Let’s say you have a true ringworld - basically a poor man’s Dyson sphere, so long that the ring completely encircles an orbit of a star. Can a ringworld have satellites - of either the natural (moon) or artificial (sputnik) variety? If so, what would the satellite’s motion be? I’m not sure that a satellite could just circle around one “point” on a ringworld - after all, there’s all that gravity from the rest of the planet pulling on it. Then again, I didn’t major in astrophysics, so maybe all the forces balance out and allow exactly that to happen…
Would a potential satellite circle nicely as above, or would it get into a helical “corkscrew” orbit around the entire length of the ring, or would it simply crash into the world?
The ring couldn’t have a satellite. The Moon orbits around the center of mass of hte Earth-Moon system, which is offset just a bit from the Earth’s center of mass (which is, unsurprisingly, at the center of the Earth). However, the center of mass of a ring is at the center of the ring. Thus, any objects in the ring’s viscintity would simply orbit the star.
I forget; did Ringworld rotate at its Keplerian orbital speed? Any way, if that were so, and the object were close to the ring, their orbital periods would be quite similar, so the object would drift slowly in the sky from the point of view of a ring-dweller.
If the orbit of the satellite was at all elliptical or inclined, it would be slightly perturbed by the ring’s gravity, so it would precess–that is, its axis of perihelion and line of nodes would turn. Other than that, it would hardly notice the ring’s gravity.
This depends on the distance of the object from the Ring. There’s a certain distance which would form a sort of L2 point, and a satellite here would orbit at the Ring’s Keplerian angular speed. Inside this distance it would drift slowly prograde, and outside it the satellite would drift slowly retrograde.
In the plane of the ringworld, there’s no gravitational effect from the mass of the ringworld inside its orbit. That’s why you have to have a method–attitude jets–to keep it centered on the sun. See The Ringworld Engineers by Niven, of course.
The center of mass description is excellent. Something could orbit the sun, inside or outside the orbit of the ringworld. That gives some interesting possibilities.
IIRC (and I hardly ever do) that works out to 770 miles per hour at the outside edge of the Ringworld.
You know, I love my kids 99.9% of the time, but when I can’t find my copy of Ringworld because they’ve randomly re-arranged my book collection again…well, I get a little testy.
The mass of the ring causes a gravitational attraction of course, but inside the ring, the gravitation of any chunk of ring is exactly cancelled out by the gravitation of another chunk or chunks on the other side of the ring. (Works the same for a hollow sphere, incidentally.)
It’s easy to see how this is the case at the very center of the ring (or sphere). At other points, the math turns out to work the same. As you move closer to any one part of the ring/sphere, the pull toward the rim grows, but a larger percentage of the ring/sphere tugs you back toward the center, also. The pulls exactly negate each other.
An intro-level college physics textbook will probably show how to prove this with calculus; I went through it once, but it was a long, long time ago.
As NoCoolUserName mentioned, though, for a ring this is only true in the plane of the ring, which is why you probably couldn’t make a satellite orbit through/around the ring itself, only around the ring/star system.
Well I’m no physicist, but that’s just what I did, and it didn’t work for me. I took a ring in the x-y plane with radius R and center (0, aR). I drew a line from the center to a point on the ring, and defined phi as the angle between this line and the y-axis in the negative direction. The ring is then defined parametrically by:[ul]x = R sin(phi)
y = aR - R cos(phi)[/ul]The distance r from the origin to a point on the ring is then given by the law of cosines:[ul]r[sup]2[/sup] = x[sup]2[/sup] + y[sup]2[/sup] = a[sup]2[/sup]R[sup]2[/sup] + R[sup]2[/sup] - 2aR[sup]2[/sup] cos(phi)[/ul]The differential gravitational acceleration at the origin in the y direction for an infinitesimal mass element dM is given by:[ul]dg[sub]y[/sub] = G y dM / r[sup]3[/sup][/ul]where y = y(phi) and r = r(phi) are given above. dM = lambda R d(phi), so this becomes an integral of one variable running over phi = 0…2pi. The value dg[sub]y[/sub] is then proporitional to the integral over:[ul](a - cos(phi)) / (a[sup]2[/sup] + 1 - 2a cos(phi))[sup]3/2[/sup][/ul]This integral is too hard for me to do, but according to my calculator it’s not zero for all values of a, and in fact it seems to diverge in the limit a -> 1.
Using the same notation, I got that the gravitational potential is proportional to the integral over 0 to 2pi of:[ul](a[sup]2[/sup] + 1 - 2a cos(phi))[sup]-1/2[/sup][/ul]This one’s a little easier, and it can be expressed in terms of an Elliptic Integral function. To make a long story short, the value of the integral does not appear to be constant with respect to a.
