I have a world inside a Dyson ring with a radius of ~1 AU, complete with humans, animals, plants, clouds, and Earth-like atmosphere. The ring is spinning at a particular angular speed, giving the inhabitants on the surface the experience of Earth's gravity at sea level.

How convincing is this pseudo-gravity as a substitute for real gravity for objects not directly touching the surface?

If a bird is flying a hundred meters above the surface of the ring, does it also experience the equivalent of Earth's gravity at a hundred meters above sea level? What about a thousand meters? What about ten thousand?

Is the atmosphere comparable to Earth's at that altitude when inside the ring?


13 Answers 13


The answer is: objects above the surface of the spinning sphere do still feel some apparent equivalent of gravity, although the higher up they go, the wonkier this apparent gravity gets.

Now, there are two ways of analyzing this type of problem: the non-inertial way and the inertial way. In the non-inertial way, we think about the problem from the perspective of a reference frame that's rotating with the shell, while in the inertial way we analyze it from a reference frame that's not accelerating (to be definite, lets say the reference frame of the star). The non-inertial way works by introducing new apparent forces that come about simply by virtue of not being in an inertial frame-- in the special case of a rotating frame, these are known as the centrifugal and Coriolis forces. This way of analyzing the problem is very useful for performing calculations, but doesn't do much in the way of physical intuition, so I'll explain from the inertial frame.

In an inertial frame, everything moves according to newton's laws-- that is, stuff moves in straight lines unless something forces it not to. So, think about the problem of a person standing on the inside of the sphere and jumping. If objects not on the surface did not feel an effective force pulling them to the sphere, we would expect our unfortunate civilian to drift off towards the sun as soon as he left the ground. However, this doesn't happen. To see why, note that our test subject isn't initially sitting still with respect to our inertial frame-- he's moving with the surface of the sphere. Thus, as soon as he stops making contact with the sphere, he'll move in a straight line in our inertial frame. Note, however, that spheres aren't straight. So, he'll collide back with the sphere eventually, a process that from his point of view looks just like an effective gravity pulling him to the sphere. This is the effect of the centrifugal force, had we instead analyzed it from the non-inertial frame.

Now, remember that earlier I said that the apparent gravity gets wonkier as you jump higher? From the inertial perspective, this has to do with the fact that circles look like parabolas if you zoom in close, but not if you zoom out far enough. Meanwhile, from a non-inertial standpoint, it's due to the Coriolis force. However, whatever way you decide to analyze it, the precise answer of how wonky gravity becomes depends on how fast the sphere is spinning, how large it is, and how high up you go. If I have more time later, I'll explain this/analyze your problem in more detail.

Hopefully that helps!

ADDENDUM: Actual Calculations

First, I'd like to clarify that how wonky your pseudo-gravity gets (ie the Coriolis force) actually depends on velocity, not on how high you go. The reason I said it depended on how high you go was that for object on a ballistic trajectory (like a person jumping), travelling higher = more speed and more time for the Coriolis force to act, thus more displacement. Now the centrifugal force does vary based on height, but 1 AU is so ridiculously large that it will be constant for all intents and purposes.

Now, for actual numbers: Based off your specs of $r = 1$ AU, $a_{centrifugal} = 9.8 m/s^2$ and the equation for centrifugal acceleration $a_{cent} = \mathbf{- \omega \times (\omega \times r)}$, we find that the magnitude of the angular acceleration needed is $$ |\omega| = 8.1 * 10^{-6} s^{-1}$$ which corresponds to a rotational period of about 9 days. Luckily, this leads to a tangential velocity of "only" $0.004c$, so we can ignore relativistic effects. Now, the equation for acceleration caused by the Coriolis effect in the non-inertial frame of the ring is $$a_{cor} = -2 \mathbf{\omega \times v}$$ where $\mathbf{v}$ is the velocity of your object as seen in the rotating frame. Note that this means $$|a_{cor}| \leq 2\mathbf{|\omega||v|}$$ Now, I'd guess that countering the Coriolis force for something powered like a bird or superman would be essentially imperceptible if $a_{cor}$ were less than $0.01 m/s^2$. Putting it all together, we find that the Coriolis force doesn't play a big role for things capable of generating their own movement as long as $$|\mathbf{v}| < 616 m/s$$ which is pretty fast. Now, for objects on a ballistic trajectory that can't maneuver to counter the Coriolis force, the effects can add up differently. For a rough idea of how much this will be, note that on earth, $$|\omega_{earth}|=7.3\times10^{-5} s^{-1}$$ So, projectiles will be affected by the coriolis force a bit less (their landing displacement will be smaller by a factor of about $\sqrt{10}$) than they are on Earth.

As a final bit of analysis/buzzkill, I just want to point out that this ring would probably tear itself apart very easily. With the angular speeds and radius you're talking about, and assuming a material tensile strength of $100GPa$, which is about as much as carbon nanotubes, we find that the necessary density of the ring material has to be less than $68 g/m^3$ in order for the ring not to tear itself apart. That's nothing. Hydrogen gas is more dense than that.

