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Say my vehicle currently in orbit had large wings that would provide the necessary lift, would it be possible to bleed off the energy so slowly and gradually lose altitude that no heat shield would be required? I was thinking that even at the low pressure at high altitude, the wings along with the high velocity would prevent the craft from falling rapidly. And thus keep the drag(thinner atmosphere at higher altitude) minimal and so the experienced heating would be manageable.

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    $\begingroup$ The space shuttle did exactly what you are describing. $\endgroup$ – David White Apr 8 at 22:27
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    $\begingroup$ @DavidWhite I am confused by the answers below. I have also thought that the space shuttle does exactly this, providing its tiles are in good shape. They aren't really a heat shield, not as much as that of Soyuz. It seems matter of placing a limit to what consider as a heat shield. Perhaps Space Exploration SE is a better place for this. $\endgroup$ – Alchimista Apr 9 at 8:39
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    $\begingroup$ @alchimista The Mercury/Geminy/Apollo/Soyuy etc return capsules had/have an ablative heat shield. The ablative heat shield evaporates. The gas layer resulting from that evaporation is an essential contributor to the heat shielding. The space shuttle used a brittle material, non-ablative. The space shuttle heat shielding was very much not robust against mechanical impact. The SpaceX dragon capsule uses an ablative heat shield. Ablative heat shield technology has progressed; a dragon capsule can do multiple reentries with the same shield, thus achieving reusability cost reduction. $\endgroup$ – Cleonis Apr 9 at 9:51
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    $\begingroup$ Kerbal Space Program says: "yes, but it takes a veeery long time" $\endgroup$ – lvella Apr 9 at 11:37
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    $\begingroup$ @Alchimista no. The shuttle absolutely had a heat shield. It wasn't an ablative heat shield, but it got very very hot on one side while keeping the occupants and systems on the other side cool. Very much a shield. And the shuttle wasn't an effective "glider" during the highest-V phase of re-entry; it was falling basically belly-first, pushing air ahead of it, much like a capsule. (It had some lift, but so do capsules; it wasn't nearly enough to do what this question is asking about). $\endgroup$ – hobbs Apr 9 at 15:46
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Being in orbit isn't about going very high, it's about going sideways very fast. In order to get to orbit, about 80% of the energy is used in achieving orbital speed, which only 20% is used in getting to orbital altitude. A re-entering craft needs to rid itself of its kinetic energy, but you can see that most of that energy actually doesn't come from its descent through the atmosphere - it already has most of that kinetic energy even before it starts to descend. Once you're in orbit at an altitude of 100km, your kinetic energy is 30x greater than your potential energy - the kinetic energy gained by falling to the ground is a mere few percent of the kinetic energy you already have.

A re-entering glider could have a slower vertical descent, but that won't really matter. It will still have a massive amount of kinetic energy due to its high horizontal orbital speed. Having a more controlled vertical descent won't really change much about re-entry, as the real issue is eliminating the horizontal velocity, which a glider won't help with at all.

You could potentially spend time in the upper atmosphere to bleed off speed before entering the thicker atmosphere which causes significant heating, but this question over at Space Exploration suggests that a wing that could provide sufficient lift in such low pressure would be impossibly large. Getting sufficient lift to remain airborne is going to be extremely difficult at high altitudes.

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    $\begingroup$ Sure, but since the wattage of the heating is a function of air pressure, if my wings could keep me at a high enough altitude so that the air pressure is kept low, wouldn't the heating be manageable enough that I don't need a heat shield? $\endgroup$ – arc31 Apr 8 at 19:53
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    $\begingroup$ I think the suggestion was rather that you could stay at an altitude where the friction with the (very thin) atmosphere is not high enough to cause thermal damage but still slowly gets rid of the kinetic energy. $\endgroup$ – noah Apr 8 at 19:54
  • $\begingroup$ @arc31 ah, beat me to it. $\endgroup$ – noah Apr 8 at 19:54
  • $\begingroup$ Most meteors burn up hundreds of miles above the earth, well above what would normally be considered the atmosphere. Most orbits are higher still. Lower It isn't quite vacuum. If it is thick enough to provide lift, it is enough to heat a glider an burn it. $\endgroup$ – mmesser314 Apr 9 at 1:15
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    $\begingroup$ @mmesser314 There is clearly some altitude where a spacecraft moving at orbital speeds does not burn, but slows down a bit. So the question is: why not keep it above that altitude as long as possible? Of course, as the craft slows down, that minimum altitude also decreases. So is there a trajectory where the craft always stays above the minimum altitude, all the way to the ground? If not, how close can it get? How much does adding lift help by reducing the rate of descent? $\endgroup$ – user253751 Apr 9 at 12:03
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The key insight here is that you have a lot of energy, and all that energy has to be turned into heat. There's no other way to lose it. Your goal is 0 m/s at 0 km height. It doesn't really matter what the original proportion of kinetic versus potential energy is, both need to be turned into heat, and that heat needs to be shed.

