Most of what follows is scientifically dubious. I'm looking to clarify one aspect and one aspect alone.

A small planet is being terraformed by an alien race who have been doing this for millennia. They proceed by wrapping the planetoid at ground level in non-permeable, initially flexible, transparent, indestructible unobtainium.

Then they inflate the balloon using gases they have brought with them in liquid form. The resulting balloon has a diameter approximately a mile more than the diameter of the planetoid. This is their standard method. They always choose small planetoids and make the balloon the same height above the ground. The balloon is pressurised and the density of the unobtainium is more than that of the atmosphere it surrounds. The pressure is such that the surface of the balloon becomes very taut like that of a rubber balloon about to burst.

Their purpose is to farm the entire surface of the planetoid. EDIT - To make farming easier they flatten any significant mountains then grind the resultant debris into topsoil and fertilise it with chemicals they bring along.

The planetoid does not have any substantial satellites - only small alien-made ones. It is circling around a Sol-like star at a Mars-like distance.


Given the conditions above, will the balloon remain concentric with the planet or will it tend to brush the ground at times?

My belief is that it will self-stabilise because atmospheric gradient* will ensure greater pressure at lower altitudes and so push the balloon away if it drifts too close to the ground. I also believe that it will end up spinning in the same direction as the planet and thus perhaps be flattened somewhat at the poles.

Am I right that it will self-stabilise or do the aliens have to have some extra mechanism to keep it from touching the surface? Might the solar wind be a destabilising factor? Would increasing the height above the ground provide a more significant pressure gradient?

If it isn't self-stabilising then suggestions for a minimal change to the system to make it stable would be appreciated.

*I've edited edited 'pressure' to 'gradient'. Apologies, it was a slip of the fingers. I did mention gradient later.


The balloon is made from non-permeable, initially flexible, transparent, indestructible unobtainium.

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    $\begingroup$ For a small (rocky?) planet, pressure gradient over 0.5 miles height will be very, very small, not sufficient to keep the balloon away from the surface. $\endgroup$
    – Alexander
    Commented Nov 28, 2018 at 1:20
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    $\begingroup$ Does it rain on your planet? Bc at least some rain comes from more than a mile high, and it will weigh the balloon down. Making the balloon several miles away from surface will trap most of water vapor inside of it. It will also enclose tall mountains. And will generate more pressure differential to keep the balloon of the ground $\endgroup$
    – Bald Bear
    Commented Nov 28, 2018 at 1:23
  • $\begingroup$ @ Bald Bear - Good point. I have to go now so I'll ponder it overnight. Thanks for pointing that out. $\endgroup$ Commented Nov 28, 2018 at 1:28
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    $\begingroup$ Potential answers may need to factor in the unequal heating of the day/night cycle, and friction from the atmosphere–balloon interface. $\endgroup$
    – rek
    Commented Nov 28, 2018 at 2:49
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    $\begingroup$ @rek: Exactly, only you would use flexible cables instead. Just take an air-supported building en.wikipedia.org/wiki/Air-supported_structure and extend it in all directions. $\endgroup$
    – jamesqf
    Commented Nov 28, 2018 at 6:30

6 Answers 6


The force that the planetoid and its atmosphere exerts on the balloon is a combination of air pressure and gravity. Assuming that the atmospheric pressure is comparable to Earth's, the pressure on the balloon is 14 lbs/square inch outwards. I have no idea what the weight of a square inch of unobtanium balloon materials is, but since the question has the balloon stay up, we can assume it's less than that, and the balloon can be kept aloft by atmospheric pressure.

Now, ignoring gravity for the moment, consider the tightly inflated balloon. The outward pressure on it is due entirely to the atmospheric pressure at the interior surface of the balloon. And that atmospheric pressure decreases with altitude. So the higher up a piece of the balloon's surface is the lower the outward force from air pressure and the lower the balloon's surface is, the higher the outwards force. This is exactly what's needed to center the balloon.

But how big is the force? That depends on the lapse rate, the rate at which atmospheric pressure declines with altitude. Here the fact that it's a planetoid works against you.

The equation for atmospheric density (or pressure) as a function of height is d = e -kgh where k is a constant, g is the surface gravity, and h is the height. A lower gravity means that density (and hence pressure) declines less rapidly with height. The height at which atmospheric pressure drops by half is inversely proportional to the gravitational binding and thus to the mass. The half-height for Earth is about 3.5 miles. Ceres (the largest asteroid in the Solar System) has a mass of 0.0002 Earth, and consequently has a half-height of 17,000 miles. Smaller planetoids would have correspondingly larger half-heights.

So we have a planetoid inside a pressurized balloon of air. The lapse rate of the planetoid is so small that it would make an infinitesimal difference in the atmospheric pressure over the mile from planetoid surface to balloon, and consequently, there would be only a small restoring force tending to keep the balloon centered on the planetoid.

