In the near future we may find cargo ships fleeing space pirates!

Given that...

  • It's a small cargo ship carrying 10 metric tons of wheat (cargo volume, approximately 13 m3). Total volume: 100 m3. Total mass (loaded): 30 metric tons.

  • Thruster technology is such that the ship can successfully maneuver through Sol's asteroid belt at an average velocity of 1,000 Km/s.

  • It's near-future, so there's no magical intertial dampening systems. Flight pressure suits we have today are OK. Any other dampening systems known and operative today or conceivably within the next 25 years are OK.

  • Knowing that today we believe all the mass in the remarkably large asteroid belt wouldn't make a body bigger than our moon...

  • And assuming at 1,000 Km/s even a baseball-sized impact would have serious consequences1...

Question: Could our astronaut survive two hours of space flight (a run of 7,200,000 Km or 180X the circumference of the Earth) inside the asteroid belt without dying from the maneuvers necessary to avoid impacts?

After two hours the pirates make a mistake, take an impact dead center of the windshield, and due to the force of explosive decompression, find themselves hurtling deeper into the belt like little human torpedoes. Our hero can slow down and avoid all future impacts.

1Obligatory XKCD, not completely relevant, but the end result gives us an idea of the urgency of the situation.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Commented Sep 3, 2018 at 22:22
  • $\begingroup$ As most of the answers have pointed out, you are not likely to hit any asteroids. However, you could try to lose the pirates by going through Saturn's rings. That would be risky just going through the plane of the rings, traveling in it is even more dangerous. I don't think you would need to go too fast, either, because there is so much debris to hit just going at the speed everything is orbiting. $\endgroup$
    – John Locke
    Commented Sep 7, 2018 at 22:47

7 Answers 7


Yes they could. Very Easily.

Contrary to myth, the asteroid belt is very nearly empty. (See https://physics.stackexchange.com/questions/26712/what-is-the-average-distance-between-objects-in-our-asteroid-belt for a related discussion.)

Planners for the Dawn mission, which spent many years moving through the asteroid belt, visiting both Vesta and Ceres (and which is still in orbit around Ceres), estimated that during that entire time, the closest it ever came to a cataloged asteroid is about a million kilometers -- over twice the distance to the Moon. Space is really, really big.

Basically, there's nothing around that's big enough to see before it's too late to maneuver to miss it. The issue will probably turn out to be devising effective micrometeor shields. (There's a good chance that the technology we use today will work even with hypervelocity impacts. Basically, there's a thin outer aluminium skin, then an air gap (well, vacuum gap!) then a very thin, very strong composite layer, then another gap then the actual hull. A small, fast projectile vaporizes on hitting the outer skin, the gasses spread and hit the middle skin which stops nearly everything and vaporizes the rest, and the hull is not touched by anything solid.

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    $\begingroup$ :-) We have a really boring asteroid belt.... $\endgroup$
    – JBH
    Commented Sep 1, 2018 at 19:20
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    $\begingroup$ To give an idea just how empty the asteroid belt really is: Its whole mass is just about one twentieth of that of our moon. $\endgroup$
    – Karl
    Commented Sep 1, 2018 at 19:22
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    $\begingroup$ @JBH The asteroid belts shown in Star Trek would collapse into a molten inferno before you could even say "Engage!" ;-)) $\endgroup$
    – Karl
    Commented Sep 1, 2018 at 19:26
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    $\begingroup$ Asteroids are the biggest incongruity between reality and movie astrophysics that I know of. When they’re not in unreasonably dense belts they’re hitting planets at laughably lethargic speeds (while nonetheless trailing fire). $\endgroup$
    – Joe Bloggs
    Commented Sep 1, 2018 at 19:36
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    $\begingroup$ @JBH, asteroid belts can't be very full. That's where plants come from. $\endgroup$ Commented Sep 2, 2018 at 8:29

If you expect to hit another asteroid, you have to work at it.

What did Alan Stern, New Horizons principle investigator, have to say?

