In a previous question I asked how Mutually Assured Destruction could be made not assured. Most of the highly rated answers focused on political methods, but one answer raised the possibility that Kessler Syndrome might make ICBMs impracticable - it's this idea that I'd like to test for practicality.

Given the Earth, with present day technology, is it technically feasible to fill LEO with enough stuff that any ICBMs fired will have only a small chance (let's define it as 5%) of making it to their target?

For the purposes of this question I'm not really interested in economics or politics - feel free to assume that the entire industrial output of the Earth goes towards achieving this, and that world peace magically breaks out for as long as it takes - I want to know if it's possible from an engineering and physics perspective.

I've tagged this as instead of because it's likely to require a bit of XKCD What If?-type guesswork, but I'd like to see answers that include calculations and that are based on empirical evidence.

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    $\begingroup$ A comment, because you want calculations: due to the realities of orbital mechanics, and the direction in which ICBMs are launched (over the poles and China<->US) it would require completely surrounding the Earth with stuff. That would block out the Sun. Since the Earth is Really, Really Big, and ICBMs are -- comparatively -- minuscule, that would require a ginormous amount of stuff. $\endgroup$
    – RonJohn
    Commented Aug 31, 2018 at 14:25
  • $\begingroup$ @RonJohn a ginormous amount of stuff is what I'm expecting; the interesting question is how much, and if it's possible to get enough stuff into orbit. $\endgroup$
    – walrus
    Commented Aug 31, 2018 at 15:05
  • $\begingroup$ Kessler syndrome is where your body doesn't properly synthesize DNA polymerase...right? The answer is yes. $\endgroup$ Commented Aug 31, 2018 at 22:59
  • $\begingroup$ @JamieClinton afraid not, Kessler Syndrome is where the density of objects in Low Earth Orbit is high enough that a single collision can cause a feedback loop, surrounding the Earth with a (relatively) dense shroud of satellite bits and other debris, causing issues for trying to get things through it into space. $\endgroup$
    – walrus
    Commented Aug 31, 2018 at 23:01
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    $\begingroup$ @kingledion XP is not caused by being unable to synthesize DNA polymerase, but rather a failure in nucleotide excision repair (which is done perpetually in cells that have high exposure to ionizing radiation). I can't think of any disorder related to DNA polymerase synthesis (or folding, etc) that has a name similar to Kessler syndrome, though. $\endgroup$
    – forest
    Commented Sep 1, 2018 at 0:24

5 Answers 5


Not possible


In order to ensure that an ICBM had only a 5% chance of surviving for 20 minutes in space, you need to launch at least 5 orders of magnitude more mass into space than has been launched since Sputnik. This is not economically feasible, so Kessler syndrome cannot protect you from ICBMs.

A note on ICBMs

First we have to define what an ICBM is going to do. It travels a ballistic course, up to 1200 km altitude, and it can be launched from any latitude. This part is important; a sub-launched missile from the North Pacific can hit Moscow while never travelling below the 50th parallel north.

What is the collision frequency of orbital objects?

From the original paper, Kessler and Cour-Palais, 1976, the collision rate for a single object in orbit is

$$ \frac{dI}{dt} = S\bar{V}_sA_c,$$

where $I$ is the cumulative number of impacts, $S$ is the spatial density of particles in orbit in km$^{-1}$, $\bar{V}_s = 7 \text{km/s}$ is the relative impact velocity (calculated in the paper), and $A_c$ is the cross-sectional are of the object in question.

Generously counting the third-stage booster, a modern Trident II missile has a 7 foot diameter and is about 12 feet long. Lets use a 7x12 foot surface as cross sectional area, which works out to $8\times10^{-6}$ km$^2$ to match units. This gives us an equation:

$$ \frac{dI}{dt} = 5.6\times10^{-5} S$$ where we will solve for $S$ in units of particles per cubic kilometer and $dI/dt$ in collisions per second.

What collision rate leads to a 5% survival rate for an ICBM?

A Poisson distribution gives the number of discrete events in a certain time period, based on a rate function, the probability that the event will happen. The probability mass function for the Poisson distribution is $$P = \frac{\lambda^k e^{-\lambda}}{k!}.$$ $P$ is the probability that an event will happen, $\lambda$ is the rate of an event happening, which is what we will solve for, and $k$ is the number of events. We want to solve for $P = 0.05$ and $k=0$; that is, there is a 5% chance that there will be zero collisions. Plugging these in, we solve for $\lambda=3.00$.

