Is there actually a way to create total orbital denial?

Question

Orbital denial is like area denial but applied to the entire orbit of a planet or even a star. I don't remember where I first heard the term but I know Max Brooks uses it in World War Z and suggests that the controlled detonation of a single space station could be sufficient to shut down operations in Low Earth Orbit indefinitely. So how would that work? The question is twofold:

1. How much material would need to be flying around that it couldn't be mapped and avoided with modern imaging and radar techniques?
2. How big would the fragments need to be for modern space stations, capsules, and satellites to be at risk at standard orbital velocities?

For those unfamiliar with the parlance area denial is a strategy used in ground combat in which a section of the battlefield, or even of a whole country, is rendered inaccessible by virtue of containing a high density of "passive" weaponry (i.e. weapons that don't require an operator). Weapons used can be as simple as stakes in the ground or as complex as motion trigger machine guns but have traditionally been things that soldiers or vehicles "trip" by either standing on them or driving over them. The example case for orbital denial would use the same principle of a dumb object you have to run into but in this case it's debris and the kinetic energies involved are much greater.

Answer 1

Ah, Kessler syndrome. Always a favorite when it comes to orbital disasters. I wrote about a similar situation, where it becomes impossible for humans to safely reach Earth orbit, based on data from space agencies. I used the equation (Eq. 2): $$N=vAn\Delta t$$ where $N$ is the number of collisions over a period of time, $A$ is the area of the spaceship or space station, $v$ is the relative velocity of the ship and the piece of debris, and $\Delta t$ is the period of time over which the encounters occur. Let's say $A=500\text{ m}^2$ (a large-ish ship/small-ish space station), $\Delta t=5600\text{ s}$ (the approximate period of the ISS), and $v=15\text{ km/s}$, twice the orbital speed of the ISS (remember, this is the relative velocity, so each object could be moving at that speed but in opposite directions).

If we set $N=1$ (one collision per orbit), we find a number density of $$n=\frac{N}{vA\Delta t}=\frac{1}{15000\cdot500\cdot5600}=2.38 \times10^{-11}\text{ m}^{-3}$$ Consider, though, a ring centered at the ISS's orbital radius (about $400\text{ km}$). If the ring has a radius of $10\text{ km}$ - enough to hit such a target, even with a non-circular orbit - then its volume is approximately $\sim10^{13}\text{ m}^3$. This means we'd need about 200 objects with random, uncontrollable paths in that area alone to make a collision likely. And that's just one orbital area!

That said, you'd need the number to be even higher, for several reasons:

• 200 objects is certainly enough to track. The US alone tracks almost 18,000 pieces; 200 would be easy.
• The objects would have to be big to cause any damage to a properly protected spacecraft - and the bigger the object, the easier it is to monitor it.
• Even if you cover this toroidal region, any ship can simply choose a different orbit (within reason). So maybe you can shut down a couple of orbital lanes, but there are plenty more available.

So, orbital denial isn't easy to manufacture in such a way that it could affect an entire planet. It's quite hard.

Answer 2

A single station blowing up would create a big headache for those planning orbital insertions but there isn't enough material in the ISS or anything reasonable to build and launch to cover more than a small area.

For one thing, you would have to shred it into confetti to cover a wide area. Good luck with that. In most explosions, most of the material would be in rather large pieces since an explosion is an expanding gas or plasma in a container that prevents it from getting out. The container generally fails along its weak points and the gas or plasma spends most of its force getting out through that opening. The only way to get confetti is for the explosion to be big enough and spread through the container evenly enough that it, essentially, breaks out of the container everywhere.

Having said that, orbital denial is easy. Have 3 or more satellites (for coverage) that launch missiles (or guided crow bars) down at anything trying to launch from the earth. It takes a lot of fuel to launch and most launching rockets tend to be big and slow. They are heavy enough that the will likely not be maneuverable enough to do much dodging. It would be like shooting clay pigeons. A launch would be like yelling "pull".

This is why there are so many treaties about limiting weapons in space. It isn't because that they are afraid that we would kill each other there. It is because they can be used to deny all access to space.

Answer 3

I think it matters greatly if this is an accident, an improvised attack or a deliberate attack. In a worst case scenario of a deliberate attack it should be easily possible. For instance:

Using a large launch vehicle (such as a Saturn V for arguments sake), launch the payload into earth orbit and send it to the moon. Loop around the moon and return in a figure of eight entering a retrograde orbit around earth at the required altitude. At this point release the payload via a small explosive charge. The payload being 40 tons of 1g steel ball bearings. 40 million of those in and around the intended orbit should effectively destroy anything in that orbit.

I doubt that such small objects could easily be tracked and I’m sure that one impact would be sufficient to cause catastrophic damage due to the energy involved – especially in a retrograde orbit. If the a space shuttle widow was damaged by the impact of a paint flake, a 1g ball bearing should be big enough and have sufficient energy to make a good sized hole in anything likely to be in orbit.

Answer 4

It's surprisingly easy. Just throw sand. Kinetic energy being proportional to $v^2$ comes in surprisingly handy for this sort of thing.

Let's say we want 1m^2 coverage per ISS-like orbit. (That is, one collision on average per orbit per 1mx1m area.) Using the same equation as @ShadoCat, i.e.

$$N=vAn\Delta t$$

$\Delta t=5600\text{ s}$ (period of the ISS), and $v=15\text{ km/s}$ (relative orbital speed, considering that the orbits are retrograde to each other).

So then, the density is

$$n=\frac{N}{vA\Delta t}=\frac{1}{15 \frac{km}{s}\cdot1 m^2\cdot5600s}=1.19 \times10^{-8}\text{ m}^{-3}$$

If we consider the volume of a spherical shell around the Earth from, say, 350 - 450 km, this works out to a volume of roughly $2\cdot10^{17}m^3$. So we'd need $~2.4\cdot10^{9}$ particles.

This sounds like a lot, but is surprisingly small. A small grain of sand might be 10mg - in which case you'd need to lift ~24000kg to orbit. A single delta IV heavy could do that.

10mg particles at 15 km/s aren't immediately catastrophic by any means (~1100J, about on par with a pistol shot), but they aren't to be sneezed at either. (Antennas, altitude jets, solar panels, etc are likely to take damage relatively rapidly. Ditto spacewalks / etc.)

This likely is an 'orbital denial' only in terms of 'no-one can stick around, or have solar panels out, etc'. A single launch through wouldn't be nearly as likely to be affected.

Note that this isn't likely to clog up the skies forever. Reentry times are going to be in months, not years. (Reentry from 450km is a lot longer than from 350, note.) (Debris caused by said cloud, however...)

(In practice, you'd do this with a number of smaller launches. It takes a lot to change orbital planes).

(I'm also glossing over a number of things here - the orbital velocity here isn't actually double on average, (although it's biased that way) etc. But this is largely counteracted by my calculations here assuming that you need to cover the globe, whereas in practice you only need to cover much less.)

Quick calculation.

Inspiration.

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Now, as to if you could do this with a controlled detonation of a single space station? No. The orbital mechanics don't allow it. It'd have to be retrograde to start, and even then the spread wouldn't be sufficient. Although if you could... the ISS is somewhere around half a million kg, or ~500cm^2 coverage all on its own.