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    $\begingroup$ If you jump, according to your explanation, you would not land at the same point you jumped from, but off due to the sphere moving. Am I correct or did I misunderstand you? $\endgroup$
    – user
    Jun 17, 2018 at 18:01
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    $\begingroup$ @user In the limit of small jumps you land where you started cause you're moving at essentially the same speed as the sphere. For larger jumps you actually do land at a different spot because your angular velocity goes up the higher you get (since your tangential velocity is constant). In the reference frame of the sphere this is caused by the Coriolis force. $\endgroup$ Jun 17, 2018 at 19:07
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    $\begingroup$ It might be good to note that if one approaches this Dyson sphere from the outside, the only gravity is due to the actual mass of the sphere. The apparent gravity only shows up if you are already spun up on it. Just like in a spinning space-station, if you fly in to dock, it doesn't matter how close you are to the spinning ring you won't feel any gravity (except the real kind). $\endgroup$
    – kηives
    Jun 18, 2018 at 3:41
  • $\begingroup$ @kηives, this depends on how the masses and speed of rotation are balanced, if it's a low mass sphere with a high speed of rotation, you'll struggle to land as you want to be stationary relative to the ground. $\endgroup$
    – Separatrix
    Jun 18, 2018 at 7:41
  • $\begingroup$ @Separatrix obviously $\endgroup$
    – kηives
    Jun 18, 2018 at 7:42

This is indistinguishable from Earth's gravity

Centrifugal acceleration on an object is $$\mathbf{a} = \left[\frac{d^2\mathbf{r}}{dt^2}\right] + \frac{d\omega}{dt}\times\mathbf{r}+2\omega\times\left[\frac{d\mathbf{r}}{dt}\right]+\omega\times\left(\omega\times\mathbf{r}\right). $$

If we assume that the particles of interest are at a contant distance from the star ($\mathbf{r}$), then the first and third terms are set to zero. If we assume that rotational velocity is contant, then the second term is set to zero. That leaves \begin{equation}\mathbf{a} = \omega\times\left(\omega\times\mathbf{r}\right). \end{equation}

In order to feel a force equivalent to Earth's surface gravity ($g = 9.8$ m/s$^2$), we solve

$$9.81 = \omega\times\left(\omega\times1 \text{ AU}\right).$$

Now, this is highly dependent on the polar angle of the location of interest. The above equation only holds at the equator. At other angles, the centripetal acceleration will be less! Let us assume that you want 1$g$ at the equator, and less towards the poles. Since $1 \text{AU} = 1.496\times10^{11}\text{ m}$, we can solve for $\omega = 8.1\times10^{-6}$ radians/sec to achieve equatorial centripetal force equivalent to Earth's gravity. This means the Dyson sphere has to rotate around the sun every 775,909 seconds; about 9 days. That is pretty fast, but since this is a Dyson sphere, I assume the material structure can handle it.

Now, to answer your specific question, what will the centripetal force be like above the surface. Lets say that we are 10 km above the equator; that is higher than Mount Everest. This affects $\mathbf{r}$; the distance from the sun. The distance from the sun of the surface is $1.496\times10^{11}\text{ m}$. Subtracting 10,000 meters from that we get....$1.496\times10^{11}\text{ m}$. Compared to 1 AU, any altitude we might be used to on Earth is insignificant.


Traveling up into the air above a Dyson sphere designed to simulate gravity with centrifugal force at the equator does not affect gravity appreciably.


The short answer is that birds would feel your pseudo-gravity.

The long answer involves making some sense of the situation with coordinate transforms.

The simplest system to explore is the inertial system, where the coordinate system is fixed. In this coordinate system, the Dyson ring appears to be spinning at a high rate. Objects travel in a straight line unless forces are put on them. One way forces can be put on them is if they are in physical contact with the ring. They can also have forces put on them by the air. But what forces are on the air itself? That question is a little difficult to answer in inertial systems. It can be done, but it'll be easier of we change things up.

Consider a rotating system, which rotates at the same rate as the surface of the Dyson ring. In this coordinate system, the Dyson ring appears to be fixed, in the exact same way the ground beneath our feet appears to be solid and not-moving. Now this is just a coordinate transform. We did no physics in this process. All we did was look at things a different way.

Why look at things this way? Well they make some calculations easier. Consider the environment you envision for your ring. You've got the "ground," with atmosphere on the inside of it. That atmosphere gets pulled along with the ground exactly the way the Earth's atmosphere is drug along with the surface of the planet. This is convenient. You presumably want a "familiar" environment, where the ground behaves like we expect ground to, air behaves like we expect air to, and birds behave like we expect birds to. Thinking in this rotating frame is very natural.

