Follow up question to: How long time would it take for a space hulk to lose orbit?

Let's assume that the spaceship crashes into the planet, I used the Impact Calculator to get some result of the crash.

Your Inputs:

  • Distance from Impact: 200.00 km ( = 124.00 miles )
  • Projectile diameter: 400.00 meters ( = 1310.00 feet )
  • Projectile Density: 25 kg/m3
  • Impact Velocity: 11.00 km per second ( = 6.83 miles per second )
  • Impact Angle: 30 degrees Target Density: 2500 kg/m3 Target Type: Sedimentary Rock


  • Energy before atmospheric entry: 5.07 x 1016 Joules = 12.1 MegaTons TNT
  • The average interval between impacts of this size somewhere on Earth during the last 4 billion years is 3.3 x 103 years Major Global Changes:

  • The Earth is not strongly disturbed by the impact and loses negligible mass.

  • The impact does not make a noticeable change in the tilt of Earth's axis (< 5 hundreths of a degree).
  • The impact does not shift the Earth's orbit noticeably.

Atmospheric Entry:

  • The projectile begins to breakup at an altitude of 104000 meters = 342000 ft
  • The projectile bursts into a cloud of fragments at an altitude of 18900 meters = 61900 ft
  • The residual velocity of the projectile fragments after the burst is 0.0382 km/s = 0.0237 miles/s
  • The energy of the airburst is 5.07 x 1016 Joules = 12.1 MegaTons.
  • No crater is formed, although large fragments may strike the surface. Air Blast:

What does this mean?

  • The air blast will arrive approximately 10.1 minutes after impact.
  • Peak Overpressure: 528 Pa = 0.00528 bars = 0.0749 psi
  • Max wind velocity: 1.24 m/s = 2.78 mph
  • Sound Intensity: 54 dB (Loud as heavy traffic)

Would it be right to assume that the calculations are correct in this matter or do the somewhat aerodynamic shape, or other features change this?

I'm looking for the effects on the planet, not the spaceship, that if the calculations are correct - evaporate before it hits the ground.

  • $\begingroup$ The outputs of the asteroid impact simulator look like Tunguska to me. $\endgroup$ – MedwedianPresident Aug 21 '16 at 13:30
  • $\begingroup$ At that size, aerodynamics simply don't count. $\endgroup$ – WhatRoughBeast Sep 13 '16 at 18:03
  • $\begingroup$ might be worth taking into considerations what happens to an energy source that is capable of punching a hole through reality is uncontrollably slammed into a planet and as a side note these things must be extremely reliable and long lasting as the empire is depicted as struggling to build new stuff from scratch so they are often working with equipment that is old on a scale of millennia which is still working perfectly (ignoring the minor personality traits) $\endgroup$ – Jason Shawcross Dec 7 '18 at 13:39
  • $\begingroup$ @JasonShawcross that was a good point! $\endgroup$ – Magic-Mouse Dec 12 '18 at 11:28

The aerodynamics matter hugely in this scenario, as does the material of the ship and it's properties.

For example: If you put the numbers in for the Apollo capsule, but assume it's made of rock, it never reaches the surface.

The biggest question here is whether or not your ship was designed for atmospheric re-entry or not. If it was, then it will probably have a shape that lends it well to entering the atmosphere shielding down. I'm making assumptions about your ship design here, it could be that it expects to use active thrust to make sure it's shielding down, but I'd design a ship to self-orient. That shielding will probably be capable of absorbing the brunt of the atmospheric impact. If the shielding is uncompromised by whatever process led to it being a derelict in the first place the ship will survive re-entry intact. At this point you have a massive kinetic impactor, which will hit the ground in one piece and will cause one heck of a crater.

If the ship isn't designed to withstand re-entry, then it's possible it might break up sooner or later. It depends on the manufacture and stresses put on the vehicle. If it has significant stress points (joins between sections etc) and little internal bracing then it will come apart sooner. If it's been designed for rigorous high G burns and is reinforced to the gills then it will have to experience a bit more burn before fragmenting. If it's covered in easy to burn off sensor blisters/guns they'll burn off and cause weak points, if it's a sleek outer hull all shiny and chrome then it might last a little longer. It's really a question of design.

Looking at the Tempest: I'd assume it would break up. It's not designed for uncontrolled re-entry, its covered in weak points (gothic architecture does not good re-entry shielding make) and it's not exactly aerodynamic.

On the other hand: There's a lot of 40K canon about imperial warships ramming... ooh... everything. So it might make it all the way to the ground.

