Exploding Stations: A Short Study

Continuing the discussion from Date of the kickstarter?:

I got tired of the rough estimates, so I went and plotted this up properly. I made a few assumptions in doing so:

  1. The planet being orbited is, for all intents and purposes, Earth. It has Earth’s mass and radius. It is an idealized, perfectly spherical Earth.

  2. The station is orbiting in a circular orbit at an altitude of 2000 km.

  3. There is no atmospheric drag.

  4. The explosion is radially symmetric, and, for our purposes, constrained to the plane of its orbit.

  5. The large chunks of the station are launched away from the explosion with a Gaussian (normal) velocity distribution. I had to make up some numbers here for the mean velocity and standard deviation, so I chose a mean of 981 m/s and a standard deviation of 327 m/s (981/3). 981 was selected from a list of velocities on Wikipedia because A) it was faster than the muzzle velocity of an M16 (so that it’s, literally, faster than a speeding bullet), and B) listed as the velocity of an actual aircraft (the SR-71 Blackbird, not, for some reason, called the “Superman”).

  6. The space station explodes into 20 large pieces. 20, because it’s a nice round number, and didn’t take too long to run the calculations. These 20 pieces are not assumed to be distributed symmetrically around the centre of the explosion, however.

Here’s the colossal mess that was spit out:

In this run, 8/20 (40%) of the space station chunks collided with the planet, while the remaining 12 (60%) were launched into a cascade of orbits. None achieved escape velocity.

I ran the calculations a few more times, to get a reasonable sample (and also to restrict the colour options of the orbits, because dear god):

Round 2:

13/20 (65%) collide, while 7/20 (35%) achieve stable orbits. None escape.

Round 3:

9/20 (45%) collide, and the remaining 11 pieces get distributed over a wide range of orbits.

Round 4:

Again, 9/20 (45%) collide with the planet.

Round 5:

14/20 (a whopping 70%) of the fragments hit the planet!

In total, I ran 10 simulations, with the number of collisions in each simulation given below:

8/20 (40%)
13/20 (65%)
9/20 (45%)
9/20 (45%)
14/20 (70%)
8/20 (40%)
11/20 (55%)
8/20 (55%)
12/20 (60%)
8/20 (40%)

The mean average of these is 100/200 (50%), so, on average, half of the space station will collide with the planet. Strangely enough, it’s not a front half/back half split (although, that does contribute the majority (~90%) of the collisions). The forward half can collide if it gets a suitably large push toward the planet from the explosion (it ends up in an orbit with a higher apoapsis, but with a periapsis beneath the planet’s surface).

Mind you, while these numbers are independent of the number of space station chunks, they are not independent of the station’s orbit. A lower orbit will result in more chunks hitting the planet, while a higher orbit will result in not only fewer chunks colliding, but more chunks escaping the planet’s hold altogether. And, naturally, these numbers are very much dependent on the mean velocity chosen for the explosion.


Nice one.
You obviously know how it works.

… the number of chunks are highly dependent on the explosive change in velocity. So for the record, no significant piece of the station would actually reach the surface of Earth (assuming the presence of an atmosphere).

I was wondering how the size of the orbit might influence these numbers, however, but didn’t have any purpose-built mathematical tools installed…

So, I quickly modded my KSP to mimic the real Sol system in terms of distances, sizes, and masses involved,

I put a small spaceship into a roughly 2000km orbit and used a manoeuvre node to find out the change in velocity required to reach the lower atmosphere (11km):

481m/s, assuming a retrograde trajectory, or approximately half of your mean change. The difference between a 1989km altitude change and a 2000km altitude change is largely irrelevant, I chose to put my target altitude at 11km because KSP doesn’t display the periapsis if it’s below the surface. By comparison, escape velocity requires an additional 2.8km/s (direct trajectory).

EDIT: Just to point it out, the only reason why the Moon appears to be on the edge of an escape orbit in the above image is the angle of the camera, and the relative inclinations of the orbit. The orbit of my spaceship is approximately equatorial, the Moon’s very much is not. If I’d bothered getting them in the same plane it’d look something like this:

Source: Kichae’s post below.

