You are on a generation ship in interstellar space, between star systems. We know that there are rocks whizzing around out there - escaped asteroids, bashed planets, we have even put a few artificial ones out there ourselves.

Unless your ship was giving off electromagnetic radiation that could be detected from it bouncing off the rock (radar or similar) how could you detect a boulder sized completely inert rock? Gravity would be negligible, it would have a very minimal heat signature, there is no incident light, nothing to warm it up. Any radiation it emitted would have long since passed its best-by date. So unless you happened to detect it when it passed between you and a star, obscuring the light just enough, how would you detect it?

I want to clarify - I want the system to be passive. I DON'T want the system to be a reflection of something from the ship.

To complicate the issue, this rock is heading straight on your axis of travel, or rather you are headed straight for it, so it does not cross in front of a star.

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    $\begingroup$ Inter-galactic means between galaxies. Inter-stellar is used for within a galaxy. Can you clarify which? It will have an impact on the answer. $\endgroup$
    – Tim
    Oct 13, 2017 at 1:25
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    $\begingroup$ Isn't this what deflector dishes are for? It's the snow plow of the future! $\endgroup$ Oct 13, 2017 at 2:01
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    $\begingroup$ I'd say the big question would be how much notice do you need? That would depend on your divert capability and how big your ship is (i.e. how far you have to move out of the way). $\endgroup$
    – Cort Ammon
    Oct 13, 2017 at 2:49
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    $\begingroup$ @Justin Thyme Depending on your drive, 0.2c may be fast enough for your craft to become an emitter, just from plowing through the interstellar medium. $\endgroup$
    – Cort Ammon
    Oct 13, 2017 at 6:11
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    $\begingroup$ Reading all the comments you've made, you've mentioned several times that you don't want to detect it in time to respond; you explicitly only want to be aware of the rock afterwards. Instead of searching (or not-searching?) for the low-energy rock, wouldn't it be easier to become aware of the rock by simply monitoring the hull and detecting the collision event? I feel like all of the answers are trying to give you a detection early enough to avoid the rock. $\endgroup$
    – Cort Ammon
    Oct 13, 2017 at 16:31

8 Answers 8


Let's suppose you are traveling at 10% of light speed, and you need at least 0.1 seconds to react to a potential collision. That means you need to be able to spot a cold rock at a distance of (0.1 x 3x108ms-1/0.1s) = 3x108m, or three hundred thousand kilometers.

The equilibrium temperature of objects in deep space is 2.7K. But for argument's sake, let's imagine the rock you're about to hit is at a relatively warm 5K. Let's also assume it has a surface area of 105m2.

Per the math on the Wikipedia page, the rock will have a black-body emission peak of 294GHz, and its power output will be 3.54W.

If you have a giant infrared microwave sensor with an area of 100m2, then at a distance of 3x108m it can detect, at most, 3.13x10-16W of power from the rock. To put it another way, at 294GHz, that means you're detecting just 1.6 million individual photons per second. Which is not impossible, but very difficult (thermal cameras have serious noise problems because the camera itself is not at absolute zero).

So, it might be just about possible to detect objects passively, if they were the size of a city block, and fairly warm, and you could destroy or avoid them in a tenth of a second.

But at 10% of light speed, you could be obliterated by an object the size of a peanut*, and if an object had been in deep space long enough it would have cooled to the temperature of the cosmic background, making it thermally invisible no matter how big it was.

Personally, if I were on an interstellar ship, I would insist on it having active sensors (radar or lidar). If that's absolutely forbidden for narrative reasons, then I don't think you could avoid all collisions, so you'd need to be able to survive them. You could have an "icebreaker" ship flying just ahead of you, basically a huge metal asteroid with engines, and if something hits it you have time to hit the brakes before you rear-end it.

