A voyage of several ships at 0.2c needs raw materials and protection from impacts with interstellar dust. For these purposes, they installed enormous thrusters on the circumference of a 10-km-diameter C-type asteroid and accelerated it along with (in front of, but not connected to) them. They keep the asteroid facing straight toward their destination, mining the back surface while keeping the front surface is untouched, other than of course the constant bombardment by relativistic trace interstellar gases. When the story takes place, the voyagers have just finished flipping the asteroid 180 degrees to prepare for deceleration as they near their destination after 30 years of travel (they remain protected by the asteroid during this phase).

What I want to know is, what is the condition of the asteroid's front surface (the surface that was facing toward their destination up until recently)? Did the bombardment make it smooth? Powdery? Is it radioactive, and if so, how far away can a human float safely near it? Are there large craters from occasional collisions with large dust particles? The more detail you can provide, the better.

I found this article helpful, since it talks about bombardment of an oriented surface at 0.2c (though in this case it's assumed to be the postage-stamp-sized Breakthrough Starshot probe). It suggests that over a 30-year journey, erosion by the ISM will be of order 1 mm deep, and that collisions with large dust particles aren't common enough to hit a tiny probe, but are very likely to hit something several square km in area (I estimate ~$10^4$ collisions with $>10 \mu m$ grains over 30 years).

So while I'm pretty sure that only the first millimeter or so of the asteroid will be affected, the article doesn't go into detail about the questions in the 2nd paragraph (consistency, presence of large impact craters, and radioactivity). I'd also like to know if the composition of the asteroid makes a big difference here.

  • $\begingroup$ I don't understand your problem: a large surface is made by many small surfaces, and you have the information for what happens to the small surface already. $\endgroup$
    – L.Dutch
    Dec 1 '21 at 14:13
  • $\begingroup$ One problem. Once you flip your ship around to decelerate, you now face your colony side to the incoming particles that will impart significant energy that your asteroid shield absorbed. depending on your rate of deceleration, this could be a very long period for the colony being bombarded until the ship is moving slow enough where the incoming particles are no longer doing too much damage. $\endgroup$
    – Sonvar
    Dec 1 '21 at 14:57
  • $\begingroup$ @Sonvar My thoughts too. It's not useful to place the impact-shield behind the spacecraft on the deceleration cycle unless you plan to re-use it on your next journey and particularly need to hang on to it. Far better to drop the shield and enjoy the fuel-savings of not having to decelerate that much mass. $\endgroup$
    – Ruadhan
    Dec 1 '21 at 15:06
  • $\begingroup$ @L.Dutch I edited the question to highlight what I still don't know (last paragraph). As you said, whatever happens to a small area will happen to a larger one, but I don't know anything about it other than that there will be about a millimeter of ablation. And of course the article didn't mention whether there will be large craters, since anything large enough to cause a crater would destroy Breakthrough Starshot $\endgroup$ Dec 1 '21 at 15:35
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    $\begingroup$ you could just rotate the thrusters. $\endgroup$
    – ths
    Dec 1 '21 at 18:11

Unexpectedly weird.



It was a good thing they had this big lump blocking for them. There was more to block than they expected, although when they saw hot asteroidstuff streaming by them they suspected things were going on up there. The trip has wildly eroded the forward facing surface of their asteroid, forming hoodoos and aeolian elements as referred to in your moniker, the fast moving cosmic medium here playing the role wind does on Earth.

Your voyagers had planned that for the deceleration they would rest on the asteroid and let its giant engines decelerate it and them together. They are less enthusiastic about hunkering in beneath the unstable looking hoodoos and spires that erosion has produced.

  • $\begingroup$ That would be pretty cool, but it would require the interstellar medium to be several orders of magnitude denser than current studies suggest. As I pointed out, ablation of the surface should be of order of a mm deep (barring larger impact craters). As much as I appreciate your reference to my username, interstellar dust is way too diffuse to carve any substantial patterns into the surface. $\endgroup$ Dec 1 '21 at 17:55
  • $\begingroup$ If they traverse a nebula, that would provide the several orders of magnitude greater than interstellar medium. $\endgroup$
    – Willk
    Dec 1 '21 at 18:23
  • $\begingroup$ Fair enough. Could certainly be cool enough to include in the story, and it makes sense that with their overkill shield, they would absolutely plow through a nebula rather than try to go around. My question is, though, would a high mass really lead to ventifacts, or would explosive collisions with large dust grains reduce them to craters? And how radioactive would the surface be, given the higher impact energies? $\endgroup$ Dec 2 '21 at 0:53
  • $\begingroup$ / how radioactive would the surface be/ I am not clear on how hitting something really hard renders it radioactive. I get that maybe on impact there is particulate radiation and maybe some EMR but that would be over fast. $\endgroup$
    – Willk
    Dec 2 '21 at 2:20

A C-type asteroid is volatile rich, low density rock (~1.7g/cm^3). This is not the most robust asteroid, but it can be large enough to provide the protection you need.

