# How tall do atmosphere retaining walls on rotating space habitats need to be?

A common, matter-efficient science-fiction habitat is a hollow cylinder or ring in space that is spun to simulate the pull of gravity on its interior surface. These habitats have been imagined as small as a spaceship, mere meters in radius, up to a ringworld, 1 AU in radius.

As these habitats increase in size there is theorized to be a point at which fully enclosing the ring is no longer necessary. Just as the Earth does not require a roof to retain its atmosphere a sufficiently large rotating habitat would not require a roof either. Tall retaining walls would suffice to hold the atmosphere indefinitely. At least, that’s the theory. In practice, can a rotating habitat retain an atmosphere without a roof? And if so how can we compute how tall a rotating habitat’s walls will need to be to prevent substantial atmosphere loss over time?

A good answer should provide the formula needed to calculate the height of the walls using the following variables:

$R$ the radius of the habitat.

$G$ the centripetal acceleration (“gravity”) felt at “sea level” on the inner surface of the ring.

$P$ the pressure at “sea level” of the atmosphere.

Additionally providing a calculation of radius vs. wall height where $G$ is equivalent to 1 gravity and $P$ is equivalent to 1 atmosphere may be generally useful.

We will assume that the atmosphere is composed of the same oxygen and nitrogen mixture as Earth’s and that the temperature of the surface of the ring is within normal Earth ranges at 25 degrees Celsius.

• As a guide: Earth's atmosphere is 300 miles (480Km) thick, and we call that at 1G human standard. As I recall, Larry Niven made his ringworld walls 500 miles high. Jul 30 '18 at 21:27
• @JBH Thank you for the reference. I anticipate the case of a ringworld will be significantly different from a planet for 2 reasons. First, the rate at which the "gravity" decreases as you increase in height will be significantly more dramatic in a reasonably (thousands of km) sized ring compared to a planet. Second, air on a planet constantly feels a gravitational pull, but on a ring, the simulated gravity is dependent on the motion of the air. If a molecule ever stops rotating there is no force holding it to the ring anymore and it will simply float away. Jul 30 '18 at 21:34
• Like a dyson sphere, a ringworld has little to no gravity of itself and should be spun to create sufficient gravity. Not only does this have the effect of forcing the air closer to the ground than a planet would require, but it creates wind where there would be none without the spin. Air motion itself has nothing to do with keeping an atmosphere in place. Jul 30 '18 at 21:59
• Also note that ringworlds are intrnsically unstable. You need circumfrential attitude jets and a good navigation system to keep the ring in place. Jul 30 '18 at 22:05
• A simple formula for what you ask is not possible. Density of atmosphere at any altitude depends upon pressure, temperature, and atmospheric composition. To model these accurately for Earth's atmosphere, the atmosphere is modeled in multiple layers, corresponding to average properties in different layers - and this all occurs under conditions where gravity is often considered constant. You should expect to model layers according and use integration for the combined result (most likely numeric integration). Atmospheric Models would be a good start. Jul 30 '18 at 23:52

This is a surprisingly complicated question. For planets there is basically only two ways for atmosphere to escape, but for a ringworld, there are also two, but they're different. And it turns out that for many configurations, the one that is unique to a rongworld dominates.

On Earth, atmosphere leaks off to space when an atom high enough up that its mean free path will take it into interplanetary space also gains enough energy to hit escape velocity. (And for any height where there's significant atmosphere, escape velocity is basically the same as on the surface -- it goes down as you rise, but not by enough to matter.)

So on Earth the fast atoms escape, which means that the light atoms escape preferentially, since for a given temperature, the lighter atoms and molecules move faster. It's the far tail of the Boltzmann distribution that may make it out and away, and it's a fairly slow process.

This process is common to both Earth and a ringworld. (Actually, the escape from the ringworld by this mechanism is somewhat faster on a ringworld because there are trajectories slower than escape velocity which cause the atom to go shooting off to the side.0. My back-of-the-envelope estimate is that this is always insignificant.)

The second mechanism is erosion by the solar wind. Basically, the high speed solar wind sometimes is blowing hard enough to push the Earth's magnetic field out of the way and it then entrains atoms in the upper atmosphere and carries them away. When this happens this is more significant than Boltzmann loss. Earth's magnetic field makes this a relatively minor effect, but a planet without a magnetic field will lose atmosphere quickly. This should be insignificant on a ringworld, because such a small fraction of the atmosphere is in a position to be eroded.

The biggie for a ringworld is over-the-wall leakage. When you get to the top of the wall on a ringworld, whatever atmosphere is up there is totally unconfined to the side. Gravity confines it from above, so that only the hot atoms escape, but to the side even 1 meter/sec will take an atom or molecule over the wall and away.

This flow will travel at a big fraction of the speed of sound at that altitude -- call it half. (A factor of two doesn't much matter, anyway!) So above the ring wall, there is a steady very fast wind at whatever atmospheric density you have at that altitude. (This effect is entirely missing on a spherical planet. Discworld may have a problem.)