Now there's a cost of viewing the world through a rotating frame of reference. The equations of motion are different. The famous equations of motion Netwon put forth (such as "all objects move in a straight line until acted upon") are only true in inertial frames. If we are in a rotating frame, the math changes. Without getting into ugly math details, the effects that appear are Coriolis and centripetal/centrifugal effects. centrifugal effects are what you are using for your gravitational force. They are called "pseudo-forces" because they aren't actually forces, but they do affect the equations of motion in the same way forces do.

The key takeaway from this is the centrifugal accelerations are always present in a rotating frame. So your birds and such are going to be affected by it, regardless of whether they are touching the surface or not. As long as they are rotating at the same rate as the ring, you'll see these effects.

These centrifugal effects are scaled by the velocity of the object in the stationary frame with respect to the rotation of the frame. In other words, for Earth, East/West velocity leads to centrifugal effects, while North/South velocities do not.

If your bird was not moving along with the ring, the same laws hold, but they get muddier. If your bird was "holding still" in the inertial frame, and watching the ring wizz by it, that bird would not experience any centrifugal forces. It's "East/West" velocity is 0. Your pseudogravity would be ineffective. However, this bird would also be subject to tremendous wind forces as the ring and the air column rotate. As those wind forces pull it "West" in the direction of rotation, its velocity will increase, and it will start to feel pseudogravity.

That same story can be viewed in the inertial frame as well. In that frame, the bird is holding still, and is being pummeled with wind forces that accelerate it.

  • $\begingroup$ I like your answer, but I have a small correction. Centrifugal forces in the rotating frame are dependent only on position, not velocity. It's the Coriolis force that depends on velocity wrt the rotating frame. In the case of a bird not moving with the ring and remaining fixed relative to a star, it still "feels" the centrifugal force. The only difference is that the Coriolis force acts even stronger in the opposite direction and pulls the bird into what looks like a circular trajectory from the rotating frame. $\endgroup$ Jun 18, 2018 at 23:09


Here are my assumptions:

  • As stated in the question, you've got a Dyson ring with a radius of 1 AU, spinning fast enough that anyone standing on the inside surface experiences 1 gee of outward centrifugal acceleration (in any frame of reference spinning with the ring, of course), thus mimicking sea-level Earth gravity.
  • There are inward-pointing walls on the edges of the ring to keep the atmosphere from spilling out, and a functioning ecosystem existing on the inside surface of the ring.

A column of the atmosphere taken from the inner surface toward the Sun would look very much like a column of Earth's atmosphere taken from the surface upward. It would have the same density profile, for sure, and probably similar composition and temperature at each altitude as well. The main difference would be in how the solar wind interacts with the ring's atmosphere and magnetic field- which depends entirely on how the ring's magnetic field is set up. It would need to be installed artificially; Earth's magnetic field comes from magnetohydronamic shenanigans in the molten outer core, while a ringworld has no core at all. So I can't say anything about the ringworld's auroras or ionosphere, except that you can probably configure those to work however you want.

Birds flying at high altitudes will actually experience more gravity than at the same altitudes on Earth- but not by any noticeable amount. Earth's gravity decreases with the square of the difference from the center of the planet; the ringworld's will decrease linearly from the inner surface to the center of the ring. However, these effects are so subtle, no actual living thing will ever be able to detect them without specialized equipment- Earth's gravity is 9.8 m/s^2 at sea level, and about 8.7 m/s^2 at the height of the International Space Station. It doesn't decrease much, and on a ringworld it'll decrease even less.

What will be affected by the subtle differences between Earth's atmosphere and the ringworld's are certain kinds of weather events. Namely, hurricanes. On Earth, hurricanes spin because the Earth spins, causing air nearer the equator to be moving eastward faster than air closer to the poles. When this air moves toward the low-pressure cell at the center of a hurricane-to-be, air coming from closer to the pole appears to be pushed westward because the ground under it is moving eastward more quickly, and air coming in from closer to the Equator appears to be pushed eastward because is has more eastward velocity than the slower-moving ground now under it. This is the Coriolis effect.

Now, because your ringworld is a cylinder, not a sphere, the Coriolis effect will not affect air moving along the inner surface in any way, and so you won't get hurricanes. Smaller-scale spinning weather events caused by wind currents moving in different directions, like tornadoes and dust devils, sure, but no hurricanes.