But would anyone really put a spacestation at a 2Mm orbit? Well, it’s not a bad place to put it as such, the transition between LEO and MEO. Atmospheric resistance is all but gone (no, the atmosphere doesn’t stop at 100km, it just grows thinner), and while you still have to deal with tidal effects these, too, are minimized with altitude.

But why not put it in high orbit? Geosynchronous or higher, where tidal forces actually help increase your orbital height rather than lower it. Well there’s the obvious reason of “fuel requirements” but I took a look at it…

Getting into a Low Earth Orbit, according to wikipedia, takes a minimum of 9.6km/s+drag. I put my spaceship into a 200km orbit around kEarth and set up manoeuvre nodes to a 2Mm orbit:

Required delta-V: ~9.6km/s (to LEO) + 457m/s (first manoeuvre node, transition burn) + 433m/s (second node, in yellow text between navball and kEarth) = 10.5km/s. This is not an entirely accurate number, I probably skipped an important 100km or so, entirely ignored drag, and wasted a bunch of fuel circularizing in LEO, but for the purpose of comparison it’ll do.

Compare this to a geosynchronous orbit of 35.7Mm or so:

~9.6km/s (to LEO) + 2455m/s + 1474m/s = 13.5km/s or about a 29% increase in delta-V. In the real world, this would be a much bigger increase in cost due to the need to carry reaction mass with you for your engines, but in the Infinity universe this is not required. Oh yeah, and the spacestation, or what it’s made of, might not even have visited the planet it orbits which would actually make a higher orbit cheaper. Oh, and keeping it in a geostationary orbit would greatly simplify flying to and from it from the surface of the planet. Oh, and…

In short, there’s a lot of reasons why you’d want a station in GSO, and not a lot for MEO. So what’s important about this? Minimum dV to get back to Earth from GSO:


Minimum dV for escape trajectory from GSO:


Of course, this balance would shift significantly if orbiting a larger, faster spinning planet (such as Jupiter). Now, for the real question: what would happen if a piece of the space station, say a crew capsule and large engine, was ejected at a vector that shifted it’s periapsis into the planet’s atmosphere?

The fall would accelerate it to over 10km/s… it would enter the atmosphere…

Decelerating at a massive rate (exceeds 15g for most of the trip) it would burn ever brighter…

Until I realized I hadn’t installed Deadly Reentry or anything similar and it calmly fell into the ocean…

(At a fatal velocity of 103.5m/s).


Ooo, that’s pretty.

Oh, I didn’t mean to suggest that chunks would survive re-entry. Just that a significant amount of the station would be available to light up the sky, as it were.

Realistically, the number of chunks are going to be dependent on how the station’s destruction model is built. Without knowing the strength of the materials used to design the station, the explosive energy of missiles in the game, or the explosive energy of whatever ridiculously hazardous materials the stations are carrying, figuring out proper numbers for this stuff is a fools errand. As I say, I chose my range of delta-v’s based off of the Rule of Cool.

That’s pretty close to what I find computationally. My calculations spit out 472.9 m/s, assuming no radial component to the velocity change.

Close! This can be solved analytically to find that an additional 2.9 km/s is needed to escape. That’s actually significantly higher than I had expected it to be. Because I think it looks cool, here’s a plot showing the orbits resulting from an extra 2.7, 2.8, and 2.9 km/s:

For reference, the green/black circle is the Moon’s orbit. (And apologies for the yellow lines; they looked so much more visible in MatPlotLib.)

Actually, in a universe where fuel requirements are a trivial matter, things like defendability (for which I’m not sure there’s a discernible difference between MEO and HEO) or detectability (probably favours a LEO, all things considered) are probably the more important considerations. I chose 2000 km because it eliminated the issues associated with LEO, and still gave something of an awe inspiring view of the planet (the angular diameter of Earth at 2000 km is about 74 degrees, or about a monitor’s width). View, of course, favours lower orbits (and an orbit at 1000 km probably avoids the LEO issues, and increases the angular size of the planet to 81 degrees). And that’s assuming the planet’s a flat disk. The fact that the planet’s a sphere actually increases the angular size (though, you no longer get to see the whole hemisphere facing you) comes into effect even at these distances, so it’s closer to 90 degrees at 2000 km, and 120 degrees at 1000 km). In geosynchronous orbit, the Earth has an angular diameter of only 20 degrees or so.

