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A Banks Orbital requires some form of unobtainium, with tensile strength much greater than any known substance. I'm trying to figure out how that substance would stand up to impacts and explosions.

To be specific, say the material in question is, to coin a word, paraneutronium, basically 'as strong as neutronium would be if it were a solid instead of liquid, and stable at zero ambient pressure'. Or, like a single layer of graphene, except with neutrons in place of carbon atoms.

Per https://en.wikipedia.org/wiki/Nuclear_binding_energy#The_nuclear_force

The nuclear force is a close-range force (it is strongly attractive at a distance of 1.0 fm and becomes extremely small beyond a distance of 2.5fm)

Assuming the neutrons are spaced 1.0 fm apart (and effectively assuming a square grid, ignoring the possibility of a hexagonal lattice like graphene; it doesn't matter for current purposes of order of magnitude estimation).

The measured binding energy of the deuteron is 2.2 MeV. Not the same thing, but a plausible proxy; say that's the bond energy we are looking at here. That's 3.5e-13 J.

Force = energy/distance = 3.5e-13/1e-15 = 350 N. That seems plausible; the above Wikipedia page gives the repulsive force between a pair of protons close to each other as 40 N, and the strong nuclear force must overcome the electromagnetic repulsion in nuclei containing tens of protons.

A square meter of paraneutronium contains 1e30 neutrons, has a thickness of ~1e-15 m, and a mass of 1675 kg.

A meter-wide strip of paraneutronium has a tensile strength of 3.5e17 N.

The maximum length of paraneutronium that could support itself in 1 g = 3.5e17 / (1675 * 9.8) = 2.13e13 m = 2.13e10 km, twenty billion kilometers. That compares favorably with the radius of a Banks Orbital = 1.8 million kilometers; we have a little over ten thousand times the minimum necessary tensile strength.

That means a single layer of paraneutronium could support, say 1 km thickness of rock ~= 3000 tons/m^2, and still have a severalfold safety margin.

All well and good when nothing goes wrong, but in a science fiction story, something always goes wrong or there wouldn't be a story. So what happens when the load-bearing shell suffers an unexpected impact?

What happens if it gets hit by a 1-meter rock at 100 km/s?

Or a 100-km asteroid at that speed?

Or a 1-megaton nuclear warhead?

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    $\begingroup$ The unit for tensile strength is generally the Pascal (Pa). You've got m in there for every tensile strength measurement. What unit are you using? $\endgroup$
    – jdunlop
    Commented Apr 27, 2022 at 20:18
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    $\begingroup$ If you're asking how something behaves when it's hit you're definitely not asking about it's ballistic behavior. Unless it's getting launched into the air ballistic armor isn't armor that is ballistic, it's armor that protects you from ballistic objects. $\endgroup$
    – sphennings
    Commented Apr 27, 2022 at 20:20
  • $\begingroup$ Neutrons do not react with anything and do not bind to each other. They cannot easily be contained either. Even so, they decay really fast. If all you have is neutrons at zero pressure, no matter which container you use, it takes a few seconds for some to escape and the rest to become protons and electrons. You can get to clump together with the gravity of a neutron star, but that adds more problems than it solves. $\endgroup$
    – The Square-Cube Law
    Commented Apr 27, 2022 at 20:24
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    $\begingroup$ To figure out what happens with an impact, it would help to know the elastic modulus of the material, and what the yield stress is. Essentially how much it can deform and return to its original shape, and what stress is when it plastically deforms. From that you can figure out how fast sound waves and shocks can move through the material. $\endgroup$
    – UVphoton
    Commented Apr 28, 2022 at 2:29
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    $\begingroup$ Magnetic Monopole Matter gets you the strength you want and a number of other interesting properties. Magnetic monopoles are theoretically possible, though they haven't yet been observed in nature. This is what you would expect thogh, as their creation requires copious amounts of energy. $\endgroup$
    – TheDyingOfLight
    Commented May 4, 2022 at 5:22

3 Answers 3

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Here's a few alternative calculations:

You will have problems when the speed of sound is insufficient to carry the information of the impact far enough to dissipate said impact before the material is moved far enough to break.

(You may have problems before this; this is an estimate of an upper bound.)

The speed of sound in a taut thin membrane is just $c = \sqrt{T/\sigma}$. $T=3.5*10^{17} N/m$. $\sigma = 1675 kg/m^2$. Overall, this works out to a speed of sound of ~14.5 km/s.

Let's say that the membrane breaks when it stretches 2.5x (from 1fm spacing to 2.5fm spacing).

Consider a massive point hitting the membrane. It ends up stretching out a cone shaped 'depression' in the membrane. The base of the cone can grow in radius at no more than 14.5km/s. The sides of the cone aren't snapping, and so can grow at no more than 2.5*14.5km/s. This then constrains the speed of the point to no more than ~33km/s.

(You can also imagine a disk hitting the membrane, and the truncated cone resulting.)

Do note that this is after the collision of the point and the membrane. 1675kg/m^2 isn't negligible.

Take your 1m rock at 100km/s. Or rather, let's call it a 0.5m radius 1m cylinder of magnetite (~5g/cm^3). If it was somehow zero thickness, or a rigid body even at these speeds, would it penetrate? Yes. 5000kg/m^2 is more than enough; it would still be traveling at ~75km/s after initial impact.

