Kablam! Satellite collision math, and a correction

By Phil Plait | February 12, 2009 10:19 am

OK, so last night I posted about satellites that collided in orbit. I mentioned that the energy created in the collision was about the same as detonating a ton of TNT. I got asked how I did that math. That’s no problem (well, a little one), but a bigger problem is that I screwed up the physics of the orbits. As commenter Marco Langbroek pointed out, the angle of impact I used was wrong. I forgot about the angle of the nodes.

Let me explain.

Both satellites were in polar orbits, more or less. One had an inclination (tilt) of about 86 degrees to the Equator — in other words, it passed 4 degrees (90 – 86 = 4) from being directly over the Earth’s poles, and the other had an inclination of 74 degrees from the Equator. I assumed that meant that the angle of approach was 12 degrees. FYI, a head on collision would be an angle of approach of 180 degrees (they are moving in opposite directions, toward each other), one catching up to the other would be 0 degrees, and a broadside "T-bone" collision is an angle of 90 degrees.

So where did I screw this up? The inclination is not the only important angle. What’s also important is what’s called the node of an orbit, or the angle around the Earth. Here’s an illustration:

Earth with two possible satellite orbits

The yellow and red tracks represent two polar satellite orbits. You can see that a satellite in either orbit will pass very close over the Earth’s pole (the south pole is seen here), so the inclinations of both orbits is high, near 90 degrees. But you can see they are rotated with respect to each other, in this case by about 60 degrees. That means that where they intersect, over the Earth’s poles, the angle between them is about 60 degrees.

That has a huge impact (har har) on the collision speed. If they had the same node and the same inclination, the collision speed would be zero; they’d be on the same orbit. But if the nodes are rotated by 90 degrees, the collision would be a broadside, one slamming directly into the side of the other.

That’s what happened with the Cosmos and Iridium satellites. The inclination difference was about 12 degrees, but the node angle difference was about 83 degrees (according to Mr. Langbroek, whose word I’ll accept here since he’s an amateur satellite tracker). So the impact angle was almost a total broadside.

How does that affect the energy of the impact? Well, it’s possible to get a pretty rough idea. What follows is basically a back-of-the-envelope calculation, meant to be pretty loose. BotE calculations aren’t supposed to be truly accurate; they’re meant to give you an idea of the resulting number. So I wouldn’t be surprised if the true conclusion I reach here as far as explosive yield is concerned is off by a factor of 2 or more, but the thing is I don’t care. We’re just trying to grasp the magnitude of the forces involved, not their exact measurements.

The collision energy depends on the relative velocities of the satellites. Imagine two cars approaching an intersection. One car is in the middle when the other slams into it at a 90 degree angle. The only important number here is the velocity of the impacting car; the velocity of the other one doesn’t matter. It could be sitting there in the intersection, or moving at 100 kph; the velocity of the collision really only depends on how fast the other car was moving when it hit.

In reality, with satellites, it’s more complicated. The actual three dimensional trigonometry of the event is a little fierce, but it turns out that an impact at 83 degrees is mathematically very close to a simple 90 degree collision (the difference in angles results in just a small percentage difference in velocity). In other words, a car hitting you at an 83 degree angle has almost the same velocity as if it were hitting you exactly broadside.

Assuming the satellites were both moving at the same speed, the impact velocity then is simply the velocity of one of the satellites, or about 8 kilometers per second.

The energy of impact depends on the mass of the satellites, too. The Iridium satellite was about 700 kg, and the Cosmos was probably about the same. I couldn’t find a good figure for Cosmos, just estimates… but it’s close enough. Remember, I’m trying to be very rough here; I don’t care if I’m off in my numbers by a factor or two or not; I just want an approximation.

The kinetic energy of an object is equal to 1/2 x its mass x its velocity2. Again, because I’m being really rough here, it doesn’t matter which satellite hits the other. We can assume they have about the same mass, so the kinetic energy (in ergs, which I’ll explain in a second) is

KE = 0.5 x 700,000 grams x (800,000 cm/sec)2 = 2 x 1017 ergs

An erg is a small unit of energy, but 200,000,000,000,000,000 is a lot of them. Blowing up a ton of TNT releases about 4 x 1016 ergs, so this collision was roughly the equivalent of lobbing 5 tons of TNT at the satellites.

Boom. Yikes.

[FWIW, my mistake in the previous post was assuming the collision angle was 12 degrees, and the velocity of collision depends on the sine of the angle between the objects. Sin(12) = .20, so my number was 1/5 as big as it should be.]

That’s why there’s a large cloud of expanding debris; each piece of shattered satellite carried away a piece of that violent energy release. The energy of collision changed the orbits of all those shards, so they are now orbiting the Earth on new paths that take them higher or lower over the surface, right into the traffic lanes of other satellites.

So eventually, some may once again find an object in their way. And because of the high velocity, the kinetic energy of impact even from a low-mass piece can be pretty fierce. A rifle bullet does a huge amount of damage when it hits something, and it has a mass of about 10 grams and moves at a paltry 1 km/sec. Now think of the damage inflicted by a small satellite chunk that masses about 1 kg (100 times as much as the bullet) and moving 8 times faster… the energy of impact is 6400 times that of the bullet. Imagine getting hit by six thousand rifle bullets, and you start to get an idea of why satellite collisions are not just catastrophic for the two birds involved, but also a danger to other objects in orbit as well.