Now bring gravity back into the picture: Since we have assume that the air pressure exerts a greater outward force on the balloon's surface than the inward force due to gravity, we can call the balloon a sphere. Newton himself proved that a uniform spherical shell of matter exerts no net force on an object inside it, and the object inside it naturally exerts not net force on the shell. So as long as the shell is inflated by air pressure, the gravity of the planetoid has no effect, and exerts no restoring force.

Bottom line: You'd have a spherical balloon of air with a planetoid drifting around inside with no particular tendency to stay in the center.


Solar wind will blow the 'wrapper' into the planet

Solar wind is a stream of charged particles blowing from the sun out into space. At a distance of 1 AU, pressure from the charged particles is in the range of 1-6 nPa. This isn't much, but integrated over the surface of a planet this will cause significant deformation of your shell. It will not remain perfectly concentric.

... also, about that solar wind ...

Solar wind is mostly composed of protons. Therefore it is positively charged. If the solar wind impacts on the shell, the shell will eventually be positively charged, since protons will strip off electrons. This will turn your entire system into a giant capacitor. Eventually, the voltage across the capacitor will be high enough and then ...

enter image description here

Does your planet have a magnetic field?

Back on topic, an electrostatically charged shell around your planet is bad for other reasons. Does your planet have a magnetic field like Earth does? Well, then you will get a Lorentz force vector based on the motion of a charged object in a magnetic field. Even if your planet's magnetic field is perfectly symmetrical, the shell won't be, thanks to solar wind, so there will be some asymmetric magnetic forces pulling on the shell.

The direction of the force will vary depending on the magnitude of the magnetic field, the distance the magnetic field extends from the planet, the distance of the shell from the planet, etc. There is also the possibility of relative motion (even if it is minor) between the shell and the planet. All of these factors will be significant, so I'm not prepared to estimate the magnitude or direction of the magnetic force on the solar-wind-charged-shell. But it won't be zero, so your shell is going to move.


The shell is going to hit the planet, one way or another.

Better idea though, what is wrong with using gravity to capture things on the planet's surface? Gravity is, after all, what kept our atmosphere attached to us for the last few billion years. A 'gravity well generator' that you put your planet into to prevent things from leaving the surface isn't any less realistic than magical handwavium shrink wrap on a planet.

  • $\begingroup$ Several comments! First, the point about solar wind moving pushing the balloon off-center is sound, given that there is no restoring force. (I'd note that you don't actually address that which is key.) Secondly, the solar wind is electrically neutral -- if it had an excess of protons, the Sun would build up an opposite charge which would retard proton emission and push electron emission making the wind neutral. What's true is that it's the protons which produce most of the pressure. But if they stick preferentially, then the balloon will develop of positive charge which will (continued) $\endgroup$
    – Mark Olson
    Commented Nov 28, 2018 at 15:12
  • $\begingroup$ (continued) repel protons and attract electrons and the charging will slow and the stop. Thirdly, the Lorenz forces felt by a stationary charged body in a magnetic field are exactly zero. Charges only 'feel' a magnetic field when they are moving relative to it, and the balloon is essentially stationary. Finally, as I noted in my answer, gravity will do nothing to bind a spherical shell (which the OP specified) to a real planetoid. (If the gravity was high enough to produce a significant change in atmospheric density in a mile of height, that effect would work. More unobtanium needed!) $\endgroup$
    – Mark Olson
    Commented Nov 28, 2018 at 15:19
  • $\begingroup$ @MarkOlson So the second point is pretty good. As to the third, if the shell builds up a charge, then it will absolutely be moving relative to magnetic field, even if it is not moving relative to the planet, since the magnetic field is itself moving. The poles shift slowly over time. Once you start getting force, then the shell will then move relative to the planet, and there is no telling what could happen then with the induced forces. My gravity suggestion is to replace the shell with 'extra gravity.' Anything that a shell keeps trapped in, an event horizon will also keep trapped in... $\endgroup$
    – kingledion
    Commented Nov 28, 2018 at 16:28
  • $\begingroup$ The Lorenz forces due to polar shift is to unutterably tiny that it's dwarfed by just about anything else you could think of. The idea that a movement once started causes more charge movement which causes more shell movement is incorrect. If you do the math, the force induced by movement is at right angles to the original movement and smaller -- there is no accelerating movement. Also, if the shell is conductive, the movement will induce eddy currents which will dissipate energy. (I fear I don't understand what you're getting at with regard to gravity.) $\endgroup$
    – Mark Olson
    Commented Nov 28, 2018 at 16:35
  • $\begingroup$ @MarkOlson wrote "... what you're getting at with regard to gravity." The remark by kingledion about gravity is not a direct answer to the question at hand. Rather, kingledion's gravity remark is to suggest to OP that the balloon not be used and that something else be used instead to keep the atmosphere on the planet. That "something else" suggested by kingledion is artificial gravity. Use artificial gravity to hold your atmosphere on the planet instead of using a huge balloon. The point was that an artificial gravity generator is just as realistic as an unobtanium balloon. $\endgroup$
    – Loduwijk
    Commented Nov 28, 2018 at 22:18