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    $\begingroup$ It's worth mentioning that the "gravity" on the ringworld it's not actual gravity but a centrifugal force, that will, in fact, vanish completely as soon as you leave the atmosphere $\endgroup$ Jun 18, 2018 at 7:53
  • $\begingroup$ No. It will only vanish if/when you cancel the transverse velocity it inherited as a consequence of being on the ring. You could - if you made a fast enough race car - fall off the ring if you drove it fast enough in a direction counter to the rings rotation. Which would be pretty damn fast if its maintaining 1G at 1AU - a Dyson ring with 1AU radius would complete a rotation ever ~10 days to generate 1G. $\endgroup$ Jun 18, 2018 at 10:45
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    $\begingroup$ @SilverCookies - in addition to the comments of Chris Becke, by Einstein's principle of equivalence, upon which the General Theory of Relativity is based, there is no local difference between gravity and "centrifugal force". The only way to distinguish between them would be to take measurements far enough apart to detect the directional difference of being attracted to a point vs an axis. $\endgroup$ Jun 18, 2018 at 16:25
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    $\begingroup$ @ChrisBecke Firstly I would appreciate if you could use a less condescending tone, since it seems this is a simple misunderstanding. Birds can't leave the atmosphere anyway, a rocket moving vertically that stops its engines after leaving the atmosphere would indeed "fall back" simply because, having a sustained sideway motion it would eventually impact the atmosphere again... $\endgroup$ Jun 20, 2018 at 8:54
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    $\begingroup$ @ChrisBecke ...think instead of a rocket moving vertically and also "northway" (in the direction of the walls) if it stops its engines after passing the walls it would simply procede in a straight line, unaffected by the gravity of the ring since, you know, there is none. I will make a thought experiment to better explain my point... $\endgroup$ Jun 20, 2018 at 8:54

Hold on. Try asking

When Earth spins, does the atmosphere spin with it?

I mean, if the atmosphere did not spin with the earth, then getting to the USA from Europe would be really fast. As you climb, the atmosphere would slow and slow (speed up and up relative to surface) and there'd be nice 800mph winds to ride all the way across the Atlantic. Returning would suck.

Well, that doesn't happen. Why not? The rotation of the Earth's surface drags the atmosphere along with it, due to ground effect. The same thing that affects airplanes that are within a wing's length of the ground, except Earth is 8000mi across.

Earth's air doesn't need ground effect to stay put. Either way; Earth would keep its atmo' due to its real gravity. Your ringworld would not. Centripetal force is all it's got, so if ground effect failed to drag the atmosphere along, the atmosphere would not be acted upon by centripetal force, and would expand infinitely into the vacuum of space. The atmosphere would be lost.

So you can take it as a fair assumption that the air is moving with the ringworld.

Aerodynamic forces are decided by your speed relative to the air. Since the ringworld's air must move with it (or be lost) and that is expected anyway from ground effect... an aircraft's airspeed will be roughly its surface speed across the ringworld (there'll be jet streams and stuff like that, but probably within 150 knots if Earth's experience maps at all.)

Several here say to generate 1.0G of centripetal force, the ringworld would have to rotate with a surface velocity of 1.2E+06 m/sec or 2,330,000 knots. So if you were in the air, you would be swept along with the air by aerodynamic forces. Which means you'd be moving nearly the speed (within hundreds of knots) of the surface.

Now what happens if the bird folds her wings? She will go in a straight line and experience zero gee. Unfortunately, the ringworld is not going in a straight line. The surface is traveling nearly parallel with her, both at 1.2 million meters per second, but slightly curving up toward her, and the paths will intersect with the differential velocity of 100 meters per second or so. Ouch. The bird would be very wise to open its wings and use them to create about 1 G of lift away from the ringworld's surface, which would hold her at altitude. Just like she would on Earth.

Airplanes don't experience gravity. They experience lift, which keeps them from hitting the ground.

The reason the ground is careening toward you is different than on Earth, but the answer is the same: lift. A Tiger Moth, Osprey, or B777-200 would fly about the same as Earth. They could even shoot a cat 3 approach if the proprietor of the ringworld were to fit the equipment.

  • $\begingroup$ @Flater - the ring, to maintain 1G, is completing a rotation every 10 days. A space station at 1AU, would take 1 year to complete a rotation. If the ring broke, then everyone and everything in it is suddenly a. experiencing 0G, b. leaving the solar system very rapidly. $\endgroup$ Jun 18, 2018 at 10:55
  • $\begingroup$ Note that since Earth spins towards the east, a stationary atmosphere would cause a strong wind blowing towards the west. It'd be the Europe-to-America trip that would go easily, and the America-to-Europe trip that would face unbeatable headwinds. $\endgroup$
    – rob
    Jun 18, 2018 at 14:59
  • $\begingroup$ Oh, thanks @Rob. My imaginary-globe method led me wrong. $\endgroup$ Jun 18, 2018 at 15:26
  • $\begingroup$ "if ground effect failed to drag the atmosphere along...." but you might as well say "if the walls were leaky, they would leak" or "if it stopped spinning, it would stop spinning". the air would certainly be spun as well, it would be impossible not to (certainly considering the million-year process of spinning up the ring). $\endgroup$
    – Fattie
    Jun 19, 2018 at 3:06
  • $\begingroup$ (In a "doubly sci-fi" sense, we could I guess imagine whole rings of this nature popping in to existence WITHOUT the air mass being spun (and of course naturally they use some ultra-hyper-lubricant, quark-smooth surface so it never does spin) - but it's kind of a "so-what"? Everyone knows that, obviously, real Dyson rings have the air spinning with them, and it would be all-but impossible to make it not spin. We'd have to unrealistically say the air was not spinning for some special reason! :) ) $\endgroup$
    – Fattie
    Jun 19, 2018 at 3:11

How Gravity Works

The most simple way to define gravity without math is to say that it is the factor of two objects mass and their distance from each other. A ring the size of 1AU is going to have a gravitational effect of it's own anyways. I'm not really sure what the other dimensions of this ring are or what it is primarily made out of, but this is easily an object that has many times more mass than that of our planet. Obviously it's a ring and not a solid sphere so there is not really any gravitational "center" to be accelerated towards like that of a sphere. This makes calculating what exactly this gravitational force due to sheer mass would actually be at any given point within the ring a bit beyond me.

