# Gravity on an Orbital Ring

Added 9-15-19: Thank you all for your comments. I will be reading them through carefully. I have quickly skimmed through the answers and I do appreciate your corrections on terminology and ideas, and my equation.

After I've had a chance to think through all the comments and answers, I will revise my questions and post more to see if what I put together holds up to your knowledge.

Angela

In my story, I have an orbital ring around Earth at geosynchronous orbit, supported by and/or connected to a series of space elevators. I have already taken care of materials, purpose, and other issues.

My question is whether the ring is in free-fall orbit, so that the conditions there are zero-G, or whether because the ring is 'attached' to Earth, there will be a gravitational pull towards Earth. If there is a gravitational pull, how strong is it? Would you use the normal gravitational equation F=GMm/r2? And if so, the mass of such a massive structure will make this an interesting calculation.

Would this also depend on how the ring is linked to the elevators?

Thank you.

• Hi Angela, welcome to Worldbuilding! You didn't mention how massive your ring would be (although this is not likely be affecting the outcome). – Alexander Sep 10 '19 at 23:31
• Gravity doesn't change because things aren't attached. Earth still pulls at the ring. If it's spinning fast enough, inertia creates a faux gravity on the inside of the ring. – 458 Sep 11 '19 at 1:04
• While the ring won't be exactly in orbit, it will be almost in orbit. (The almost is because the ring is a rigid-ish solid-ish object with non-zero width and thickness.) The gravitational acceleration (with respect to the ring) experienced by an object near the ring will be minuscule. (Cannot say "on" the ring, because the gravitational acceleration will be so very very small.) (And the word you are looking for is "anchored" by those space elevators.) – AlexP Sep 11 '19 at 1:06
• @alex If the forces are equal it's coincidental, as the ring's outer pull force is relative to how far you are from earth's center, and the inner pull force is relative to the rings angular velocity. – 458 Sep 11 '19 at 1:36
• Welcome to Worldbuilding. It seems like you (and many more in the comments/answers) don't have a good idea of what an Orbital Ring is. An OR can exist at any orbits (even some impossible for normal satellites), and they don't need to be at geosinchronous. Most OR in fiction are at lower orbit and the elevator that hang from them aren't actual space elevators, but simply cable that hang from the ring – SilverCookies Sep 11 '19 at 11:40

You described the ring as "orbital" and "at geosynchronous orbit"; therefore, it's in freefall. That's what "orbit" means- in freefall, but moving sideways so fast you miss the planet completely.

The fact that your orbital ring is attached to Earth simply means that if it moves relative to Earth, it'll break the tether and cause problems. The best way to avoid these problems is to put the ring in a circular orbit over the equator at geosynchronous height moving in the same direction that Earth spins- in other words, a geostationary orbit, which you've already done.

The formula for Newtonian gravitational acceleration is

$$a_g = {GM \over r^2}$$

Plugging in Earth's mass and the radius of geostationary orbit, we get

$$a_g = {\left(6.674 \times 10^{-11} {N m^2 \over kg^2} \right) \times \left( 5.972 \times 10^{24} kg \right) \over \left(42164000 m\right)^2} = 0.224 {m \over s^2}$$

So people on the ring will experience about 0.02 gees of acceleration from Earth's gravity (1 gee = 9.8 m/s^2), but the ring itself (being in freefall) will be accelerating in exactly the same way. As such, people on the ring will feel weightless- they'll float around in it, just like astronauts on the ISS (or any other spacecraft in orbit).

Now that I've said that, I will note that you've specifically characterized this space station as a "ring". This suggests that you'll be spinning it for artificial gravity. Just to get you started, the formula for centripetal acceleration is

$$a_c = {v^2 \over r} = {v \over t} = {r \omega \over t}$$

where $$a_c$$ = centripetal acceleration (i.e. the "gravity" that people standing on the inside rim of the ring will feel), $$r$$ is the radius of the ring, $$v$$ is the linear speed of the edge of the ring, $$\omega$$ is the angular speed of the ring (in radians per second), and $$t$$ is the period of the ring's rotation (i.e. the time it takes to make one full revolution). You want $$a_c$$ to be equal to one gee (9.8 meters per second squared); if you know how big you want the ring to be, this will tell you how fast it should spin.

