# Imperial Star Destroyer as a meteorite

Fictional situation, looking for a factual answer.

IIRC, an Imperial Star Destroyer from Star Wars is about a mile long. In the movies, we see them wrecked on planets. But if an ISD were to be ‘shot down’ in real life, and assuming they don’t burn up like the ones in the movies didn’t, and it becomes a mile-long meteorite, what would be the effect on the planet if it hit a continent? What would the effects be if it hit an ocean? (Let’s use Earth as the planet.)

I don’t have the math here, but one thing to point out is that if a star destroyer was “shot down” over a planet, it’s essentially starting out a 0 velocity and then accelerating toward the planet. As opposed to meteors which are traveling at great speeds when their orbit intersects with the Earth.

I thought of that after I posted. If it were stationary, it would have an initial velocity of zero. If it was orbiting in free fall, it would be going about 17,500 mph. If it was approaching the planet from some distance, there’s no telling how fast it would be going when it hit the atmosphere.

I don’t know the answer, but I suspect the material composition may matter. And the overall density.

Don’t forget that, unlike a meteorite, an Imperial Star Destroyer has a lot of empty space inside it. The Chelyabinsk meteor made a much bigger boom than Skylab even though Skylab measured pretty damn big.

Also, whatever it is that’s holding them up (it’s not just orbital dynamics, or they wouldn’t “fall” when damaged at all), it might not fail all at once. If a ship weighs a million tons, and battle damage leaves it with only 950 thousand tons worth of lift, it’s going to go down, but fairly slowly.

Antigrav systems in Star Wars are known to be small and cheap, so they’re probably distributed through the entire hull of a star destroyer, and so wouldn’t fail all at once.

From this website,

I’d guess a density of 750 kg/m³ and an ISD mass of 40,000,000 tonnes

As far as the physics goes, I don’t know how to proceed any further than this, and that’s ignoring air resistance:

ETA: I did find an online impact effects calculator. Here is a link where I plugged in some numbers, but I really don’t know how it works.

https://impact.ese.ic.ac.uk/ImpactEarth/cgi-bin/crater.cgi?dist=480&distanceUnits=1&diam=1&diameterUnits=2&pdens=750&pdens_select=0&vel=2.8288&velocityUnits=1&theta=90&wdepth=&wdepthUnits=1&tdens=2750

The projectile begins to breakup at an altitude of 56800 meters = 186000 ft

The projectile reaches the ground in a broken condition. The mass of projectile strikes the surface at velocity 2.77 km/s = 1.72 miles/s

The energy lost in the atmosphere is 6.48 x 1016 Joules = 1.55 x 10^1 MegaTons .

The impact energy is 1.51 x 1018 Joules = 3.60 x 10^2 MegaTons .

The larger of these two energies is used to estimate the airblast damage.

The broken projectile fragments strike the ground in an ellipse of dimension 1.28 km by 1.28 km

Transient Crater Diameter: 3.26 km ( = 2.03 miles )

Transient Crater Depth: 1.15 km ( = 0.716 miles )

Final Crater Diameter: 3.83 km ( = 2.38 miles )

Final Crater Depth: 443 meters ( = 1450 feet )

The crater formed is a complex crater.

At this impact velocity ( < 12 km/s), little shock melting of the target occurs.

At this impact velocity ( < 15 km/s), little vaporization occurs; no fireball is created, therefore, there is no thermal radiation damage.

The major seismic shaking will arrive approximately 1.6 minutes after impact.

Richter Scale Magnitude: 6.3

Mercalli Scale Intensity at a distance of 480 km:

I. Not felt except by a very few under especially favorable conditions.

II. Felt only by a few persons at rest, especially on upper floors of buildings.

The ejecta will arrive approximately 5.45 minutes after the impact.

At your position there is a fine dusting of ejecta with occasional larger fragments

Average Ejecta Thickness: 9.14 microns ( = 0.36 thousandths of an inch )

Mean Fragment Diameter: 367 microns ( = 14.5 thousandths of an inch )

The air blast will arrive approximately 24.2 minutes after impact.