Fortunately, the asteroid belt is so huge that, despite its large population of small bodies, the chance of running into one is almost vanishingly small - far less than one in a billion. That means if you want to come close enough to an asteroid to make detailed studies of it, you have to aim for one.

Given a 2 hour journey, vs. with months required for the New Horizons probe to traverse the belt, your chance of problems is more likely closer to 1 in trillion.

You are far more likely to die of a unsuspected heart attack when in "perfect health" during any given 2-hour period.

Don't base your expectations on collision chances based on the advice of a golden-hued protocol droid.

  • $\begingroup$ New Horizons doesn't impress me and isn't much of a comparison. Yes, it set a speed record of 58,000 kph, but that's only is a scant 16 kps. My ship is travelling 62.5X or 1,000kps. Perhaps given the vast lack of density there's still not an issue. I can't find exact numbers for its traverse, but let's say 3 months. My ship would traverse the same extent in only 34.6 hours. No, NH isn't much of a justification. $\endgroup$
    – JBH
    Commented Sep 2, 2018 at 0:31
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    $\begingroup$ @JBH, the example of a probe which casually cruised through the asteroid belt with no significant maneuvering capability or even method of detecting if it was going to hit something because no one was seriously concerned about the possibility of it hitting anything isn't a relevant example? $\endgroup$ Commented Sep 2, 2018 at 8:04
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    $\begingroup$ @JBH, and what about Cassini? Voyager 1 and 2? Pioneer 10 and 11? Galileo? Ulysses? Juno? Juno is even more of an example showing there's not a problem because it's solar powered; the wings have a span of roughly 20 meters. How big is the nosecone of your spacecraft? $\endgroup$ Commented Sep 2, 2018 at 21:13
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    $\begingroup$ @JBH Speed is irrelevant, only distance covered. $\endgroup$ Commented Sep 2, 2018 at 23:08
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    $\begingroup$ @JBH. You actually have it backwards, the more quickly you traverse the distance, the less likely you are to have a collision. This is precisely equivalent to the question of running or walking in the rain. You get wetter if you walk than when you run. Minute Physics has an excellent explanation. $\endgroup$ Commented Sep 2, 2018 at 23:28

Lets do math and assumptions

Assume the asteroid belt is $3\times10^{21}$ kg. Now assume that the asteroid belt is entirely composed of 100 g ball bearings, the smallest objects that can penetrate your Whipple shield and kill your spacecraft. The asteroid belt then consists of $3\times10^{22}$ tiny ship killers.

The asteroid belt is roughly a torus, extending from 2.3 to 3.3 AU from the sun. That means the 'torus' has a radius of 0.5 AU. At a distance of 2.8 AU, this 0.5 AU distance is $\arcsin{\frac{0.5}{2.8}} = 10.3$ degrees above the plane of rotation. Looking at this graphic, you can see that is an underestimate of the actual vertical separation of asteroids. So the following volume estimate is an under-estimatation of the space that the asteroid belt occupies.

The volume of a torus with radius 2.8 AU, and tube radius 0.5 AU is $$2\pi^2 (2.8)(0.5)^2 = 13.8 AU^3.$$ More usefully, that is $4.6\times10^{25} \text{ km}^3$. This means the particle density of these ball bearings is 0.00065 km$^{-3}$. That means, there is one ball bearing every 1530 km$^3$.

If your spaceship has a cross section of 100 m$^2$ and needs to move a distance of 1 AU to cross the asteroid belt, then the volume of its path is 14960 km$^3$. Given the particle density above, this means there are about 10 ship killers in the average path through the heart of the asteroid belt.

At 1000 kilometers per second, you would traverse the asteroid belt in about 42 hours, so you would have to avoid one object every 4 hours. You would only have to deviate your course by less than 100 meters to miss each object, since the average distance between objects in the ball bearing scenario is around 23 km. Changing your course by 100 meters every 14,440,000 km is...well the angle isn't really important. Vibration can get you off course by that much, the thrust needed for a course change is trivial.


