## Wednesday, December 4, 2013

### The Specific Stability Requirement of Shell World

Gravity's shell theorem is extremely neat, but there's some heavy critical thinking that needs to be applied when we start going into the ideas of mega-structures.  The shell theorem states that when something (the object) is placed inside a spherically symmetric distribution of matter (the shell), then neither the object or the shell will experience a net force.  However, "net" is the key.

The shell world is a proposal to surround a small body in our solar system in a shell that covers the entire planet/moon/asteroid and holds in the atmosphere.  In other cases, it may be called worldhouses, worldsheets or anything else reflecting the general concept.  If you look at this recent conception of the idea, you'll notice that the advocates are specifically advocating relying on the weight of the shell as an alternative to a pressure vessel.  That means that, in fact, this is a particular type of gravity balloon.
Mars or perhaps a moon in another solar system could be encased in a shell of dirt, steel and Kevlar fiber.

Key word here is "dirt".  This reflects a desire to "weigh down" the sheet so that it doesn't have to rely on material strength.  This is obviously possible from a physics standpoint.  The structure would be initially balanced.  Obviously, the pressure of the atmosphere pushing up would be equal to the weight of the shell pushing down.

The question of stability is "what happens when the planet moves inside the shell?"  This is actually surprisingly easily answerable, and it's a straightforward stability analysis.  But only for specific assumptions - namely if there is no rigidity.  You have two different cases you can entertain as limit cases:
1. the shell is so rigid it never experiences deformations
2. the shell is completely non-rigid

In the first case, the structure is always stable.  The reason is that the net force between the planet and the shell is always zero.  So even if the planet moves, it will not affect the shell.  Only the atmospheric pressure at different locations will change - after all, the planet still has some gravity.  The atmospheric forces are restorative, so this means the structure is always stable.

The case with no rigidity is harder to appreciate well, but it's somewhat of a fundamental assumption I want to keep in mind when writing about gravity balloons.  You would certainly prefer to place no requirement on the materials.  Then the problem is only one of gathering the materials - not doing anything with them.  So for the shell world we can apply this sort of assumption.  This means that any acceleration on one segment on the shell is not taken to influence other parts of the shell.  You still have some obvious problems.  In order for this to be useful in any way, there would have to be a flexible membrane that maintains a pressure barrier even when the shape is deformed.  And yes, this is precisely what I have in mind for the nature of the shell.

We will also have to neglect the gravitational effects from the shell itself.  This could be significant in some cases, but for something large like Mars, it's irrelevant.  It can be ignored with no major ill-effects.  In fact, if the shell starts to gain significant mass relative to the planet itself, then that starts to look a lot like a gravity balloon.

That's a lot of qualifiers, so let's get into the specifics.  I couldn't find an expression for atmosphere pressure including the full form of Newtonian gravity, so I made one.  With that, we can set the stability requirement.  To do this, we imagine that the sheet lies at some radius from the center of the planet, and at this point, the upward force from the pressure exactly equals the downward force from gravity.  Because of that, we can formulate a simple stability requirement.  To have stability, the fraction change in pressure with a change in altitude needs to be larger than the fraction change in gravity.  Following that logic, and using the form I've developed, you can get a very clear stability requirement.

Calculation of the Stability Requirement
for a Shell World

This stability requirement is formulated in terms of radius.  This is to say that below that radius, a world sheet would be stable, but above it, it would not.  What does it mean to not be stable?  It means that pressure will "outrun" gravity.  If a deformation of the sheet develops, the pressure will cause it to expand until the sheet hits the surface of the planet, or until it breaks.  The system obviously isn't differentiated to begin with, so failure is always a possibility.  Even if it's stable, a leak can destroy it - meaning that it will eventually fall to the ground.  This is to be expected, since we don't normally expect things to float in the air without additional support.  The worldsheet can, but only if it fits this criteria.

Now, the H' value needs a calculation associated with it.  Take that from my Physics Stack Exchange post.  The expression has mass of the planet in it.  This is the dimension over which we want to investigate the system.  Here is the requirement restated:

Stability Requirement in Terms of Planet's Mass
and Other Universal and Gas Parameters

This is a useful form, because it can directly be applied to find out which bodies can have a stable world sheet.  I did that for a few interesting bodies, here is the table:

Table of Viability of Worldsheets for Various
Interesting Bodies in the Solar System
M (kg)R0 (km)Stable r (km)r / R0
Earth5.97E+246,371.0 2,370,438 372.1
Mars6.42E+233,390.0 254,698 75.1
Moon7.35E+221,737.1 29,166 16.8
Ceres9.50E+20476.03770.79
Vesta2.59E+20262.7102.80.39
Sylvia1.48E+19143.05.90.041
Phobos1.07E+1611.30.00420.00038

