I may not have been totally clear about what I was defining r, y and phi as. At least in my notation, phi is not the polar coordinate; it’s the angle around the off-center ring. I’m afraid, then that this simplification is only true for a = 0 (the ring is not off-center), in which case the integral is trivial anyway.
Incidentally, the actual potential that I get (though I may have erred) is:[ul]V = (GM/pi R) × 1/(1+a) × K(2a[sup]1/2[/sup]/(1+a)), where K(k) is the complete elliptic integral of the first kind[/ul]
For an additional z-offset of bR, it’s[ul]V = (GM/pi R) ((1+a)[sup]2[/sup] + b[sup]2[/sup])[sup]-1/2[/sup] K(2a[sup]1/2[/sup]/((1+a)[sup]2[/sup] + b[sup]2[/sup])[sup]1/2[/sup])[/ul]
Yeah, you’re right. If I’m reading this correctly, then K(k) can be approximated as ln(4/(1-k[sup]2[/sup])[sup]1/2[/sup]) for k close to 1. Using my expression for V(a) above, and x = 1-a, I get:[ul]V = (GM/pi R) × 2/(1+a) ln((1-a)/(1+a)) = (GM/pi R) × ln(4/x)[/ul]If this is correct, then g = GM / pi R[sup]2[/sup] x. Using values for M and R from this page, it’s equal to 1g at xR=29.7km, which is apparently small compared to the width of the ring, so I think it’s not important…
I’ve rethought this (and gotten rid of my headache) and you’re right, there is a net gravitational force for a ring. Otherwise, duh, there would be no perturbation. However, I am not sure if there are any Lagrangian points in analogy to the three-body problem. Sounds like an excellent homework assignment for an advanced mechanics class, though, and I’m tempted to take a whack at it, if work is slow today.
Yes, you’re right about the y/r. I drew the diagram your way, decided I didn’t like it, and redrew it with the origin at the center of the circle, then went on to hybridize the two.
BTW, just to provide some motivation for my confused babbling, when considering the long-term behavior of an orbit, it’s a common celestial mechanics trick to “average” the other moons in the system into rings, which then just create a J[sub]2[/sub] perturbation, like the oblateness of the planet.
For the idealized case of a one-dimensional ring, as scr4 says above, there will be a synchronous orbit outside but close to the ring. However, the synchronous orbit inside the ring, for a light ring spinning faster than its Keplerian speed, will be far inside the ring. For the Ringworld’s rotation period of ~10 days, the synchronous orbits inside the ring would have to be well inside the orbit of Mercury (year ~90 days).
The actual Ringworld, though, is more like a section of cylinder, with a radius about 100 times its width. Close enough to the surface, its gravitation no longer looks like the idealized ring; instead it looks more like a (somewhat) uniform flat sheet. The difference in gravitational acceleration across a uniform flat sheet is 4 pi G (areal mass density); using the parameters from the site Achernar links to I get an areal mass density of about 8.6 x 10[sup]6[/sup] kg/m[sup]2[/sup] and a delta-g of only 7x10[sup]-3[/sup] m/s[sup]2[/sup]. This is not nearly enough for a synchronous orbit outside the Ringworld: The sun’s gravity at the Ringworld radius is about 6x10[sup]-3[/sup] m/s[sup]2[/sup], so the radial gravitational acceleration toward the sun can vary from about 2.5x10[sup]-3[/sup] m/s[sup]2[/sup] (just inside the Ringworld) to 9.5x10[sup]-3[/sup] m/s[sup]2[/sup] (just outside), with corresponding Keplerian-orbit periods from about 0.75 to 1.5 Earth years. (So there are neither Ringworld-stationary nor sidereal-stationary orbits near the Ringworld.)
For an idealized ring, there should be close orbits circling the ring (i.e. in orbital planes perpendicular to the ring plane, threading the loop of the ring). I think these can have nearly arbitrary average sidereal period, as long as stability is not an issue (for an orbit with a short period, let it approach the ring closely on the side opposite the sun; for an orbit with a long period, let it approach closely on the sun-side). For the Ringworld, the limited acceleration provided by the ring’s gravity will place limits on these periods: The threading orbits close to the Ringworld should be somewhere between the Keplerian extremes (0.75-1.5 yr[sub]Earth[/sub]) since they see both the inside and the outside of the Ringworld.
The potential is constant in both the rotating and the nonrotating frame, so this is a little simpler than the restricted 3-body problem. If we restrict motion to the plane of the Ring, angular momentum is conserved. Now, I’m sure you’re familiar with the process of setting up an “effective” potential - the potential in the rotating frame - with an L[sup]2[/sup]/2ma[sup]2[/sup] term. This makes it a 1-dimensional problem. This doesn’t allow for the threading orbits that Omphaloskeptic mentions, but it’s something.
For an ideal (1-dimensional) Ring, the potential diverges to negative infinity at a = 0 (the sun) and a = 1 (the Ring). There will, in general, be one minimum in the effective potential, inside or outside the Ring depending on the angular momentum. The frequency of perturbation will in general not simply relate to the orbital frequency, both of them the solutions to equations involving nasty elliptic functions. So, noncircular orbits won’t be closed, but they’ll be stable to perturbations in the plane.
For a Ribbon (2-dimensional cylinder), I tried to do the math but couldn’t. However, I believe that the potential has a cusp at a = 1. (The derivative is a step function.) I also believe that there will again be a single minimum in the effective potential where resides a stable orbit.
