I guess it has something to do with their being both a high horizontal and a vertical velocity components during re-entry. But again, wouldn that mean there is a better reentry maneuver that the one in use?


Re-entry velocity from LEO is ~ 7800m/s, from lunar space it is as high as ~11,000m/s[1].

Different books give the terminal velocity of a skydiver as about 56m/s[2] or 75m/s[3].

The difference is that in order to orbit, you need to go very fast sideways. You essentially fall so fast sideways that you miss the ground when falling towards it (see related xkcd). A skydiver, whether he dives from a plane or baloon, has only marginal horizontal velocity and almost zero vertical velocity to start with. The skydiver then accelerates to terminal velocity, which is quite slow (see above).

In comparsion, the Apollo capsule had a terminal velocity of 150m/s at 7300m altitude$^4$. It is from that moment on that Apollo behaves like a skydiver. Drogue chutes are pulled that slow down the craft to 80m/s, and then finally the main chutes that slow down the craft to 8.5m/s[4]

But thats only the very last phase of the flight. You need to somehow slow down from 7800m/s to 150m/s first and descent from space deep into the atmosphere. Creating chutes that can both withstand that and are big enough to slow the craft down enough that high up in the atmosphere is simply not feasible from an engineering point of view, and even if it were, it would probably prohibitive from a weight/delta-v point of view.

In comparison, the Falcon 9 first stage does not have problems with re-entry heating, although it also reaches space. But that stage does not achieve orbital velocity. It only goes about 2000m/s at separation, which is slow enough that heating is not an issue when it comes down (see this question on Space StackExchange).

1: Atmospheric Entry. Wikipedia, the free encyclopedia.
2: Tipler, Paul A. College Physics. New York: Worth, 1987: 105.
3: Bueche, Fredrick. Principles of Physics. New York: McGraw Hill, 1977: 64.
4: W. David Woods. How Apollo Flew to the Moon. Springer, 2008: 371.

  • $\begingroup$ How can "the terminal velocity of a skydiver" be $56\,\rm m/s^2$? That's an acceleration, not a velocity. And $75\,\rm m/s^3$ is even stranger. $\endgroup$ – Henning Makholm Jan 14 '17 at 10:12
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    $\begingroup$ @HenningMakholm urgh, I see why you are confused. those are the footnotes, I will edit that to be clearer $\endgroup$ – Polygnome Jan 14 '17 at 10:18
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    $\begingroup$ This is the better answer. It's all about the initial speed. All that energy has to go somewhere. $\endgroup$ – candied_orange Jan 14 '17 at 19:27

The distances and speeds involved are materially different. On the scale of a parachute dive, the atmospheric density doesn't change much (and is relatively high). A parachutist quickly reaches a terminal velocity where the drag from the air matches the pull of gravity.

In a re-entry, you're approaching in a much less dense atmosphere, and you're going much faster. At these speeds, drag warms you up much faster. Also, you're plowing into the atmosphere, and that means you're increasing the drag. Between these effects, you see substantially more heating. A parachutist dropping from orbit would have the same issues with burning up.

There are some interesting things that are done regarding reentry maneuvers. The Chinese had one lunar orbiter which skipped off the atmosphere. The idea was simple. If the orbiter were to re-enter our atmosphere directly, it would receive too much heating. Instead, it was allowed to just enter the rarified upper fringes of our atmosphere, bleed off some of its velocity (into heat) before skipping off the atmosphere similar to a stone on a pond. This gave it time to get rid of some of that heat before a second re-entry brought it down safely.

