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A person at rest 500 km above the Earth falls straight downwards. She has a snug magical force field around her that is totally rigid and completely protects her from outside heat. The force field does not change the shape of her body and DOES NOT protect her internal organs from rapid deceleration. Can anyone give me a rough estimate of the following:

  1. The time taken to fall the first 400 km to the top of the atmosphere.
  2. The time after that to reach the ground.
  3. The maximum deceleration she would experience.
  4. Whether she would survive the fall (ignoring the final impact).

I need this for a surprise book I am writing for my daughter, which I am hoping to keep generally within the bounds of believable science.

Steve Pemberton
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CapIsland
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    Terminal velocity will also depend on the drag coefficient, mass and size of the magical field.... unless it magically has no drag. But if it has no drag and there will be no deceleration. Can you stuff her in a pickle barrel instead? That would make the problem easier. /;O) – Woody Nov 19 '23 at 23:38
  • @Woody, I'm busy struggling with some discontinuities in my atmospheric model, but for a rough estimate, a skydiver isn't any harder to model than a pickle barrel (or a spherical forcefield). – Mark Nov 19 '23 at 23:46
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    @Mark ... maybe a spherical pickle barrel ? – Woody Nov 19 '23 at 23:50
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    It's a cool question and I would like to answer, but now that you've said "the force field can take on any aerodynamic shape" it makes the problem a lot harder. I would have opted for a ~2 m diameter spherical shape so the solution is fairly straight-forward with a density scale height of 7 or 8 km up to the "knee" and about 14 or 15 km above. Now that it's a shape-shifting field, of course it would change to a space-plane-like reentry vehicle with a huge wingspan and therefore tiny loading (kg/m^2) and use aerodynamic lift to spend hours or days slowly descending. – uhoh Nov 20 '23 at 02:51
  • If you want your subject to survive, give them some initial velocity. Falling straight down makes you slam into the atmosphere but deorbiting with a magic heatproof wingsuit might just be survivable – Sanchises Nov 20 '23 at 06:46
  • Let's add that one way to make it survivable is to do what all spacecraft do: add a bunch of horizontal velocity, and glide, achieving quite acceptable 4-6g at the peak. It's only plunging straight down that causes the risk of splat. – SF. Nov 20 '23 at 10:35
  • Yes, I did realise that if the forcefield could be manipulated greatly, then it could just turn into a parachute at the end. But assume the forcefield hugs the body at an inch out but could, for the sake of making the woman survive, alter slightly without becoming wings.

    @Sanchises, I did consider giving the woman horizontal velocity, but this makes the next part of the novel unworkable. She has to come straight down to enable her rescuers to predict where to position something 'sci-fi' to save her.

     – CapIsland Nov 20 '23 at 11:53
  • I have also added another part of the question here: https://space.stackexchange.com/questions/64809/person-flying-upwards-at-high-speed-after-falling-downwards – CapIsland Nov 20 '23 at 13:06
  • "The maximum deceleration she would experience." would depend on the time she should come to a stop. If momentarily, the deceleration will be infinity, obviously. – U. Windl Nov 20 '23 at 13:46
  • I meant the maximum acceleration she would experience EITHER during the fall in which she dies OR during the fall in which she survives (until she hits the ground). – CapIsland Nov 20 '23 at 14:22
  • Can someone point me to a gravity and air resistance simulator so that I can do all this myself? I just need to know how air resistance works if you're shooting upwards. – CapIsland Nov 20 '23 at 14:29
  • My other question was closed because stupidly I used the word 'wormhole'. I am just interested in the physics, nothing else. I can write the novel: I not as good with the physics. – CapIsland Nov 20 '23 at 16:35
  • "at rest" relative to what? The gravitational center of the Earth? The point on the surface of the Earth on the line between them and the gravitational center of the Earth? the Moon? the Sun? the Milky Way galaxy? something else? (all assumed that you mean at some instant in time) – Makyen Nov 20 '23 at 17:53
  • At rest relative to the Earth. – CapIsland Nov 20 '23 at 18:10
  • It's a minor point, but the thermosphere can extend 500-1000km above the surface of the Earth. Starting at 500km up, you don't have to fall anywhere to be within the atmosphere.

    Yep, technically the ISS orbits inside the atmosphere!

     – Brian Nov 20 '23 at 19:16
  • @Makyen, presumably at rest in an Earth-centered Earth-fixed coordinate system. Earth-centered inertial is the only other coordinate system that's even remotely sensible, and that only adds about 1% to the speeds and forces involved (and zero difference after aerobraking). – Mark Nov 21 '23 at 02:17
  • Thank you everyone for helping me so much with this question. I have no idea what I have done wrong with my other question and no one else seems to know either, or is willing to tell me. So I will stop here and wish you all the best. – CapIsland Nov 21 '23 at 09:15
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    For what it's worth, you might get more appropriate answers on [worldbuilding.se]. That community is good about not just checking the "truthiness" of fictional devices, but also at suggesting alternatives. Just make sure to use the "science based" tag (or whatever it is called). –  Nov 21 '23 at 17:02
  • I'm just waiting for someone to model this problem with a spherical cow. – candied_orange Nov 21 '23 at 17:32
  • They thought this through in the 60's: see MOOSE – user71659 Nov 22 '23 at 03:07
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    @user71659, MOOSE was intended for orbital re-entry, not vertical re-entry. Very different load profiles. – Mark Nov 22 '23 at 03:31
  • 'Magical anything' means you must either migrate such Questions over to SE Worldbuilding, or explain how 'magic' belongs in real-world Space Exploration

    If you do switch over to SE Worldbuilding, you'll still have to explain to your readers whether she has a magical force field, or not, and if she does, how you justify putting those particular limits on that.

