Is aerodynamic control in an exosphere possible, in case velocity is high enough?

Exospheres are different than the lower parts of atmospheres (if there are any): the molecules don't collide anymore and it doesn't behave like a gas. However, it still causes some tiny little drag on satellites.

If a fast enough spaceplane were to approach the Earth and pass through its exosphere1, could they have aerodynamic control in the exosphere? If so, how fast would such a craft have to be?

1 The Exosphere begins at the Thermopause which is usually between 500 km (310 mi) and 1000 km (620 mi) above the Earth's surface depending on solar activity.

• For answering this question 2 clarifications are needed: 1.) how important are this "4000km" for you? The Exosphere starts somewhere between 500 -1000 km and the answer will be could be different for 500 km or 4000 km. 2.) do you want a theoretical answer or a practical answer? ...finaly: SOMEHOW RELATETD: en.wikipedia.org/wiki/… – CallMeTom Aug 30 '20 at 7:49
• @CallMeTom and uhoh of course I'd like to know it for all: lower exosphere (440 mi / 710 km), the example I provide (2,500 mi) and the upper exosphere (6,000 mi / 9,700 km). My example is just that, an example. – Giovanni Aug 30 '20 at 9:47
• @CallMeTom I'd like both a theoretical and a practical one, if they must be separated (like, if the required velocity is so high that you'll have little chance to use your craft's yoke in the exosphere of Earth). – Giovanni Aug 30 '20 at 9:53
• @Giovanni okay thanks! I've made an edit to your question to reflect those clarifications and to include a bit of "prior research." I also added the lift tag since in spaceflight that's any component of aerodynamic force not parallel to the velocity vector. – uhoh Aug 30 '20 at 10:43
• Relevant: what-if.xkcd.com/30. Quote: "if you turn, your plane rotates, but keeps moving in the original direction. The X-Plane author compared piloting Martian aircraft to flying a supersonic ocean liner." – John Dvorak Aug 30 '20 at 10:44

Yes. The craft is a solid body moving through, well, not air, but something that at least exerts a drag force on it. So if the craft adjusts that drag by adjusting giant vanes or electromagnets or something, then it's guilty of committing aerodynamics in the general sense of that term.

But the aerodynamics is rudimentary because there's no way to generate lift (a force orthogonal to the craft's direction, due to a pressure differential) because there's no pressure, only the individual impacts of molecules on the craft (drag). The concept of airfoil applies only when the fluid's particles push against each other, not just against the solid body.

The craft could slow down by increasing total drag (think releasing a big parachute), or change orientation (excess drag on one side). It couldn't change the direction of its trajectory, except by the indirect method of increasing drag during certain intervals.

How fast the craft has to go to maneuver like that depends on the reach of the drag-producing devices, and on how dense the molecules are at that moment.

• Well there's still pressure but an extremely tiny one. Otherwise it would be absolute vacuum. – Giovanni Sep 1 '20 at 6:39
• @Giovanni It's a matter of median path between collisions of the particles. If a particle can bounce off other particles and back into the craft, the other particles into others so they bounce back to bounce the previous one back again, etc, so you have a pressure wave, and the pressure change caused by the craft acting back on the craft, you have lift. If by the time the particle bounces into another the craft is already a kilometer away and the influence of craft on pressure doesn't reflect back on the craft, a lot general gas properties cease to apply. – SF. Sep 1 '20 at 8:16

Practically it is not possible to control an space/aircraft aerodynamically above about around 80 km altitude. That's why the US (Forces) define the needed altitude to get the Astronaut Badge as above 80.4672 km*.

*(strictly speaking it is defined in imperial units (50 multiples of 1.609344 km), as the Author of this answer I don't want to share a distance in a non-SI-Unit).

[..::EDIT::.. Yes, you can construct a situation, where drag influences an object, but that is far away from "controlling" it!]

Some spacecraft are designed for a special aerodynamic behavior (see GOCE), but that is not about controlling it but to minimize drag.

Reentry vehicles are often designed to be stable during reentry, but that works for much lower altitudes thus higher atmospheric pressures.

..::EDIT::..

As you mention, you could maximize the effects of the control surfaces by being faster, but in orbital mechanics being faster results in being in another orbit. To be low but fast (needed to control something aerodynamically) would mean: you are in the periapsis of a highly eccentric orbit or you just pass Earth's sphere of influence. Even then the effect of aerodynamic control would be very small.

• Yeah I knew that at orbital velocity you can't have aerodynamic control above 50 miles (80 km). But nowhere did I write about orbits. I meant what you're describing in your penultimate sentence: passing Earth's sphere of influence. I suggest you need solar escape velocity or more to have control in the exosphere. My question is whether it's possible at all, and if so at what speed. But otherwise your answer is very good (I can't upvote since I'm unregistered). – Giovanni Aug 30 '20 at 11:31
• of particular interest: How did the X-15 control attitude above the Kármán line? – uhoh Aug 30 '20 at 12:43
• I imagine that at some degree of speed (probably way up near the dozens of km/s) the sheer explosive energy released from bombardment with tenuous ions would impart a non-negligible momentum to a traveling craft, making an aerofoil do at least something to trajectory. – BMF Aug 30 '20 at 16:17
• Thats why I started my answer with "Practically...": YES, in THEORY it is possible to construct a situation where an object is influenced by the exosphere. But the Question is about "CONTROLLING" it! – CallMeTom Sep 1 '20 at 4:35
• 1.) Hours of converting Lockheed Martin documents (including explaining my old boss the difference between Pound-Mass and Pound-Force... he still does not belive something stupid can be true and is wondering were this 0.981-factor error comes from), and days of rebuilding CAD-models because LM build a model using Inches in a mm-environment. (oh I miss my old colleague swearing about LM ) 2.) as long this forum tries to be professional and the community downvotes flippant answers, I will not support even more stupid things. 3.) My contribution to a better world. 4.) Mars Climate Orbiter – CallMeTom Sep 1 '20 at 7:02