It’s been WAY too many years since I took physics and calc. In the intro to Ringworld Engineers, Niven says it took Dan Alderson “several years to quantify the instability.” Anything that takes Dan Alderson several years is outta my league–although he may have been quantifying how much the ringworld would move with respect to the sun given a particular amount of interstellar dust or galaxial gravitation or something.
The way it was explained to me in physics class is:
From a viewpoint perpendicular to the ring (“looking down” on it) pick a point inside the circle–don’t pick the center, that’s the trivial case. Draw two lines that intersect at the point. The area between the lines on one side of the ring is not equal to the area on the other side, because we didn’t pick the center, but the gravitational attraction cancels out because we’re farther away from the side that has a greater area. I wish I could draw the picture, it’s almost intuitively obvious.
IANA maths wizard, but it seems intuitive that what is true inside a hollow sphere is also true inside a great circle ring slice of that sphere.
The great thing about living on a ringworld is that spaceflight is almost impossibly difficult; you have to accelerate a space probe until it is in orbit around the star, and this speed is very different to the speed of the ring floor; a satellite of the local star would have an orbital period of about a year at the distance of the ringworld, while the ringworld rotates in a falan, approximately 94 Earth days; about four times as fast as orbital speed.
So you would have a satellite whipping past every four months or so depending on the distance from the ring floor.
You could lob a rocket over the rim wall, but it would fall outward, and you would never see it again.
No, Achernar’s right; even in the plane of (and within) a ring (approximated by a circle with nonzero linear mass density) the force on a particle is nonzero. The integral is not elementary (it can be expressed, as Achernar found, using elliptic integrals), but near the ring it scales approximately as ~1/x (x the distance to the ring), something like the force near a long straight wire.
NoCoolUserName’s explanation is a nice way of thinking about things, though the conclusion is wrong. When you draw the two lines intersecting within the ring, the amount of mass is larger for the side farther away, but only by a factor of the distance (there’s also an extra secant factor which we can’t easily handle with this level of explanation). The gravitational force, however, decreases with the square of the distance, so it’s smaller for the far-away parts of the ring: the particle will be accelerated toward the nearest point on the ring. For the case of a sphere, however, the amount of mass is proportional to the distance squared (again with a secant factor), since you’re looking at a surface instead of a line: the two factors cancel in this case. (That they cancel exactly for the case of a sphere, and not for a cube or some other surface, requires a little more work to show.)
Omphaloskeptic, the explaination I quoted from my physics class was, of course, for a sphere. I had generalized to a ring without knowing that it was, in fact, correct.
The fact that a non-central point has a gravitational attraction to the ring makes the ringworld more unstable than just drift. Once the sun moves off-center, it is attracted to the ring!
The final answer (from “Larger than Worlds”) is to start the ringworld moving in a lateral direction (90 degrees to the plane) and use electormagnetic effects to create a Bussard jet, with the sun to prime it. That’s traveling in style!
Archenar, it’s a bit more tractable if you note that y/r=cos(phi), so the integrand is just cos(phi)/r[sup]2[/sup], or:
1/[(1+a[sup]2[/sup]) sec (phi) - 2 a R ]
The rest is left as an exercise for the reader, as my Schaum’s Mathematical Outline is at the office, and I have a splitting headache that will not accommodate integration by parts or any other such shenanigans.
Anyway, no net force means no L2 point (or any other Lagrangian points, for that matter). The ring just profides a J2 perturbation in addition to the gravity of the star.
Actually, Niven’s Ringworld rotated faster than Keplerian speed. If it was exactly Keplerian speed, the surface would be weightless (free fall). If it were slower, there will be gravity on the outisde surface of the ringworld.
Which means you can have a synchronous satellite “above” (inside) the ring. You can also have a “low orbit” satellite very close to the ring, which won’t be synchronous.
Crap. The integrand should be 1/[(1+a[sup]2[/sup]) sec (phi) - 2 a ], obviously.
And I dug out my calculator, but it’s puking on anything other than a=0 (for which, reassuringly, the integrand collapses to cos(phi), which integrates to 0 over 0,2*pi, as we would expect.)
The plates also rotated at non-Keplerian (Can we use a word like that? It’s almost sinful.) speeds. They were held in place by wires.
Picture a dozen playing cards in orbit around the sun. They drift, they turn, they wobble. Now thread them all together and give them a bit more than orbital velocity. They stretch the thread and stay stable–just what we want!