Ouch. That’s not too far off from the estimated 14 km/s that the Tunguska asteroid/comet is thought to hit the atmosphere at, and roughly half the ~19 km/s that the Chelyabinsk meteor was moving at. Given that the bits of space station are considerably less durable than solid chunks of space rock, I think there’s a pretty good chance that they’d not just burn up, but explode, in the atmosphere at that kind of speed.

It’d be like fireworks!

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A point I felt might be a good idea to reiterate in this topic.

It just occurred to me: the only way to make all of a spacestation change it’s velocity by nearly a km/s is with a very large explosion at the centre of mass.

While putting hazardous materials under a few layers of armour would probably be a good idea, to make sure the first piloting error wont cause a chain reaction, a Death Star-like “lets put the explosive bit where it’ll do the most damage to us!”-approach seems… contrary to common sense. Put 2 layers of armour on the outside of your explody bits, and 3 on the inside - something makes it go “boom” at least most of the boom will be going away from you.

Which, with a roughly even distribution of boom containers and self-propelled booms would result in a Gaussian velocity distribution with a mean velocity of 0, relative to the station’s original orbit, would it not?

I may have forgotten to put RCS on my ‘spaceship’ (the crew capsule and large engine in the reentry pictures was literally all there was to it, Infinite Fuel Cheat ftw), or install any piloting assistance addons (I didn’t even bother with Kerbal Engineer), so I’m honestly surprised I got that close to the optimal numbers.

If you don’t mind, I think I’ll edit that plot into my first post, btw. The orbital inclination of the in-game Moon is rather extreme and as I tried to keep a roughly horizontal orbit… actually, even if you do mind I’ll probably have already done it :stuck_out_tongue:

I agree.

Both that fuel is largely irrelevant, and that LEO would probably be better for both - if you had some sort of ground-based defence (large number of armed shuttles with short-term lifesupport?) it’d get to you faster.

I guess the real question is whether there are any altitudes that offer a particular benefit to a station? Geostationary simplifies navigation (important if computers suddenly cease to exist), LEO holds tactical advantages and limits the distance smaller shuttles have to go, twice the Moon’s orbit would be easier to access from space, and somewhere between the Van Allen belts would probably make hardware last a little longer.

2Mm probably has some sort of significance behind it, seeing as it’s the cut-off line for LEO and doesn’t have a particularly interesting orbital period (127 minutes), but couldn’t tell you what it is or if it’d be useful to a space station.

Nice, thanks people.
Though if they are tight on budget, I wouldn’t be surprised if they either had the entire station disappearing in a giant ball of awesomness, or if they chose the station’s orbit and the chunk speeds so none (of the big ones, at least) would collide with celestial bodies - it would save them money on shiny but not gameplay-mandatory effects.

Also, I’ve tried to run numbers to calculate what an actual exploding 30-km station would do to the planet, energy wise. Assuming that only a fraction of the station is vaporised and that only an equivalent energy is emitted during the explosion, I’ve found that the energy needed to vaporise 1 km3 of iron is around a 2 000 megaton eq. TNT, or 40 Tzar Bomba nuclear tests.

Math :
With a density of 7, a 1km3 cube of iron has a mass of 7e9 metric tons, or 7e12 kg. Iron vaporise at 2750°C, so I take that as a difference of temperature (in Kelvin). Its specific heat capacity when solid is about 452 J(kg K), and its specific heat of vaporisation is 6.2e3 J/g.
Its specific heat of fusion is around 0.2e3 J/g, so I’ve ignored it, and didn’t find a different specific heat capacity for liquid iron, so I assumed it was the same as solid.
So the total energy is : (specific heat capacity * mass * temperature difference) + (specific heat of vaporisation * mass)
Which gives (452 * 7e12 * 2750) + (6.2e3 * 7e12) = about 8.7e18 J

Assuming an orbit at 1000 km and such energy emitted in 1min (the time to have a cinematic explosion) and negligible atmospheric absorption, the epicentre would receive around 11.5 kW/m² during that minute. Which is about a ten times the 1 kW/m² we get from the Sun (before that pesky atmosphere gets in the way, in both cases).