* Edit re: relativistic collisions

Why do I assume a micrometeoroid will not pass right through the ship, as it would at orbital speeds? For objects in Earth orbit, we're talking about rail-gun speeds (~105ms-1), but for interstellar travel we're talking about speeds a thousand times greater, closer to what you'd see in a particle accelerator.

Suppose a pebble passes through a spacecraft at 10km/s, and its temperature increases by 100K (which I think is a low estimate). It might melt and fly apart, but at that speed it'll be out the other side of the spacecraft by the time that happens. Now, suppose the same pebble was traveling a thousand times faster. It hits the same number of molecules on the way, but each of those collisions is a thousand times more energetic. Loosely speaking, by the time it passed through the spacecraft its temperature would have increased by 100,000K. At that temperature, it's not held together by chemical bonds; it's a collection of highly energetic charged particles radiating in all directions, so most of the (considerable) kinetic energy is going to end up transferred to your spacecraft as heat.

In some ways that is good news; if you did see a large obstacle in advance, you could pretty much blow it to atoms with a .22 rifle. Although, given the constraints above, the only way that'd happen is if you were constantly accelerating towards a big bright nebula or something, and could see objects occluding it.

  • $\begingroup$ You will NOT be obliterated by something as small as a peanut. it will go right through you. The spaceship is basically hollow. The outer hull will have a hole in it, the particle keeps going, Just like a bullet keeps going through a piece of paper. The hull of the ship won't stop it. $\endgroup$ Oct 13, 2017 at 14:28
  • $\begingroup$ Like I said, the temperature would be negligible. I have already eliminated a temperature trail. $\endgroup$ Oct 13, 2017 at 14:29
  • $\begingroup$ No object colliding ay relatvistic speed would "just keep going." Upon impact you would have a spray of plasma that proceeeds to incinerate the rest of your ship. $\endgroup$ Oct 13, 2017 at 15:39
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    $\begingroup$ @Justin Thyme - not exactly the answer, but interesting enough: (dailymail.co.uk/sciencetech/article-3757314/…) Your problem is that you need to do something about whole spectrum of objects - from hydrogen gas to small pebbles. If your make the spaceship flimsy enough that pebbles will go through - then hydrogen will erode the hull very quickly. $\endgroup$
    – Alexander
    Oct 13, 2017 at 17:42
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    $\begingroup$ One common idea is to spray a fluid in front of you and recapture it (as you're accelerating). You do (slowly) lose fluid over time, though. Bonus points if it doubles as the ship's radiator. $\endgroup$
    – TLW
    Oct 15, 2017 at 0:14

Cosmic Microwave background radiation has a known pattern. If a rock passes in front of the microwave camera, it will distort and block this pattern, even if it has the same temperature and no radiation originating from it. To increase the chance of spotting the particle, there should be multiple cameras that scans the same region. By comparing the camera outputs, one can map out the 3D shape of the surroundings, including oncoming doom.

However, I doubt that evasive maneuvers would be the solution. Generation ships are for the long run and over the course of centuries, these evasive actions will use way too much fuel. Without evasive maneuvers, running a ship for centuries requires no more fuel (excluding energy used by people) than a ship that is designed to run for few months. Thus, instead of changing the course of the ship, they should try to deflect the particle. Which will decrease both the amount of energy required and the time frame to notice the object.

  • $\begingroup$ I am tending towards the 'not what you see, but what you DON'T see' solution. That is, the rock blocking out the background. seeing a shadow and not the rock itself. $\endgroup$ Oct 14, 2017 at 5:16

Gravitational detector.

Rocks have gravity. Negligible gravity, you disdainfully assert. But gravity nonetheless. Can the gravity of a rock in space be used to detect it? Interstellar space is a nice place for this sort of thing because there is nothing to get in the way and no other gravity sources except you, and the rock.

One measures gravity with a gravimeter. These are nifty, and sensitive, and used commercially to measure variations in surface gravity of the earth. The Cavendish experiments famously measured the gravitational attraction between 2 non-earth objects.

For reference.