Unfortunately, these speeds are well above the speed of sound in mundane materials, so you can't just consider the impactor a rigid body. And when you do the estimate for non-rigid bodies, you're fine.

That being said, this does indicate that you have to worry about collisions with other things made of neutronium.

Alternatively, consider high-speed lead colliding with said membrane. Each individual lead nucleus is ~5.5fm in radius. This means that each lead nucleus should collide with ~1 neutron.

Breaking 4 of the bonds in said neutronium requires 1.4e-12 J. Corresponding velocity of a lead-208 nucleus is 2,847 km/s.

This puts an upper bound on things - lead traveling at 1% of c or faster or the equivalent will destroy it (or at least poke holes in it).

(That being said, there 'should' be all sorts of interesting nuclear reactions that also occur...)

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    $\begingroup$ Good analysis, thanks! Though the example of the lead nucleus reminds me to be concerned about cosmic rays. What is the density of cosmic rays with energy on the order of 10 MeV or greater? I know it is small on an everyday scale, but I have sudden visions of it making femtoscopic lace curtain of the shell of an Orbital over a few million years. $\endgroup$
    – rwallace
    Commented May 4, 2022 at 10:11
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    $\begingroup$ Very rough estimate: thenakedscientists.com/forum/index.php?topic=81359.0 you get an upper bound on total cosmic ray incident energy per unit area of ~4.8 x 10^-5 W/m^2. And your neutronium has ~1.4x10^18 J / m^2 of bond energy. Dividing gives ~10^15 years or so. So call it 10^12 years before 1-in-100 bonds are broken, at a conservative estimate. Not a problem. $\endgroup$
    – TLW
    Commented May 4, 2022 at 17:24
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    $\begingroup$ I wouldn't worry about bulk effects. Destroying said neutronium requires ~300MT TNT / m^2 even assuming that said energy was perfectly coupled into said neutronium. This is a fair bit of energy. I'd be more worried about 1d effects - you 'only' require ~350J/m to break a 1d line, best-case. Which is a huge amount compared to other 2d materials (graphite is something like 1nJ/m), but still relatively tiny. $\endgroup$
    – TLW
    Commented May 4, 2022 at 17:59
  • $\begingroup$ graphene*, rather $\endgroup$
    – TLW
    Commented May 4, 2022 at 20:57
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Boom.

The biggest problem I see with your paraneutronium is that it has no repulsive force. Neutron is stuck to neutron somehow, every femtometer. If a piece of matter hits it, the electrons sail through the barrier, and the nuclei aren't far behind. If they're more than a centibarn in cross-sectional area, they're guaranteed to "interact" with the neutrons. For many nuclei that would be energetically unfavorable and I suppose nothing would happen ... but what if forming the higher isotope releases more than 2.2 MeV? Then a lot of energy gets loose, even as a hole is formed in the pattern. Now this might not add up quickly, if the violence of interaction drives back anything that would strike... but it seems like it should do something eventually.

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  • $\begingroup$ Heh, fair enough. But that would apply also to normal matter being placed gently in contact with it, which would make it useless. To postulate structures like Banks Orbitals, we have to postulate some unobtainium that can support loads made of normal matter, so we have to assume that paraneutrons (or whatever) repel ordinary nucleons without otherwise interacting with them (not an unreasonable stretch, given that we are postulating such material in the first place). $\endgroup$
    – rwallace
    Commented Apr 27, 2022 at 21:31
  • $\begingroup$ Maybe, but that ties everything up in a bow. If the substance uses an unspecified force to repel normal matter, then what normal matter can't it repel? $\endgroup$
    – Mike Serfas
    Commented Apr 27, 2022 at 21:36
  • $\begingroup$ Right, but the substance has a specified, finite tensile strength and thickness, and while it disobeys the laws of particle physics, it obeys other things, like Newton's laws. It seems to me this should be enough to calculate what sort of impact it would take to punch a hole in it. $\endgroup$
    – rwallace
    Commented Apr 27, 2022 at 22:07
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    $\begingroup$ A lot depends on what happens when e.g. an individual lead nucleus hits said neutronium, @rwallace . Does it elastically collide with one neutron? Does it elastically collide with ~100 million neutrons? Does it inelastically react with one neutron? Etc. $\endgroup$
    – TLW
    Commented May 3, 2022 at 23:33
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Okay, it seems to me we can reason roughly as follows.

Take the case of a 1 m object impacting at 100 km/s (around the high end of expected natural impact speeds, assuming an object coming from interstellar space). That's on the order of 1e3 kg, that must be decelerated from 1e5 m/s in 1 m (being the size of the impacting object), in other words in 1e-5 s. So for about ten microseconds, the exerted force would be on the order of 1e13 N.

A meter-wide strip of paraneutronium has a tensile strength of 3.5e17 N.

On the face of it, that's four and a half orders of magnitude safety margin. In practice somewhat less because the material has limited elasticity etc. Say it's three orders of magnitude.

A quick dimensional analysis suggests, and experience with guns versus armor confirms, that the amount of armor a projectile can penetrate, is a linear function of its diameter. That means at that speed, a kilometer-sized asteroid would punch a hole in the structure.

That sounds plausible, and means the active meteor defense system only needs to deal with objects that can feasibly be spotted at interplanetary range.

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