  • $\begingroup$ Sorry forgot to link to the ship wh40k.lexicanum.com/wiki/Tempest_Frigate it was in the previous post but both should of cause be independent, $\endgroup$ – Magic-Mouse Nov 3 '15 at 13:06
  • $\begingroup$ I'm not familiar with the 40K canon, but ramming things and uncontrolled (presumably) atmospheric re-entry are very different. That ship does not look like it was meant to land, to me it looks as a strictly deep space vessel and from the shape of it, I can not conceive how it can survive re-entry without breaking up. $\endgroup$ – ventsyv Nov 3 '15 at 20:20
  • $\begingroup$ @ventsyv: That was more a comment on how ridiculous 40k canon is than any effort to say it'll survive re-entry. There's at least one instance of a 40K warship ramming (entering?) a pocket universe, so uncontrolled re-entry should be a doddle. $\endgroup$ – Joe Bloggs Nov 4 '15 at 9:37
  • $\begingroup$ Well the interesting part was not to make the ship survive, the interesting was to make the people on the ground survive. $\endgroup$ – Magic-Mouse Nov 4 '15 at 12:05
  • $\begingroup$ The re-entry survivability also depends on the angle. 30 degrees is pretty steep. $\endgroup$ – John Dvorak Mar 4 '17 at 13:04

Working from the facts I can see from your links on the spacecraft it has no chance at all of entering the atmosphere in one piece without power. From the illustration http://wh40k.lexicanum.com/wiki/Tempest_Frigate:- The armored prow will immediately initiate a pitch "down" tumbling the ship into a top or bottom exposure to re-entry pressures. All of the bits sticking out will break off but not before tumbling the ship more leading to a catastrophic break up of the hull.

The lighter sections will either burn up or impact without any real impact to the planet.

The worry is the heavy dense bits within the ship. Bits of armor, reactors and shielding, heavy pressure tanks etc. These can reach the ground as has been evidenced from deorbiting earth satellites. See http://news.nationalgeographic.com/news/2011/09/110909-nasa-space-debris-uars-satellite-top-five-science/ . You are going to get a "shotgun" blast of these producing small impact events scattered over a wide area. Total impact to the planet as a whole, about zero. However you would not want to be standing where one of these bits hit.


It's absurdly low density & thus terminal velocity, coupled with the fact that it isn't ice only held together by convenience, but rather incredibly tough and built for battles that involve using relativistic weaponry and ramming space orks...

It would certainly go into a spin(though not likely nose down, a more dense object might go nose down, but here the thermal effect from the leading edge and the widened nature of the prow would create lift in the forward section), which could potentially (stupidly enough) increase it's chances of surviving orbital entry (at least until splashdown) because of the increased drag (slowing it down) and the dispersion of heat across the entire body.

That is, until we consider shearing forces, which the calculator doesn't deal with.

I have to question the mass, I don't think johnson was too worried about it tho. ???///??? ///???///

Hopefully some clarification for you

So, imperial college's calculator assumes the object is an asteroid which it's not.

For the purposes of the calculator.

So first we approximate the volume of the ship, it's pretty much just a cuboid, getting fancy with the spires and stuff is :S

Then we cut a cross section to derive an approximation of 'diameter' for the calculator, for whatever it does with it, and to determine density(combined with the provided mass) and with those we find it's terminal velocity.(it's not under power, it's a hulk)

I just left the angle of attack as I found it.. I guess I should check that link of yours to see if there was any retained momentum and direction.

We don't need to shift the surface density at the point of impact because..well, it could be anything.

1.5km depth, approx 200m w & 200m h (the max beam is listed at .4km at the grossest point, 200m is a total guess to the average.)

The mass is listed at 6.1mt (or 6100000000kg) Which gives us a density of 101.6666kg/m3

I used a drag coefficient of 1.5, arbitrarily.

So we have a impact speed of 1.62246 km/s

I'm well aware that this is entirely different to what I posted as a link earlier, and have no idea why (except probably I was disgusted with such a low density.)


But I'm not sure it makes a material difference to the answer, because the results of the calculator presume that it is a natural body that formed, essentially, out of convenience, whilst the Imperium invested the power of magitech into forming fabulously strong synthetic compounds.

This ships has an accel rating of what, 41m/s2, which of course doesn't mean it's hull or internals can survive terminal velocity even without an impact at the end.

Problem with that are it's nature as an object of war that survives impacts from relativistic objects of significant dimensions and continues fighting. Judging it's ability to remain intact is all but impossible without a rulebook assertion, drawing analogs from armor/hitpoints whatever the system uses.

If we assume it survives to the surface due to magitech(which I would, throwing synthetic materials is not the same as throwing clay) the total energy release is similar anyway, and the scatter of debris can just be more earth thrown into the air instead of more ship.

  • $\begingroup$ I'm confused about this answer, the question was "can i assume the calculations are correct, because it is a space ship made of steel bolted together, and not a rock hold together by gravity" and "would the posted effect be correct, considering the planet not the space ship" but the answer contains a link to a different calculation than i posted and comments about the effects on the ship and density. Would you be nice to clarify it for me ? $\endgroup$ – Magic-Mouse Dec 7 '18 at 11:49
  • $\begingroup$ I would blame any earlier confusion on fatigue, but it would be even more acute now. Essentially, I don't see a problem with using the calculator to determine the ground impact. If you don't like the result you can say the ratio of inelastic to elastic collisions is lesser or greater and have the ship 'absorb' more or less energy intact, I don't think Games Workshop release the chemical specifications of materials and systems their ships are made of, so it's up to you. $\endgroup$ – Giu Piete Dec 7 '18 at 13:12

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