Comments (88)

  1. “Kablam!” is the sound my head makes when trying to understand all this math. Luckily, I don’t need much math in my work (writing tax software — eek!).

  2. Ah, a new comment on the previous post gives a collision angle that is actually about 100 degrees or so. That means, if it’s accurate, the relative velocity is a little over 9 km/sec, so the energy would be even higher, about 3 x 1017 ergs, or 7 tons of TNT. Ouch.

  3. Off-topic but a court in Washington decided Vaccines don’t cause autism click on name for link.

  4. Erin

    That’s a different kind of Big Bang… O_o

    Oh, Phil, they made a vaccine ruling this morning in a special court:

    Another point for science! :D

  5. Erin

    @Jamie Mueller

    LOL, yay for dual informants!

  6. DrFlimmer

    UUAAAHHHH, ergs…. argh. Phil, couldn’t you use Joule in this case? All your numbers are in SI, so why did you make that transfortmation? (yeah, it’s not too difficult in this case, but please, come on…)

    I just couldn’t resist! But indeed, that is impressive and dangerous. Hopefully they don’t do any harm to the next shuttle missions (especially that one that goes to Hubble…) or the ISS. And even harm to other satallites could be a problem, depending on which satallites get hurt.

  7. B-but…you’re a scientist! Scientists don’t correct their mistakes! They are part of the vast global conspiracy to protect their flawed “theories”…! Obviously, you are doing this as part of a cover-up, to hide the real nature of those satellites, which as we all know were spreading vapor trails in the plasma currents coursing up from the hollow pits at the poles (The satellites were in polar orbits, don’t you know…you know what that means!) in order to hide the trilateral commission’s secret tunnels down into the center of the earth where the Lemurians are plotting with Putin to take over control of the universe from the good people of Uranus, as written in verse 666 of the book of whozeewhatsis the elder (of zion).

    Now, why don’t you fess up to the real reason you haven’t disclosed your secret identity?

    I rest my case.

  8. Tomas

    @DrFlimmer unfortunatelly, erg is a unit quite frequently used in astronomy (texts, books) despite almost everything being done in SI.

    But you’re right and I would like to IMPLORE Phil, please, try to use SI units more often, if for nothing else, than just so more americans get used to seeing them in use (which can be only beneficial).

  9. For a biologist, it is surprising that I actually appreciate the mathematical expoundings that Phil gave here.

    What’s the chance all this debris and potential subsequent debris will leave orbit and fall onto unsuspecting ecosystems?

    Things colliding, despite the loss to the corporations, is still pretty cool, and the fact that it was such a huge chance it couldn’t happen, makes it a little exciting. Anyone for the lottery today?

  10. I’m sad there would not have been roaring flames as it happened. Silly vacuums, always ruining the fun.

  11. Ed

    It seems to me that timing is going to make a big difference too. The idea that this collision was the equivalent of a T-bone accident (that is that only the velocity of the impacting body is important) is based on the impact occurring after the front part of the impacted satellite has passed. If, instead, the impact occurred at the front corners of the two satellites; wouldn’t we have to take into account the energy from the velocities of both objects?

  12. Nigel

    Your calculations assume an *optimal* collision, with the centre of mass of each object passing through the same point at the same time, whereas statistically it was most likely a glancing contact, with less energy release. Tracking of the debris will give some idea of what really happened.

  13. Tom Woolf


    I don’t believe the debris will become a danger to the ecosystem or individual members of it due to terminal velocities. Granted, I have no idea if there are dangerous materials on the communications satellites, but the air should slow or burn up anything that makes it back down to earth’s atmosphere.

    As to vacuums ruining all the fun… You’ve been talking to my cats, haven’t you?

  14. mike

    Phil, your correction has a mistake, though it’s probably OK for the BotE.

    One evaluates collisions in the center of mass. The orbits are about the same radius, so the speed is about the same. This means you’re high by a factor of 2 in energy, assuming the two satellites are roughly equal in mass; the center of mass is not stationary WRT the earth for a “broadside.”

    Your analysis is correct in the limit that one satellite is much bigger than the other.

    Sciencegoddess, it’s a certainty that some parts of the satellites will fall to earth and probably burn up in the atmosphere. But I wouldn’t expect it to affect ecosystems at all. An intact satellite — even Skylab and Columbia (both of which were much bigger) — doesn’t make a significant impact (so to speak); a satellite blown into little tiny bits will burn up much more easily due to the vastly increased surface area.

    It’s not like a several-mile-wide asteroid. That’s a whole ‘nother ball of wax^H^H^H iron/rock/carbon/ice/whatever. Despite the fact that this satellite was named Iridium, it won’t be wiping out dinosaurs.

  15. Phil, for clarity: the 83.5 degree angle comes from a calculation with Rob Matson’s ALLCOLA software, which is software specifically to find these kind of close approaches/collision scenario’s.

    The ALLCOLA output:

    TLE file #1: botsing1.tle
    TLE file #2: botsing2.tle

    Target satellite: #24946
    Name: IRIDIUM 33
    Perigee: 776.05 km Apogee: 779.28 km
    Velocity at perigee: 7.4651 km/sec
    Satellites in range: 1

    Date UTC Time Norad Name Range/OrbSep NdAng EphAge

    2/10/2009 16:55:59.81 22675 COSMOS 2251 0.8/ 0.1 83.5 -0.05

  16. Fergus Gallagher

    You need to do the calculation in the Zero Momentum (Centre of Momentum) frame. One way to look at this is that, since momentum is always conserved, a certain amount of energy is needed to keep the bits after collision moving.