Action and reaction

You omit to give a mechanism for construction other than a semi plausible "inflation". As young engineers we joked about "sky hooks" but as with all things formerly implausible, such as space elevators, these are now a reality see wiki. So tethers during construction would aid stability on completion and provide a temporary framework for supporting the outer shell. If the planetoid is mined for obtainium to use for the tethers and shell then the orbit will not change as the total mass is unchanged.

How many tethers are required? Mathematically four (tetrahedral) would be the absolute minimum, diametrically that becomes eight, say a safety factor of three in pairs for redundancy /maintenance then 12 pairs of obtainium to the core (similar to those shown in photo) and 30 obtainium (green) for the lattice should be enough for a blue inflated translucent unobtanium Icosasphere. Which tethered together rotate together Click to see animation by Illustr CC BY-SA 4.0 Note I have truncated top and bottom as more economic and less problematic to produce 2 level platforms

enter image description here enter image description here

Concept of spherical objects with one suspended inside another**

If inflation during construction is uniform on opposite sides the system will stay self centric.

kingledion raised the issue of solar winds but earth is not pushed into outer space, any additional mass will pull the planetoid inwards etc. Equally if containing conductive strands the lattice frame acts more like a faraday cage (with it own electromagnetic issues, but a wide open lattice will not cause hardship).

kingledion, Mark Olson et al. Have in general pointed out that there are similarities to Newtons shell theory and there would be nothing to stop the planet and shell moving at different rates thus on completion over time the core and case could drift into one another. Using diametrically opposite tethers will stop any such drift.

The "space anchors" would provide an interface for space elevator occupants to transfer to the 12 orbital spacepads, and the lattice could be used for monorail links between them. Given the layout any failure/maintenance would still provide a choice of back-up routes. However due to the spin the shell shape is likely to be squished top and bottom, so 2 polar spaceports would make more sense. To counteract the inward pull at the poles a pillar or stanchion form support is needed as proposed by rek ? these could be triangular in nature to align with the truncated oblate (flattened) spheroid.

What could create a calamitous wobble with a catastrophic collision is significantly uneven internal weather, or a drastic external collision event. Both of these would be mitigated by use of tethers.

In either case with enough force the "skin" could rupture or exceed the tethers strength and the whole system instantly collapse. But then unobtainium is indestructible, per your specification !

  • $\begingroup$ I like the engineering solution. You rightly point out that some way of getting to the outside is necessary for the export of produce from the farm. If I understand you, the tethers are under tension. I suppose there would have to be some damping to prevent oscillation. External collision is just a risk they take. They have many of these farms which are tended by robots. Ships drop by now and again to collect the harvest which could be lifted via the tethers. $\endgroup$ Commented Nov 28, 2018 at 22:41
  • $\begingroup$ KJO - Ah helium - now you're talking. Maybe introducing helium will help things along. The helium will float towards the 'roof' and make the pressure gradient steeper with denser gases staying near the ground. (Or will it?) I may make this a separate question. $\endgroup$ Commented Nov 29, 2018 at 17:20

I think you will need some kind of alien stabilization technology. Imagine a tennis ball inside a large inflated balloon. The pressure of the air inside will not keep the tennis ball from bouncing around, even in zero gravity. On a planetary scale the rocky planet still needs to spin and orbit its star, so the planet-sized balloon would have to keep up with the orbiting planet, using some kind of propulsion system.

I think a better technique for terraforming is to understand how Earth keeps its own warm atmosphere from leaking out into space. It's really quite simple. It's just gravity. Gravity keeps the air molecules on our planet. Our planet also has a spinning core of molten iron which creates an electromagnetic field around our planet, protecting it from solar winds which might strip away our ozone layer and atmosphere. The aliens would only have to pick a sufficiently massive planet, fill it with their preferred atmosphere, and generate a protective electromagnetic shield. No balloons needed.