Also, if you spin the ring, you'll probably want it to be in the same plane as Earth's equator. A spinning ring is like a top or a gyroscope- its axis of rotation will want to point in the same direction all the time. If you mount the ring perpendicular to the tether, spin it up, and wait six hours, you'll find that the Earth (and the tether) will have rotated under it, while the ring itself has not. To keep it from crashing into the tether, you'd need to transfer all of its angular momentum somewhere else (no easy task, but doable if you have two identical rings stacked on top of each other, spinning in opposite directions- if the structure connecting them is sufficiently rigid, their angular momenta cancel out, and you can spin the whole assembly any way you want), or abandon the wheel-on-a-stick visual and mount the ring in the same plane as the tether and the Equator, like a unicycle.

Finally, the whole assembly will not have any noticeable effect on the Earth's orbit unless it's a significant fraction of the mass of the Moon- and the Moon is much bigger than most asteroids. Getting something that big into orbit will not be at all feasible any time in the foreseeable future, whether it's lifted up from Earth or dragged down from interplanetary space.

Edit: I should clarify that the above paragraphs were written under the assumption that your "orbital ring" was nothing more than a ring-shaped space station with some space elevators attached. The sort of thing that we could plausibly build once we figure out space elevators. However, I've been informed that this may not be the case, and that instead you may be going for a megastructure of mind-boggling scale, encircling the entire planet at geostationary altitude.

In the case of a megastructure at geosynchronous altitude spinning in time with the Earth, the answer is quite simple: The ring is in freefall, and therefore anything inside will float around like in any normal-sized space station without artificial gravity.

However, if you push the ring outward (by lengthening the tethers) without changing its angular speed (i.e. keeping it at 1 revolution per day), you'll start to get centrifugal gravity. Things inside the ring will "fall" toward its outside edge. We can compute how large the ring needs to be in order to get one gee of artificial gravity this way.

Here's the basic equation:

$$a_c = a_g + g$$

where $$a_c$$ is the centripetal acceleration at the ring's edge, $$a_g$$ is the acceleration due to Earth's gravity at the altitude of the ring, and $$g$$ i our goal: standard Earth surface gravity, 9.8 m/s^2.

The centripetal acceleration is the acceleration required to put an object on a circular path. Part of that, for a person standing on the ring in in our scenario, will be supplied by Earth's gravity, and the rest by the outside rim of the ring. Its that "the rest" bit that we want to be one gee.

Looked at another way (specifically, from a frame of reference spinning at the same rate as Earth), $$a_c$$ is the centrifugal acceleration pushing the ring's occupants outward, $$a_g$$ is the gravitational acceleration pulling them inward, and once again, the difference is what they'll actually feel.

So, plugging in the formulae for $$a_c$$ and $$a_g$$:

$${r \omega \over t} = {GM \over r^2} + g$$

Rearranging things a bit, trying to solve for r:

$${r^3 \omega \over t} = GM + gr^2$$ $${r^3 \omega \over t} - gr^2 - GM = 0$$

Aaaand that's a cubic equation that I don't know how to solve.

Wolfram|Alpha gives a fantastically complicated solution for r, as well as a couple of equally-complicated solutions with imaginary numbers that can safely be ignored. So I'll just let WolframAlpha puzzle through the numbers on its own... and the solution turns out to be r≈1.15989×10^10 meters. That's about 17 times the radius of the Sun, 30 times farther out than the Moon, and 375 times the radius of geostationary orbits.

Good luck building that!