    Two cases:

    1) two identical balls, travelling towards each other in straight line, at the same speed v, collide. The total momentum before and after the collision is zero and so all the energy is available for an explosion, or whatever. E=2*1/2mv^2 = mv^2

    2) A stationary ball is hit my an identical ball (travelling at 2v – so the relative speed is the same as above). After the collision, the centre of mass of the remnants has to move at (total momentum)/(total mass) = m2v/2m = v. So energy of CM frame is 1/2(2m)v^2 = mv^2. But the total initial energy is 1/2(m)(2v)^2 = 2mv^2. So the energy for the explosion is the difference, mv^2. The same! But this is only HALF the initial energy.

    This is (one reason) why the LHC has has 2 counter-rotating beams – the maximum energy can go into the collision. SLAC (linear accelerator) can get more collisions, as one side of the collision is a solid block with a much better chance of collisions, but lots of the energy is “wasted” in conserving momentum. (Another good reason is that the centre of collisions in LHC are stationary inside, e.g., ATLAS – at SLAC they shoot off at near the speed of light)

  17. Rob

    Please, please, please use SI units. Ergs have been deprecated since 1948! I don’t think you’re that old.

  18. Copernic

    Speaking of bullets, the likelyhood of these two hitting each other is somewhere similar to that of two hunter’s bullets hitting each other out in the woods.

  19. Mike, I have a cat where all you have to do is make a deep whirring noise with your mouth and he’s gone!

    I figured most of the debris would burn up, but one never knows, especially with toxic stuff.

    As for the SI issue: At the lab, I instinctively do everything SI, but when I get home, I couldn’t tell you if that was 100mls of milk…I would automatically tell you how many ounces it was. It is so location related and I think if I thought in mls at home, I would assume whatever liquid it was would probably be toxic or less than palatable! lol!

    I remember being in elementary school in the 70s when they had that BIG push for the metric system and yet it still didn’t catch on.

  20. Oops, I meant Tom Woolf, not Mike, for the cats.

  21. Argon

    Five tons of TNT or one pound of TNT: Hollywood would still depict the collision with a loud bang in the soundtrack.

  22. mike


    Believe it or not, I have a (rather nice, actually) E&M textbook written by Purcell that is entirely in cgs units. Yes, I had to calculate electric potential in statvolts and charges in esu. It’s not nearly that old; the copyright date is 1985 (1st ed is probably late-60s like the rest of the Berkeley Physics Series).

    It turns out electrostatics “looks” nicer that way (none of those pesky 4*pi*e0 factors), but IMO it’s not worth it.

  23. hale_bopp

    I saw your post and the image they had of the orbits on spaceweather.com and thought, hmmm…something doesn’t match here.


    Just eyeballing the spaceweather diagram, 100 degrees looks about right…good enough for a back of the envelope calculation!

  24. Wayne

    Arg, CGS units! Say it isn’t so! Why must Astronomers, who arguably study the largest and most massive objects of any science, measure things in grams and centimeters??? I thought it was nuts in grad school and I still avoid them like the plague today. It frustrated me and my students no end to discover the book I adopted for Solar System Physics (Planetary Sciences, by de Pater and Lissauer) was full of CGS confusion, ESPECIALLY when you get into electromagnetism. I can sort-of see why theorists like them (fewer 4pi’s running around etc.) but why-o-why do astronomers/astrophysicists think they’re so great?

    SI all the way man, it’s the only decent thing to do.

  25. CaptXpendable

    Sorry for another off-topic post, but I just saw a new video on YouTube promotining 365 days of astronomy, and Phils book has a cameo.

  26. Mike, you’re right about the procedure, but in this case, it doesn’t matter. Doing it your way yields the same answer as Phil’s method.

    From a center-of-mass perspective, the satellites are approaching each other at

    800,000 cm/s * sin(45) = 560,000 cm/s

    That’s the velocity of each satellite toward their center of mass, so the total kinetic energy for both satellites would be

    2 * (.5 * 700,000 g * (560,000 cm/s)^2)

    Which is equal to 2*10^17 ergs, the same as Phil’s answer. (In fact, it’s exactly the same, because sin(45) is exactly equal to the square root of 0.5.)

    Phil’s method works, however, only because the satellites in question are (estimated to be) of equal mass. If the satellites hadn’t been of equal mass, Phil’s method of determing the kinetic energy of one satellite relative to another wouldn’t have worked. But with the colliding objects being of equal mass, Phil’s cheat to make the math a little easier is not only accurate, but extremely common.

  27. Sili

    I too have absolutely no feel for the cgs units. Joule (or even calories), but just how big is an erg?

  28. Brian B.

    Did you leave the apogee and perigee of each satellite out of the calculation because at the point of impact, those figures are identical?

    I’m trying to get my head around this. At first I thought it was an omission on your part, but after thinking about it, it makes sense that the actual ground speeds of both satellites would be nearly identical at the point of impact because their altitudes would be identical.

    It is my understanding that actual ground speed changes as the orbital altitude changes, but for all orbits, the speeds are identical at identical altitudes. I’m a PR guy, not a scientist, but I have just enough understanding of orbital mechanics to be dangerous.

    Thanks in advance for clarifying.

  29. Earl: right. I drewa bunch of diagrams, trying to remember kinetics from like 20 years ago, and realized it didn’t really matter since the satellites are moving at roughly the same speed and have the same mass. The velocity is the big player, since it gets squared, so the angle is actually important.