Centrifugally Created Pseudo-Gravity

In pseudo-gravity the acceleration is due to centrifugal forces instead of the warping effects that mass has upon space-time. Lets say there is no atmosphere for a thought experiment, an observer drifting from the center towards the inner surface of the ring would be like standing inside a giant rotating hula hoop. The ring would be spinning by at great speed while you moved in a straight line towards it. Add in atmosphere and things get..... messy. You would be moving forward at a set speed and begin experiencing drag from the atmosphere at a 90 degree tangent to your forward travel that was moving several thousand miles per hour. Its mostly advisable to math the speed and angle of the ring as you land, or better yet, match its speed and rotation from the outside and just not deal with atmospheric reentry. Until the object moving towards the ring from the center connected with the surface and was brought up to the same speed the ring was rotating at it would not experience the same pseudo-gravity everything else is. This effect does not apply to objects already on the ring and rotating with it, and the atmosphere rotating with the ring would play a big role in boosting said object rotational speed as well. Think of a juggler on a train, when he begins juggling his objects they don't suddenly move towards the rear of the train as soon as they leave his hand. This is because they retain the same velocity imparted to them by the train and are moving at the same speed relative to the train. They don't just suddenly stop having velocity because they are not touching the train right? This is why you can toss or drop something on a jet airplane and not have it impact the rear of the cabin at 340 miles per hour.

So a guy jumping up and down, or tossing a ball, or a bird flying would not really notice. "But what if they went in the opposite direction of the spin and lost velocity?" you might ask. Well, since the ring has to be spinning at thousands of kilometers per hour to adequately simulate a -9.8 m/s acceleration uniformly across the entire ring you would have to move in the opposite direction at least that speed to cancel out the centrifugal effects. Jumping upwards or even operating smaller slower aircraft would not cause enough of a differential in this velocity to really matter much beyond having to possibly account for it in extremely fast vehicles.

But what about moving towards the center of the ring? Doesn't a circular objects rotational velocity increase as you move towards the center? Wouldn't this effect the feeling of the pseudo-gravity? Again, since this is such a truly huge object. At 1 AU in radius you would need to move enough towards the center that you leave the atmosphere before you begin to experience any noticeable difference that what is experienced at the surface. Birds, bees, people and baseballs would not be in danger of flying off into space unless you found a really entertaining way to accelerate them to incredible speeds and great distances.


This is a bit tricky, since atmosphere isn't a solid object under acceleration. The easiest way to do it would be to have the entire thing enclosed and not rely simply upon centrifugal forces to hold it in. A pressurized tube like a giant inner tube would be best since then you can guarantee that the atmospheric pressure was more or less uniform throughout. You also need an enclosure to counteract the effects of solar wind. Somehow overcoming the fact that drag effects at the surface and at the top of the atmosphere would not be the same you also have to worry about solar wind stripping your atmosphere away. Frankly, I think a pressurized totally enclosed ring shaped tube would be all around a better idea. Maybe less dramatic than a Halo style open wheel, but Halo is also using made up, misused, or fake words to describe most of what is happening on their ring world anyways.


So, imagine an object which is at an altitude of 100 meters on this world.

The simple confusion here is between

  • (A) objects that "teleport in". in your thought experiment, imagine a rock "just appears" at that point (and let's say for now there is no atmosphere).

  • (B) objects that are simply an ongoing part of the system. So, the bird was born on the ring, learned to fly, and happened to fly 100 meters upwards.

In case "A", as the OP suspects, there is absolutely no "like-gravitation-on-Earth" effect. Nothing, zero.

In case "B", there IS a "like-gravitation-on-Earth" effect. For the bird, it's exactly like being on Earth. No difference whatsoever.

Note: all of what I say has absolutely no relation to an atmosphere. Picture it as a total vacuum for clarity.

Note that a comment under the answer perfectly explains it:

"the deeper point here is that not every person intuitively realizes that it is having been part of the rotating system (as in, interacted with its components) that puts you into a state where things feel like gravity as we normally experience it"

That is the nub of OP's question.

A point of confusion regarding the atmosphere. Note that you cannot teleport in anyway - there's no such thing as magically instantly displacing your momentum.

But let's say on Earth, someone teleported in, to a location 500m above the surface, and the person was moving at a great speed, say 1000 km/h in one direction (North, whatever, doesn't matter).