     – Robbie Goodwin Nov 27 '23 at 22:01
  • @RobbieGoodwin, "magical anything" is a useful technique in science to focus on one phenomenon while ignoring irrelevant factors. For example, unless you're specifically dealing with relativity or mechanical stress, it's common to assume that the objects you're looking at are perfectly rigid. Fluid dynamics, for example, get far more complicated when you add deformation of solid objects to your simulation, while not generally changing the results much. – Mark Nov 27 '23 at 23:12
  • @Mark Really? Where is that method taught, and at what level? 'Focusing the attention' is obvious; using 'magic' to do that sounds about as scientific as… uh… magic! – Robbie Goodwin Nov 28 '23 at 21:12

1 Answers1

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Your question is under-specified (you don't give the size or posture of your subject), so I'm assuming an average-sized woman falling in the classic face-down skydiver posture. I'm also modeling this using the "perfect gas model" of reentry, which is bordering on incorrect -- there's likely to be substantial compression heating at the point of peak deceleration.

Falling the first hundred kilometers takes about 153 seconds. Nothing really interesting happens here, or for the next few minutes. Look around and enjoy the view -- if you're falling near the poles, keep an eye out for the aurora.

After about 305 seconds, you reach the Karman Line, a hundred kilometers above the surface. At this point, you're moving about 2675 meters per second, and the atmosphere is thick enough that you're experiencing about a milli-g of drag.

Acceleration builds up quickly as the atmosphere thickens, and 16 seconds later, about 57 kilometers up, you hit your peak velocity of 2800 meters per second, with drag exactly balancing gravity. Acceleration is still building, though.

Twelve seconds later, at roughly 28 kilometers up, you hit peak deceleration, a crushing 25 times the force of gravity. Wikipedia's acceleration tolerance chart suggests that this is probably not survivable falling back-first (five seconds above 20 gs, where the chart gives a limit of one second). It's definitely not survivable in any other posture. If you didn't have the forcefield, your arms and legs would be dislocated and broken by the forces involved.

The forces fall off nearly as fast as they grew, and you'll be down to a tolerable 4 gs just thirteen seconds after peak, and under two gs just eight seconds later, having shed 2600 m/s in 32 seconds. Peak heat dissipation was roughly 25 megawatts, which Wikipedia says is similar to the reactor output of a Los Angeles-class nuclear submarine.

From here, you drift down, slowing as the atmosphere gets thicker. You hit the ground 560 seconds after you started falling, at a leisurely 50 meters per second. You probably won't even leave a significant crater.

If you're looking for survivability, your best bet is to spread the forcefield out into a circular plate at least two meters across. The fall to the Karman Line isn't changed much, but deceleration starts six seconds earlier, peak velocity is 2750 meters per second, and peak acceleration is only 21 gs. Time spent above 20 gs is roughly a second, right at the chart's limit.

The real change is the descent after slowing down: it really is a drift, lasting 860 seconds after peak deceleration, for a total fall time of 1190 seconds, landing at just 16 meters per second.

Mark
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    Thank you very much for that wonderful write up. So the woman would not survive the five seconds above 20gs even in the force field. The characters in the novel would have 333 seconds to say goodbye.

    I assume there is no way the woman could reduce the deceleration to survivable levels even with a head down streamlined forcefield?