The Universal Gravitation Equation is: F = GMm/R2 where • F is the force of attraction between two objects in newtons (N) • G is the Universal Gravitational Constant = 6.674*10−11 N-m2/kg2 • M and m are the masses of the two objects in kilograms (kg) • R is the separation in meters (m) between the objects, as measured from their centers of mass

What is the smallest force that has been measured today? Let us assume in this spacefaring future, measurements this sensitive are commonplace.

These folks http://newscenter.lbl.gov/2014/06/26/smallest-force-ever-measured/. claim to have measured a force of 42 yoctonewtons, each yoctonewton being 1 septillionth of a newton, or 10-24 (1 E-25 in excel).

Let us see what sort of body at what distance would exert a force of 42 yoctonewtons on our 1000 kg detector mass. Maybe a 1 kg (m) rock 1 km (R=1000) away from our detector (M=1000)? hmm.. how to paste excel rows and keep formatting. I will do it as an image. Any better ways, please suggest in comments.

excel rows pasted as a pdf

A 1 kg rock 1 km away from the detector mass produces a force many orders of magnitude greater than 42 yoctonewtons. It is almost silly, like detecting a pie in the face! Negligible indeed!

The distance to produce the 42 yoctonewton gravitational attraction between a 1 kg rock and our 1000 kg test mass is 126057432 m or 126,057.432 km. Not even the distance from Earth to the moon, but a fair distance. And a larger mass could be detected farther away.

Of course with 1 gravitational measurement it might be a small rock close up or a big rock far away. That is the thing about gravity. My intuition told me that with 3 consecutive measurements, and assuming stable mass, velocity and trajectory of the detected rock, there can be only one mass and velocity of the detected rock that fits. But on testing that idea I found it not to be the case, as follows:

This rock is moving directly at the detector so we do not need to worry about angles. Let us first have a 1 kg rock moving at 10 m /second, and 3 consecutive measurements 1 second apart.

more excel rows as jpg

Fine. Now for a 100kg rock can we find some velocity that mimics those force readings at each time point?

still more excel rows as jpg

Hmm. Yes we can mimic the small rock with a large one. A small close rock approaching slowly is very different from a large distant rock approaching fast, but gravitationally they look the same.

Can we work around this? So far I have not come up with a workaround. I thought putting an additional detectors on 1 km booms at right angles might help, which should demonstrate how little intuition for math I have. Booms did not help in my excel model of different perspectives on the same approaching mass.

In sum:

  • All mass has gravity
  • Gravity can be detected
  • Rocks in space can be detected because of the gravitational force they exert on a detector.
  • It was difficult for me to use gravity to characterize rocks in space by mass and velocity. But I am thrilled at the thought that someone here might be able to do it.
    • Math corrections always welcome.
  • $\begingroup$ Thank you sincerely for doing the math, It is quite illustrative. You have assumed the mass of the ship to be 1000 kg. My ship is much more massive than this, so the numbers would be much higher. The difficulty is that R^2 thing, It's okay when the rock gets close, and I have the rock coming from behind, so the relative velocities between the rock and the ship are not that great. So maybe... a gravimeter used for another purpose that happened to have a data record... $\endgroup$ Oct 15, 2017 at 13:50
  • $\begingroup$ I envisioned a 1000 kg dedicated homogenous mass used as the detector. Perhaps an ingot of osmium. 1000 kg is just the detector. More mass = easier to detect rocks. I think it would be harder to detect inflections on an inhomogeneous mass (like a ship) because its internal gravity would be shifting as things moved. I think once you are flying spaceships and detecting yoctonewtons, one distance is not intrinsically more implausible than another. $\endgroup$
    – Willk
    Oct 15, 2017 at 14:40
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    $\begingroup$ You could have a two part system. If passive detector finds anything interesting, active one will fire to check. $\endgroup$ Oct 16, 2017 at 17:47
  • $\begingroup$ @Will I now get the part where the 1000 kg mass was the detector. I should have realized that part from the math. It might work. The ship is no longer under acceleration, just coasting. The engines are shut off. If the detector were at the back of the ship (the rock comes from behind) it just night detect the difference in gravitational pull from a rock coming from behind. The instrument would have to be very sensitive. It could even give a sense of the closing speed. Interesting. Now I need a reason for the detector to be there in the first place. Maybe for research in the engines. $\endgroup$ Oct 16, 2017 at 18:09
  • $\begingroup$ You could have it to detect rocks so you don't get hit, which is what I assumed your interest was. Or maybe because you might want to go visit bodies you discover in the interstellar void. Possibly retrieve small ones for study. The cool thing about your mass overtaking the ship is that (as I learned by doing here!) there is a continuum of solutions for a given gravitational signal, from something tiny and close approaching very slowly to something incredibly massive and very far away, but approaching really fast. $\endgroup$
    – Willk
    Oct 16, 2017 at 21:19