• Reading the second part of this answer ("[...] two identical rings stacked on top of each other, spinning in opposite directions- if the structure connecting them is sufficiently rigid[...]", "abandon the wheel-on-a-stick visual and mount the ring in the same plane as the tether and the Equator, like a unicycle") it seems to treat the "ring" as a space station which happens to be shaped like a ring - and not as a circular mega-structure surrounding the entire planet, which I think is what OP aims for. – G0BLiN Sep 11 '19 at 13:31
• Note that if treated as a mega-structure, more of the parameters of the orbital ring become known (e.g. ω and t must equal Earth's ω and t so that the tethers are geostationary, r must be big enough to contain the Earth and then some, etc.) - in fact, if you aim for a to equal one gee, you can find an exact r - maybe add this? – G0BLiN Sep 11 '19 at 14:31
• @G0BLiN You are correct; I did interpret "orbital ring" in the question as "a ring in orbit that could plausibly be built in the near future, minus the space elevators", rather than a mind-bogglingly huge megastructure encircling the entire planet. In the geostationary megastructure case, the answer is quite simple: You're in freefall and so is the ring; therefore, you feel weightless. – Someone Else 37 Sep 12 '19 at 2:44
• You could, however, push the ring farther out without changing its angular speed, and get artificial gravity that way. I'll do the math. – Someone Else 37 Sep 12 '19 at 2:45
• Thank you! It's just been too many years since I went through physics classes, I was fuzzy on the details. I can do the math however, and will follow through with that. Angela – Angela Sep 16 '19 at 0:15

There would be some 'gravity' on board the station, but it would actually be in the opposite direction - away from Earth.

This is because a station attached to a rotating body (Earth) by a space elevator would have to be at a distance somewhat beyond that of a geostationary orbit. The stations are being held in place by the space elevators; and if the elevators were to snap, the stations would, I believe, zoom off into an elongated elliptical orbit. Since you have one solid structure, a ring, orbiting earth, it wouldn't zoom off. Rather, it would act like a centrifuge space station... a centrifuge space station 85,000 km in diameter.

The equation for determining the acceleration you get from a centrifuge is:

$$a = R(\frac{2\pi}{T})^2,$$ where 'a' is the acceleration, R is the radius, and T is the rotation period.

At geostationary orbit, a ring station with a rotational period exactly equal to Earth's day period will find that its acceleration outward is exactly equal to the pull of Earth's gravity down, so there will be no net pull up or down. If the rotation period is constant, however (which it would be, since it's tethered to Earth by the space elevators), then increasing the orbital radius would increase the 'gravity' pulling outward, away from Earth. This is a linear relationship, so doubling the radius doubles the outward gravity. If you went and made the ring station 170,000 km in diameter, you'd have an outward gravity of about 0.45 meters per second squared, while the downward gravity would be about 0.05 meters per second squared, for a net outward gravity of about 0.4 meters per second squared, about one twenty-fourth of what you get on Earth.

So, unless you wanted to make the ring absurdly huge (like, ten times the distance from the Earth to the moon huge), getting rotational gravity from a ring station tethered to Earth isn't really feasible. If you were closer than standard geostationary orbital distance to Earth, you would get downward gravity from Earth... but then the station would need to be a rigid body, holding itself up against Earth's gravity; which is, as far as we know, impossible from an engineering standpoint on account of the sheer size and mass of the station.

So, long story short: The station is effectively in free-fall orbit, and while there might be some outward pull, depending on how far past geostationary orbital distance the station is, it won't be very much.

