Say that a heavy (150ish kg) android was involved in some kind of incident in orbit and came crashing back down to earth (traveling at orbital-ish speeds upon re-entry).

Would they be going fast enough to "arrive" with a bang or would atmospheric drag have slowed them to more of an anticlimactic soft thud? (for the sake of argument, assume they don't burn up on the way down)

Playing around with this crater size calculator, sizeable (i.e. violent) craters only occur with objects traveling at several km/s at impact as opposed to at re-entry.

The idea is that the impact is violent enough to be heard as a thunderous boom in several nearby towns, but sadly my rudimentary google-math suggests otherwise.

  • $\begingroup$ Looking at the satellites and space stations falling, I guess it'll burn in the atmosphere, not much mass on impact. But if it has some heat shield? Shape matters a lot, too. $\endgroup$ – Mołot Jan 2 '17 at 8:03
  • $\begingroup$ @Sam, thanks much for the crater calculator link, useful to me for a writing in progress. I owe you beer. $\endgroup$ – Catalyst Jan 2 '17 at 10:24
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    $\begingroup$ Depends, did it nose dive or do a belly flop? In any cases there will be sonic boom followed by a short cry... $\endgroup$ – user6760 Jan 2 '17 at 10:26
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    $\begingroup$ Side note: It's actually pretty difficult to de-orbit. I.e. if somebody jumped off the ISS, they wouldn't fall to Earth - rather they'd just end up being stuck in a similar orbit. They'd have to find a way to lose most of the 7.7km/s 'sideways' velocity first (e.g. maybe it gets hit by a counter-orbit satellite). In any case, it wouldn't ever be traveling at orbital-ish speeds on re-entry - not unless it actively accelerated itself. $\endgroup$ – Luke Briggs Jan 2 '17 at 21:41
  • $\begingroup$ @user6760 based on a humanoid shape, I'd say wind resistance would make it hit feet first. $\endgroup$ – Xandar The Zenon Jan 3 '17 at 3:30

So, as your impact crater calculator is showing, no, there wouldn't be an exceptionally large boom or crater from a 150kg robot hitting the ground from orbit. Unless the android was exceptionally streamlined (think "bullet-shaped"), it would be slowed to a terminal velocity of only a few hundred meters per second before impact, not enough to make a sizeable crater. There'd be a "thud", but not a "thunderous boom".

However, there would be a boom which might be heard in nearby towns: the sonic boom as it passes through the upper atmosphere. That boom would occur pretty high up, though, (several km) so it's not going to be shattering windows or anything like that.

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    $\begingroup$ if anything hit the ground that's larger than a few solidified drops of molten metal and plastic :) $\endgroup$ – jwenting Jan 2 '17 at 13:26
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    $\begingroup$ Ehhh, the OP specified that they don't burn up. It's entirely possible that an android fighting orbital battles would be made of materials and be shaped so that it could survive an otherwise-unprotected entry. $\endgroup$ – Salda007 Jan 2 '17 at 18:04
  • $\begingroup$ I seriously doubt it won't burn up, unlike in battle where you can have radiators to disperse energy, during reentry you have none such, you'd need a heat shield. And it doesn't have to burn up, what I describe would have it break up and melt due to aerodynamic forces :)\ $\endgroup$ – jwenting Jan 3 '17 at 7:34
  • $\begingroup$ There's any number of explanations @Sam could give as to how the droid doesn't burn/break up during reentry (I can think of four or five off the top of my head), but for the purposes of answering their question, it doesn't actually matter. The question specifically stated, "for the sake of argument, assume they don't burn up on the way down." "break up and melt due to aerodynamic forces" is, in fact, what is colloquially referred to as "burning up". $\endgroup$ – Salda007 Jan 4 '17 at 10:09

No crater. No crash.

Android with enough dV to deorbit would be designed to fall to pieces upon reentry. And small scattered pieces would easily burn in the atmosphere. Even if we assume they do not burn nor melt, pieces would be small and scattered and of low terminal velocity.

Even without special design, such android wouldn't have aerodynamic heat shield (why would it?) and a lot of his mass would be in manipulators. Then there is reaction mass. To deorbit he would use most of it, making it lighter. Or, if you count without fuel, at least it would have empty tanks. These are making it less dense, lowering terminal velocity. And if some leftovers explode, the better. You have even more, even smaller parts. No threat.