Peak Overpressure: 846 Pa = 0.00846 bars = 0.12 psi

Max wind velocity: 1.99 m/s = 4.45 mph

Sound Intensity: 59 dB (Loud as heavy traffic)

Damage Description:

Glass windows may shatter.

~Max

Impressive. Thanks.

So a bit over 6,300 mph (ignoring drag) at impact. This page says:

The velocities at which small meteorites have impacted the Earth range from 4 to 40 km/sec. Larger objects would not be slowed down much by the friction associated with passage through the atmosphere, and thus would impact the Earth with high velocity. Calculations show that a meteorite with a diameter of 30 m, weighing about 300,000 tons, traveling at a velocity of 15 km/sec (33,500 miles/hour) would release energy equivalent to about 20 million tons of TNT.

That’s almost 90,000 mph at the higher end of the velocity range – about 14 times faster than the Star Destroyer in the diagram.

Except the meteors most likely to strike the Earth are going to be in a crossing orbit that is most closely aligned with that of Earth because any meteor coming laterally or head-on has a very low chance of hitting the Earth or coming sufficiently within its sphere of influence to have its trajectory altered to impact the surface. Earth’s average solar orbital speed is 29.78 km/s so assume that an impacting body is probably going to have a speed relative to Earth of somewhere around 20% of that, or about 6 km/s, and almost certainly not greater than escape speed (from the surface of the Earth is 11.2 km/s). By comparison orbital speed at LEO is ~7 to 8 km/s. A large capital ship somehow ‘falling’ from orbit is going to be at least in close order of magnitude to a meteorite, and probably more resilient than a natural body even if it is mostly hollow space. One should expect a large capital ship falling from orbit will do at least catastrophic regional damage notwithstanding of the hazardous chemicals and volatile fuels that it will disperse upon impact.

Of course, a ‘starship’ capable of moving at relativistic speeds is a weapon of incomparable destructive power. Even a small shuttlecraft from the Enterprise moving at fast sublight speeds could devastate an inhabited world or shatter a decent sized moon. Who needs phasers or a fancy “Death Star” with “turbolasers” when you have real physics?

Stranger

750kg/m^3 ? I would have thought most spacecraft are more air than anything else; even in the “unlimited energy” world(s) of Star Wars, it’s got to be more practical to build a lighter vessel to make it easier to propel. Armour is of limited value when the objects encountered are as a matter of course in the km/sec speed range.

IIRC, a meter cubed of water would be 1,000kg (density of 1). Perhaps a better comparison to start with would maybe be a battleship - while excessively armoured, and over-built with thick steel, they still float. Maybe half or more of the ship is above water, suggesting an overall density half that of water (and it’s using ballast). But a Deathstar or Destroyer does not need internal structures of half-inch steel, so say 1/4 or less the density of water, 250kg/m^3.

However, relatively intact suggests it did not “fall from the sky” so much as the guidance systems failed and it misdirected itself to the ground at a relatively low velocity to simply crumple the half or so of the lower structure… (for assorted values of “relatively low velocity”) There must have been a component of the anti-gravity/guidance drive still running.) Otherwise, it would be a pancake and a debris field. Aircraft that impact vertically tend to consist of a crater and a mess of tinfoil. Admittedly, structural strength on impact is not in a aircraft’s design parameters - but no metal structure is going to retain shape in a km/sec crash. The best one could hope for is extreme flattening, the equivalent of putting the item in a giant hydraulic press, aided by high-temperature softening from reentry heat.

(I found it humorous in what was it, Airport 77 ? Where a jumbo jet actually sank and then managed to sustain its overall structure at depth.)