Heat shields are very effective ways of getting rid of the heat. Ablative heat shields do so by burning up, other heat shields take advantage of the fact that radiative cooling increases with the fourth power of the temperature.

The idea of a glider is that you'd shed the heat slowly, by taking more time for the descent. You still need to shed the same amount of heat energy, but the power is inversely proportional to the available time. That's only a first power, though.

The engineering problem with the idea is that you need to generate lift at hypersonic speeds, without generating too much drag. That drag generates friction and thus heat, which was exactly what we're trying to reduce. But without lift, you'd fall too fast.

That lift needs to be fairly constant (slightly smaller than your weight), despite the deceleration. Else, you'd accelerate further or bounce back into space. And the Centre of Lift also needs to remain fairly stable despite the deceleration, or you'd tip over. You can't pump fuel around like Concorde did to shift the Centre of Gravity to match the CoL.

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    $\begingroup$ Bottomline is: maybe? $\endgroup$ – lvella Apr 9 at 11:39
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    $\begingroup$ @lvella: Bottomline: it's hard to prove a negative. $\endgroup$ – MSalters Apr 9 at 11:43
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    $\begingroup$ The heat explanation seems redundant. All the energy has to be turned into heat, yes, but if you can turn it into heat slowly enough, you can radiate it off and avoid melting, which is the point. $\endgroup$ – user253751 Apr 9 at 12:05
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    $\begingroup$ @user253751: The "slowly" is exactly the point of the third paragraph. But I then explain what it requires: a slow descent, which requires lift, which causes drag, which causes heat. $\endgroup$ – MSalters Apr 9 at 12:21
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    $\begingroup$ @user253751: the SR-71 had to be carefully designed to handle heating, and this was at a max speed of 3000km/h. Orbital speed is 8x this. Drag is proportional to the square of the speed, so the heat that the "glider" needs to overcome is 64 times the heat that caused problems to the SR-71. $\endgroup$ – Martin Argerami Apr 10 at 4:04
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I hope I didn't make a mistake in this quick back-of-the-envelope calculation, but:

A vehicle mass of $7.8*10^4 kg$ and an orbital velocity of $2.8*10^4 km/h (\stackrel{\wedge}{=} 7,78*10^3 m/s)$ result in a kinetic energy of $2.35*10^{12} J$, i.e. more than two terajoule. That's quite a bit of energy if you consider that a nuclear power plant's output is measured in gigawatt and hence would need to run for many minutes to produce this energy (and, incidentally, corroborates the anecdote that the first stage of a Saturn V burned more fuel than all of Great Britain).

The potential energy of a 500 km orbit should be around $3.8*10^{11}J,$ that is about 1/8 or so of the kinetic energy. The combined energy that must be shed is $2.74*10^{12}J.$

The average heat dispersal rate during a 30 minute (= 1800s) descent then is $1.52*10^9W$, which is the output of your standard average nuclear power plant block. The air cooling during re-entry works really well (the word "air cooling" never before entered my mind when I imagined re-entry).

If you want to bring the power down to an order of magnitude which can be handled without a heat shield, say, 1 MW = $1*10^6W$ (but doesn't that still sound hot?), you need to stretch the descent by a factor of 1,000.

The descent would no longer take 30 minutes but 30,000 minutes, or about 20 days.

It's not surprising that landing the second stage proved harder than envisioned.

Perhaps the 1 MW assumption is too pessimistic though: After all, Falcon 9's first stage descends without a real heat shield and only some half-arsed entry burn. Differences:

  • The initial velocity is only around 6000 - 8,000km/h, that is, less than 1/3 of an orbiting vehicle; the kinetic energy, which grows quadratic, is therefore less than 1/10 for the same mass, which makes it less than 1/20, assuming a weight of $3*10^4kg$ for an almost-empty first stage.
  • The entry burn reduces the velocity to about 600 m/s. Together with the lower mass of perhaps 30,000 kg (including landing-burn fuel) this results in a kinetic energy of $5.4*10^9J$, and with 80 seconds to go after that we arrive at a heat dispersal rate of $6.75*10^7W$, or 67 MW. Still, good air-cooling.

If we take that as an indication for what a largely unshielded vehicle can sustain we may de-orbit not in 20 days but instead within 1/60 of that, i.e. 8 hours or so. But it may well be that a vehicle that can take a certain degree of heat on its outer shell for a minute would overheat when exposed to it for many hours. On the other hand, large wings or canopies — constructive and mass issues aside — would distribute the heat load more. The truth may be in the middle.

From pureley energetic considerations, a de-orbiting time of many hours is a likely scenario.