What would happen? This teleporting person would experience incredible forces from the air and would quickly "slow down" (in fact matching the speed of the surface of the Earth), and then fall to Earth.

Interestingly exactly the same thing happens on the Ring if someone weirdly teleports in with a non-matching speed.

  • $\begingroup$ That poor bird teleporting in, would definitely find himself above Vne and would get his feathers plucked. In fact I'm not entirely sure how anyone would get on or off that ring without getting really, really plucked. $\endgroup$ Jun 19, 2018 at 3:24
  • $\begingroup$ Heh :) (As it says, "picture it as a vacuum for clarity") $\endgroup$
    – Fattie
    Jun 19, 2018 at 12:43

The short answer:

Ironically in a system like this the higher the bird goes the stronger the force will be pushing it outwards.

Why? The Dyson ring is a rigid structure that, at 1au, needs to have a velocity of 1200km/s to achieve 1g. At 1au, the acceleration due to the suns gravity is only 0.006m/s², much MUCH smaller than 1g which is 9.8m/s².

So, the ring is NOT in orbit. If broken into pieces it would fly out and away from the sun - although it would fall back again in an elliptical orbit as the escape velocity from the sun at 1au is about 40000km/s.

So the ring, the air in the ring, the bird and everything else in the ring is travelling at 1200km/s and consequently trying to fly away from the sun, only the rigid structure of the ring prevents that. This is literally the source of the apparent gravity: Everything on the ring should follow a highly elliptical orbit from that point, but the rigid structure keeps forcing it into a circular motion around the sun.

In order to escape this effect, objects would have to decelerate in the opposite direction to the rings motion: If the bird left the atmosphere - and then used rockets to accelerate backwards to 1170km/s (relative to the ring (and air) below) it would, ironically, merely be in a stable orbit around the sun at that radius as the orbital speed of the earth around our sun is 30km/s. Any speed between that and 1200km/s and it would start dropping towards the sun - because relative to the sun those speeds would be slower than 30km/s and hence insufficient to maintain a circular orbit at that distance.


If you stood up on the surface of the ring, you would experience gravity -- or at least it's equivalent in acceleration, which you could not readily distinguish it from.

If you then jumped straight up, you would most likely continue to experience that gravity because (a) since you were originally rotating with the ring, you would simply continue to do so, and (b) the ring makes traction with the atmosphere, causing that and everything in it to rotate also.

The exception is if you run counter to the direction of rotation and then jumped up -- you would negate the direct effects of the ring, but there's still that atmosphere to worry about. (However, if you could run fast enough, you could negate the force of "gravity" without ever leaving the surface.)

To an observer seated in a chair on the ring, if you jumped straight up, you would just come straight down again (from his viewpoint). But if you could jump up after countering the effects of the spin, and somehow get the atmosphere to pass by you without effect, you would just keep going straight up from the viewpoint of a sidereal observer. By sidereal observer, I mean someone in a spaceship who is not moving along with the ring. (You could argue that this is your "real" motion as viewed from the "side" -- "real" + "side" = "sidereal".)

Of course, if you could do this, you'd be headed straight for the sun with no way to turn around. Oops.

  • $\begingroup$ This is incorrect. Air is not involved. If you "jump up" you make a parabola exactly "as if" you were on Earth, $\endgroup$
    – Fattie
    Jun 18, 2018 at 20:19
  • $\begingroup$ @Fattie -- Actually, it is correct. The motion of the air, being in sync with the rotation of the ring, would effect any angular motion you make while jumping up (either with or against the rotation). And a Dyson ring without any atmosphere (or sidewalls or whatever to hold in that atmosphere) would be pretty useless. $\endgroup$
    – Jennifer
    Jun 25, 2018 at 6:38
  • $\begingroup$ Hi @jennifer: I believe there are 2 things you don't understand. Point 1: Say you are on Earth. (Earth has atmosphere.) You jump up and down, say one meter. Now say you are on Earth, but zero atmosphere (simply, you could be in a vacuum room in a research center.) Again, you jump up and down 1 meter. (Note too that the surface of the Earth is, in fact, spinning very rapidly.) In both those cases (atmosphere or vacuum), exactly the same thing happens. You go up and down on the same spot, exactly as per normal experience. The atmosphere is utterly, totally, completely unrelated. $\endgroup$
    – Fattie
    Jun 25, 2018 at 13:27
  • $\begingroup$ Point 2: Now say you're on the Dyson ring. You jump up and down 1 meter. Again, you try it in both the normal atmosphere of the Dyson ring, and you try it in a vacuum. (Simply, you could be in a vacuum room in a research center.) Once again, on the Ring, those two jumps are absolutely identical - absolutely no difference. The atmosphere is utterly, totally, completely unrelated to the "virtual gravity" issue you experience in a Dyson Ring. (And indeed, in Point 1, The atmosphere is utterly, totally, completely unrelated to the issue of the fact that the Earth's surface is spinning.) $\endgroup$
    – Fattie
    Jun 25, 2018 at 13:29
  • $\begingroup$ (Footnote - of course, there are absolutely tiny aerodynamic effects of the air as you jump up and down, ie, you will be slowed down a tiny bit depending on what sort of clothes you are wearing; this is utterly unrelated to the discussion at hand.) $\endgroup$
    – Fattie
    Jun 25, 2018 at 13:31