I don’t think coherent and consistent physics is really a feature of Star Wars. In fact, at one point I hypothesized that their “galaxy far, far away” is actually in a small pocket universe or inside the event horizon of a black hole where physics are altered and distances are much less, which would explain how they can fly across the galaxy in hours through ‘hyperspace’ and destroy a planet with a laser virtually instantaneously instead of the many hours it would take for thermal and shock effects to propagate through an Earth-sized planet. Until The Last Jedi it wasn’t even clear that kinetic energy weapons could be used to damage a starship even though it is obvious that any material object propelled to relativistic speeds is fundamentally a superweapon capable of shattering large spacecraft and pulverizing moons.

Stranger

Plot point of John Varley’s “Red Lightning” - free-ish energy source let’s someone accelerate out of the solar system, then come screaming back in at something like .9c. Good aim - wanted to hit the US, actually impacted slightly off the Atlantic coast. Much havoc ensues.

My point is that, Star Wars-wise, ships don’t orbit. They basically use their engines to stay in place. Which is why they fall when the engines fail.

The video below makes some comparisons (and goes into all sorts of other details…mostly how we track them) about meteors:

The meteor that made Berringer Crater in Arizona was about 0.8 miles in diameter and solid iron (so, probably massed more than a Star Destroyer). About 10 megatons of TNT, 600x the energy of Hiroshima, was what occurred on impact. So, ballpark for a Star Destroyer. We have built nukes bigger than this (although I don’t think we have those anymore).

A crashing Star Destroyer would annihilate a large city (e.g. New York City). (This assumes dropping from orbit, IIRC there are examples in Star Wars lore of hitting a planet while in hyperspace and the planet is destroyed.)

My question is why a Star Destroyer would hit the planet at that speed? We see them hovering the same way bricks don’t over cities. Presumably if they were crashing from orbit they would be working to not crash and maybe live and not hit the planet at 17,500 mph. My point being, seeing a crashed Star Destroyer on a planet that is not in a big hole doesn’t mean much. Lots of ways it might be less catastrophic.

It should be added that the the person who figured the Berringer Crater was made from an iron meteorite did not count on it being vaporized on impact (which it was). He spent a lot of time and money looking for that iron but it wasn’t there.

The same would happen to a Star Destroyer hitting at those speeds. There’d be nothing left to see except a big hole in the ground (and probably a few bibs and bobs tossed out).

In one of Larry Niven’s War Against the Kzin a force-fielded protected ship at a significant fraction of c hits an asteroid in the Centauri system, effectively destroying it (the asteroid, that is).

As I said, I envision the impact where the star destroyer ends up a partly pancaked wreck where scavengers can climb through (or fly through) is less a result of free fall and more a result that a few of the anti-gravity devices failed or their guidance fails, and the ship proceeds to alter course for the surface at a moderately lethal speed.

Perhaps a hint of this is that there is at least one bit in one movie, IIRC, where a destroyer horizontal wrt the planet below sustains damage and stars to nose down toward the surface. Maybe the anti-gravity or whatever in the nose fails, and the remainder is not sufficient to counter all the effects of gravity.

In an old video Kyle Hill did the math on what would happen if the Raddus from the movie The Last Jedi hit the earth at 0.9999c. It would vaporize the earth’s surface down to the mantle (complete destruction).

Nice.

Someone cleverer than me might like to work out how conservation of momentum works with ships that use reactionless drive. They don’t expel any propellant, so they don’t work like anything we are familiar with.

If the laws of action/reaction don’t apply to these vessels, would they necessarily possess vast quantities of momentum when they impact a planet? The energy release might be very small, or even be negative in quantity (which might manifest as an implosion or something). !gnaB

Totally fiction but in the Star Wars universe they use repulsorlifts (basically anti-gravity) drives. That’s why we see Star Destroyers hovering effortlessly over cities. They do not feel the pull of gravity. We see their use on small scales like Luke’s speeder and small cargo sleds to the biggest starships.

Presumably if you start wrecking the ship those drives may give out and not be enough to hold them up from the planet and they will fall.

Needless to say we have no such tech and no idea how one would be made even in theory.