• "This is because a station attached to a rotating body (Earth) by a space elevator would have to be at a distance somewhat beyond that of a geostationary orbit" Can you explain why? Are you assuming the station is also acting as the counterweight for the space elevator? If so, what if the elevator extended beyond the station and there was a separate counterweight further out? – Hypnosifl Sep 11 '19 at 1:49
• @Hypnosifl Yes, I am assuming that the station is acting as a counterweight. You COULD set up a separate counterweight further out. If you did, you could set up the station closer to Earth, such that it got substantial gravity from Earth, but that would likely require even more advanced engineering, as well as leaving the station itself in a precariously unstable situation. If the cables snap on a station sitting beyond geostationary orbit snap, the station goes flying... if the cables on the counterweights of a station INSIDE stationary orbit snap, it falls. – FlyingLemmingSoup Sep 11 '19 at 2:04
• it's a ring around earth. you can make it any diameter you want and suspend the elevators from it, which will serve as anchors to keep it centered. no need for a conuterweight. that said, a solid ring at GEO boggles the mind. – ths Sep 11 '19 at 10:55
• The most stable setup would be to put the ring exactly in geostationary orbit, and provide a counterweight extending outwards at each attachment point of a space elevator. That way, a failing space elevator would simply mean that the cable to the counterweight needs to be cut in order to keep the station safe. It could actually be the same cable extending through the ring station, but some means to cut the cable in an emergency would likely be a good idea to ensure that space elevator debris cannot damage the station. – cmaster - reinstate monica Sep 11 '19 at 12:32
• If the counterweight is positioned far enough out of the ring, it already has escape velocity in the event that its cable needs to be cut, so it will just fly away from earth, never to return. So, there would be exactly zero danger from counterweight debris to the station. Plus, such a counterweight tether would provide excellent launch points for deep space missions. The situation for the inner part of the space elevator is not so easy. – cmaster - reinstate monica Sep 11 '19 at 12:37

It's microgravity (free-fall)

First I would like to clarify that geosynchronous orbit and geostationary orbit are not quite the same thing, and that considering the ring is tethered by space elevators you must mean geostationary. The difference is that geostationary is a specific geosynchronous orbit that sits on the equator and rotates at a rate equal to the Earth so that any point on the orbit is always directly (or nearly) above the same point on the ground, which is necessary for a tethered space station of any kind. The only alternative would require some sort of force continuously adjusting the station's acceleration to avoid it falling to Earth.

That said, anything in geostationary orbit will have basically negligible gravity. There is of course gravity, but it is just enough to keep the station in place, without pulling it downward. As massive (pun intended) as a ring structure around the Earth might be, I don't imagine it could be built large enough to actually have a strong affect on gravity, or you would end up having to adjust all the equations about the force of gravity anywhere on the planet.

• Not a criticism of what you wrote, but I despise the term "microgravity", because it suggests that what's relevant is the weakness of the gravitational pull, when in fact it's "free fall" that really matters. (In this specific case, the gravity really is weak, at a couple percent of normal surface gravity, but still far stronger than the one-millionth of normal that "micro-" in the SI context would indicate.) – Monty Harder Sep 11 '19 at 18:56

As Chronocidal observed, solid rings don't orbit well... they're intrinsically unstable, and likely to crash (because as soon as they get off-centre, the nearest bit gets pulled harder by gravity and the further bit less hard and it'll accelerate towards the planet). You could fix this if your space elevators were orbital towers, but making compression members 36000km long sounds like an even harder engineering challenge than making space elevators which are already pretty crazy hard. Possibly the counterweights on your elevators can provide a correcting outward force without the need for compression members, but as your ring won't be unbendably stiff (it'll be a quarter of a million km in circumference, and good luck making something that size really stiff!) you'll need a hell of a lot of space elevators to hold it still. Bicycle wheel earth!

What you could use instead is a Dynamic Orbital Ring. These use much shorter tethers than space elevators (maybe a mere 1000km tall, and a ring circumference of 46000km... child's play if you can make a space elevator!), and maintain their position by being spun so it isn't in orbit and isn't held up by towers, but held down by the tethers. They can also be placed in various orientations and altitudes, unlike your geostationary ring which could only be built in one orbit. You build stuff on the tethers and supported by the ring, but you don't attach anything to the spinning ring itself because its sole purpose is to hold everything else up.

Gravity on a ring of this type is indeed computed in the same way as you'd compute force due to gravity under normal circumstances. Thus, at 1000km altitude on one of these rings you'd experience about 75% of Earth's gravity, for example.

You could build a dynamic ring of this type at geostationary altitude if you really wanted and had access to materials of implausible strength, but I'm not really sure why you'd want to.