I see three options:

  • The bird is in a stable orbit with about the same diameter as the ring. The surface would be passing under it quite rapidly. (This won't work if there is an atmosphere "in" the ring, e.g. held be side walls.) It feels weightless.
  • The bird is at the diameter, but not in a stable orbit. It feels weightless.
  • The bird is moving at more than orbital velocity, like the rest of the ring and the atmosphere above it. It does not feel weightless.

When you stand on a rotating surface (green ball in the picture) to simulate gravity, the perceived apparent gravity is due to the centripetal acceleration allowing the circular motion combined with the inertia of your body. circle with green dot just inside circumference and red dot nearer the centre

Now imagine putting an object within the circle, but in contact with the circle walls and at rest with respect to the translational velocity of the ring: will it experience any centripetal force? No, because there is nothing to transmit it. Therefore the object will not perceive any apparent gravity.

By Newton's first law then it follows that the red object will keep its motion until a force acts on it.

  • 8
    $\begingroup$ Unfortunately, this isn't correct. The intuitive reason is that once you leave the surface, you travel in a straight line, so you eventually impact the curved surface of the sphere again. From the perspective of the person jumping, this would seem basically identical to a gravitational force pulling you back to the surface of the sphere. See my answer for more detail. $\endgroup$ Jun 17, 2018 at 16:39
  • 3
    $\begingroup$ @L.Dutch from the inertial frame's perspective, the problem is that when you jump you don't go straight up. You travel at a diagonal since you're already moving sideways due to the rotation of the ring. $\endgroup$
    – Carmeister
    Jun 17, 2018 at 17:27
  • 2
    $\begingroup$ You never "feel" gravity, even on Earth. The only force you feel is the ground pushing up beneath you. The feeling of jumping would be the same on Earth or on the Dyson ring (as long as you didn't jump too high). $\endgroup$
    – Carmeister
    Jun 17, 2018 at 17:39
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    $\begingroup$ This is incorrect, unless, when you jumped, you were moving horizontally with sufficient velocity to cancel the rotation of the ring. Being in physical contact with the ring has nothing to do with it. What matters is whether you are moving along with the ring when you jump. $\endgroup$ Jun 17, 2018 at 19:34
  • 3
    $\begingroup$ @L.Dutch The only way you can feel yourself being pulled to the ground when you jump is by interaction with other stuff (like the air) and by visually looking at your position relative to earth. If you were falling towards Earth in a (small, to ignore tidal effects) opaque box, you would not be able to tell you were falling vs just floating in space. This is the essence of the weak equivalence principle, which helps form the backbone of general relativity. $\endgroup$ Jun 17, 2018 at 20:20

I think there is some missing principles here. When we are stationary on earth, we are in fact not stationary at all.... only in relation to the surface but what is 0 units of velocity on the planet earth is really equal to the rotational velocity of the planet rotation. This is around 460 m/s at the equator. So, if Earth were to stop instantly like in 10-60 seconds or absolute 0 seconds, you would be flung away from the planet at the rotational speed. When the time to stop is so short, you wouldn't be slow down. It is like being on a merry-go-round and it got spun up really really fast, you'll be thrown off. If the planet stopped rotating instantly and gravity stopped instantly like that then in theory you would be flung off. Now in space, there won't be any meaningful friction to slow you down. No air drag of any kind. Of course, Earth's gravity isn't quite like it would be in a rotating ring like artificial gravity by means of centrifugal/centripetal force but by mass but even then if the planet and its core had a sudden seize, it would be ugly because the mere gravity of Earth without rotation would be near zero, anyway... not enough to keep you from be flung off the surface into outer space. Being on the rotating ring would be like the merry-go-round. Not, the mass of the ring would need to be considered.

A rotating Dyson Sphere shell would have potentially massive mass to actually have gravity and if it even rotates, it would potentially have a gravity field. It is unclear if the ring world would have enough mass but in theory, it could if you have the material to make it. Practicality aside, if there was the material collected, refined, etc. for an advanced civilization (and the energy to go along), it would be theoretically possible for a ring world with a mass per 10,000 cubic kilometers being comparable to that of Mars or Earth, it would be theoretical that you'll have a gravitational force but you may need, in this case, the ring to rotate around the orbit of the star in the middle at least somewhere in the realm of the rotational speed of Earth. It might need to be a little faster but can't think of exactly the amount but you're looking at a rotational speed at this scale being earth like. At least, we are talking about a ring that has a 1 AU radius of 2 AU diameter. The problem with the ring or possibly even a basic classic Dyson Sphere would be the outer side may have very little gravity because you begin to be in the situation where you maybe flung off if you let go of the merry go round.