Math :
A 1 000 km, or 1 000 000 m sphere has a surface of 4 * π * radius², or 4 * π * 1 000 000², which gives 1.26e13m²
Given that, surface illumination (in W/m²) is energy * fraction of surface / time, or 8.7e18 * (1/1.26e13) / 60, or about 11.5e3

Note that it drops with the square of the distance, though. So it falls to one Sun at 10 000 km, and not much at Moon orbit. On the other hand, it rises at about 50 Suns at a low 250 km orbit.

And this is about as low-end as you can get, the end-result may have a few zeros more. So it is safe to assume that such a actual station have good chances to cause an Endor holocaust to whatever unfortunate world it orbits, even before km-sized chunks begin to hit it.
As such, I would advise against putting them around inhabited or fragile worlds, and suggest having interstellar treaties to ban such irresponsible orbit placement.

On the other hand, given that ships are supposed to be comfortable in the corona of a star, they probably have little to fear from the radiation alone, as long as they avoid other potentially hazardous effects like warp field disjunction.

Also, at 1000km, such a station (and I assume, the resulting fireball) would have an apparent length of three times that of the Sun. That would make for an impressive sight, right before your retinas fry.

It’s kind of hinted at in the Wikipedia article that LEO extends up to the inner radiation belt, but Wikipedia also claims that the inner belt starts at an altitude of 1000 km, so I have no idea what’s going on there. Every source I’ve found so far just gives 2000 km as the upper limit without justification or cause. It seems kind of arbitrary from where I’m sitting right now.

That would seem to be at the high end of the requirements here, wouldn’t it? The station doesn’t need to be very nearly vapourized in order to explode, after all. All it needs to do is lose structural integrity in an explosive fashion, no?

A fireball 10 times brighter and 3.5 times larger than the Sun would be a hell of view, though. Hell of a view!

Perhaps it’s the height where someone once said the inner edge of the Van Allen belt might be over some part of Earth and a bureaucrat at NASA put it in an official report? I guess it’s as good a place to put it as any - certainly wouldn’t make sense to use it’s actual minimum height (according to wikipedia) of 200km for the top of LEO :stuck_out_tongue:

I’d say.

Oh and ThornEel… a better way to describe 8.7e18 J?

“Approximately 50 kg of antimatter”. EDIT: Typo, oops.

Well, 1km3 would be 1/(303), or 1/27 000 of the 30 km-sided cube containing the station. Now, said cube wouldn’t be filled, obviously, but even assuming that only 1/270 of the cube is filled, that’s still 100 km3. While 1/270 sounds reasonable to me (for example, a long block of 3033 km), I assume that 1% is more on the low-end of what is needed for it to loose integrity.
The station would probably not be made of iron. I don’t know how much that would affect the energy thrown around, but it’s probably safe to assume that it would be something stronger than iron, so it would be higher. It would also probably not be as dense, but even a 4-times lower density wouldn’t change the magnitude of those numbers much.
Also, I assumed that most of the energy would go to vaporise it, and little more. So if it’s turning it into superheated plasma instead, or if only a small fraction is actually used “efficiently” at heating (instead of being wasted on radiation), it means that we can start adding zeros here and there.

OTOH, the station may have some highly heat-absorbent stuff lying around, like metallic hydrogen ready to go off and refrigerator around, so I guess we could also have something lower.

Also, a billion-ton chunk of iron hitting the ground at 45° at 10 km/s would apparently have an impact of 13 000 megaton eq. TNT. And this time, it’s on the surface.
According to Wolfram Alpha (you tell me if it’s accurate), Earth gets that kind of impact every 200 000 years. So not a dino-killer, but that should punch quite the hole in the landscape and the ecosystem.

Strange, I find around 100 kg (so 50 kg of antimatter).
That said, I had made quite a few mistakes initially before writing the steps down a bit more rigorously (there should none left), so take my results with a grain of salt

Finally, I realise that I made a low-end estimation, but not a high-end. So assuming a 30 km-sided cube station with a density of 100 g/cm3 entirely turning into energy (actual matter/antimatter reactions are only 2/3 efficient because some of the energy turn back into neutrinos and other useless ghost particles, but let’s not bother with that), I find 2.427e35 joules, which is 20 years worth of Sun energy.
Assuming that this energy is released in 1s, this gives us, for this one second, the energy output of a type II supernova.