  • $\begingroup$ Clever, and you're probably right, but the OP says to assume a mostly intact android is falling, and this opposes that assumption by adding new information to the context. $\endgroup$ – Zxyrra Jan 2 '17 at 9:01
  • $\begingroup$ @Zxyrra he asked to assume they would not burn, and that does not oppose main idea of my answer (falling to pieces). Edited to make it more clear. $\endgroup$ – Mołot Jan 2 '17 at 9:09
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    $\begingroup$ @Zxyrra pointint that the scenario assumed in Q is unfeasible is also our job. +1'd $\endgroup$ – Mindwin Jan 2 '17 at 12:07
  • $\begingroup$ @Mindwin Can you repost your comment, but reworded to make sense. $\endgroup$ – a4android Jan 3 '17 at 1:00
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    $\begingroup$ @a4android that language barrier, omg. Sorry about weird grammer[sic] $\endgroup$ – Mindwin Jan 3 '17 at 9:50

You have to look at the right sort of calculator. Avoid impact calculators as they assume your space android can read the ground intact. Be assured it won't.

For your in-answer enjoyment, please refer to this xkcd what if question. You're welcome!

The Wikipedia entry on atmospheric entry has this useful information.

Of satellites that reenter, approximately 10-40% of the mass of the object is likely to reach the surface of the Earth.

Assuming that only 10-40% of the space android survives re-entry, its velocity will be greatly reduced.

An aerodynamic heating calculator can be found here.

Research hasn't been able to discover a calculator for the amount of deceleration a deorbiting space android would experience. This would be useful as it would determine its velocity on impact. With various estimates on its terminal mass, ranging from 10% to 40%, it would be possible calculate its impact crater.

Hopefully other answers can provide the velocity change re-entry.

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    $\begingroup$ @Zxyrra A slip of the fingers. Fortunately, fixed by Michael Karnefors. Thanks Michael! $\endgroup$ – a4android Jan 2 '17 at 11:50
  • $\begingroup$ Satellites typically designed to be rather flimsy, as the owners don't want to pay for launching more weight than needed. They're also not terribly aerodynamic - think solar panel "wings" &c. An android, by definition, is more-or-less human shaped, and probably quite dense. Depending on composition (e.g. volatiles which explode from friction heating) it's likely to survive atmospheric reentry fairly intact. $\endgroup$ – jamesqf Jan 2 '17 at 18:10
  • $\begingroup$ @jamesqf The robustness of the android was an unknown. I stuck with what was known about satellite & spacecraft ere-entry. I contemplated the android could be covered Shuttle-type tiles or similar re-entry safe materials, but didn't have enough information for an answer. Otherwise we're in agreement about its likely survival from re-entry. $\endgroup$ – a4android Jan 3 '17 at 1:04
  • $\begingroup$ But we can deduce a lot about it just from the fact that it IS an android, and so "man-like" by definition. $\endgroup$ – jamesqf Jan 3 '17 at 4:29

We are the space androids. We are here to protect you. Here to protect you from the terrible secret of space.

Normally a space android will not crash, as we are perfectly constructed for your safety and happiness. However some space androids of the "shover robot" class may be sufficiently malfunctioning to do so. Do not trust the shover robots. In this case, those towns nearby may hear a sonic boom during re-entry, but the actual ground collision would be likely to be unspectacular unless you were very close. Space debris such as spent fuel tanks, rocket parts etc. in this size range comes down rather frequently (1-2 a week on average), and often manages to do so completely unnoticed even by people living nearby.

As a decent comparison, NASA's UARS satellite crashed in 2005, and the largest impactor fragment after burn-up was 158kg, very close to the size of your space android. The energetic yield of the impact was estimated at 153 KJoules. You'd get a similar impact from crashing a car into a brick wall at 36mph.

Thank you for your attention. Now you will be protected, please go stand by your stairs.

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    $\begingroup$ Your satellite crash paragraph would be stronger with citations or calculations for its numbers. In particular, 158 kg, 153 KJoules, and 36 mph. $\endgroup$ – Brythan Jan 3 '17 at 3:35
  • $\begingroup$ Certainly, human friend. This link gives the details of the largest impactor from the UARS re-entry. A simple application of the kinetic energy formula 0.5m*v^2 gives 44m/s, or 99mph as the terminal velocity of the object. wikipedia is in accord. $\endgroup$ – The Pusher Robot Jan 4 '17 at 0:39
  • $\begingroup$ As for the car: for the same KE, solving for v and assuming a 1200kg car, 1200*0.5*x^2=152994, x=16m/s. Or about 36mph. I trust that answer is satisfactory. Now please go stand by your stairs, human. You will be protected. Thank you. $\endgroup$ – The Pusher Robot Jan 4 '17 at 0:54

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