You can estimate how long the atmosphere will last fairly easily. First of all, you can neglect the ring entirely and just look at a cross section of the ring. You have gas confined by the ring wall on two sides and by gravity on the top. It's 1G (and the temperature's the same, also) so the ring world's atmospheric decreases in density with altitude the same rate as Earth's does. (See barometric formula.) The density drops by a factor of two roughly every 6 km. (We're ignoring temperature effects here, which complicate it a lot but don't make a big change in the results.)

I don't know how to do pretty equations, so I'll try to be very clear.

Assume that the density is e-z/a where z is the height above the ground and a is the 1/e height. (The height by which the density drops by a factor of e ~2.718.) The amount of atmosphere contained by the wall is the integral of the density from ground to height h, the height of the wall times the width of the ringworld. Call this A0 = 1-e-h/a which is pretty close to 1 for any reasonable height h. The amount of atmosphere above the wall and thus able to escape to the side is the same integral evaluated from height h to infinity. Call this dA = 1/e-h/a.

The time T it takes for about half the atmosphere to leak away is T=w*A0/dA*v, where w is the width of the ringworld, and v is the speed of sound. Since A0 is roughly 1, this reduces to T = w/dA*v.

For w = 1 million km, and v=1000kph, we get T=1000/dA hours. There are about 9000 hours in a year, so to get the time to 100 years we need 1/dA to be 900. since 1/dA is e-h/a, it's easy to solve for this and see that we want h/a to be about 7., So if the walls are about 7 times the 1/e height, the leakage will be about half the atmosphere every century. Since the 1/e height is about 10 km, this suggests that for a million-km wide ringworld, we need 70 km walls for a century's stability.

To get the stability to a million years it needs to be about 16 1/e heights high or about 160 km.

Given all the approximations I made, I think I can say that 1000 km walls will be entirely adequate for long-term atmospheric stability.

Just a reminder that the radius of the ring world does not figure in because air pours over every bit of wall at the same rate. I assume that it was Earth-normal temperature, gravity and pressure at the surface, but you really don't have to. The impact of temperature is on the velocity of the molecules at wall height which has little to do with surface temperature. The surface pressure doesn't figure in because if you double the pressure you also double the loss rate and it cancels out. Surface gravity matters: lower gravity means a proportionally larger 1/e height and thus requires proportionally higher walls.

The easy answer is that you can choose your wall height however you like. The diameter of the ring and the angular velocity of the ring will determine your gravity at the edge of the ring. The density of the air at the edge of your ring will depend on your centrifugal force at that point, and the weight of the air above it. I would say that the size of your wall depends on what kinds of clouds you want to have visible from the surface of your ring, how tall your nimbus clouds are allowed to get, and how high up you want them to be from the surface of the ring. You can adjust the speed and diameter of the ring until it's just right for your world.

Here's a calculator you can use to get angular velocity (rotations per minute) of your ring for a given acceleration (gravity) and radius. https://www.artificial-gravity.com/sw/SpinCalc/

Remember that earth's gravity (that's your Centripetal Acceleration) is 9.8 meters/second^2. So if you adjust your angular velocity and radius until you get something similar to earth, it will give you a good starting point.

The rate of change of gravity at earth is calculated in an easy way here: https://www.mansfieldct.org/Schools/MMS/staff/hand/lawsgravaltitude.htm

The rate of change in gravity as you leave a planet is exponential, but from what I can tell, the rate of change of gravity as you approach the center of the ring is linear. If I'm not mistaken, this means that you will need to have a significantly thinner atmosphere to get the same air density at the surface.

Here's another calculator: http://www.calctool.org/CALC/phys/newtonian/centrifugal

So you can use SpinCalc to get your angular and tangential velocity at the edge of the ring, and then use the centrifugal force calculator with those values and different radii to find out what the gravity will feel like as you approach the center of the ring.

Earth has pretty evenly spaced cloud layers. https://en.wikipedia.org/wiki/Cloud#/media/File:Cloud_types.jpg

Your ring's upper cloud layers will be much wider relative to the lower layers, because its gravity is tapering much more slowly as you get farther from the surface. I'm making an assumption that the clouds will follow a similar pattern at all, but since I've never seen a ringworld in person, and I'm not a meteorologist, this is my best guess.

The upper end of earth's atmosphere (about 300 miles) has a gravity of about 8.9m/s2, so for example, if you've decided that the gravity you want at the edge of the ring is 9.8m/s2 (earth), you can play with the diameter on calctool to figure out which diameter has 8.9m/s2 and then take the difference to get the size of your wall (ignoring the weight of air in the upper atmosphere, which isn't that much anyway.)

Sorry I didn't include a formula. Centrifugal force is not a simple thing; you're gonna end up doing integrals if you try to do it on paper, I'm pretty sure.

On Earth, the boundary between the atmosphere as space is the Kármán line, at 100 km altitude.

https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line

The Kármán line, or Karman line, lies at an altitude of 100 km (62 mi; 330,000 ft) above Earth's sea level and commonly represents the boundary between Earth's atmosphere and outer space.[2] This definition is accepted by the Fédération aéronautique internationale (FAI), which is an international standard-setting and record-keeping body for aeronautics and astronautics.