At 1 parsec, people around the neighbour star would see a flash for one second, some 3 years later, who would have 1/50 of the Sun’s glare.
Given that the I-Novae Engine is probably unable to manage such light effects from one star system to another, I think it’s safe to assume this won’t be the effect chosen for destroyed stations.

Here’s what I’m talking about. The station doesn’t need to be vapourized to catastrophically explode. It will certainly guarantee that result, but so will melting the iron. So will deformation of weak points caused by a shock wave. Presumably, this stations been under attack for some time, and is going to have some structural issues due to a barrage of weapon fire. So destruction due to vapourization is… well, let’s call it a worse case scenario.

I’ll trust Wolfram on those statistics, but the damage done would be significantly less. Since the space station won’t be a solid mass, as it enters the atmosphere it’s going to have both a very large mass-to-cross-section ratio, and is going to experience an exceptionally large amount of drag. I may work the numbers out tomorrow, but I’d say the shock of impacting the atmosphere would cause each colliding section of station to undergo a catastrophic deceleration. Most of its energy will go into heating the atmosphere and tearing the bits of station apart.

As I say above, the bits of station that do collide with the planet’s atmosphere could essentially be used in-game as fireworks. Bits come down, the presumably already in-game atmospheric re-entry effects are used, and then the chunk of station explodes when it slams into the troposphere.

I like how you don’t classify KSP as a purpose-built mathematical tool. :slight_smile:

Ah, typo. Cheers. Underlying meaning was “how the hell would you produce this sort of energy?” I mean the station would pretty much have to be made of enriched uranium to even come close.

i think in reality if you need to destroy spaceship or station you have no need to literally vaporise it. damaging core systems are enough to transform spaceship or station to a chunk of metal and plastic scrap. huge loss of momentum or mass are not required.

The station isn’t vaporized, only 1% is - and that 1% is vaporised efficiently, at that.
I find it hard to blow the station up in multiple chunks (not simply disabble or cripple it) without vaporising a significant part of it.

What will break it will be either the material ceasing to be solid or being mechanically torn off. I doubt that we will see Homeworld Cataclysm’s ram frigates. I also suppose that it will blow up once its integrity is low enough, instead of being ripped apart bit by bit (which may be more realistic, but less cinematic and probably less satisfying gameplay-wise).

So while it has been previously weakened, the blunt of the movement and loss of integrity comes from the station itself exploding. And unless there are loads of high-pressure stuff lying around (like metallic hydrogen) or some more exotic (read : handwavium) effect like a warp disjunction, this comes from things suddently and violently releasing lots of energy around the station, causing it to suddenly expand, compromising integrity and transmitting lots of kinetic energy around.

I also thought that it would be more logical for an effective detonation to vaporise stuff instead of simply melting it. Particularly if you have only some dozens of seconds and a 30km-long object, as efficiently melting it without vaporisation would ask to heat it from lots of places at the same time.
Given that, what percentage of a 30km-long station do you need to vaporise to begin throwing bits around? 1% sounds pretty reasonable to me.

Otherwise, how would you calculate the energy needed to blow a station up? I guess we could go for the energy needed to rip it apart and to move chunks around at a reasonable speed, but I don’t know how to calculate the “rip apart” energy, particularly for advanced materials and a structure can’t guess.

Another reason to have vaporised stuff is for the giant fireball. How would you make the giant fireball otherwise?
Interestingly, a billion tons of vaporised material would probably make for quite the sight even without Hollywood flammable-aether fireballs.
But I’m not sure hoping for realistic space explosions in a space combat game (or movie) is more reasonable than hoping for commercial fusion power in the decade.

So we could go the easiest way : the station collapse into hyperspace in a minute-long blinding flash of cold white light. But that would be boring, wouldn’t it?

Yeah, that conveys a sense of how much (more than we can imagine) energy this is - though without using billions of tons of TNT, you could get the same amount of energy with “only” a thousand tons of Tzar Bombas.
Which makes me think, should I start to give energy in kilotons eq. Tzar Bombas?