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  • $\begingroup$ Could you add to your answer regarding the following: the SpaceX Falcon rocket jettisons the fairing halves seconds after second stage engine ignition. On the last launch stream I saw the second stage velocity at that moment was around 8000 km/h. Other rocket companies allow the fairing halves to burn up on reentry. SpaceX reorients the fairing halves so that they 'surf' on the atmosphere, and the fairing halves survive their reentry. The fairing halves have a lot of surface area compared to weight. Is it conceivable that even starting from orbital velocity fairing halves can survive reentry? $\endgroup$ – Cleonis Apr 9 at 15:15
  • $\begingroup$ @Cleonis Interesting idea; but you'll notice that I carefully avoided all specifics of rocket engines or aerodynamics. All I did is find out how much energy needs to be shed over which time. But yes, they are large and light, only 1000 kg each, I think, making the kinetic energy . I cannot immediately find information about how long they need to slow down to parachute speed. If it's 10 minutes, then the power is about 3MW. Oh, there is an interesting video on space.com showing the re-entry conditions: space.com/spacex-falcon-heavy-fairing-video.html $\endgroup$ – Peter - Reinstate Monica Apr 9 at 15:44
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    $\begingroup$ You're missing one thing about the Falcon 9 first stage: one of the purposes of the entry burn is to act as a heat shield. Yes, rocket exhaust is cooler than the compression-heated air around the rocket. $\endgroup$ – Mark Apr 9 at 20:06
  • $\begingroup$ Expanding on my previous comment: even if it is possible to perform a gliding reentry from orbit; only an object similar to a fairing half can do it; for gliding reentry the object must be very lightweight for its surface area. Even if a non-shielded glider can return payload from orbit it will be a very small payload only. By contrast, for the weight of an ablative heat shield you get a far, far bigger payload-from-orbit capability. So: in terms of bang for your buck an ablative heat shield is far superior to gliding reentry. $\endgroup$ – Cleonis Apr 9 at 20:22
  • $\begingroup$ @Mark Interesting point. Yes, I never thought of rocket exhaust as a cooling agent either. $\endgroup$ – Peter - Reinstate Monica Apr 9 at 21:24
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Accepting for the sake of argument that it's possible to build a glider that can generate meaningful lift in a near-vacuum, a slow gliding reentry is actually worse for heat buildup than a fast ballistic reentry.

One of the key discoveries of the early 1950s was that a blunt reentry vehicle is superior to a streamlined one. During a blunt-body reentry, most of the vehicle's orbital energy is shed as adiabatic compression of the air in a shock wave slightly ahead of the vehicle; very little goes into friction heating of the vehicle. The main source of heating is radiative heating from the shock wave, and the main purpose of the heat shield is to protect the vehicle from this heat for a few minutes.

Since the vehicle's orbital energy is mostly going in to the air around it, this means that the faster you can stop, the less heat you need to deal with. The Mercury capsules with their crushing 12 g reentry profiles needed far less heat shielding than the Apollo capsules, which entered with a more comfortable 4 g peak acceleration.

A slow, gliding reentry goes to the other extreme: the hypersonic shock wave is formed on the surface of the vehicle, maximizing heat transfer. You need to make your reentry really gradual to keep the heat load reasonable; Peter's answer suggests that you're looking at multi-day reentry periods.

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  • $\begingroup$ That the heat emerges not at the vehicle skin is an incredibly interesting piece of information explaining how 1 GW heat power is survivable for several minutes. $\endgroup$ – Peter - Reinstate Monica Apr 10 at 13:51
  • $\begingroup$ So you want a blunt braking (basically a cobra position) until you are at Mach... maybe 3 so that your titanium can bear the heat produced by the friction of standard aerodynamic flight with laminar flow. And probably several skimming dives at shallow angles with bounces before main re-entry. $\endgroup$ – Peter - Reinstate Monica Apr 10 at 13:53
  • $\begingroup$ @Peter-ReinstateMonica, a skip reentry would only be worth it when coming in quite fast, eg. a Moon return. There's less than a 1% speed difference between orbiting at the ISS's altitude and orbiting at the Karman line. $\endgroup$ – Mark May 3 at 22:41
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You have to bear some fundamental physical principles in mind here:

An orbiting satellite that is only experiencing atmospheric drag (like a spherical or irregularly shaped object) will gradually lose height but increase its speed in the process (closer orbits have higher speeds). This is of course the last thing you want in this case. In order to reduce the speed you have to prevent the spacecraft from losing much height by also having an aerodynamic lift. For this you need a suitably shaped object or wing orientated such that the lift force is comparable to the drag force. For a plane surface the two forces are equal for an angle of 45 deg relative to the direction of motion. The space shuttle for instance used to enter the atmosphere with the nose pitched up by 40 deg, so almost equal lift and drag forces. This resulted then only in a rather slow height decrease (the space shuttle took about 8000 km horizontal distance to drop just 130 km).