Thus, 100 km (62 mile) high walls should be adequate.

• It would be interesting to reconcile my link with your link from the perspective of atmospheric loss in relation to a ringworld, which must be spun to create sufficient gravity. This is important since the OP asked for hard-science. Jul 30 '18 at 22:02
• @jbh 300 miles is well above the ISS and most other LEO satellites. IOW, while yes there's some atmosphere at 300 miles, it's so thin that spaceships are up there. Jul 30 '18 at 22:34

Suppose that you want to use a space habitat to duplicate a natural environment for Earth animals and plants. In a society which is building thousands of space habitats for humans to live in - partially to have a larger human population than can live on earth, and partially to have survivors in case a disaster wipes out life on Earth - having copies in outer space of all Earth's ecosystems to preserve members of every species so they can be reseeded on Earth if need be would be a good idea. And if the future humans building space habitats have any use for domestic plants and animals, they would want to have a source for breeding replacement domestic plants and animals if something wipes out the domestic ones on Earth and in other space habitats.

And if a cylindrical space habitat is, for example, 100 kilometers long and 10 kilometers in diameter, it will have a surface area of 3,141.59 square kilometers, between Mansel Island in Canada and Yamdena Island in Indonesia in area - https://en.wikipedia.org/wiki/List_of_islands_by_area - and also between Lake Tana in Ethiopia and Lake Amadjuak in Canada in area - https://en.wikipedia.org/wiki/List_of_lakes_by_area1.

If the dimensions are increased by ten times to 1,000 kilometers by 100 kilometers, the surface area will increase by one hundred times to 314,159 square kilometers, between Sumatra in Indonesia and Honshu in Japan in area - https://en.wikipedia.org/wiki/List_of_islands_by_area - and also between the Caspian Sea and Lake Superior in the USA in area - https://en.wikipedia.org/wiki/List_of_lakes_by_area1

So what area would be necessary for a more or less self sustaining ecosystem that requires relatively little human intervention to maintain? I'm not sure how certain the answer is, but I suspect that the answer varies greatly between different types of Earthly ecosystems.

And another question is what is the required volume for a more or less self sustaining ecosystem that requires relatively little human intervention to maintain? I'm also not sure how certain the answer is, but I also suspect that the answer varies greatly between different types of Earthly ecosystems.

Suppose that an African savannah ecosystem required a height of 10 meters above the ground to function well. In that case a roof could be build over the ecosystem with a height of 10 meters, supported by many evenly spaced pillars that woulds seem more or less like trees to the animals. And more ground could be laid down over the roof, converting it into another ground level and doubling the total area of the savannah in the space habitat. And another roof could be built over the upper level at a height of thirty meters to make a third ground level, and so on and so on, level after level.

Of course most multi-celled Earth plants and animals would need gravity like that of Earth to flourish, and every level closer to the center of the rotating space habitat will decrease the gravity slightly. So eventually a limit would be reached.

If one arbitrarily assumes that a ten percent increase or decrease from Earth's gravity would be acceptable, a rotating cylindrical space habitat with a radius of 5 kilometers should have a zone of acceptable gravity for levels between about 4,500 and 5,500 meters from the central axis. A rotating cylindrical space habitat with a radius of 50 kilometers should have a zone of acceptable gravity for levels between about 45 and 55 kilometers from the central axis.

If heights between 10 meters and 1,000 meters are necessary for various Earth ecosystems, a 10 kilometer diameter and 100 kilometer long space habitat could have between 1 and 100 levels, depending on the ecosystem chosen, and thus between 3,141.59 and 314,159 square kilometers of surface area. A 100 kilometer by 1,000 kilometer space habitat could have between 10 and 1,000 levels and thus between 3,141,590 and 3,141,590,000 square kilometers of surface area.

Note that the total surface area of Earth, land and sea, with all its different ecosystems, is about 510,000,000 square kilometers, meaning that a space habitat with 3,141,590,000 square kilometers of surface area would have more than six times the surface area of Earth.

If between 10 and 1,000 meters of air above the surface are hypothetically needed for various Earth ecosystems, why should the topmost level of the Earth ecosystem have an open sky extending 5 kilometers or 50 kilometers to the center axis of the space habitat, depending on side walls to keep the air in. Why not build another roof on top of the top level of the Earth ecosystem? and have above it an open area of vacuum extending to the center axis of the space habitat, with maybe a pressurized zero g central cylinder?

And of course such a design for space habitats to duplicate Earth ecosystems for the sake of Earth planets and animals can also be used for human inhabited space habitats, since humans are much more comfortable living in artificial indoors environments than most animals are.

Since it is probably impossible to make materials strong enough to build cylindrical space habitats with diameters so great that walls hundreds or thousands of kilometers high to hold in atmosphere would not reach all the way to the central axis, I see no point in discussing how high such walls would have to be. And I have an intense dislike for wasting air made with material shipped in from comets and icy bodies in the outer solar system by letting some of it escape very slowly over gigantic atmospheric retaining walls, to say nothing of the wasted material to construct those walls.