From this, we conclude that Ceres could not have a world sheet without additional stabilizing features on it.  There is physically no radius at which it could work with air at normal conditions.  Now, there are ways we could relax this.  Factors that affect the air properties will affect the conclusions here.  Refer to the mass implication equation above to see how to apply these.  My conclusions are:
• Lower temperature -> more stable
• Pressure at surface -> no effect on stability
• Higher molecular weight -> more stable

Even considering these, it would be difficult to imagine how a Ceres worldsheet could be made to be stable.  I used the temperature of 293 Kelvin, because I'm assuming that the world sheet habitat would be at room temperature.  That's not very easy to change if you're going to have humans living there.  Moving on, the pressure won't affect stability because the worldsheet's thickness is designed to exactly balance the pressure with gravity.  A higher molecular weight is an interesting proposition, but what would you use?  Nitrogen has a higher molecular weight than Oxygen gas, so maybe with a higher Nitrogen partial pressure a Ceres worldsheet would be more viable.  Otherwise, more Argon could do the trick.  But that will only buy you a factor of 2 or so.  For smaller bodies, it just could not be made stable period.

Given these results, I would say that some of the illustrations of the shell world can't really be practical.  You wouldn't just need the shell to have tensile strength.  That wouldn't be enough.  It would have to be fully rigid, meaning that it tolerates torsional stresses, and that just doesn't sound practical.  In reality, it would make far more sense to have it tethered to the surface of the planet.  Without that detail, it seems very dubious that it could ever work on a Ceres size object.  On Mars it could work without any tethers to the surface.

With some active tensioners in the tethers, it wouldn't be that big of a deal, actually.  Tethers are easy to shorten or loosen.  You could have some active control for the shell's altitude at different points over the planet, as well as ways of sensing how far it is from a balanced point.

This result comes with an interesting implication.  This means that there is a range of masses (roughly spanning from Anahita to Ceres) that are both too large to host a habitable pressure at their center as a gravity balloon, but also too small to host a stable world sheet.  Of course you could just abandon the world sheet and instead use an interconnected network of caves.   That replaces the "tensioners" that I was referring to with more-or-less a big pile of rock.  I suppose I'm still left wondering why anyone would be happy with gravity 2 or 3% of Earth's, which is about where the stability point for world sheets comes up.

1. Your Pressure equation for varying gravity is incorrect. The scale height is when the potential energy of the gas at a given altitude is equal to its thermal energy (kT) - thus one needs the ratio of gravitational potential of the radius variable and the scale height. If we call the surface Ro, the scale height radius is (Ro + H), and the radius variable is R. The ratio of their potentials is thus [(R-Ro)/(H)]*[(Ro+H)/R].

The inequality becomes (I think):

(Ro+H).Ro/(2H) > R

...which implies that so long as the scale height is smaller than the body's radius the stability condition will be met.

2. BTW if you want an online discussion of the barometric equation in a non-homogeneous gravitational field, there's this paper: http://web.ist.utl.pt/~berberan/data/43.pdf

1. It's not different, it only looks so. You're using z, which is the height above the surface radius (the altitude). I'm using z itself. Those are just convention:

z = r - r0

Plug this into my equation's form.

P(r) = P0 exp( H' (1/r - 1/r0) = P0 exp( H' (1/(r0+z) - 1/r0 )

Now there is some algebra. I won't clutter the screen up with that. If you combine the fractions you should find the following form:

P(r) = exp( -(H'/r0) * (z/(r0+z) )

I believe this is the same expression for pressure that your link has. I am reasonably sure of that. However, I may have mucked up the stability requirement a bit. For a low altitude to radius ratio, your stability requirement says r > 2 H, whereas mine says r < H' / 2.

It's not immediately obvious to me if this is because of the definition of H, or if it's because of my algebra. I might come back to that, since a factor of 4 is fairly important. However, I do not believe your pressure profile is different from mine.

2. Hi Alan
Not quite. The exponent should be (H'/z)*(r0/(r0+z). Using the expression you've derived, it's (H'*z)*[1/(r0*(r0+z))] which inverts the z*(r0+z) product.