Now, if you tried to fly a shuttle and slowing your shuttle to 0 velocity difference to that of the rotation of the Earth, the problem is, as you are flying towards the planet, you're intended landing spot would have already moved (rotated). Yes, you'll hit the planet or potentially land or crash. The question is where? This might not be the landing strip so you'll then have to kind of fly the shuttle to a suitable landing spot. If your plan was to jump onto a rapidly rotating merry go round at the spot of you originally was aiming for, the good change is you'll miss your mark.

Getting into the doorway / passage way of a rotating Dyson Sphere from outside to the inner space, can be a tricky business. It can be tricky to land at a specific spot on the Dyson ring as well because you'll need to approach at near parallel and gradually converge but you'll need the correct matching velocity which means you'll need to be a little faster and begin gradually slowing down to match or otherwise be able to land without being too much faster or slower. A m/s faster or slower won't necessarily kill you but say... a 100 m/s faster or slower can mean catastrophic crash. 100 m/s is basically 224 mph crash.... either you crash into something in front of you or something behind you crashes into you. A jet can make such a landing at that speed if you are a little faster as long as you have a clear runway to land. However, if there is something stationary in front of you, then it would be like landing a jet on a road that has a dead end and a big structure like a building in your way right at the end of the road. If you don't make the stop in time, you crash.

I am not sure how air will be contained in a rotating ring to form an atmosphere. Not sure it will be quite like on Earth. The means in how Earth retains its atmosphere would need to be replicated in some suitable manner. IIRC: The magnetosphere of Earth is at least partly responsible for how Earth retains its atmosphere. It's a big magnetic field. We would have to produce the equivalent around and along the ring through a series of magnetic field generators but there is no current feasible means in current human technological and industrial capacity to do anything of this sort. This is something a race that would be described to be of a Type II civilization on the Kardashev scale or somewhere along that line to even be able to do such an endeavor.

  • $\begingroup$ This answer is problematic, it's long and doesn't get to the point clearly. It also makes the assumption that the net gravitational vector of a heavy ring whilst on the inside would be towards the adjacent surface - this is false. The sum of all vectors of gravity at any point inside the ring would be the center of mass - ie. the dead center of the ring. Could you edit down to be more succinct and readable, else this answer will likely get deleted as low quality. $\endgroup$ May 18, 2021 at 8:04

This is a good question, that I have no idea how to answer, specifically I have now idea what the answer actually is. I do know that Niven's Ringworld design calls for walls to hold the atmosphere from spilling out into space.

Here's my best guess at an answer:

  • fact, the "gravity" experienced by any object subject to rotation depends on the speed with which it is travelling

  • fact, the lower atmosphere will spin with the ring due to friction with it's surface

  • fact, the energy picked up through friction isn't the going to propagate very far vertically through the air column, on Earth we only see frictional effects in the bottom 300 or so metres of the atmoshpere

Supposition, the lower atmosphere, and everything in it, will experience substantially the same pseudo-gravity as the surface itself but because this gravity is governed by velocity it will dissipate relatively rapidly with altitude. There will also be an atmospheric density below which the gravitational "drop-off" effect accelerates noticeably.

In short something a hundred metres up will probably experience as much if not slightly more gravity than we might expect from a planet but a kilometer or more up it would slow down enough, stellar relative, that it stops experiencing gravitational forces faster than it would on a planet. As such the atmosphere will thin out more rapidly at altitude on a rotating megastructure than on a planet.

I advise anyone looking at building mega-structures to read Larry Niven's essay, Bigger Than Worlds in full as a primer, it has a lot of useful notes.

  • 1
    $\begingroup$ Re :"energy picked up through friction isn't the going to propagate very far vertically through the air column". This is false. $\endgroup$
    – TLW
    Jun 17, 2018 at 18:08
  • $\begingroup$ @TLW Ground friction effects on earth top out at about 300m above mean topography. $\endgroup$
    – Ash
    Jun 17, 2018 at 18:29
  • $\begingroup$ Earth's atmosphere is still spinning at 100km height. See e.g. adsabs.harvard.edu/full/1966SSRv....6..248K. I'm not sure where you got that number. $\endgroup$
    – TLW
    Jun 17, 2018 at 18:30
  • $\begingroup$ @TLW Yes it is but not because of surface friction, as the article you have cited points out we don't actually understand exactly why the upper atmosphere goes as fast as it does, which is considerably faster than it should given the Earth's surface speed and our atmospheric viscosity, although they suggest that in part it is because of gravitational coupling and in part because of differential heating. $\endgroup$
    – Ash
    Jun 17, 2018 at 18:42
  • $\begingroup$ The totally overwhelming point in relation to this actual QA is that, whatever happens physically to our air on Earth, would be exactly the same in the DRing. This has been discussed to death now. $\endgroup$
    – Fattie
    Jun 30, 2018 at 13:12

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