  30. And ergs, Joules, who cares? A unit is a unit, and once you compare it to something familiar (TNT) then it doesn’t matter. Astronomers tend to use CGS, and I have the TNT scale factor memorized (1 megaton of TNT = 4 x 1022 ergs). I could start with Joules, but it doesn’t matter: 90%+ of my readers don’t know a Joule either. :)

  31. Joe Meils

    This is EXACTLY why I never try to write “hard” science fiction. Too many math and engineering geeks out there with their T.I. calculators at the ready, just waiting for a chance to pounce…

  32. Surprised none of the commenters on this particular thread have so far touched on how big of a disaster this is for Low Earth Orbit satellites, and polar orbits in particular. When the first article was published it was not clear that these satellites were in a polar orbit. Though people familiar with Iridium would probably have known that.

    If you look at Phils picture above you’ll note one key feature: polar orbits all congregate, surprise, at the poles. What is left of the two satellites is now presumably circling the earth in orbits roughly the same as those followed by the original . Except that rather then being in a tight, easy to miss, solid object the debris is in a nice wide cloud much like a shotgun blast. So every satellite in polar orbit now stands a serious risk running of into a cloud, even the smallest components of which are more than capable of disabling it, every time it passes over either of the poles.

    The initial press-release said that 600 separate pieces were being tracked. But that was for pieces 10cm or larger. How many <10cm pieces were created by the collision? For reference NASA's Orbital Debris Program Office estimates that the destruction of the Chinese FENGYUN 1C generated more than 150,000 pieces of debris larger than 1 cm. Any reason to think that the most recent incident would generate any less than that?

    In addition as the debris cloud spreads, and descends due to orbital decay, those pieces of ex-satellite are going to be running almost straight across the equatorial type orbits used by the ISS and the shuttle amongst others.

  33. Sundance

    Phil, the whole description about the impact being side-on, and cars colliding in intersections is just plain wrong. Sorry. The fact that both objects are moving is important, no matter what the angle they hit at is. Here’s how it should have gone…

    The energy of the impact is determined by the _difference_ betwen the velocities of the two satellites (or cars, or whatever). We want to ask “how fast does satellite B see satellite A approaching it?”. We can answer this by subtracting the vB, the velocity of satellite B, from Va, the velocity of satellite A. So if vI is the velocity of impact, then vI = vA – vB. Since velocity is a vector quantity, it has both magnitude and direction, so when we figure out the difference of their velocities, we need to take direction into account. If they meet at an angle of 83 degrees (let’s call it 90 degrees to keep things simple! This is a BotE calculation after all), both travelling at 8km/sec, we can figure out their impact velocity by constructing a vector diagram. Draw two arrows of equal length, representing the velocities of the satellites, with the “tip” of the first arrow touching the “tail” of the second arrow. They form a right-angled triangle, with the total length from the tail of the first vector to the tip of the second being the combined velocity of impact. This will be longer than either of the satellite velocity vectors by a factor of the square root of two, approximately 1.414. So the impact velocity should have a magnitude of 1.414 x 8 km/sec, from the frame-of-reference of satellite B.

    Notice that this construction of subtracting one vector (arrow) from another works no matter what the impact angle is. If the impact angle was zero degrees, we’d draw an arrow, then draw another one, starting at the tip of the first arrow, running back in the opposite direction (because we’re subtracting the velocities, if we were adding velocities we’d have both arrows pointing in the same direction). That’s part of the beauty of physics, you use the same ideas and techniques to solve a huge range of problems. It’s engineers who use a different formula in every situation!

    Note also that since there are only two impacting bodies involved, the problem reduces to a two-dimensional one. We need to know the 3D components of their velocities to figure out what angle they hit at, but once we know that we can solve the problem in a reference frame where the problem becomes two-dimensional, so the comment about “three dimensional trigonometry” is a bit misleading. Since two vectors define a triangle, which must lie in a plane (i.e. it’s 2D, not 3D) we can figure this out with 2-component vector maths. Let vA = (8,0). That is, satellite A is moving “east” at 8 km/sec, and “north” at zero km/sec (i.e. moving due east). Let vB = (0,8), so satelite B is moving due north. Then vI = vA-vB = (0,8) – (8,0) = (-8,8). The magnitude of vI is given by Pythagoras’ theorem, vI = sqrt{(-8)^2 + 8^2} = sqrt{64 + 64} = sqrt{128} = 1.414 x 8 km/sec

    Now, let’s assume both satellites are the same mass. What’s the energy in the centre-of-mass reference frame? Why does that matter, you ask? Well, think about a meteorite hitting the Earth at 11km/sec. As far as the Earth is concerned, the meteorite is moving at 11km/sec, so it’s energy is 0.5 x (mass of meteorite) x (11000 metres/sec)^2. But there’s a discrepancy if we view things from the meteorite’s point of view, because the Earth is moving towards it at 11km/sec, but is about a gazillion times more massive, so the meteorite should see the Earth approaching it with kinetic energy of 0.5 x (mass of meteorite x gazillion) x (11000 metres/sec)^2. What we need to do is calculate the kinetic energy of both Earth and meteorite in the centre-of-mass (CoM) frame and add them together. The Earth is moving a gazillion times slower than the meteorite in the CoM frame, so the energy it contributes to the collision is 0.5 x (mass of meteorite x gazillion) x (11000/gazillion metres/sec)^2 = (energy of meteorite)/gazillion (since the gazillions cancel out :) ) and it looks like all the impact energy comes from the meteorite. Phew!