3. But the one you cite as correct is not the exponent in the link you gave:

http://web.ist.utl.pt/~berberan/data/43.pdf

Page 410, equation 37. It has the independent variable z in the numerator, consistent with my form. You are saying that the correct form should only have the z variable in the denominator, and I don't think this matches the paper. I don't see it either. I outlined the conversion between the two forms in more detail here:

http://physics.stackexchange.com/questions/89013/expression-for-atmospheric-pressure-with-altitude-including-tidal-forces/90664#90664

A point the paper made is that the pressure should limit to a constant value, because you're essentially out of the gravity well. This reflects a prediction that all atmospheres have some non-zero leak rate, which is logical, but pedantic. I agree with that statement. An exponent of the form z*(r0+z) can not exhibit this behavior because it goes to zero as z goes to infinity.

3. Hi Alan

(z/H)*(r0+H)/(r0+z)

I hope that makes more sense now. Think of it as the exponent of the homogeneous gravity field, with a correction factor for the finite spherical size of the gravitational source.

Of course the negligible gas-mass assumption that this approximation makes means the function isn't valid when taken to infinity. At some point the total gas mass would cross its virial mass and collapse. But as a description of an atmosphere under a ceiling, it should be valid.

1. The constant I've used for H is different from the usual definition, and this is indicated by H'. If your expression here uses only H, then I can explain most of the differences. To put one in terms of other:

H' = r0^2 / H

You can plug this into the expression I have, the result is the following exponent:

- (z/H) * (r0/(r0+z))

You can see this is similar to your expression. On Earth, H is small (about 8 km) compared to the radius. That means that our expressions will return the same values for Earth. Small bodies, like Ceres, will have larger H values due to lower surface gravity, putting it in the same range as its radius. Our formulas will disagree for such small bodies.

I'm not sure where that difference comes from. I used the gravity form of g=GM/r^2, and so I thought that it was exact for the case of a homogenous gas.

4. I used the potential for a spherical mass, GM(1/r-1/r0), then defined GM as g0r0^2. Like I said you need to find the ratio of the potential energy V of the two different altitudes. So V(H) = g0r0^2*(H/((H+r0)*r0)) and V(z) = g0r0^2*(z/((z+r0)*r0) and V(z)/V(H) = (z/H)*((H+r0)/(z+r0)) ...QED

5. Though my source doesn't bother defining H' in the spherical gravity field - they just leave it as kT/(m.g0). I am not sure why, but they seem to know what they're doing. Either way it's a constant and not much different in most cases. Intriguingly for this discussion, the pressure profile they compute is explicitly for an atmosphere bounded by a shell. Exactly what we're discussing here.

6. After all that, I did the same analysis and got exactly the same stability conditions. A World-Sheet would need anchors or supports of some sort, to guard against perturbations. Mountains of Atlas, one might be tempted to say.

1. I was hung up on your statement "you need to find the ratio of the potential energy V of the two different altitudes". That statement is obviously wrong, because potential has an arbitrary constant of integration. Dividing two absolute potentials can't possibly be meaningful. However, I lacked a precise statement to offer as a replacement. While I was sitting on that, it seems that you've made additional corrections. Nonetheless, I think this form of equations will help make things clear.

My replacement statement:
Pressure falls exponentially according to the change in gravitational potential divided by a characteristic gas-specific potential. Mathematically:

P / P0 = exp( - Delta_U M0 / (RT) ) = [def] exp( - Delta_U / U_gas)

This is no longer arbitrary because it relies on the difference in potential between two points, and allows you to calculate the pressure ratio between those two points.

I introduced H' myself, so it's not surprising that we don't see it within your reference. I'll recap the definitions here. In my notation, M0 is the molecular mass and g0 is the surface gravity.

U_gas = RT / M0 = 84,087 m2/s2

H = RT / (M0 g0) = U_gas / g0 = 8.6 km (as expected)
H' = M0 GM / (RT) = GM / U_gas

In my Stack Exchange question, I wrote down two expressions for the pressure given radius / altitude. The form I've written in this comment is the "one equation to rule them all". You can obtain the expression for either case with that, using the specific definition for gravitational potential difference. Take r0 to be the surface for the purposes of shell-world.

Delta_U = U(r) - U(r0)
Newtonian gravity, U(r) = -GM/r:
Delta_U = GM * (1/r0-1/r)
Constant gravity U(r) = g r:
Delta_U = g * (r - r0)

If you plug these into the governing equation along with the definition of U_gas, you can obtain the expressions in the Stack Exchange question:

P / P_0 = exp( M0 g / (RT) * (r0-r) )
P / P_0 = exp( M0 GM / (RT) * ( 1/r - 1/r0 ) )

I have a high level of confidence in these expressions, and now I can articulate what I think they mean. It is an exponential of a ratio of potentials, but that is the gravitational potential difference divided by a m2/s2 quantity that you can isolate from the ideal gas equation.