You wouldn’t have a single solid piece of granite circling the Ringworld.
Even on Earth, soil expands and contracts due to temperature changes, and (even more so, I believe) due to changing water content. Also, I don’t believe the soil layer is that thick. I always envisioned a few meters thick, but even a kilometer thickness of soil wouldn’t do much.
If the Ringworld were instantaneously split into thousands of segments, would those segments have escape velocity form their Sun, or would they be in an elliptical orbit? If the later, you could construct all the segments in orbit at their aphelion. Then you just slow them down and let them all fall inwards, and simultaneously connect them when they are all at their perihelion. Instant spinning ring.
ETA: You don’t need to construct the entire width of the Ringworld at once. Once it’s spinning, you can drop in other parts to attach to the existing ring.
Gotta think like Pak.
According to Niven’s book All the Myriad Ways, Niven was greeted at Boskone (the Boston Area SfFcon) by members of MITSFS* chanting
“Ringworld is Unstable! Ringworld is Unstable!”
A bit more dramtic than just letter-writing.
*The MIT Science Fiction Society, which Niven says can be pronounced various ways, including “Misfits”. He couldn’t have disliked them all, despite the jibe – he married Marilyn “Fuzzy” Pink, who was an officer (President?), and creator of the eponymous Pinkdex computerized index of the MITSFS holdings.
They would have escape velocity many times over - the Ring rotates at about 700 miles per second as I recall, while the escape velocity would be only about 25 miles per second.
I forget my Niven-physics.
Why wouldn’t something as massive as the ring have it’s own gravitational pull? After all, full scale (1:1) maps of various planets’ continents (including Earth) were found included in the topography, and they lay well within the boundaries of the inside surface of the ring…
It does, but the gravitational pull of a non-rotating ringworld is a.) much too small and b.) directed the wrong way (most of the mass is on the sunward side of you, and will tend to pull you off the Ringworld. Fortunately, not very rapidly. (I calculated the effect back when I first read the novel. It’s a pretty straightforward calculus problem, but I’m not about to work it out right now)
So you really do need to rotate the Ringworld to get that artificial gravity
I have to admit that it’s not at all clear to me that a flexible Ringworld will be stable. I imagine the OP is thinking that it will act like a lot of individual particles in orbit together, so the force from one part won’t strongly pull on another. But the mere fact that the ringworld is a linked whole is going to mean that forces on one part will affect another part, even if it’s flexible, or even “stretchy”. The particles in Saturn’s rings aren’t connected at all, so they’re a bad model.
I think if you’re standing on the Ringworld, the gravitational pull from the material underfoot would cancel out the gravitational pull from the rest of the ring, and the net effect is zero.
Anyway, the Ringworld is only “unstable” in the sense that if its position were perturbed, gravity would not pull it back to the original position. It’ll just stay at the offset position. In that same sense, a series of unlinked objects circling a star would be just as “unstable” - if you push one of the pieces away from the original orbit, it would not try to return to the original orbit.
Actually not. While a solid dyson sphere would stay in place if perturbed, a Ringworld would rapidly fall into the Sun. I did the calculations a few years back, and determined that when the offset is small, the offset would double every couple of months (and when the offset is large the offset will grow even faster).
A ringworld is dynamically unstable, from what the things I read way back when discussed. The analogy would be a ball on top of a hill, or a pencil balanced on a point. The least deviation from optimum would pull it further and further off center.
The gravity inside a uniform body (a ring, thin cylinder, sphere) is effectively zero. Ringworld itself has no net gravitational effect on occupants of the internal surface. It’s all determined by the central sun and the centrifugal “force” from the spin.
If you consider the ringworld as a collection of discrete pieces in the same orbit, the result is still dynamically unstable. The least perturbation, and two pieces will fall into each other - thus creating a larger gravitational effect that will grab some more pieces - the typical planetary accretion model. Hence the need for magical rigid scrith. Again, the only stable “same orbit” situation is using trojan points, but requires two large bodies, the central one substantially larger than the second one. (Or a situation where the bodies are so small and apart that they have effectively zero gravitational influence on each other).
