5

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.

Ventifacts and Yardangs
  • 4,176
  • 13
  • 30
  • 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. – L.Dutch Dec 01 '21 at 14:13
  • 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. – Sonvar Dec 01 '21 at 14:57
  • @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. – Ruadhan Dec 01 '21 at 15:06
  • @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 – Ventifacts and Yardangs Dec 01 '21 at 15:35
  • @Sonvar I've edited the question to be more clear. The asteroid has been rotated (so that its rockets face ahead, allowing it to decelerate). The ship, however, is still behind the asteroid. The ship and asteroid are not physically connected. Any ISM will now hit the back of the asteroid during the deceleration phase, not the ship. – Ventifacts and Yardangs Dec 01 '21 at 15:37
  • 1
    you could just rotate the thrusters. – ths Dec 01 '21 at 18:11
  • I assume that the energy requirements to accelerate and decelerate this asteroid are hanwaved away? – Alexander Dec 01 '21 at 18:26

3 Answers3

2

Unexpectedly weird.

hoodoos

https://www.worldatlas.com/articles/what-are-badlands-and-where-do-they-occur.html

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.

Willk
  • 304,738
  • 59
  • 504
  • 1,237
  • 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. – Ventifacts and Yardangs Dec 01 '21 at 17:55
  • If they traverse a nebula, that would provide the several orders of magnitude greater than interstellar medium. – Willk Dec 01 '21 at 18:23
  • 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? – Ventifacts and Yardangs Dec 02 '21 at 00:53
  • / 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. – Willk Dec 02 '21 at 02:20
1

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.

Sonvar
  • 3,207
  • 6
  • 14
  • Isn't 60 AMU way too high? Only 0.1% of the interstellar medium is heavier than Helium. – Ventifacts and Yardangs Dec 01 '21 at 17:58
  • 1
    "Due to not losing energy to space" - why not? As the rock would heat up, it will radiate away more energy. – Alexander Dec 01 '21 at 18:24
  • @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 – Sonvar Dec 01 '21 at 19:27
  • @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. – Sonvar Dec 01 '21 at 19:29
  • 1
    @Sonvar so, when calculating "build up of ~333.5K" over 30 years, did you factor in radiant heat loss? – Alexander Dec 01 '21 at 19:31
  • Radiant heat loss would be fairly minimal and strongly dependent on the actual chemical make up, and thus heat transfer coefficients. I used more conservatives numbers and assumed radiant heat transfer is negligible. – Sonvar Dec 01 '21 at 19:33
  • @Sonvar here is a task for you. A square meter of surface receives an amount of energy E per second. Emissivity of the surface is ε. Using Stefan–Boltzmann law, calculate equilibrium temperature of this surface. Assume that the only outgoing energy transfer is radiant emission from one side of the surface. – Alexander Dec 01 '21 at 19:51
  • @Alexander that equation is very dependent to energy received verse initial temperature. based on a energy input of only 335k over 30 years, in the absence of a solar flux, the radiant heat loss fraction is in the ^10-8 magnitude, thus really not impacting the calculation by very much. Plus, I'm basically doing RADCON math, Making assumptions on many different unknown variable. My math can be +/- 20% off and still be in the ball park, and thus such variables such as radiant heat loss can be negligible. If anything, the heat would be less than what I calculated, just barely – Sonvar Dec 01 '21 at 20:09
  • @Sonvar "that equation is very dependent to energy received verse initial temperature" No. Equilibrium temperature is not dependent on initial temperature. – Alexander Dec 01 '21 at 21:09
  • "The Stefan–Boltzmann law describes the power radiated from a black body in terms of its temperature." Yes, initial temperature does contribute to how much energy an object will radiate. And still, the delta in K is still too insignificant to be accounted in a black body thermodynamic temperature heat transfer. – Sonvar Dec 01 '21 at 21:46
  • "how much energy an object will radiate" - initially yes, eventually no. Notice that I'm asking about equilibrium temperature. Think about electric range, for example. When on, its surface will reach certain temperature (let's say 300C). And it does not matter whether before turning on, you took this range out of freezer or applied blowtorch to heat it above 300C - it will still come to same 300C after some time. – Alexander Dec 01 '21 at 22:49
  • I still think this is still a moot point. My calculations are based on heating of the material of the first centimeter. True equilibrium temperature of the asteroid itself would not change considerably. I calculated energy deposited into a few grams of material, verses millions of tons of asteroid material. In reality, vitrification would happen in mere milliseconds then heat is transferred to the rest of the asteroid in the form of conduction heat transfer. Radiant heat is still negligible. – Sonvar Dec 02 '21 at 14:34
0

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.

L.Dutch
  • 286,075
  • 58
  • 587
  • 1,230
  • 1
    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? – Ventifacts and Yardangs Dec 01 '21 at 15:43