If the mass of an asteroid belt were deployed as a minefield, then it would require a small number of course corrections of nearly negligible angle to avoid hitting the mines. But the asteroid belt is not a minefield, and there are far less than $3\times10^{22}$ objects to avoid. If you drop the number of objects by an order of magnitude, you need to make one course correction on average. If you drop it by two more orders of magnitude, then you have 1% change of striking anything, etc.

I don't have a count of objects of ship-endangering size available. But based on the worse case estimate, we can see that the density of such objects is very low. In the actual asteroid belt, a path plotted straight through is very unlikely to hit something.


You bring up a bunch of issues, so this answer has multiple parts.

Could an astronaut in a near-future space ship survive transit through our asteroid belt?

The image of a field full of rock is an out and out myth. The Asteroid Belt is mostly empty space.

It is 4% of the Moon's mass spread over 50 trillion trillion cubic km. Of that mass, most is locked up within only a handful of objects. About one third is accounted for by Ceres, while Vesta, Pallas, and Hygiea make up 9%, 7%, and 3% (respectively). That leaves just under half of the Asteroid Belt's mass (1.9% of the Moon's mass) for the rest of that 50 trillion trillion cubic km.

There are only an estimated 800 trillion asteroids larger than a metre within the Belt. That may sound like a lot at first blush, but 800 trillion asteroids divided by 50 trillion trillion cubic km is just under one asteroid per 50 billion cubic km.

Worrying about hitting one of those while traveling through the Belt would be akin to driving over salt flats the area of the United States with only 2000 people scattered across them and worrying about hitting someone.

You could well say "what about smaller objects?", but the situation doesn't change very much. 4% of the Moon's mass over 50 trillion trillion cubic km is only 60 mg per cubic km (60 picograms per cubic metre). There is no worst case distribution where there'll always be something dangerous to hit, and since most of that mass is actually concentrated in large objects, which themselves tend to be concentrated in several orbital groups, it's extremely easy to not hit anything in the Asteroid Belt.

This shouldn't be shocking. If there was any significant concentration of mass scattered throughout the Belt, it would quickly collapse into a planet. That's why much of what little mass is there collapsed into Ceres.

TL;DR: In fiction, ships traveling through the Belt must swerve around asteroids. In reality, the many space probes we've sent through the Belt paid minimal attention to collision risks. This won't change much even with higher speed travel.

In the near future we may find cargo ships fleeing space pirates!


Thruster technology is such that the ship can successfully maneuver through Sol's asteroid belt at an average velocity of 1,000 Km/s.

Keep in mind that ships traveling at a constant velocity, by definition, aren't using their engines/thrusters. In space, you don't need to push to stay in motion. You only need to push to speed up and to slow down.

A single Merlin engine (currently used on SpaceX rockets) can already generate nearly 1000 kN of thrust. That means 8 or 9 modern engines could accelerate 30 metric tons (30000 kg) to 1000 km/s if fuel wasn't the problem that it currently is. Unless people in your near future setting are still struggling with fuel mass and power sources like we are today, there's no reason to imagine any ship in your universe would need more than minutes to get up to speed. In other words, the vast majority of their travel time should be unpowered.

  1. This raises a problem for pirates. How do they detect ships with no drive plumes in a region (literally) many trillions of times larger than the volume of the Earth? Hell, even if the ships ran hot for their entire journeys, spotting them in such a large zone would still be next to impossible without high powered ground-based tracking.