So basically, what you want here is a speed reduction whilst more or less staying at the same height. In principle, you could reduce the horizontal speed to zero this way. With an airplane this would be called stalling (a situation you usually want to avoid), but in this case it would be the solution of your problem, provided you do this at a height where the drag (and lift) is not so strong that you need a heat shield. You would then of course fall right down, but the speed you achieve falling from 200 km or so would be quite moderate and hardly require any heat shield either. And when you have picked up some speed, you could re-orientate the spacecraft and land it like an airplane (like the space shuttle did).

Having said this, your suggested glider re-entry without heat shield has actually already been done almost 20 years ago with the SpaceShipOne using a shape-changing airfoil design, but this was a suborbital (ballistic) trajectory, reaching about 100 km height but only a speed of about Mach 3.

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  • $\begingroup$ Glider re-entry without a heat shield was done 41 years before that, with the first suborbital flight of the X-15. $\endgroup$ – Mark Apr 10 at 3:03
  • $\begingroup$ @Mark: the X-15 was close to (the lowest of the lowest) orbital height, but exceedingly slower than orbital speed (around 10% of orbital speed at best). $\endgroup$ – Martin Argerami Apr 10 at 4:20
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    $\begingroup$ @MartinArgerami Your point stands, but IIRC X-15 reached a top speed of about 7300 km/h, or Mach 6, or 2 kilometers per second. That is about one quarter of the orbital speed of 8 kilometers per second. I'm afraid I don't know whether Mach 6 was the speed in level flight, which would make a more meaningful comparison here. $\endgroup$ – Jyrki Lahtonen Apr 10 at 7:22
  • $\begingroup$ @JyrkiLahtonen: a quarter of the speed means we're talking about a 1/16th of kinetic energy. Other suborbital flights are also undertaken without heat shield: check eg en.wikipedia.org/wiki/SpaceShipOne.. In an extreme case, even a zero-g airplane flight is following a suborbital trajectory --> and here you have your gliders actually working :) $\endgroup$ – Carl Berger Apr 10 at 12:22
  • $\begingroup$ @JyrkiLahtonen, IIRC it was during the dives.. but, why would it make a difference whether the top speed was reached in level flight or during a dive? I can see a difference depending on air density (hence altitude).. $\endgroup$ – Carl Berger Apr 10 at 12:48
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Nope, can't be done. The problem being there are two things you have to come to terms with, earths gravitational field and its atmosphere. The first forces you to obtain an orbit. From there, every way down is in an angle. The usable angle of reentry is relatively small. Dig in too deep and you'll burn up. Come in too shallow and you'll just bounce off, back into space.

An orbiting shuttle slows down its orbiting speed to start a fall in that angle. That fall continues until its airspeed is slow enough to allow it to fly, gliding down to the surface. The choice of the direction of that angle aims to minimize the friction during the fall. Going from burning up on reentry to 'just' needing a heat shield is already quite an achievement. The idea that it would be possible to enter earths gravitational field and then just gently glide all the way down to the surface without ever getting hot is not realistic.

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    $\begingroup$ Why would it "bounce back" into space if you come in shallow? The system as a whole (K.E + P.E) is losing energy due to some friction. So it gets into lower and lower orbits and not back into space... $\endgroup$ – arc31 Apr 8 at 20:51
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    $\begingroup$ Think of it as a skimming stone on the surface of water. Its not exactly like that, but gets close enough. In theory it goes much smoother than in practice. The 'top' of the atmosphere is not a flat surface. Like a skimming stone, you may very well dip at first contact, depending on what you hit , but chances are you'll skim. It doesn't really matter, because the thing that makes the heat shield a necessity is the fact you're not just coming in from outer space, but from an orbit, which puts you at escape velocity. $\endgroup$ – Berend Apr 8 at 21:28
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    $\begingroup$ @arc31, you're not bouncing back into orbit, but you are bouncing back into space, and when you come back down, you'll be entering the atmosphere at a steeper angle. $\endgroup$ – Mark Apr 9 at 3:43
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    $\begingroup$ @arc31, Strictly speaking there is a flight path that could bring you down without needing a heat shield, but not as a glider. Getting on that glide slope and staying there takes a lot of propulsion to stop the fall. Controlled reentry as a glider, which the shuttle does, can not get you there. $\endgroup$ – Berend Apr 9 at 5:47
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    $\begingroup$ In your last sentence you write "to enter earths gravitational field". That is very misleading, maybe you meant "to enter the athmosphere". (Gravity at 100 km altitude is still 97% of that on the surface.) $\endgroup$ – g.kertesz Apr 9 at 6:52

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