• Things in orbit don't go "off-centre" from gravitational pull. Orbit implies freefall, so the gravity felt by any part is effectively zero. You also can't "fall" out of orbit. As your perigee altitude decreases, your apogee altitude increases. – Fax Sep 12 '19 at 14:12
• @Fax you'll note that I didn't say that gravity would pull the ring off centre, merely that once off-centring has occurred, gravity will amplify the problem, inevitably resulting in a crash. You'll also note that I didn't say that the ring was in orbit, only that they "don't orbit well". Do feel free to propose how a solid planet-encircling ring can be in a stable orbit around that planet, though. – Starfish Prime Sep 12 '19 at 14:18

From the question and some of the answers I read there seems to bit of a misunderstanding of the subject at hand. I will try to clarify.

An Orbital Ring (OR from now on) is NOT some kind of giant ring that somehow occupy an entire orbit around a planet. Such a structure would be orbitally unstable and would require active correction to maintain it's position, it would also be impossible to build without exotic materials.

An OR is a massive active structure that is composed mainly by a large set of co-orbital objects (we'll call them the core). OR are powered by electricity and require no loss of material to be maintained (no remass needed).

More refined versions of the concept have the entire orbit surrounded by a (magnetically suspended) shield, and more facilities can be built on top of the shield. The shield can protect the core from things like an atmosphere, meaning your OR can, in theory, be in an orbit with a lower point just a few meters above ground. An OR can be built with modern technology (although it would be a massive project, 1000s of times more expensive than anything ever built).

In some versions, long cable sometimes hung from the OR down to the surface of the planet; usually to ease material transport from and to the OR. These cables are usually just a few hundreds Km long and aren't true space elevators in the sense that they aren't supported by gravitational nor centrifugal forces and require no counterwheight like normal space elevators; for all purposes they are just a cable hanging from a very tall structure.

In theory you can build an OR at any orbit, however building one at GEO would make little sense. OR are usually built to easy access to space, you can lift material to the ring above the atmosphere and then magnetically accelerate the material up to any orbit you desire. If you build your OR at GEO then might as well build a normal space elevator instead.

The gravity. Here's the thing: the only part of the OR that is actually "in orbit" is the core, the rest (the shield, the cables, the city, whatever...) is not in orbit at all; it's just standing there sitting on top of the shield. If you have an OR at GEO then the gravitational attraction will be toward the Earth and have a value of about 0.23 m/s2 (less than the Moon). At a much lower orbit the gravity will be just a bit less that the surface gravity of the Earth.

Start with a simple beanstalk made out of carbon nanotubes or better. A specific strength of 100 or more -- that is MPa / (kg/m^3).

Park an artificial asteroid in geosynchronous orbit. Spin a cable of this stuff down towards the Earth (with a small weight on the end), and another outwards (with a much heavier weight on its end). Attach to ground. Solve all of the instability problems.

Repeat a dozen or more times.

Each of these have a "station" at geosynchronous orbit (what is left of the artificial asteroid), a cable going down to Earth, and a tail going out. To launch something to space, you send it out the tail then drop it (this causes the cable to whip some, so there are limits to how much you can use it this way).

At the geosynchronous station, spin and Earth gravity are equal. At the tail-station, there is going to be more spin gravity than Earth gravity. Below the geosynchronous station, there is going to be more Earth gravity than spin-gravity.

Now, do something crazy. Spin a fiber from one beanstalk to another and hook them up with carbon nanotubes. Repeat until you have a ring.

These carbon nanotubes are theoretically zero stress, but the entire orbit is not stable. You can, however, wheel in or out the tails to fix instability without ejecting mass, and transport force over the ring of nanotubes, and it is plausible that you could make that entire thing dynamically stable.

Along the nanotube ring, there is going be zero net gravity -- Earth and Spin gravity add to 0. The ring will be 35,786 km above the Earth, or just over 40,000 km above the center of the Earth. This makes it about 1/4 of a million km long, or about 7 times as far around as the Earth is.