    So back to the satellites. Their impact speed is 1.414 x 8000 m/s, so if they have the same mass, in the CoM frame they’re both moving together at 0.5 x 1.414 x 8000 m/s each. Hence one satellite has a kinetic energy in the CoM frame of (0.5) x 700 kg x (0.5 x 1.414 x 8000 m/s)^2 = 1.1 x 10^10 Joules. Therefore the total energy of the impact should be twice that, or 2.2 x 10^10 J.

    Now 1 ton of TNT is equivalent to about 4.2 x 10^9 Joules, so the impact energy is about 2.2 / 0.42 = 5.2 tons of TNT! So, Phil’s answer was basically correct, even though the calculation used to obtain it was questionable.

  34. Phil Plait, FYI, 1 gram TNT = 4184 J* (exactly).

    *Source: Wikipedia — TNT equivalent. (Click on my name for the link.)

  35. Fergus Gallagher

    Phil@12:51. If the orbits are circular, then speeds must be [so near as makes no difference] EXACTLY the same: v^2 = GM/r. The two radii must be the same or they’d never collide!!

  36. mike

    Ehh, you’re right Phil. Cerebral flatulence. It’s been a long time since I’ve done collisions as well.

    It’s true units are arbitrary, but these days more folks have intuition with SI units. And many of the handy-dandy conversion tables relate to SI units. And as NASA guys, we both know what happens when people mix up their units….having said that, I’m doing most of my calculations in knots, nautical miles, degrees and sidereal hours. Take THAT SI!

    Astronomers don’t use cgs units. They use all kinds of wacky units. Like AU, Mpc/h (well, not with the h much anymore), Jansky, milliarcseconds, seconds per year (really! it’s a unit of “proper motion”), sidereal days (which aren’t quite the same as the days other folks use) etc.

  37. ShaneC

    Sigh – awright I wasn’t going to say anything bout a few posters have mentioned some of it, specifically, elastic vs. inelastic collisions, conservation of momentum, and “energy release”.

    First: was there, in fact, a “release” of energy here? Answer: probably not; the only energy lost in the collision would have gone to deformations and separations of physical components, and was probably fairly small. The vector sum of the momenta would have been passed to the two bodies and their pieces. While E=0.5 x em vee squared is a nice big number it is not really as significant as momentum (the first derivative). If we could sum over the resulting masses and velocities we could see exactly how much kinetic energy was lost in the collision, indicating how much was “released”.

    Second: the assumption of intersection of centers of mass. As anyone who has fooled with a billiard table knows, the ways that two bodies (nearly fully elastic in the case of billiard balls) can collide in 2D is nearly infinite. If we assume two irregular shapes with lots of doodads hanging off them, it gets really complicated. Chance favors a collision that is not CoM to CoM, with the resulting collision dynamics getting very difficult indeed, so it’s likely that we have the result of two main clouds of debris in nearly the same orbits as the original satellites plus another cloud whose CoM would be roughly in an orbit corresponding to the satellites’ vector sum, plus pieces radiating out all over the place, some of them going up, some down, some back, etc.

    Finally, of course, there’s the “kablam”. Huh? Did you hear something? Sorry, I didn’t catch that. Oh, no wonder, my radio was off.

  38. Sundance, I think you’ve got it right; it is the difference in the two velocities that matter. I’ve computed the impact velocity (I obtained the two spacecraft orbits, and hence pos. and velocities at the time of impact) and obtained ~11.6 km/s.

  39. TaoMacGuy

    Now why am I thinking of the “Satellites in Orbit” version of the “Nuclear Fission” game? What would the “critical mass” of satellites in orbit need to be to create a self-sustaining “reaction?”

    Let’s see… Collision #1 releases 600 smaller pieces (e.g. satellite “neutrons”), some of which impact other satellites, releasing yet more satellite neutrons…

    I have this Simpson-esque image of satellite fission is *super* slow motion.


  40. MadScientist

    Just to complicate things, were the centers of mass of each satellite headed for the same point, or did one satellite bump the other like a billiard ball grazing another billiard ball?

    OK, enough speculation – both satellites are toast.

  41. Sundance

    *Sigh* I didn’t really want to get this nerdy, but since people have been wondering about glancing vs. CoM-to-CoM collisions, and elastic vs inelastic collisions,

    First, assume a spherical satellite… :)

  42. TDoc

    Agree with NoAstronomer on “Surprised none of the commenters on this particular thread have so far touched on how big of a disaster this is for Low Earth Orbit satellites.”

    Iridium satellites have a lower orbit and this accident should’ve shed light to other satellites that share the same path.

    On a different note, the story is grabbing attention from the media in several aspects… http://www.newsy.com/videos/russia_and_u_s_collide_in_space

  43. Tomas

    @Phil, I don’t think it’s about what you readers know; if you’re using speeds in m s-1 and masses in kg, then using Joules is straightforward as J = kg m2 s-2, and once you get the factor of 10-7 in, it just makes more sense to go with J rather than ergs (I regret you underestimate your reader base) as most people are more familiar with J (and SI) than ergs, or Å that matter (which I always hated and astronomers keep insisting on using them – I mean, c’mon we have µm). And, i gotta say it again, using non-SI units was really the only thing that bothered me about your book (if you wanna keep it simple for US readers, provide imperial values in brackets AFTER values in SI).

