Observed rotation rates of small asteroids point to a limit on the material strength of these bodies. This limit is not very high

But that's only part of the picture, because an upper limit on rotation rate only tells you about the global mechanics of the body. There's also the question of whether we expect collisions to cause high rotation rates which stress the bodies in the first place.

Given the fact that we have direct observations of asteroid material (falling from the sky all the time), what material quality could we expect? It's simple to look up values for ordinary Earth rocks. Compare the two, there is 3 orders of magnitude difference.

  • Asteroid rotation rate based limit: 20 kPa
  • Granite, Basalt, Quartzite, etc tensile strength: around 20 MPa

The most obvious use of asteroids seems like it could be cutting it into beams, and using those for station construction, like what we do with certain kinds of lumber or granite counter-tops on Earth. If we did that, would these materials be worth a darn? Do we have a measured range of asteroid material strengths? Would the rock be too fragmented to use?

Also note that the target for NASA's Asteroid Redirect Initiative is intended to be soft rock. This eliminates some dangers of losing its (unstable) orbital location and falling to Earth, although I do not know if it's the reason for the requirement.


Not entirely answer for your question but answer of what would be used as the construction materials in space, instead of raw asteroid rock.

Considering that the considerable number of meteorites are iron, while the asteroid rock structural endurance itself is poor, with abundance of solar energy forging them into iron is totally viable, and there, Wikipedia provides abundance of information on mechanical properties of various iron alloys. I mean, why settle for brittle rock when good iron is abundant?


I may be misunderstanding your derivation of asteroidal material strength, but I believe your analysis doesn't measure something that is, say, related to von Mises yield strength. Instead, you are measuring the cohesion of the asteroid's rubble.

I think we can do a fairly good analysis of various asteroids' composition and then do terrestrial tests to determine material properties of the raw or alloyed material.

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    $\begingroup$ The equatorial stress was taken to be uniaxial. The referenced book I got the equation used a pretty unsophisticated model of body forces, such that it's the same as if a hula-hoop was spinning in space. I put in values from asteroids discovered since publication of that book. Obviously it doesn't give the asteroid rock strength but it still sets a lower limit on that value. $\endgroup$ – AlanSE Jul 21 '13 at 20:45
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    $\begingroup$ There's one more thing I worry I've not communicated well. Small asteroids with high measured rotation rates are not rubble piles because any rubble on the equator will fly off. Their sizes are several meters or several 10s of meters so gravity does pretty much nothing. $\endgroup$ – AlanSE Jul 21 '13 at 21:33
  • $\begingroup$ Some good points. I upvoted your physics question too. $\endgroup$ – Erik Jul 21 '13 at 22:08

Asteroids larger than ~200m in size are expected to be rubble piles, an amalgamation of dust, sand, rocks and boulders held together mainly by chemical forces (and at larger sizes, with help from gravity). So as you rightly point out, what's of interest for building materials is not the asteroid's macro tensile strength, but that of its constituent pieces.

The range of strengths for asteroid materials can be deduced from the values of meteorites collected on Earth. Really, why travel to an asteroid to collect a sample when the Universe delivers samples to Earth for free? ;-)

There are some issues with this approach:

  • We don't have a precise one-to-one mapping of asteroid types to meteorite types, composition wise.
  • Meteorites get beaten up in their passage through the atmosphere and subsequent impact on Earth.
  • Meteorites that lie around on the ground for significant periods of time get weathered and can be chemically and physically altered by exposure to the elements.

With that in mind, you can find some compressive strength values for a few meteorites here. Compare those values to that of concrete, which is about 3,000 psi. I haven't checked all those meteorites but, unless noted, I believe they're all non-iron, and probably chondritic (non carbonaceous), which would correspond to the materials in S-type asteroids.

Can they be cut into beams and used for construction, as has been suggested? That would depend on whether there are large enough pieces in the asteroid to provide long enough beams, and also whether or not there are cracks due to past impacts. I'm completely ignoring the difficulties of cutting beams from an asteroid, in space. I would ask, however, if beams are useful. They probably are not the best way to build in free space, and if building on the Moon or some other location with surface gravity, the cost of getting them there would be very high (and technological hurdles would have to be jumped over).

The current in-space fabrication movement (for example, Made in Space) is moving in the direction of 3D printing, which requires printer stock in the form of powder. So it might be that what's useful from asteroids are not the boulders, but the dust- and sand-sized particles, which can be mixed in with a binder brought up from Earth and used to 3D print habitats and other structures.

  • $\begingroup$ First sentence you might want to edit "larger"->"smaller" $\endgroup$ – AlanSE Feb 19 '18 at 17:06
  • $\begingroup$ Nope, "larger" is correct. Asteroids smaller than ~200m are still going to be mostly rubble piles, but we believe there will be a percentage that are not, that will be solid (or cracked/fractured) monoliths. What that percentage is, however, we don't know. Asteroids larger than ~200m are almost certainly practically all rubble piles (unless they're large enough to have become differentiated or are the leftover iron-nickel nucleus of a differentiated asteroid). $\endgroup$ – J.L. Galache Feb 22 '18 at 9:13

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