Suppose we take that ring, and we detach it from the beanstalks. Then we spin it up.

f = m v^2/r -- to hit 10 m/s^2 we need v^2/40,000 km = 10 m/s^2, or v^2 = 400,000,000 m^2/s^2, or v = 20000 m/s; roughly 6 times geosynchronous orbital speed. At that speed, things attached to the ring will experience roughly 1 gravity.

For every m^2 of cross section the ring has 1/4 million km * m^2 * 1300 kg/m^3 = 3 * 10^14 kg of weight. This is also under the effects of 1 gravity; so the carbon nanotubes have to support that weight.

Carbon nanotubes can handle 130,000 MPa in tension, or 1.3 *10^11 N per m^3. The ring is under 10^15 N of tension, so the tubes can hold less than 1/20,000th of the ring's length without support.

In theory you could relieve tension at beanstalks, but 20,000+ beanstalks seems like a bit much. Even at 1/3 G you'd still need 6000 beanstalks.

Now, you could reduce the height of the ring; it is spinning and supported by beanstalks, so it isn't really in orbit at this point. At 1/3 the height you get 1/3 of the mass, together with 1/3 G ... still requires 2000 beanstalks.

In short, spin-gravity around an entire planet is a large step beyond building a beanstalk.

On the other hand, theoretical materials that are ridiculously stronger than carbon nanotubes could be used. But barring that, there will be no spin gravity of any kind.

• Would you need 20,000 beanstalks, or just 2 beanstalks made of cables comprised of 10,000 carbon nanotube wires each? – gbjbaanb Sep 11 '19 at 23:04
• @gbjbaanb You need roughly 20000 anchor points (that support the ring) to relieve enough tension on the ring to prevent it from tearing itself to pieces. The problem with a rotating ring is each piece has to hold up every other piece. The "beanstalks" could in theory provide support, reducing tension, but it requires a lot of them. Now, you might be able to do something where you have a thick "channel" with something rotating in a "groove" held up by the channel; that might work with fewer than 1000s of beanstalks – Yakk Sep 12 '19 at 0:45

The people in the Ring would, essentially, be in zero G.

Being attached to the Earth by elevators would have no effect on this. Gravity is a warping of space/time around a mass. It is not transmitted like heat but is a warping of the universe.

If the ring is in a geostationary orbit (being far enough out to rotate just fast enough to keep up with the ground) then they are in free fall on the ring.

Now, if the ring is lower or higher it would need to continually thrust to maintain position. That would add a sideways force that would behave as gravity.

The mass of the ring wouldn't really cause gravity unless it was massive enough to cause problems for Earth. Even then the gravity would only really be felt on or near its surface.

• It would not need to thrust to remain in position due to being higher or lower than a geostationary orbit. A ring in any orbit would need some sort of active feedback to maintain position because such an object is not stable in any orbit. Being higher or lower than geostationary orbit (while maintaining geostationary rotation velocity) would only result in tension (higher orbit) or compression (lower orbit) forces experienced as hoop stress in the ring which it would need to be able to withstand (without breaking itself apart). – J... Sep 11 '19 at 17:14
• @J..., I understand what you are saying and I was a bit off on that last bit but constant thrust over a wide area could reduce that stress. The constant outward (or inward) thrust on all the parts would do it but one wrong move would make for a very bad day. – ShadoCat Sep 11 '19 at 17:26
• You would also need an impractical amount of fuel. And the station would be instantly destroyed when you ran out. – J... Sep 11 '19 at 17:42
• @J..., yep. That's why Geostationary is the best. though you might be able to throw some wires out and steal enough energy from the Earth's magnetic field to keep it going (so long as nothing went wrong). It's not like we're using that magnetic field for anything... ...(joking). – ShadoCat Sep 11 '19 at 17:50
• Solar panels would work much better. You'd still need reaction mass, in any case. – J... Sep 11 '19 at 17:52