     – CapIsland Nov 20 '23 at 01:19
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    @CapIsland Head-first or feet-first is worse: you get deeper into the atmosphere before starting to slow down, so the forces are higher (but briefer), while your tolerance is lower. If you can spread the forcefield out into a sphere (or even better, a plate), deceleration starts earlier, and forces drop to (barely) survivable levels. – Mark Nov 20 '23 at 01:51
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    "just 16 m/s" is still 57 km/h so guaranteed injury and and over 40% risk of death based on pedestrian/car impacts at 50 km/h. – Criggie Nov 20 '23 at 05:11
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    @Criggie Surface impact was explicitly excluded in the question. – gerrit Nov 20 '23 at 07:53
  • @Mark, if it's not too much trouble, could you give me the advice the characters in the book would give to the woman on how to change her forcefield shape/cross-section, and how this would change what happens, stage by stage? – CapIsland Nov 20 '23 at 10:31
  • I have also added another part of the question here: https://space.stackexchange.com/questions/64809/person-flying-upwards-at-high-speed-after-falling-downwards – CapIsland Nov 20 '23 at 13:07
  • Can someone point me to a gravity and air resistance simulator so that I can do all this myself? I just need to know how air resistance works if you're shooting upwards. – CapIsland Nov 20 '23 at 14:29
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    If you can modify the shape in real-time, you can start with a disc for maximum drag in the less dense regions (though of course this would be unstable) and go progressively smaller in cross-section to keep the drag tolerable, before going wide again to decelerate – Chris H Nov 20 '23 at 14:48
  • Thank you for the idea, @ChrisH. – CapIsland Nov 20 '23 at 16:33
  • My other question was closed because stupidly I used the word 'wormhole'. I am just interested in the physics, nothing else. I can write the novel: I not as good with the physics. – CapIsland Nov 20 '23 at 16:35
  • Seems like the kind of analysis where jerk (rate of change of acceleration) would be relevant? I know that as you regress in to higher and higher derivative orders they become harder to predict/model (and where should you stop). But I think this is one of those cases where at least jerk could at least enter the discussion? – Lamar Latrell Nov 20 '23 at 19:49
  • @CapIsland : just barely over 20 g might be survivable if she is very fit and lucky. The world record (https://en.wikipedia.org/wiki/John_Stapp#Work_on_effects_of_deceleration) is above that. – vsz Nov 20 '23 at 22:17
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    @LamarLatrell, maximum jerk in all scenarios is under 120 g/s, which Wikipedia suggests is tolerable in any orientation. – Mark Nov 20 '23 at 23:35
  • I often find my intuition around jerk fails me. That number in other words is over 1km/s^3. One kilometre.. . Thanks for looking it up – Lamar Latrell Nov 21 '23 at 01:21
  • Thank you everyone for helping me so much with this question. I have no idea what I have done wrong with my other question and no one else seems to know either, or is willing to tell me. So I will stop here and wish you all the best. – CapIsland Nov 21 '23 at 09:14
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    @CapIsland You could try going to worldbuilding and ask there instead - I believe they do science-based answers... – eirikdaude Nov 21 '23 at 09:23
  • @eirikdaude, Thanks, but I am not a member there. I am a member here. The other question is no different from this one. Either this one should have been closed or that one should be opened. – CapIsland Nov 21 '23 at 09:28
  • Awesome answer. You could add a little final info - what kind of speed does a real-world (military) skydiver hit the ground at, if trying to get down as quickly as possible without breaking anything? For comparison's sake? – AnoE Nov 21 '23 at 09:44
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    @Mark "Head-first or feet-first is worse: you get deeper into the atmosphere before starting to slow down," implies a static posture. Maybe a change in posture at critical times to reduce the deceleration? – chux - Reinstate Monica Nov 21 '23 at 12:33
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    @Mark "you hit your peak velocity of 2800 meters per second, with drag exactly balancing gravity. Acceleration is still building, though." Hmm, at this point, is not acceleration zero? and then one slows down later (deceleration, not building acceleration)? – chux - Reinstate Monica Nov 21 '23 at 12:46
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    @chux-ReinstateMonica before that point, the acceleration is leisurely <1 g. At that exact point it's zero. Soon after you'll be going pretty fast through some thick atmosphere and your accelecation will go way up. If you're headed back-first when you hit those g's, you'll have a hard time. If you're oriented in any other way, you die. – John Dvorak Nov 21 '23 at 16:52
  • @chux-ReinstateMonica, before, you're accelerating downwards. At the point of balance, you feel exactly 1 g of drag (you don't feel gravity, since it's acting uniformly on your body), with no net acceleration. After, you're accelerating upwards (but still moving downwards because of the velocity you've built up). – Mark Nov 21 '23 at 20:08
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    @Mark and John, Agree the magnitude of the acceleration builds after that peak velocity point. I tend to think of acceleration as a vector and building implies an increase (more positive), when the scenario you describe tells of a rapid negative increase in acceleration from 0. IAC, now it is clear what is meant - it increases in the opposite direction - and quite rapidly. – chux - Reinstate Monica Nov 21 '23 at 20:24
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    As for modifying the force field, the bigger the better--scale it up enough and she could land with it (landing with a big flat thing is common behavior--see "parachute". Nobody does it with fixed surfaces because it would weigh too much.) It can't be a plate, though, because that is unstable and will most likely flip. A gentle bowl would be optimum. – Loren Pechtel Nov 22 '23 at 04:53
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    Randall Munroe would be proud of an answer like this! Add more stick people though :P – scotty3785 Nov 22 '23 at 09:11
  • @LorenPechtel, bigger is not uniformly better. Turns out there's an optimum around 10 square meters (depending on the person's mass). Below that, increasing speed at the start of aerobraking makes for a higher acceleration, above it, increased drag makes for a higher acceleration. – Mark Nov 22 '23 at 23:03
  • It is worth noting that the "two meter diameter flat plate" is a pretty good approximation of the MOOSE one-man reentry equipment – ikrase Nov 26 '23 at 09:10