If a spacecraft travels at 10% the speed of light will it be destroyed by collisions with interstellar dust and particles?

The spacecraft will be traveling to nearby stars, not going through a nebula.

  • $\begingroup$ Why specifically 0.1xc? Or is it simply: Is there a practical limit on the speed of a space journey imposed by potential, but inevitable, collisions with interstellar dust and particles? $\endgroup$
    – Ng Ph
    Commented Jun 30, 2021 at 14:07
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    $\begingroup$ It's from a comment by Zubrin that fusion rockets could reach 0.1c $\endgroup$
    – sno
    Commented Jun 30, 2021 at 16:17
  • $\begingroup$ Thx! I didn't know this. Your question makes sense. $\endgroup$
    – Ng Ph
    Commented Jun 30, 2021 at 17:11
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    $\begingroup$ Given the huge ten orders of magnitudes of variations in density, it would help if you would clarify your question. Are you interested in sending a spacecraft to a nearby star. At 10% c, a one-way journey to the closest would take over 40 years. Are you interested in sending a spacecraft through the heart of the Orion Nebula at 10% c? That would be a one-way journey lasting 13400 years. Note well: In 13400 years, humanity (if it survives that long) will probably have slightly improved technology. $\endgroup$ Commented Jul 1, 2021 at 21:07
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    $\begingroup$ You mentioned Zubrin in a comment. He has said, multiple times, that if we are planning something in space that would take over 50 years to accomplish it would be best not to undertake that venture because technology will overtake in 50 years. That means once we can make a vehicle that can go at 10% c, we should only think of sending it to the nearest stars -- and that means having to plow through a interstellar medium with a rather low density of 0.3 atoms/cc as opposed to a million atoms/cc as in the current answer. $\endgroup$ Commented Jul 1, 2021 at 21:26

1 Answer 1


Based on this answer:

$$E \approx \frac{1}{2} m v^2 = \frac{1}{2} m c^2 \left(\frac{v}{c}\right)^2$$

The mass of a proton $m_P c^2$ is about 938 MeV, so if an innocent atom of hydrogen or bare proton in space were hit by a spacecraft at 0.1 c, in the spacecraft's frame it would look like a speedy proton. Of what energy?

$$E = \frac{938}{2} 0.1^2 = 4.7 \text{ MeV}$$

From here I clicked proton projected range and found out it will be about $1 \times 10^{-2}$ g/cm$^3$, or about 5-10 microns in some low-density material.

Cold interstellar medium is about $1 \times 10^{12}$ protons/m$^3$ or $1 \times 10^{6}$ protons/cm$^3$, but comments below indicate that typical values are way lower, only 0.05 to 0.3 protons/cm$^3$. So I will use 0.1 /cm$^3$ ($1 \times 10^{5}$ /m$^3$) going forward.

At 0.1 c each square meter of spacecraft intercepts $3 \times 10^{7}$ m$^3$/s or $1 \times 10^{12}$ protons/sec.

This is not a lot of heat

The power deposited in the 10 micron thick layer can be gotten from multiplying 4.7 million volts by $1.5 \times 10^{-7}$ amps (coulombs/second) per square meter, or 0.7 watts. In order to calculate what temperature can re-radiate that power use the Stefan–Boltzmann law with $\sigma = 5.67 \times 10^{-8}$:

$$P = \sigma T^4$$

which gives us a skin temperature of about 60 Kelvin, not even enough to keep standard electronics working.

Sputtering is not a problem

Our 10 micron thick square meter has of order 0.1 mole or about $10^{23}$ of atoms.

From random links 1, 2, and Figure 1 in 3 I will ballpark the sputter rate at $10^{-4}$ atom/atom which is conservative.

So at $10^{12}$ protons per second we'll sputter $10^{8}$ per second, meaning we'll loose 10 microns every 30 million years.

Note that the sputter rate may be a few orders of magnitude faster because my estimate was conservative and at 3300 K the atoms are already pretty energetic and predisposed to leave.

But what about things bigger than protons, like cosmic dust?

Damage by things bigger than a proton is going to be a big problem, but I don't know how big. I will leave that aspect to another answer

projected range of a 4.7 MeV proton

  • $\begingroup$ "Shields up, Scotty!" problem solved :-). Certainly doing what you can by way of shaping the spacecraft so impacts are at a shallow angle will help a very little bit, too. $\endgroup$ Commented Jul 1, 2021 at 13:57
  • $\begingroup$ Downvoted because you used 10^6 atoms per cm^3, the upper density of a molecular gas cloud. That's a protostar. A better value is 0.3 atoms per cm^3, the density of the Local Interstellar Cloud. $\endgroup$ Commented Jul 1, 2021 at 14:06
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    $\begingroup$ The Local Interstellar Cloud, and the nearby G Cloud (with a similar density) extend to several nearby stars. The Local Bubble, in which the Local Interstellar Cloud is embedded, has an even lower density of 0.05 atoms per cc, and extends for over a hundred light years. A better answer would address these local densities that are over six orders of magnitude smaller than the value you used. $\endgroup$ Commented Jul 1, 2021 at 21:02
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    $\begingroup$ @DavidHammen should be much happier with this version. :) $\endgroup$
    – PM 2Ring
    Commented Jul 26, 2021 at 14:00
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    $\begingroup$ @PM2Ring Much happier. $\endgroup$ Commented Jul 26, 2021 at 14:51

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