For our Trident missile, let us assume 20 minutes (1200 s) are spent in the orbital regions of space. Therefore $\lambda$ is in units of events per 1200 seconds. Converting to units of events per second, we get $\lambda = 0.0025$. This number is equivalent to the $dI/dt$ that we want.

How much space do we need to fill with particles?

Mentioned in the introductory note is the point that ICBMs can get into all sorts of unusual parts of space, because they are sub-orbital. They can fly right across the north pole. Orbits that take our anti-ICBM particles into high latitudes will also inevitably cross the equator, where there will also be anti-ICBM particles that are orbiting at low latitude. Thus, there will be a distribution of satellite orbits that peaks at the equator and reduces in density near the poles.

I'm going to ignore that; instead I'm just going to say we need a constant density debris field over the Earth. We'll see why in a minute. In order to have a good chance of stopping an ICBM, we need density $S$ of particles in all parts of space that an ICBM might reach over its 20 minute stay. This is a shell from 300km to 1200km from the surface of the Earth, or at a radius of 6670 to 7570 km.

$$ V = \frac{4}{3}\pi(7570)^3 - \frac{4}{3}\pi(6670)^3 = 5.7\times10^{11} \text{ km}^3.$$

How much mass of particles do we need?

Plugging $dI/dt$ into our collision rate equation, we get S = 44.6 objects per cubic kilometer of space. Multiply this by the space we need to fill and we need 25 trillion particles. As you can see, increasing the particle density near the equator is not really necessary; this is plenty of ball bearings to launch.

Generously assuming a 100g particle can kill a ballistic missile (doubtful!), we need to put 2.5 trillion kg of particulate mass into orbit to allow only a 5% chance of an ICBM to get through. The total amount of spacecraft launched since the dawn of the space age is roughly 13 million kg.


Kessler syndrome is dangerous on the timescale of weeks and months. Once there are enough particles, your spacecraft won't last out a year, and then it is hard to justify launching something into orbit. But the particle density required to stop an ICBM, with a fast, suborbital path through space is immense. Especially with mobile ICBM platforms (i.e. submarines) nearly any trajectory for a ballistic missile is possible, so you would have to cover everything to be safe.

Even dropping the requirement to a 50% kill-rate and redoing all the calculations only drops the mass requirement by a factor of about four. You still need roughly five orders of magnitude more mass in space than has been put there in the past 50+ years to form an effective defense. Kessler syndrome cannot effectively guard against ballistic missiles.

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    $\begingroup$ This is a fantastic answer; 45 units per cm3 seems pretty reasonable until you consider just how big space is. $\endgroup$
    – walrus
    Commented Aug 31, 2018 at 15:25
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    $\begingroup$ I believe that you're overestimating the necessary size of a particle. ICBMs are not heavily armored, and a 5.56mm bullet is approximately 4 g. Otherwise, great math, much better than my ballpark figure. $\endgroup$
    – o.m.
    Commented Aug 31, 2018 at 15:57
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    $\begingroup$ Fantastic! May I add a detail which makes things even more unlikely? In LEO, atmospheric drag causes orbits to decay rather quickly. TIangong-1's orbit decayed in about a year, starting from 350km. So not only do you need 2.5 trillion kg of stuff up there, you also need to have 2.5 trillion kg of stuff whose orbits have not yet decayed. $\endgroup$
    – Cort Ammon
    Commented Aug 31, 2018 at 15:58
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    $\begingroup$ @kingledion, those heat shields are designed to resist heat during a very specific reentry trajectory. Merely tumbling the missile might be enough to prevent a clean separation of the warhead. $\endgroup$
    – o.m.
    Commented Aug 31, 2018 at 16:27
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    $\begingroup$ @o.m. I wondered about the 100g size too, but when you're talking 25 trillion particles the extra couple of orders of magnitude are probably unimportant. $\endgroup$
    – walrus
    Commented Aug 31, 2018 at 16:29

Given the awesome destructive power of nukes, a 5% survival rate isn't an effective defense, because just one warhead getting through will take out an entire city. (Even an authoritarian government isn't going to be happy about trading away entire cities at a stroke, particularly since you can't effectively counterstrike--the same space junk that blocks the other side's warheads will block yours just as effectively.) You need something more like a 5% chance of even one warhead slipping through; if the attack contains (say) 5000 warheads, that requires a survival rate of 0.01%. The Kessler effect just isn't going to give you that kind of shield.


