# Could you skim along the Karman line using a parafoil?

Imagine a person just wearing a spacesuit equipped with a parafoil returning from low orbit. Could they maintain altitude starting at 7.8 km/s above or along the Karman line slowly losing speed—maybe it takes a whole orbit or three—until subsonic without getting cooked inside their suit?

I'm looking for reasoning as to why it couldn't possibly work if the foil was large enough. The Karman line defined as the boundary where aerodynamics and orbital mechanics swap dominance implies lift doesn't stop there at all so we might imagine a very large & light parafoil like terrestrial sporting kites or paragliders slowing below orbital velocity without losing altitude while in layers of atmosphere thin enough to not let a space suited human overheat.

OR

Ideas about how to approach the calculation of approximate size for a aerodynamic surface to generate enough lift for a ~150 kg payload to stay high enough in the lower thermosphere to not get too warm at orbital velocity and then slowing down dipping into lower layers that then generate little friction heat at lower velocities.

• No, the parafoil would get shredded. Even if it didn't they'll get cooked inside their suit going at 7.8km/s at 100 km altitude. Inflatable re-entry devices are close to your idea
– JCRM
Jul 21 at 15:42
• no, you want to push the balloon, so it protects you from the atmosphere
– JCRM
Jul 21 at 16:25
• "The Karman line as defined as the boundary beyond where you couldn't obtain aerodynamic lift seems quite a bit above where serious reentry heating would start.": Meteors become brightly incandescent around 75-120 km altitude. An astronaut at the minimal orbital velocity for that altitude wouldn't experience such extreme heating, but...likely still far too much. You will either want to have braked to a much lower velocity by the time you reach that altitude, or you'll want some thermal protection. Jul 21 at 19:14
• I'd say sort of. The only effective way to generate lift at 100km would be a massive, ram-wing glider. This would need to be pretty huge too. This also depends on whether you mean the person is on a re-entry trajectory, or on an orbital trajectory at 100km, and encountering particles of the atmosphere as they orbit. Jul 22 at 6:20
• generating lift at the notional Karman line isn't particularly hard, it's defined at the point where lift and orbital mechanics swap dominance after all
– JCRM
Jul 22 at 11:10

Using a naive lift calculation at 100km runs into several issues. As you go above about 50km your mean free path for the atoms in the atmosphere is becoming long enough that you no longer have a 'fluid' but instead encounter each atom individually in a series of collisions.

At 7 kilometers a second fluid flow assumptions also break down as you encounter the atmosphere at faster than the local speed of sound with shock waves forming.

In terms of low speed re-entry, each kg of vehicle+payload has 24 M joules of kinetic energy that needs to be dissipated, turning a Kg of frozen water into vapor only needs ~5 M joules, so default math says our pilot gets boiled. We can dissipate that 24 Mj over a longer period, but need to radiate that energy away and optimal radiation needs higher than human comfortable temperatures.

These combine to mean a parafoil with pilot hanging underneath is not the right hammer for the job.

At this speed/pressure combination it is possible to get compression lift, but this requires a rigid body, so the paperplane from orbit project is interesting, though note that they designed to 230 degrees C so were using non standard paper.

This suggests that scaling up an oversize hang-glider like vehicle optimised for compression lift using metal foil might be able to make a 'slow' reentry, but it will still be averaging higher temperatures than a human in a basic space suit can handle for an extended period, and be problematic getting the required area at low enough mass.

For many of the use cases for this fragile several hundred meter hang glider something like the MOOSE project is possibly a better option, putting the pilot in a high drag ablative shield that can dump as much energy into the atmosphere as possible, and using expanding to allow for high temperatures in the heat shield without melting the pilot.

• @GremlinWranger—love the paperplane, MOOSE & compression lift pointers, thanks! Jul 23 at 9:29

So you want to re-enter without getting heated to above 300K. Sure.

Radiation heat loss = tlarea × temperature⁴ × 5.67/10⁸

Since the area of an average man is 1.9m², and an average woman is 1.6m², this gives us 873-735W. Minus 100W for the body, and round down, and you get 700-600W. For convenience, we shall use the latter value.

Moving at 7,800m/s, and at an energy dissipation of 600W, you can only tolerate a force of 77mN, giving a deceleration of 0.5mm/s²...not impressive. In the whole of one orbit, you decelerate less than 3m/s!

But this isn't what concerns me. What concerns me is what happens at 5,500m/s. At this speed, you have regained half of your weight, meaning you need 736N of lift, but lost only 30% of speed. So you can only tolerate 110mN. That means you need a lift to drag ratio of 6,700. Impossible.

It was suggested below that perhaps an extremely large vehicle could dissipate the heat. At 300k, the radiative flux is 459Wlm². This is way less than the radiation the sun, ar 1373W/m², and is approximately equal to the reflection from the earth at midday (412W/m²). Add to that 239W/m² from radiation. So you will probably overheat at noon.

If we assume that the descent occurs at night, then we get an average of 330W/m². Assuming a more realistic lift to drag ratio of 2, you have to dissipate 2.2MW. Thus 6667m² surface area, or 3333m² wing area, 4 times the A380 wing area.

• I think your calculation assumes that the astronaut's body does the breaking and heat absorption/dissipation; however most of the braking and heating down to a probably negligible fraction can be presumed to get done by the chute/parafoil which we might easily make very large, say 100x10 metres = 1000 sqm area to dissipate heat over. PS You don't really explain how you get from energy dissipation to deceleration force—this would involve the astronaut's mass to derive kinetic energy at least? Jul 25 at 18:09
• given the reenteree is orbiting, they will be doing so both at noon and at night at various points
– JCRM
Jul 26 at 8:08
• @JCRM if you time correctly, you could get the worst heating on the night side Jul 26 at 14:46
• @Prototypist force times speed equals power Jul 27 at 1:00