    • If pirates can see cargo ships over these distances, why wouldn't the cargo ships be able to see those pirate ships. Remember, they'll have to burn towards the cargo ship after spotting it. That would give the cargo ship plenty of warning (and time to start increasing their own speed). I mean it's not like pirate ships can just sit next to "shipping lanes" and wait. (There's no sitting in space. There's only falling around the Sun, planets, moons, etc.) The only way to stay in a lane is to actually travel along its orbital path. What's worse, the paths cargo ships must take change as the planets orbit the Sun. Even worse than that, the lane for any given orbital configuration is highly speed dependent. That means one ship traveling at 950 km/s would have a vastly different "shipping lane" from a ship traveling at 1300 km/s. So, there's no easy way for a pirate to travel along a shipping lane at a slower pace and wait for other ships to catch up.
  2. If a ship requires no power to stay at 1000 km/s, of course it can maneuver at that speed. You have to define what kind of course correction constitutes "successful". Success is only a matter of warning time and a ship's ability to accelerate. If you're saying that these ships are able to detect obstacles with enough lead time to briefly use their engines for course corrections, then you've already answered your own question.

Question: Could our astronaut survive two hours of space flight (a run of 7,200,000 Km or 180X the circumference of the Earth) inside the asteroid belt without dying from the maneuvers necessary to avoid impacts?

Given that the odds of needing to maneuver are nearly non-existent, the astronaut wouldn't have to worry. However, even when maneuvers are needed, unless you've somehow accidentally ended up right on top of a planet, minor nudges are all that will be required.

After two hours the pirates make a mistake, take an impact dead center of the windshield, and due to the force of explosive decompression, find themselves hurtling deeper into the belt like little human torpedoes. Our hero can slow down and avoid all future impacts.

Maybe you want to set this in Saturn's rings? That is the only location in the Solar System with the kind of debris density you're imagining. Of course, you'd have to find a reason for two ships to want to stay inside the rings for so long.

I'm sorry. I don't mean to rip holes throughout your entire idea. Things in space just don't work like things on Earth. If you want input based on the facts, than the fact is that most traditional forms of conflict don't work in space.

  • Fuel dictates travel speed, not travel distance.
  • The speeds you can get up to and slow down from determine what destinations you can travel to.
  • The visitability of destinations (for your speed capabilities) depends on orbit and mass.
  • Space is big and bumping into people or things is hard.
  • If you are close enough to see other people, everyone can see everyone else.
  • Running after ships (or running away) is like shooting a gun. You can have a significant effect in the short term, but you can't do it for very long. In the case of space ships, you'll run out of fuel.
  • Etc, etc.
  • $\begingroup$ All exceptionally good points and well written. +1. Though I'd like to clarify that Fuel does in fact very much dictate distance, it dictates how much you can change your orbit. Merely nudging yourself and waiting an eternity will not get you very far. Once you have enough dV to get somewhere, any further fuel used will elongate your trajectory and get you there faster, but ultimately it's all about distance. I'm sure you knew all this given your informed answer, but the bullet-point seemed misleading to me. $\endgroup$
    – Ruadhan
    Commented Sep 3, 2018 at 11:30
  • $\begingroup$ "Though I'd like to clarify that Fuel does in fact very much dictate distance" Yes-ish. I was trying to describe delta-v without doubling the length of my already long post. $\endgroup$ Commented Sep 7, 2018 at 5:41
  • $\begingroup$ "Fuel dictates travel speed, not travel distance" Yes-ish. However, it's not so much about speed per se (which will always be in relation to something, to begin with, and is a scalar quantity) as it dictates your ability to change your velocity (a vector quantity). This is why fuel budgets in rocketry are often measured in delta-v, as the delta-v very directly tells you how much a spacecraft can maneuver (and already incorporates both the properties of the engine as well as the mass of the spacecraft). Any given maneuver will also often be expressed in terms of delta-v expenditure. $\endgroup$
    – user
    Commented Sep 7, 2018 at 6:05
  • $\begingroup$ Regarding shipping lanes, see also In space, do “shipping lanes” make sense? $\endgroup$
    – user
    Commented Sep 7, 2018 at 6:06
  • $\begingroup$ "... it's not so much about speed per se ... as it dictates your ability to change your velocity ..." Indeed. That's why I said "the speeds you can get up to and slow down from...". As I just commented. I was trying to describe delta-v in layman terms. If we're not going to teach the reader vector addition, multiplication, and other essentials for understanding orbital dynamics, then there's only so much detail we can go into. $\endgroup$ Commented Sep 7, 2018 at 6:10

My gut feeling is that, as stated in other answers, your astronaut is more likely to die from boredom than from an impact.