  44. Ben

    If I understand this correctly (and I’m far from being a rocket scientist), all these calculations seem to assume a collision where the center of masses of the two satellites are in a 2D plane relative to their trajectories at that moment. But was this really the case? Is there anything known about this yet? Having so many debris particles (600+?) seems to point at this, but do we know for sure? Chances are what they are, wouldn’t it be even more amazing if the satellites were at exactly the same height? (Say within 1m or so.) Anybody know?

  45. Brian H

    I just dropped in to say, as a theorist, units are for the weak, hbar=1, c=1, k_B=1! Do everything in GeV and powers there of!

  46. Brian H

    @Ben: Any two vectors define a plane, and since they did collide, they were moving in the plane defined by their velocities at the instant of collision. But yeah, lucky shot with them being at the same altitude at the same time at the same place.

  47. Brian H

    Oops, correction: any two vectors define a family of planes, and the point of collision picks one out uniquely. It’s the plane containing the point of collision, perpendicular to the cross product of the velocities at the moment of collision (or, more properly, just a tiiiiiiny bit before)

  48. Sundance

    @Brian H

    Yes, for the benefit of anyone who doesn’t know what a cross product is, the family of planes specified by the velocity vectors are planes that are parallel to each other – like the pages in a closed book. The point of impact picks out one, like selecting a single page.

    And why stop at “units are for the weak”? A real theorist knows _numbers_ are for the weak! :) Once you’ve solved the relevant equations for the variable you’re interested in, who cares what numbers you plug into it?

  49. Sundance, I think you and I said essentially the same thing, but you determined the velocity of one satellite relative to another, while I determined the velocity of each satellite relative to their combined center of mass. (2 * sin(45) = sqrt(2) ) Even so, Phil’s BotE calculation is reliable since it’s been simplified by the assumption that the satellites are of equal mass (and spherical :) )

    I appreciate the comments about the potential fallout from the debris cloud.

  50. Lawrence

    Question for those who know: Is this a significant event in the general proliferation of space junk? What signal traffic is carried by Irridium satttelites? Was there an interruption of service to the users of that signal? Does the debris field have a deleterious effect on signal traffic to sattelites in higher orbit?

  51. cm? g? erg? it’s m kg and J=Nm
    As others said before please use SI unites until you have a good reasons not to use them. (For example eV for energies at atomic scale)

    But for satelites kg are more useful than g and m are more useful than cm so the resulting energy should be in J.
    the difference between them is just a shift of the decimal point so it shount not make any difficults to converte tons of TNT to ergs (as you know the value) and then to J.
    And cgs units are gnerally shitty because they have different incarnations with respect to electrical phonmena.

    And do not underestimate your readership I think more then 90% know what a Joule is. I have a good feeling what a J is (approx the 10th of the energy to lift one kg one m up) but not what an erg or a ton of TNT is. (I mean how many of us have witnessed an explosion in the range of tons of TNT?). Do not forget that europeans grow up with the m kg s system.

    kind regards !

  52. It seems to me a couple of things haven’t gotten much exposure in these comments yet.
    It all focuses on the paths that the debris cloud takes. (1) The debris will probably NOT be limited to the plane defined by the vectors of the velocities and the impact point, and in fact could have MANY possible orbital velocities, some quite a bit more elliptical than the parent objects. (2) Tracking as many pieces of the collision as possible will give a good estimate of the elasticity (or lack thereof) of the collision, comparing the momentum of the pieces with their energies. Postulating that the small pieces will be in statistical clouds going in similar directions to the bigger pieces, one could (using the right orbital simulations, something for which i have no source) project into the future which satellites will be at risk over time.
    What an interesting idea, “satellite fission”. Perhaps we are growing the end of the exploration of space? How about sending up garbage collectors to clean up the bits? I see a novel in the works….

  53. Sundance

    Earl, yes we did, I was just being more pedagogical (or long-winded, depending on yor reference frame :) ) because I felt Phil’s discussion was misleading to anyone who didn’t know how to do vector addition. The statement that “[the car that gets hit] could be sitting there in the intersection, or moving at 100 kph; the velocity of the collision really only depends on how fast the other car was moving when it hit” is just plain wrong. No offence Phil, the blog is great, but we all make mistakes from time to time. The velocities relative to the CoM are what matters for calculating the impact energy, and that certainly does depend on how both cars (or satellites) are moving. So Earl, you and I did the same thing, I just went to greater lengths to explain where the factor of sin 45 comes from ;-)

  54. @Lawrence

    It’s really to early to tell how bad the debris from this event is. But I have to think it’s at least on a par with the Chinese ASAT test from last year.

    The Iridium satellites are part of a network of 66, now 65, satellites that provide satellite telephone service. Some service problems were reported in the media, but the service provider has an in-orbit spare that they are moving into position. AFAIK the debris field wouldn’t have any significant impact on the signals.


    I would expect the majority of the debris field to form two rings around the earth, largely following the original orbits of the two satellites. That is pretty much what the debris from the Chinese ASAT test did. That satellite was in a fairly similar orbit, polar and at about the same altitude.

  55. MadScientist


    Hollywood will not only put in a big ‘bang!’, but there will be a toroidal (or in some instances spherical) propagation of illuminated particles at time of impact and natural laws will be violated with bits scattering in every direction rather than primarily in the directions travelled by the two satellites.