Possibly, with a Slight Change

The suggestion as written will not work, for two main reasons:

a) Near Earth Space is just too big, as kingledion outlines in his excellent answer.

b) LEO orbits are affected by atmospheric drag, so the debris field will rapidly de-orbit itself.

The Change

Use "smart debris" instead of "dumb debris".

Launch a series of satellites in much higher - and therefore more stable - orbits. These satellites use radar / IR / whatever other sensors make sense. On detecting anything in LEO or that looks likely to enter LEO, they perform a kinetic attack.

This is basically the Star Wars plan from the Reagan years, but completely automated, and utterly denying space access to all nations, including the launching nation. This lack of command and control should make it much simpler to implement.

For bonus points, rival powers can contribute satellites to the system, so there is no worry about a "back door" that will allow one states weapons to get through.

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    $\begingroup$ How do you contribute satellites without getting them shot down too? $\endgroup$
    – Joshua
    Commented Aug 31, 2018 at 20:19
  • $\begingroup$ @Joshua - fun engineering problem! Probably each nation has a 'turn off' code, and if all nations partner to the MAD treaty issue their code at the same time, the system is disabled for a brief period. $\endgroup$
    – codeMonkey
    Commented Aug 31, 2018 at 20:43
  • $\begingroup$ @Joshua make them start on a timer. Launch the first and declare that space access is denied in 2 years period. Or however long you need to launch sufficient numbers of satellites $\endgroup$
    – Hobbamok
    Commented Jul 19, 2021 at 8:04

It is possible but difficult.

For commercial or even military spaceflight, it is usually unacceptable if one mission in twenty fails within the first half hour. What you are calling for is that nineteen missions in twenty fail within the first half hour. The amount of debris that would be required would be awesome.

Assume that each piece of debris has an independent chance to hit a missile. To go from a 95% survival rate to a 5% survival rate, you need 58 times the amount of debris (0.95^58 is about 0.05).

The ISS has to maneuver every now and then. Call it a 0.01% chance of having to maneuver in a half hour, and you would need a 30,000-fold increase of debris over present rates.

On the long run, these debris particles would collide with each other. Some would deorbit, some would go to higher orbit, some would break down into flecks too small to harm a missile.

Fortunately for your scheme, you can replenish the debris cloud by sending up rockets with enough velocity to make most their fragments reach orbit. If they get hit on the way up, well, that helps with the dispersal ...

  • $\begingroup$ "For commercial or even military spaceflight, it is usually unacceptable if one mission in twenty fails within the first half hour." 100% the wrong assumption. We're talking about NUKE DEFENSE here. It is unacceptable if any one of up to 5000 makes it through. And then you're assuming uniform distribution of junk which is not the case (and if such a "defense" were being deployed it would be quite easy to launch the ICBMs on such a trajectory that theyd avoid the high concentration orbits anyway $\endgroup$
    – Hobbamok
    Commented Jul 19, 2021 at 8:03
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    $\begingroup$ @Hobbamok, I think you are miss-estimating how Cold War era planners thought about acceptable exchange rates. A 95% damage mitigation would have been thought the greatest thing since sliced bread. $\endgroup$
    – o.m.
    Commented Jul 19, 2021 at 17:22

No. Not gonna work. If I was facing this problem I see an immediate solution:

I will deliberately launch my nukes into an orbit with an AP so low that it will decay in two hours. The nuke will deorbit itself into target.

Kessler Syndrome can't reach that low because the deorbit time is two hours.

If you managed to get it that low, then I will pick an "orbit" so low that it requires continuous thrust to avoid decaying in a quarter of an orbit.

  • $\begingroup$ A decaying orbit can't be targeted with any precision. The final re-entry prediction for Tiangong-1 was "plus or minus two hours", with a potential impact zone that covered a substantial portion of the planet. Controllers attempted to target Skylab's re-entry, and still missed by several thousand miles. If you want any sort of targeting ability, you need a powered re-entry. $\endgroup$
    – Mark
    Commented Aug 31, 2018 at 19:52
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    $\begingroup$ @Mark: I specifically said "nuke will deorbit itself" which definitely implies powered re-entry. $\endgroup$
    – Joshua
    Commented Aug 31, 2018 at 20:18

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