Now about how the surface should look.

By what I can read, interstellar space has a density of about 1 atom cm^3. By my calculations, at the peak speed of 0.2c, you should have ~6 billion interactions with particles per 1m^2 per second.

Each atom the asteroid interacts with will impart a great deal of energy. Using the Omnicalculator at https://www.omnicalculator.com/physics/relativistic-ke I assumed an average mass of atoms striking the surface as 60 AMU. With that, it calculated that each impact deposited 1.86*10^-10 Joules or 1.15 GeV. This doesn't sound like much, but its enough to liberate about 10's or 100's of thousands of other atoms. Some of these atoms may gain escape velocities of the asteroids, but much of them could stay in the vicinity of the crater. Further impacts will erode some of the loose debris, but would continue to liberate more material.

Ove the course of 30 years, for ease of calculations, I calculated a total 5.68*10^16 atomic strikes per square meter. even if the liberate upwards of a million other atoms, that would still only liberate a small fraction of a mol of material per square meter, resulting in a slight dust accumulation.

Another part is the friction heating. 1 eV is equal to 11600 kelvin in material. 1.15 GeV will equate to 1.33410^13 kelvin. That sounds like a lot, but if the energy received over a full square meter was deposited uniformly into 1 mol of material, it would equate to 1.2810^-4 K average temperature. Due to not losing energy to space and some conduction heating to material below in the asteroid, the temperature will build up over 30 years. Now, my math might be off, but I am only calculating a temperature build up of ~333.5K. I would assume the top few mm's would be evaporated away and the next few cm's to be vitrified due low heat transfer coefficient of the asteroid material.

So overall, the peaks and plains facing direction of travel would be covered in a glass material. Many of the peaks would be eroded into the crates and valleys. In the edges and shadows of the craters and mountains would be protected from vitrification, so dust and debris would build up here.

  • $\begingroup$ Isn't 60 AMU way too high? Only 0.1% of the interstellar medium is heavier than Helium. $\endgroup$ Dec 1 '21 at 17:58
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    $\begingroup$ "Due to not losing energy to space" - why not? As the rock would heat up, it will radiate away more energy. $\endgroup$
    – Alexander
    Dec 1 '21 at 18:24
  • $\begingroup$ @VentifactsandYardangs I was using the highest mass of particles that are naturally produced outside of a super nova. I would agree, my numbers may be high, but I was leaning on the conservative side. Average mass of particles would probably be in the order of 5 or so, but I am unsure of the actual number $\endgroup$
    – Sonvar
    Dec 1 '21 at 19:27
  • $\begingroup$ @Alexander Dissipation of heat is a real issue in space as convection heat transfer does not really occur. Most of the heat loss will come from radiant heat and atoms obtaining escape velocity from energy they gain from heating. $\endgroup$
    – Sonvar
    Dec 1 '21 at 19:29
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    $\begingroup$ @Sonvar so, when calculating "build up of ~333.5K" over 30 years, did you factor in radiant heat loss? $\endgroup$
    – Alexander
    Dec 1 '21 at 19:31

The impacts per unit surface won't change with the size of the shield. The figures that you have can be applied also to the large one.

In other words, with a larger surface you intercept more particles, but you have also more surface to erode, therefore the two effect balance each other out.

You might expect the relativistic impact between the space medium and the surface to produce some radioactive byproduct as a consequence of the merging of the atomic nuclei of the shield with the nuclei found in the space medium, but the shield itself will protect the ship from the radiation.

Not so lucky would be those looking at the ship from the destination point.

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    $\begingroup$ I commented and edited my question in response to your comment, which this seems to basically be saying the same thing as. Could you give more specific answers to my questions (surface qualities of the asteroid and whether craters will be visible)? And as for radiation, the ship will be protected, sure, but how risky would it be for a person in a suit to physically go to the asteroid's surface? Are we talking eat-a-banana levels of radiation, or coughing-up-blood levels? $\endgroup$ Dec 1 '21 at 15:43

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