    Chunks of rock/iron fall to earth every day; except for the occasional very bright flash, hardly anyone notices these things these days. Odds that pieces would hit any human or animal are fairly remote (but certainly not impossible). The most hazardous stuff in large amounts on spacecraft tend to be propellants (mercury, hydrazine, etc), beryllium parts, and in some instances only, the radioactive material in a Radioisotope Thermal Generator (RTG). The RTGs are incredibly tough; they’re meant to survive reentry and also survive if something goes wrong with the launch and the rocket has to be destroyed. So the chance of radioactive material being scattered are also incredibly low – then again I guess an RTG could be smashed open if it hits something solid and traveling at a high speed. Anyone want to find out if either the Cosmos or the Iridium has a set of RTGs? So as far as hazards go, the propellants will be scattered over too large an area to do any harm, beryllium parts will probably come down somewhat intact and not spread toxic stuff everywhere, and the RTGs (if any) – well, some small chance of problems there – but nothing like the hazards of a nuke set off in the atmosphere.

  56. Does the analysis assume that all the kinetic energy is released at the impact ? In my opinion, the kinetic energy of the fragments should be taken into consideration.

  57. Brian Fane

    “Astronomers tend to use CGS”? So people that look at distances of millions, or billions of light years, masses of millions of suns, and times spanning billions of years work in centimeter-gram-seconds? You guys trying to really inflate your numbers?

  58. Nigel Depledge

    Dr Flimmer said:

    UUAAAHHHH, ergs…. argh. Phil, couldn’t you use Joule in this case? All your numbers are in SI, so why did you make that transfortmation? (yeah, it’s not too difficult in this case, but please, come on…)

    Hear, hear.

    It’s high time you joined the SI, Phil.

    For 1/2 mv2, where m = 700 (the kg is the SI unit of mass) and v = 8000 m/s, I get a KE of 2.24 x 1010 J

  59. Nigel Depledge

    The BA said:

    And ergs, Joules, who cares? A unit is a unit, and once you compare it to something familiar (TNT) then it doesn’t matter. Astronomers tend to use CGS, and I have the TNT scale factor memorized (1 megaton of TNT = 4 x 1022 ergs). I could start with Joules, but it doesn’t matter: 90%+ of my readers don’t know a Joule either.

    Because, Phil, some of your readers are from out-of-town, i.e. not the USA.

    The SI is international. Until I started reading your blog, I had never heard of an erg, except as a kind of desert. How can you measure energy in deserts? However, I am sure that all educated Americans have at least heard of the SI and thus of the Joule. And even if they haven’t heard of the SI, they should be aware of it, so you are performing a valuable educational service. Whereas the erg as a unit of energy is an anachronism.

    So there.

  60. Nigel Depledge

    Mike said:

    Astronomers don’t use cgs units. They use all kinds of wacky units. Like AU, Mpc/h (well, not with the h much anymore), Jansky, milliarcseconds, seconds per year (really! it’s a unit of “proper motion”), sidereal days (which aren’t quite the same as the days other folks use) etc.

    Not to mention the coolest unit in the whole of science:


  61. Nigel Depledge

    Tomas said:

    …or Å that matter (which I always hated and astronomers keep insisting on using them – I mean, c’mon we have µm)

    Or, indeed, nm, which are more useful for wavelengths of visible and UV light than either Å or µm (and are easier to type, too!).

  62. I’ll bet this isn’t covered in Dr. Plait’s “Death”.

    What nobody seems to be concerned about is that this collision WILL bring on the end of the world.

    All that debris will continue to knock out satellites which will then become debris that knocks out satellites.

    In a short time we will lose almost all capability to communicate. No more cell phones, no military communication, no targeting or tracking systems, etc.

    Our way of life will come to a screeching halt. Mass anarchy and chaos.

    And it’s just my luck to be really low on plastic and duct tape. :(

  63. Nigel Depledge

    Brian H said:

    I just dropped in to say, as a theorist, units are for the weak, hbar=1, c=1, k_B=1! Do everything in GeV and powers there of!

    Erm … so you quote a load of unit quantities?

    Also, can you tell me what is this week’s Unit of the Weak please?

  64. Nigel Depledge

    Brian Fane said:

    “Astronomers tend to use CGS”? So people that look at distances of millions, or billions of light years, masses of millions of suns, and times spanning billions of years work in centimeter-gram-seconds? You guys trying to really inflate your numbers?

    Obviously, they’re trying to compensate for something. Have you ever read about barnacles? ;-)

  65. gerwen

    A little more perspective on the energy of the collision:

    Rifle bullet – Springfield 30-06 – 4 x 10^3 Joules
    One of the largest naval guns ever built – 4 x 10^8 Joules
    Satellite collision 2 x 10^17 ergs, or 2 x 10^10 Joules

  66. Now would be a good time to start a satellite collision insurance company. The skies have been pre-jinxed!

    (If you believe in that sort of thing. ;) )

  67. Sundance

    Okay, to be slightly optimistic, maybe there’s a bright side to this whole “debris threatening other spacecraft” business.

    Any interstellar probe or eventual manned mission is probably going to have to deal with impacts from high-velocity dust particles, over a period of several years/decades, while on its way to another star system. So learning how to clear the debris from satellite collisions out of the way of other satellites (including the ISS) may spur us into developing some of the technology necessary to make interstellar travel a reality.

  68. Nigel Depledge

    CafeenMan said:

    In a short time we will lose almost all capability to communicate. No more cell phones, no military communication, no targeting or tracking systems, etc.

