 1:40 PM on November 1, 2007


mce: Perhaps... but that would mean the center of mass model is wrong, and I don't see why that would be the case. I'll try to work out a "proof" later.
posted by phrontist at 1:42 PM on November 1, 2007


And there's the heating/cooling effects of the sun. One half of the ring would be at the night side of the Earth, whilst the other would be facing the sun.

It wouldn't stay ring shaped for long.
posted by popcassady at 2:00 PM on November 1, 2007


phrontist: You can't always deal with extended objects as a point mass. This is one case were that doesn't work. Consider a thought experiment: Imagine a massless ring with two point masses along a diameter. Now put that ring with its centre on another point mass (earth) - it won't move, because the forces are balanced (so we have an equilibrium point). Now place the ring so that one mass is slightly closer than the other to the earth. The force on the closer mass will be greater than the force on the further mass (because of the inverse square nature of gravitational force). The ring will move away from the equilibrium point because of this, so the equilibrium is unstable. Now consider adding two more masses on a diameter perpendicular to the first. This doesn't change the behavior, because we have the exact same thing as before, just on two axes now. You can continue doubling the number of masses and this will remain true. Double the number of masses an infinite number of times and you have a ring (and you've done the integration that this problem would call for).
posted by ssg at 2:27 PM on November 1, 2007


ssg: You convinced me at first, but now I'm doubting that reasoning again.

Now consider adding two more masses on a diameter perpendicular to the first. This doesn't change the behavior, because we have the exact same thing as before, just on two axes now.

Except that the two perpendicular masses are now shifted... so it's not symmetrical.

Someone in #physics on freenode is working on the integral right now.
posted by phrontist at 2:38 PM on November 1, 2007


It might hula-hoop in sync with the moon's revolutions, crushing or displacing whatever's on the opposite side of the planet.

I was thinking the same thing, except not hula-huping, rather moving like this. It's probably completely wrong, but the moon would have to have some kind of an effect I think.
posted by kisch mokusch at 2:43 PM on November 1, 2007


The fact that a ring is unstable is basic geometry.
posted by vacapinta at 2:52 PM on November 1, 2007 [1 favorite]


For those who are curious, some helpful dude in #physics worked this out. I'm not going to solve it though, because the reasoning in vacapinta's link convinces me.
posted by phrontist at 2:56 PM on November 1, 2007


TWF: I am going to suggest to my son that he call his band "The Oblate Spheroids"
posted by nax at 2:56 PM on November 1, 2007 [1 favorite]


Assuming uniform spherical non-magnetic earth, no atmosphere, etc., Steven C. Den Beste is exactly right: the ring is effectively an arch. Any cross-section of the ring is under compression equal to the weight of the entire ring. If it were made of some unusual material that could support that compression, it would hover at uniform height (unstably). But the slightest deviation from a perfect circle at uniform height would cause any material to immediately bend or break. The slightest nudge from an air molecule or micro-meteorite would probably be enough.

What would be interesting would be to put the ring into a spin. If it were spinning at orbital velocity, there would be no compressive forces because each piece would be in free-fall. And there would be no tensile forces for the same reason. But this is unstable, so any perturbation into an ellipse would again subject the ring to immense tensile and compressive forces. It would probably shatter spectacularly.
posted by Araucaria at 3:36 PM on November 1, 2007


Not sure if this helps, but there is no net gravitational force anywhere inside a hollow spherical shell.
posted by neuron at 3:49 PM on November 1, 2007


Yes, there's no force inside a spherical shell (and equivalently, no net force on a spherical shell from a mass inside it), but a ring is different; once it becomes slightly offcenter there will be a force acting to pull it farther offcenter.

I remember from Ringworld discussions that you can stabilize it by bouncing it up and down along the third axis — dunno why.

If the ring isn't spinning or anything when you put it up there, it'll be under compression. It's a very long thin object, so it'll presumably fail by buckling. If you spin it fast enough to put it under tension, then it wouldn't buckle.
posted by hattifattener at 5:51 PM on November 1, 2007


Let's make a couple assumptions:
-Get rid of the Earth and replace it with a gravity source of equivalent magnitude that is infinitely small.
-The band is a perfect circle with its center exactly coinciding with the point gravity source
-The gravity of the band is negligible compared to the gravity source.
-The entire universe is comprised of the gravity source, the band, and nothing else.
I don't see why the one about the relative masses of the two objects is at all relevant.
posted by Flunkie at 7:53 PM on November 1, 2007


ssg, your argument from the thought experiment is very elegant and convincing, but I think it may have a problem, even though it leads to a correct qualitative result in the case of the ring; namely, it seems to work equally well for a spherical shell (imagine a massless spherical shell with two point masses at the antipodes, and etc.), and we know that the spherical shell will not be drawn toward the Earth if we move it off center.

The difficulty seems to be one of dimension. Points don't necessarily add up, even infinitely, to a sphere-- or a ring. Vacapinta's link is careful to frame the argument in terms of adding up (integrating) balanced segments whose length approaches zero in the case of the ring, and little balanced caps whose area approaches zero in the case of spherical shells.

If the force of gravity were to vary by the inverse of the distance rather than the inverse square, your argument would predict that the nearer part of the ring would still be pulled toward the Earth with any displacement, but the argument in Vacapinta's link shows that the ring would still feel no net force when displaced. The spherical shell on the other hand, would resist any displacement of the center of the shell from the center of the Earth.
posted by jamjam at 8:12 PM on November 1, 2007


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