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    $\begingroup$ You probably need to emphasize that the parachutist would need to be in low earth orbit (i.e. with tangential speed of approx 7km/s, said implicitly)as opposed to simply dropping from just beyond the von Kármán line to have the same issues with burning up -to differentiate your statement from the steak drop situation. Also, correct me if I'm wrong, I thought many re-entering spacecraft use some variation on the skip re-entry described in your last paragraph- Apollo did this if I'm not mistaken as I recall Frank Borman explicitly describing it in an interview. $\endgroup$ – WetSavannaAnimal Jan 14 '17 at 3:45
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    $\begingroup$ @WetSavannaAnimalakaRodVance I know its done, though I don't know how often. From what I can gather it's a bit more demanding on the controls to ensure you finally re-enter where you intend. $\endgroup$ – Cort Ammon Jan 14 '17 at 3:57
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    $\begingroup$ @WetSavannaAnimalakaRodVance, Apollo specifically tried to avoid skipping -- a skip on one of the Earth-orbital missions would bring you down in entirely the wrong place, while a skip on a Moon-return orbit would take days before the second re-entry. $\endgroup$ – Mark Jan 14 '17 at 8:16
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    $\begingroup$ "Also, you're plowing into the atmosphere, and that means you're increasing the drag." Except that it's not drag, that is, friction, that causes the heat - it's compression. A parachutist just doesn't go fast enough to create much compression . $\endgroup$ – Don Branson Jan 15 '17 at 0:43
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    $\begingroup$ @TylerH Yes. A human body would not survive that. the whole thing is unrealitsic as hell. See what-if.xkcd.com/28 for doing it with a steak. Not exactly a human body, but close enough. $\endgroup$ – Polygnome Jan 15 '17 at 10:30

A human parachuting from $h=4000\,\mathrm m$ (cf. http://adventure.howstuffworks.com/skydiving1.htm) needs to get rid of their potential energy $mgh$. If we assume that all this energy is used to evenly heat up the skydiver, who essentially consists of water with its well-known specific heat capacity $c_{H_2O}=4182\frac{\mathrm J}{\mathrm{kg}\cdot \mathrm{K}}$, the temperature rises by $$ \frac{gh}{c_{H_2O}}= \frac{9.81\cdot 4000}{4182}\,\mathrm K<10\,\mathrm K,$$ not enough to incinerate. One might object that the heating would occur mainly on the front instead of evenly, but the skydive is long enough for much heat to be transported across the body or even convected away form it by the surrounding air.

Note that the mass of the skydiver did not enter into the above simple estimate, only the staring altitude and the specific heat.

A re-entering spacecraft, on the other hand, not only starts form a higher altitude (more than 100 kilometers), but additionally has to get rid of its substantial kinetic energy. If we don't want to look up the numbers for orbital speed, let's try from memory. One always hears that one orbit takes about 90 minutes, hence the velocity must be at least (using the slightly smaller Earth circumference) $v\ge \frac{40000\,\mathrm{km}}{5400\,\mathrm s}\approx 7400\,\frac{\mathrm m}{\mathrm s} $, so the energy per kilogramm of mass is $\frac12v^2\approx 55\,\frac{\mathrm{MJ}}{\mathrm{kg}}$. We see that this is a lot, lot more than the mere $40\,\frac{\mathrm{kJ}}{\mathrm{kg}}$ of our skydiver.

The spaceship might get rid of much of the kinetic energy by using its rocket engine, but that would be unwise: It took a whole lot of fuel to bring the rocket into orbit in the first place (or rather: into orbital speed, after all the potential energy is small compared to the kinetic energy in low earth orbit); hence it takes (almost) as much fuel to achieve the same $\Delta v$ upon re-entry. But in order to have so much fuel available in orbit, it must have been transported there in the first place. This is infeasible because of the fuel-to-payload ratio.


As others explained, the maximum speed of a parachutist is much smaller than that of orbital vehicles. But this is because parachutists jump from a relatively small height: the record jumps by Eustace (https://en.wikipedia.org/wiki/Alan_Eustace) and Baumgartner (https://en.wikipedia.org/wiki/Felix_Baumgartner) were performed from the height of about 40 km, and the maximum speed was about 1.3 km/s, whereas the orbital speed is about 7.9 km/s. @Cort Ammon said that "A parachutist dropping from orbit would have the same issues with burning up." That is correct assuming the parachutist dropping from orbit has an orbital speed. However, the low orbit height (say, 150-200 km) is comparable to the record height of parachute jumps. If a parachutist were dropped from such orbital height with zero initial speed, his/her maximum speed would be much smaller than the orbital speed. To achieve speed comparable to the orbital speed, a parachutist would need to be dropped from a height comparable to the Earth radius (about 6000 km).


protected by Qmechanic Jan 14 '17 at 9:58

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