I ask about a glider without any special thermal protection (pilot in the space suit), so both answers to the other question do not cover the topic.

When the glider enters the atmosphere, it starts generating lift. Could this lift keep the glider high enough, in low density air, and this way prevent the heat damage? The glider then would fly a long distance, losing speed and altitude slowly, until reaching speed and elevation at which it can safely land. Seems that lots of lift could be generated at speed and air density where the metal burns already.

Is there any analysis ever been done if such landing is possible?

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    $\begingroup$ No. It doesn't matter what the vehicle is, it must enter the atmosphere at high speed, I believe in the order of 2 to 3 km per second! That is very fast and there is no way to slow it down without generating a lot of heat. You have to convert kinetic energy into something. $\endgroup$
    – Simon
    Commented Sep 11, 2016 at 19:18
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    $\begingroup$ "When the glider enters the atmosphere, it starts generating lift": True but it also compresses air forward to an extreme pressure. Air heats, becomes a magma and the aircraft burns. If you remember radio blackout when a capsule re-enters atmosphere, the reason is the presence of the ionized magma. $\endgroup$
    – mins
    Commented Sep 11, 2016 at 20:15
  • $\begingroup$ The speed is high but the atmosphere is very sparse there. Somewhere very close to the upper boundary, it could potentially generate enough lift to stay high enough to avoid the overheating. $\endgroup$
    – h22
    Commented Sep 11, 2016 at 20:27
  • $\begingroup$ "the atmosphere is very sparse there", hence air doesn't do much to absorb kinetic energy / decrease speed, then air starts to be denser and heats under pressure while energy is transferred. That's why most of the shooting stars burn in atmosphere. Add to this that the range for the entry angle of attack is very small. $\endgroup$
    – mins
    Commented Sep 11, 2016 at 22:29
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    $\begingroup$ @mins: you mean plasma. There is no magma in space. $\endgroup$
    – TonyK
    Commented Sep 11, 2016 at 22:56

4 Answers 4



The stagnation point temperature goes up with the square of true air speed.

Temperature dissipation is proportional to true air speed and density.

Lift is proportional to the square of airspeed and density.

The lower wing loading of a glider (compared to the Shuttle, for example) means that all of the reentry will be slower, but that is not necessarily an advantage. Note that the Shuttle needed external cooling to prevent heat stored in the heat shield to dissipate into the Shuttle's structure. If the glider flies a longer reentry, it will be closer to thermal equilibrium. Also, the aerodynamic shape of a glider is a disadvantage here because the lowest heat load is possible with a blunt object - this is why reentry vehicles look like they do.

Just to give you an idea what temperatures are involved:

If we express speed as Mach number, a typical reentry speed would be Mach 25. This makes the reentry temperature at the stagnation point $$T_s = T_{\infty}\cdot \left(1 + \frac{\gamma-1}{2}\cdot Ma^2 \right) = 126\cdot T_{\infty}$$ at Mach 25. If we assume 195K at the edge of space, this comes out to 24,570K - theoretically, because ionization effects will cause the eventual temperature to be lower and the ratio of specific heats $\gamma = c_p/c_v$ is no longer constant. To calculate the real value needs a non-equilibrium gas model since the nose radius of typical glider wings and tail surfaces is in the order of one centimeter. The distance between the detached shock wave and the structure is too small for the gas molecules to reach equilibrium.

On the other hand, the epoxy resins used in glider construction cure at room temperature first and are then tempered at about 60°C. The glass transition temperature $\text{T}_g$ of such resins is at best a couple of degrees above the curing temperature. $\text{T}_g$ marks the onset of weakening of the composite matrix, and heating the glider above it will permanently damage it. Given that the glider will stay for dozens of minutes in air of a temperature of several thousand degrees and has no thermal protection will make sure that all which reaches the ground is a charred lump of material.

  • $\begingroup$ Can you please clarify this sentence that I cannot understand "Note that the Shuttle needed external cooling to prevent heat stored in the heat shield to dissipate into the Shuttle's structure". What external cooling would be available during the descent other that using the rest of the shuttle as a heat sink? $\endgroup$
    – AdrianHHH
    Commented Sep 13, 2016 at 11:15
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    $\begingroup$ @AdrianHHH: Right after landing the shuttle had to be plugged into an external cooling unit that would feed Freon lines running through the shuttle's avionics bays and hydraulics. The heat source were both the internal systems and the heat shield. $\endgroup$ Commented Sep 13, 2016 at 12:20
  • $\begingroup$ Compare What are the top temperatures occurring during reentry? on Space Exploration. $\endgroup$
    – user
    Commented Dec 5, 2016 at 13:28
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    $\begingroup$ Reportedly this is why SpaceX's Starship abandoned a composite hull for a steel one. Carbon fiber layup is much worse than steel when it's very cold or very hot. $\endgroup$ Commented Dec 29, 2020 at 5:23

Remember that temperature of a gas is related to the speed of its molecules. When the molecules hit an object moving 25000 feet per second and bounce, that gives them a temperature of many, many thousands of degrees (C or F). A balsa object would char and be destroyed. A glider would have to be designed to have good lift to drag at Mach 22+, not a minor trick.

And when it loses only 30% of its speed, it will need to support half its weight as it will no longer have orbital speed. That equates to significant lift, thus significant drag and since power is speed times force, very large dissipated power that will appear as heating.

So, sorry, but heat shielding is necessary to prevent thermal destruction.


Yes. With enough wing span-to-weight ratio it should be able to shed the speed in the most upper atmosphere, skipping without losing altitude for considerably longer than it would take the space shuttle to shed speed. Would a balloon pop if the entry was shallow enough? With a pressure of .00001 atmosphere no one has tried it; yes it is just possible. A wing shaped blimp that weighs much lighter than air at ground level should work. Getting it to orbit would be impossible, but constructing it in space for entry is possible.

This was the closest picture to what I am talking about showing that there is a material capable of withstanding entry faster than orbital speeds. The one I am thinking of would be much bigger and wing shaped and can land without a parachute like a stalled landing. It would not need parachutes or retrorockets because of the surface-to-weight ratio.

enter image description here



For small objects, there is probably be a 'critical aircraft density' below which a return from orbit is possible. A balsa wood model glider would probably land unscathed after a very long re-entry... The same may be expected from paper plane. The small mass will mean that little energy would be dissipated, and the large drag and long re-entry would ensure that its dissipation rate ('power at reentry') is kept low enough...

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    $\begingroup$ this is a bunch of hypothesis with no actual data. this should have been a comment. $\endgroup$
    – Federico
    Commented Apr 21, 2017 at 12:20

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