    Our way of life will come to a screeching halt. Mass anarchy and chaos.

    We’ll be allright. Most cellphones work through land-based base stations (there’s one at the core of each cell), not through satellites. Satellite phones are hideously expensive, and only worth getting if you intend to go where there are neither landlines nor cellphone base stations.

    Plus, global radio communication by shortwave radio will not be the slightest bit affected by the presence or absence of satellites, except in terms of traffic density.

  69. Gary Ansorge

    Dang. Where’s a deflector dish when you need one?

    Hah! debris impact at 8 km/sec is nothing. Try that at warp 6,,,or, for that matter, a “mere” 1/10th light speed(or 30,000 km/sec). We really need to be thinking of some kind of extensive magnetic field sweeper/deflector. Or we could just wait for global warming to cause the Earths atmosphere to “balloon” outward and decelerate the debris fields.
    ,,,and what will we do when our space tethers are in place, stationary with respect to all the debris/satellites in the equatorial plane?

    Yes, we definitely need some creative problem solving here.


    GAry 7

  70. Sarcastro

    For the folks that dig this kind of stuff, I recommend going out and downloading the fansubs for the anime Planates (go buy it too, but the commercial translation is garbage so get the fansubs as well). It’s about a crew of an orbital garbage scow; their job is to clean up stuff like this if they can’t prevent it. The physics of it are done quite well.

  71. Sundance

    @ Sarcastro

    Anyone remember the show “Quark”? :-)

  72. I vaguely remember reading about ergs in old university books my father had when I was a kid. He got his bachelor’s degree in physics in 1970. We need to adapt. Joules FTW.

  73. I heard that the debris will effect the space station. How does that work? I saw this video about the crash that had a lot of information. I think if you click my name it pops up. According to the story, these were personal communications satellites.

  74. Rob Pendragon

    Can someone explain why it is 83degrees. Sorry, but a large number of sites seem to be using 102degrees.

    I understand how (roughly), you can get from one to the other (360-(83)*2))/2=102ish

    I presume that the source of this error is the directions in which the satellites were moving. I did a quick calculation and I get the angle between the vectors of the two satellites to be around 100degrees. Where have I gone wrong?

    I based it on:
    Iridium-Inclination=86dgrees, RA 119.6degrees.
    Kosmos-Inclination=74dgrees, RA 8.74degrees.

  75. Emily

    What’s all this I keep hearing about wanting to use Jules for these complicated satellite collision calculations?

    Doesn’t everyone know that he was a science fiction author back in the late 19th century??? What would he know about orbital velocities? He was mostly interested in balloons and submarines, and long walks under ground.

    Wells had a much better handle on things flying through space, but even he probably could not have calculated the energy released by the satellite collision, even with the help of Jules.

    But, on the other hand, I completely agree that urges should not be discussed in polite company.

  76. Gerwen, you can find even more perspective on the relative joules of various explosions with my painfully accumulated Boom Table:


  77. dumb ass

    The energy of the collision is not that relevant. The number of new satellites or space junk in orbit is however. The new orbits of this space junk pose a serious potential for additional collisons. How long until we lose another satellite. The collision created aproximately 5% more space junk than previously existed. Can anyone calculate the when the next collision will occur. The time frame between future collisions should decrease at an exponential rate. How long until there are no more working satellites and only space junk?

    For “sicence goddess” There is a possibility that one or more of the current satellites in orbit carry a Plutonium or Uranium power source. Let see how that interacts with the atmosphere.

  78. Fry

    So this explains those clanking sounds that occur right after Leela takes off!

  79. tussock

    Units? Late comment?

    8.3 MWh (megawatt-hours), or about how much electric power an average western house uses in a year.

    But yes, SI please (or kilo, mega, giga, tera, or peta SI, as people know from their computer’s Bytes), unless you’re using c=1 and such to simplify relativistic calculations.

    If your readers really don’t know it, teach them. 30 GJ works just fine, and is no harder to understand than any other very large number we will never experience.

    Tons of TNT equivalence is a bit misleading even for nukes, better IMO to reference that one tonne of TNT releases 4.2 GJ for comparison (1 short ton = 3.8 GJ), or that the Hiroshima nuke was around 120 PJ, or that the dinosaur killer might have been 400 ZJ as you need larger numbers.

    Hmm. Two larger cars hitting at 100 degrees at fast highway speed would potentially release 1 MJ, so these satellites had thirty thousand times that impact energy, and were carrying nearly 100 times the momentum.

    How much energy does it take to vaporise satellites? 1600 kg of aluminium takes 21.7 MJ to melt and boil from 0K, so if there’s much sizable debris left we can say almost all of the energy (99.5%+) was not lost to the collision and they just clipped each other. That loss of energy would at most slow the resulting parts by 20 m/s at those speeds, or knock at most 0.5% off their average orbital height (1% at the south pole, down to 680 km altitude).

  80. Tim

    What I want to know is about SOCRATES, the collision detection system
    on Kelso’s website. SOCRATES = Satellite Orbital Conjunction Reports Assessing Threatening Encounters in Space. Right after the collision, I looked there for a model run using the actual orbital parameter
    and didn’t find anything.

    Has anyone run any of the several collision detection programs using the best orbital data and seen
    if this could have been detected beforehand? The links on Kelso’s site pointed to STK,
    who made all the cool animations that we saw on the news the next few days, but there was
    a conspicuous absence of any mention of their prediction software.
    If it predicted it, that’s news and if it didn’t, thats news, too



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