I have to assume that the ship is of a substantial size if it is to have generations of people, agriculture, animals etc.

I also have to assume that you have artificial gravity, for people and animals not to fly around, and crops to grow.

We don't know the size of the rock. But if it is large enough to make a dent but not large enough. It could be pulled out of the path downwards towards the gravity drive. And if the acceleration is large enough it will simply be pulled out of the way.

I don't know enough about space, or your world to confirm this would work in reality but in space engineers it works and is even weaponized.

This is also done by Jupiter, with asteroids going in to our solar system, the asteroids is inside Jupiters gravity well and pulled out of the way and kicked out of the solar system again

  • $\begingroup$ No artificial gravity. using rotation. $\endgroup$ Oct 13, 2017 at 14:46

I'm not sure this totaly meets your requirement for a passive system, but I think its at least part way there and is a novel approach so worth a mention

What you need is a “vanguard” Think a very wide ultra-thin circular membrane together with minimal stabilisation and communications equipment traveling a long way ahead of the generation ship on exactly the same heading. If the vanguard is sufficiently far ahead there would be time for it to detect any rock passing through its membrane and give sufficient warning to the generation ship to fire emergency lateral thrusters to adjust course slightly and avoid the object.

A variation on this would be a small fleet of small probes fitted with wide angle laser projectors and detectors plus minimal communication and stabilization equipment. These could also fly far ahead of the generation ship in a wide circle and create a net of laser light that could detect any rock passing through and warn the generation ship in a similar way. If any of the probes were hit themselves the others could take over giving multiple redundancy. The other advantage would be the capacity to have a much wider net giving a greater safety margin.

Extended answer

Assuming a hole in the pressure vessel which is survived but leaves the ship damaged and the crew trying to figure out what happened.

First I would assume they would find the hole quickly as there would be emergency procedures in place to detect and deal with such an eventuality. After the immediate aftermath the hole would be visually examined and that would give them a lot of information.

They would know if it had been an explosion from within the ship outwards, or a hit on the ship inward and in cases of external impacts the location of the hit on the ship and the exact time would give away the direction of approach. The size of the hole would also tell them something about the energy of the impact and possibly even the mass and velocity of the projectile. There is an outside chance that they might even be able to find fragments. Considering it approached from behind it can impact at relatively low or relatively high velocities.

If you need further sensor confirmation there might have been cameras on the hull which would have detected the impact flash. The other possibility (if approaching from behind) might be interference with communications from Earth at a very low level or general scientific sensors measuring anything you want on an ongoing basis in all directions continuously. If it was metallic it might have made some record in magnetometer readings during its approach.