However, to back up the gut feeling with some math, let's try the approach of the average free path, like your astronaut and the asteroid belt were a gas.

We have that

$\lambda =$$ 1 \over \sqrt{2}\pi\sigma ^2 n$

where $\lambda$ is the average free path, $\sigma$ the collision diameter, twice the particle size (astronaut being about 2 meters), which we can round up to 5 meters or $5 \times 10^{-3} km$, and $n$ the number of particles per unit volume, for which we can reuse kingledion's value of 0.00065 $km^{-3}$

Putting those values in the formula above, we get $\lambda = 14 \times 10^6 km $, which is about double the distance you expect to travel in those two hours.

Which confirm my gut feeling, and gives your astronaut the reasonable safety to just worry about slowing down.


In addition to the answers posted by others (the asteroid belt is almost entirely empty space), the entire asteroid belt is moving at pretty close to the same speed, in the same direction.

If the astronaut is also travelling at the same speed, in the same orbit, the entire asteroid belt would appear (almost) stationary. This gives lots of time to react, as well as low impact speeds.

  • $\begingroup$ The average speed of an asteroid is 25 Kps. I'm travelling 1,000 Kps. $\endgroup$
    – JBH
    Commented Sep 3, 2018 at 15:16
  • $\begingroup$ @JBH "The average speed of an asteroid is 25 Kps." (I assume you mean 25 km/s.) In what reference frame? Earth's sun-fixed average orbital speed is about 30 km/s, and Mars' is about 24 km/s. Thus 20 km/s sounds like a reasonable approximation for average orbital speed in the asteroid belt. However, if you're moving through the Belt, you're also likely to move laterally at a not insignificant fraction of orbital speed. So while there will still be some (possibly still significant) relative motion, you can't just take some random "average speed" and assume that that's equal to relative speed. $\endgroup$
    – user
    Commented Sep 7, 2018 at 6:12

You can probably get away with trusting in luck, since (as mentioned by others) the asteroid belt is mostly empty space.

That aside, there are a few more pieces of information we need to answer this:

  1. Since you are travelling "fast" compared to the asteroid belt, the volume of your ship is irrelevant. All your impacts are going to be on the front of your ship. You would need to specify (or decide) the surface area of the front of your ship. A ship with a front surface of 10m x 10m, 1m long, would have 100 times the chances of hitting an asteroid than one 1mx1m front surface, 100m long (both are 100 cubic meters).
  2. You need to specify how far ahead you can see a potentially damaging object.
  3. You need to specify how fast you can change course. Sure, a couple of G (for short periods, in life-and-death situations) would probably be ok, but what is your sensor response time, reaction time, warning time (for people to strap in)? Does your ship need to turn, then boost, or can it move laterally? Despite what space opera tells us, it will almost certainly be done by a computer. It is not complicated - if you're going to hit something, move in any direction.
  4. You mention moving at 1000km/s, but don't forget that to the occupant, actual speed is meaningless - it's just acceleration/deceleration that they can tell. At 1000km/s, occupants will think they are floating. If you have enough reaction mass, the quickest way from A to B is usually to accelerate at maximum power, for exactly half the time, flip the ship around, then decelerate at maximum power for the the rest of the trip.

For example, if you have a front surface area of 1 square meter, and you can see a minimum-dangerous object at a range of 1000KM, and you are travelling at 1000KM/s, you would have as little as 1 second to move up to half a meter (assuming a circular front surface) + the width of the object.

Conversely, if you can detect the minimum dangerous object at 1 AU (the width of the belt), and it takes 2 hours to traverse the belt, then you would have 2 hours to react & turn.

I suspect that, all else being equal, an object twice the surface area would be seen twice as far away, and would require less than twice as far to turn; turning earlier also means you have to turn less. Based on this, the larger the object is, the easier it is to miss.


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