  • $\begingroup$ Thanks for your reply - However I'm assuming that the 'tennis ball' has an appreciable gravitational field and this will create a pressure gradient within the balloon. I hypothesise that it is this pressure gradient that centralises the balloon. I don't understand why propulsion is needed, the balloon is also in orbit around its star. We don't need propulsion to make our atmosphere keep up with us. Note: They pick planetoids for story reasons that I won't go into. $\endgroup$ Commented Nov 28, 2018 at 9:05
  • $\begingroup$ A planet size balloon would not need any propulsion to keep up with the planet, as the balloon would be orbiting with the planet. The balloon is also orbiting around the star as well. Otherwise, the international space station would need constant acceleration to "keep up" with the astronauts inside of it. But it doesn't: the ISS and the people inside it are somewhat independently orbiting Earth. $\endgroup$
    – Loduwijk
    Commented Nov 28, 2018 at 22:25

Have you ever wrapped a gift for your girlfriend/boyfriend in a balloon? If you have done it, or have at least seen the thing, you should have noticed that the pressure in the balloon doesn't keep the gift in the middle of the balloon.

You simply have a body immersed in a fluid, which will then experience the usual buoyancy force. Since the planetoid will be subject to the gravitational attraction of the main body (I assume it is the central star), that won't displace the planetoid with respect to the wrapping.

However, your aliens, in order to wrap the planetoid, must have matched its orbital velocity with the velocity of the wrapping. So it is safe to assume that the center of mass of the planetoid and the center of mass of the wrapping are moving in the same way. If they didn't do it, the planetoid will simply burst through the wrap at some km/s. So my suggestion to the alien is: match the orbital velocity of the planetoid! Then orbital mechanics will take care of the rest.

If they didn't bother in setting up a rotational regime coherent with that of the planetoid, what will happen? The atmosphere in the wrap will be initially at rest with respect to the wrapping, except for the layer in contact with the planetoid surface, which will be drag around. This drag will then transfer to the outer layers, until the entire wrap and the enclosed atmosphere spin coherently with the planetoid.

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    $\begingroup$ I think it's implied the introduced liquid evapourates to produce the atmosphere. $\endgroup$
    – rek
    Commented Nov 28, 2018 at 2:46
  • $\begingroup$ @rek, I miss how your comment relates to my answer $\endgroup$
    – L.Dutch
    Commented Nov 28, 2018 at 2:48
  • $\begingroup$ "You simply have a body immersed in a fluid...." $\endgroup$
    – rek
    Commented Nov 28, 2018 at 3:20
  • $\begingroup$ @rek fluid is either a liquid or a gas. $\endgroup$
    – L.Dutch
    Commented Nov 28, 2018 at 3:22
  • $\begingroup$ L.Dutch - (1) The present isn't likely to have appreciable gravity so the pressure gradient of the 'atmosphere' will be nil. Also the present isn't orbiting the Sun and therefore experiencing microgravity. (2) Correct me if I'm wrong but I think I covered every one of your other points in my original question. $\endgroup$ Commented Nov 28, 2018 at 8:55

Space.com has an article on that subject.

The idea that small worlds may have insufficient gravity to hold-on to an atmosphere and therefore a shell is necessary. The article did not mention, however, stabilizing the shell around the planet. For that, I have done a thought experiment and I hope you find that clear to follow.

The thought experiment starts with taking a steel ingot. Put it on the water, and it will sink. Now, take that ingot and flatten it, then turn it into a concave plate. Put it on the water, and it will float. A ship made of steel will remain afloat as long as it is not filled with water. Pull it up a bit and leave it, it will drop back to its original height. Push it down a bit and leave it, it will pop-up again to its original height. Now, fill it with water and it will sink. This effect is the result of Archimedes law. Going back to the planet, replace water with an atmosphere, and the air above it with a vacuum. The shell is right between the two.

We now cover the atmosphere with a shell. it is atop the atmosphere. The surface atmospheric pressure is the weight of the atmosphere under the planet's gravity, and as you go up the atmosphere, the pressure goes down. The weight of the shell adds-up to that weight. It presses the air underneath as it tends to go down, until it cannot press it any further. At this point, the upwards pressure counteracts the downward weight of the shell and the shell should stay afloat in balance.

Let's say we pushed the shell on one side. It is like we made a dent on the side we pushed (Remember the concave plate?), and a bulge on the opposite side. If the shell was artificially held in this off-center position, the atmosphere will be moved as well. It will shift to a new equilibrium state and the surface pressure will once again become uniform all-over the planet's surface.

The pressure would go down as you go up, so the lowered section (dent) would experience a stronger upward force than the higher section (bulge) on its Antipodes. The shell will move back to its central position. All the "air columns" of the atmosphere behave like a series of communicating vessels. The shorter air columns (under the dent) will push up the shell at a force, greater than that of the taller columns (under the bulge)


Solar wind would exert some pressure and will drift the shell a bit off-center, but not noticeably when viewed from the surface.

large and small asteroids will not have sufficient gravity to stabilize the shell around the center. It is not necessary to install pillars, cables will be just fine, because the bubble cannot collapse without venting the atmosphere.


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