  • $\begingroup$ This is a really neat trick. But I don't want to detect until AFTER the collision. I want it to show up in a replay of the recorded data. $\endgroup$ Oct 13, 2017 at 14:44
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    $\begingroup$ Correct me if I’m wrong but it seems to me that you need it to be completely undetected ahead of the impact? Perhaps we should be thinking along those lines instead? You want a “what was that!” moment, partial depressurisation / or whatever and then, let’s look at the cameras and sensor records to see what happened. One option they wouldn’t be expecting would be a very high velocity rock approaching from an angle other than dead ahead (might make an interesting question in its own right). I'm sure I could come up with a few reasons. $\endgroup$
    – Slarty
    Oct 13, 2017 at 16:43
  • $\begingroup$ for my particular plot line, yes. For a general answer, it shouldn't matter. In my case, it came from behind. The only data is the archival recorded sensor data. But I can have ANY sensor data recorded. What I need to know is what that data should be. But it would probably apply to ANY general case, not just my plot line. $\endgroup$ Oct 13, 2017 at 18:12

Black-body radiation

You ask:

Unless your ship was giving off electromagnetic radiation that could be detected from it bouncing off the rock (radar or similar) how could you detect a boulder sized completely inert rock?

By the electromagnetic radiation that the rock gives off: its Black-body Radiation.

Now this may sound fancy but it really is not. The simple fact is that everything that has a temperature above absolute zero "glows". This glow depends on the square of the square of the temperature, that is to say T4.

For temperatures that we are accustomed to, the radiation is in the infrared.

enter image description here

As things get hotter the radiation goes up into the visible part of the electromagnetic spectrum, i.e. "red hot" actually means what it says: that it so hot that it does become red.

enter image description here

So this rock would simply be detected by automatically monitoring the surroundings with infrared sensors.

You said: "it would have a very minimal heat signature". But no matter how minimal, that is still detectable.

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    $\begingroup$ minimal heat is fine as long as there is sufficient difference to the temperature of your background, in this case you background is space at ~4K but I have no idea what temperature interstellar rock is likely to be at, it could be similar $\endgroup$
    – jk.
    Oct 13, 2017 at 12:35
  • $\begingroup$ @jk it will be similar. Rock will cool by raising and be heated by background radiation. If it is cooler than 4K, it will be heated faster than it cools. If it is hotter, it will cool faster than it heats. Equilibrium will be at the temperature of radiation. $\endgroup$
    – Mołot
    Oct 13, 2017 at 14:54
  • $\begingroup$ Even in interstellar space, the stars provide lots of background noise when it comes to black box radiation and everything else. This rock has been in space for billions of years. It has cooled down to pretty much background temperature. Since the rock is on the axis of travel, there would be no fluctuation of the background due to movement. I thought about the masking effect of the rock on background noise - detecting what is NOT there. $\endgroup$ Oct 13, 2017 at 14:54

If the close miss was close enough, the rock could be detected by occulting background stars. At the moment, the anomaly is filed away for later, the most probable case being small sensor malfunction, to be checked by maintenance. After all, what are the chances of something actually being there? (You there who said "astronomical", let's see if your humour is improved by a nice 8 hours EVA shift!)

It is only later, while reviewing the logs (and finding nothing wrong with the sensors) that some technician, IA or expert system adds two and two, and realises that there actually was a rock out there.

Note that for the same reason lasers beams always end up diverging and as such weakening, occultation can only be visible up to a certain distance depending on the size of the rock. Like lasers, who diverge less the bigger the aperture is, bigger occulting objects will be visible from further.

Anything small enough to only be called a rock will have to be insanely close for an interstellar object to be visible that way.

  • $\begingroup$ The inhabitants of the ship know for certain there was a rock out there. The tricky part is to determine its previous path after the fact from recorded data. Was it an intentional 'shot' (enemy action) or was it natural? But that is plot, the question answer should not be specific to the plot. $\endgroup$ Oct 16, 2017 at 17:57

The black body radiation answer that someone else provided is the best option you have in this scenario. I just want to add that It would also be quite easy to detect as deep space has a relatively uniformed energy signature . The object, no matter how minimal it's radiation, would be noticeable on the background.


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