By the way, this reasoning led Maxwell to deduce that the Rings of Saturn were composed of countless small objects. http://www-history.mcs.st-andrews.ac.uk/~history/Extras/Maxwell_Saturn.html
"The final result, therefore, of the mechanical theory is, that the only system of rings which can exist is one composed of an indefinite number of unconnected particles, revolving round the planet with different velocities according to their respective distances. These particles may be arranged in series of narrow rings, or they may move through each other irregularly. In the first case, the destruction of the system will be very slow, in the second case, it will be more rapid, but there may be a tendency towards an arrangement in narrow rings, which may retard the process. "
Not true. If you do the math, the net pull is away from the surface. You’ve got some stuff near you, but you’ve got a LOT of stuff that’s relatively nearby , but forward of you.
With the earth, everything is to the “bottom” of you, so it’s all yanking you down. Not true in a Ringworld.
of course, there’s an awful lot of stuff that’s far away from you, but distance makes its effect negligible.
Oops, you are right, I didn’t think through. The reason is: from inside a 2-dimensional ring, the amount of mass within a given solid angle is proportional to distance, but since gravitational force is proportional to distance^(-2), the gravitational pull from a unit solid angle is inversely proportional to distance. So the pull from the closer part of the ring is stronger than the far side.
And if it were a sphere, the amount of mass within a solid angle goes as distance^2, so the gravitational pull is the same in all directions anywhere inside the sphere.
This also means that gravity from the ring does act to pull things down onto the inside surface of the ring.
I think the net pull is towards the surface, for the same reason the ring is dynamically unstable. If the ring/star system deviates from symmetry and the star gets closer to one side of the ring than the other, the star is pulled towards the closer side. If the net pull were away from the surface, the ring/star system would be stable.
I remember doing the calculus in my college physics class. The short answer is that “the gravitational force inside a hollow sphere shell of uniform areal mass density is everywhere equal to zero.” Everything cancels out beautifully. The mass that is closest to you exerts a greater force, but that is balanced by the fact that there is more mass further away.
http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/grvtysp.htm
If the Ring were non-spinning, the gravitational field on the inner surface would be pointing outwards, towards the “ground”. The simplest way to show this, if you have the physics background, is via Gauss’s Law. Make your Gaussian surface a cylinder fit inside the ring, with a very small height. No mass is enclosed, so the net gravitational flux through the cylinder is zero. The flux is inwards through the top and bottom of the cylinder, so it must be outwards through the edge.
That’s only true for a sphere, not for a ring. And in fact, finding the gravitational field due to an enclosure is exactly equivalent to determining the stability of the enclosure: The net force of a Dyson sphere on an off-center sun is zero, and hence by Newton’s 3rd law the net force of the sun on an off-center Dyson sphere is also zero, and hence the Dyson sphere has neutral stability. Meanwhile, the net force of a ring on an off-center sun is such as to increase the offset, and so the force of the sun on the ring is as well, and so the ring is unstable.
It’s not clear to me, either, that a flexible Ringworld would be stable, but it also wasn’t clear that it wouldn’t be stable. The only argument that it was unstable that I had seen implicitly assumed that it was a rigid ring.
We don’t know in what way the Ringworld is flexible, or what its properties are, but we do at least know that it is flexible. It’s 16 or 17 light minutes across. When Fist-of-God formed, it took that long for the far side to even see it hit. It couldn’t have maintained a circular shape during the impact, since the far side couldn’t react until 16 or 17 minutes after the impact.
Yeah, I was surprised when I figured that out - especially since the gravity inside an infinitely long cylinder is zero.
I don’t know what you’re talking about here. The mass is all to the sunward side of the ring. If you add up the contributions from all the masses, divided by the square of the distance, the net force is toward the center of the ring. If you’re standing on the outside of the ring, this is very clear. If you’re on the inside, the pull from the nearest portion away from the center isn’t enough to cancel the inward pull.
I can’t visualize the Gaussian envelope you’re drawing.
Just land *me *on the thing, and it’ll be safe.
Maybe it is, maybe it isn’t.
Gravity on “outside” of ring. One unit of gravity towards the sun due to the stuff close by that is under you. Plus one unit of of gravity also towards the sun due to all the stuff far away. Total equals two units towards the sun.
Inside the ring. Stuff under your feet (and away from the sun). One unit of gravity away from the sun. Rest of the stuff one unit towards the sun. Units cancel.
You can’t say it does or doesn’t work till you do the math. And my gut/poor memory says its zero inside the ring and non-zero outside the ring.