# When do aircraft become solarcraft?

Any body travelling through particles undergoes drag. Any body able to generate lift (for instance spheres cannot generate lift) can generate lift if it undergoes drag.

First by assuming one body in a circular orbit around Earth at 250km altitude. Assuming this body has a non-zero lift to drag ratio, is able to control its attitude on three axis, and is able to provide constant thrust for multiple orbits.

If thrust is used to push this body in the same direction as its velocity vector, and its lift is steered to counter Earth's gravity, then it should be able to orbit slower than its orbital velocity, which means it's flying, rather than being in an ever missing Earth free fall / orbit.

Does this make it an aircraft? (deliberately ignoring Karman line/altitude definitions)

Secondly assuming the same body in a higher circular orbit in very thin exosphere: When does atmospheric drag and lift become less influent than drag and lift generated by solar radiation pressure? Is there an altitude at which this body cannot be an aircraft anymore, because dominant force acting on it is SRP, therefore it becomes a solarcraft?

Shouldn't this very region of high atmosphere become the boundary between flight and spaceflight?

Edit: This question is not about redefining Karman line or redefining what an aircraft, or a spacecraft should be characterized by.

This question is more about lift in its very essence, in the first place. Everything else are consequences.

I made the premise that every medium generating drag or pressure to some body moving into it, allows this body (if designed and oriented to do so) to generate lift. Is this correct?

Why can solar radiation pressure generate lift, but very thin atmosphere couldn't? why can particles travelling at light speed generate lift by hitting some slanted surface, when ~5-10km/s thin atmospheric particles produce drag and only drag?

I remember this comment to an answer to this question :

"Indeed it is true that "aerodynamics of gliding works about the same at ... 0.1 bar and 0.001 bar" but somewhere before you get to ~10E-6 bar that all changes: the gas becomes non-collisional, i.e the mean free path of the molecules or atoms is much larger than the scale of the vehicle. Then continuum aerodynamics doesn't work anymore. Notably, the mechanism for generating lift is very different, relying on momentum exchange from molecules or atoms impacting the surface of the vehicle. The usual mechanism of impact-adsorption-reemission at thermal velocities is very inefficient"

So why couldn't some spacecraft orbiting circular in leo, and having wings oriented with a positive angle of attack relative to Earth (like an airplane's wing relative to horizon) "fly" slower* than its intended orbital velocity, thanks to thrust provided by some engine, for instance air-breathing electric propulsion, or whatever kind of long duration, very low thrust engine?

*even if it flies at 7.123km/s without decaying, and should theoretically orbit unpowered at 7.124km/s

ADDENDUM: my interpretation of "lift" may be wrong, but it implies any kind of thrust generated by deflecting particles. For instance, a deeply stalled airfoil in dense atmosphere at 50deg angle of attack still produces some residual "lift" which should be defined by "thrust".

• The Karman line was pretty much invented to answer this question.
– user20636
Commented Apr 29, 2019 at 13:18
• @JCRM what definition of Karman line are you referring to? It would be great to have a link towards the latest and widely accepted one.
– user19132
Commented Apr 29, 2019 at 13:23
• @qqjkztd Karman is 100 km, a nice round number because it is somewhat arbitrary and not very scientific. There is a serious scientific, evidence-based proposal to lower it to 80 km. Why is FAI considering lowering the Karman Line to 80km?. Relevant: Have spacecraft ever dipped below the Karman line and then safely continued spaceflight? and also What would a “Kármán plane” look like, a bird, or a plane?
– uhoh
Commented Apr 29, 2019 at 14:42
• WaterMolecule's answer to this question explains well why lift only works in continuum flow / low Knudsen number regimes: space.stackexchange.com/questions/31925/… In fact, this question is probably a duplicate of that one, since you state that it is "about lift in its very essence". Commented Apr 29, 2019 at 21:03
• @OrganicMarble do greater than 10 Knudsen number ballistic collisions with thin atmosphere at 7km/s (or "impact lift" as mentioned by Joshua) behave the same as solar radiation pressure? (Are both kind of particles changing momentum in the same way when being deflected at some angle?)
– user19132
Commented Apr 29, 2019 at 21:42

I think there are a few misconceptions to clarify here:

• Rotating bodies can generate lift. This is known as the Magnus effect.
• Lift is a hydrodynamical phenomenon: Differences in flow velocity above and below a moving body translate into pressure differences which heave the body up. However in very rarified gases this mechanism stops working. This is because when the gas becomes thin enough, the continuum approximation breaks down, particles become uncoupled ballistic singles and pressure is not translated from one point to another anymore. The Karman line is approximately this height, thus above this limit, Lift doesn't exist.
• Drag is a different phenomenon, it is the integral of all bumps with particles going another direction and thus well-defined and existent even in rarified gases.
• Orbiting slower than orbital velocity: There are not many cases where the force balance $$mv^2/r = GM/r^2$$ that defines the orbital velocity is upset by other forces. Only then this can lead to a different orbiting speed. In astrophysical contexts this would not be very exotic (gas in dense protoplanetary discs or black hole accretion discs does feel its own pressure gradient and magnetic torques), but in a spaceflight context I think this would be rather exotic.
• With the same reasoning we know that there is no lift generated from the solar wind, only drag. The drag can be either in the linear or quadratic regime, depending on the mean-free path of the respective gas. With the density of $$\sim 10$$ protons per $$cm^3$$, I estimated the mean-free path inside the Solar Wind to be $$10^{12}$$m, so we're in the linear regime, and then the drag friction time computes as $$t_f = \frac{r \rho_0}{c_s \rho_{gas}}$$, with $$r$$ being the object size, $$\rho_0$$ the density of the object, $$c_s$$ is the speed of sound of the gas (used to parameterize the temperature which gives bumps, it's still a non-continuum gas) and $$\rho_{gas}$$ is the density of the gas. Now we can search for the intersection of the two friction timescales, $$t_f^S$$ of the solar wind and $$t_f^A$$ in the atmosphere, so we require $$t_f^S = t_f^A$$, which translates into $$c_s^A \rho^A = c_s^S \rho^S$$. The speed of sound depends on temperature $$T$$ and mean molecular weight $$\mu$$ through $$c_s = \frac{k_B T}{\mu}$$ and I take $$T^A = 10^3K, \, T^S = 3.6\cdot 10^4 K, \, \mu^A=30, \, \mu^S=1, \rho^S= 10^{-23}g\,cm^{-3}$$. Then this approximately reduces this criterion to $$\rho^A \approx 10^3 \rho^S$$, which is fulfilled at an atmospheric density of $$\rho^A = 10^{-20} g \; cm^{-3}$$, which we reach only way beyond the exosphere. Interestingly this value would be reached radially before $$\rho^A = \rho^S$$ is reached, because the solar wind is so much hotter than the exosphere. I also wrongly assumed that $$\mu^A$$ doesn't change, but taking that into account wouldn't change the final conclusion.

To wrap up, with your proposal there would be quite some space.. until you reach space.

Edit:
To support the drag time computation as stated, I reference Weidenschilling (1977) and references therein, which state and use the friction times for different regimes of Knudsen and Reynolds numbers.

• How comes srp drag can be translated into lift? how would you control attitude on yaw roll pitch, on a body spinning to produce lift from magnus effect?
– user19132
Commented Apr 29, 2019 at 15:16
• @qqjkztd: Sorry I don't understand your vocabulary. I'm a hydrodynamicist, unfortunately not a pilot. Can you post a link to the srp drag? Commented Apr 29, 2019 at 15:18
• here it is, drag is pressure since it's not converted into lift link
– user19132
Commented Apr 29, 2019 at 15:19
• page 542 "Airplane-like attitude and flight control by shifting and tilting four sail panels". which means srp can produce lift. Why couldn't very thin atmosphere do the same?
– user19132
Commented Apr 29, 2019 at 15:21
• It's a misconception that aircraft generates lift by forcing air to move faster above the wing. Any aircraft generates lift by redirecting part of airflow downwards. This works for ballistic model too. If a gas molecule hits the wind and reflected slightly downwards, it pushes wing slightly upwards. That's a direct consequence of law of conservation of momentum. Everything else is merely a convenient computational model. And ballistic model ("Newtonian approximation") is actually one of the best approximations for highly supersonic lift. Commented Feb 8, 2021 at 16:24

Some issues with this:

1. Drag is dependent on cross-sectional area, and so is solar radiation pressure. This would mean that different spacecraft would have a different definition of space. We might get around this by designing a "standard spacecraft", but that's exactly what defining the Karman line was for in the first place.

2. The exosphere is complicated and its boundary is based on where the solar radiation pressure exceeds Earth's gravitational pull on a Hydrogen atom. Why hydrogen? Why not atomic oxygen or something else?

3. Solar radiation is not constant, so again your definition of space would change with the sun

As a result, we needed to make a single definition. Thus a spacecraft is a craft that operates in space - that is, above the Karman line - at some point in its flight.

The Kármán line definition is not ignorable.

Wikipedia:

The Kármán line is therefore the highest altitude at which orbital speed provides sufficient aerodynamic lift to fly in a straight line that doesn't follow the curvature of the Earth's surface.

If you can remain above the Kármán line you are a spacecraft. Full stop.

In fact it's a bit worse. It is not possible for an aircraft to remain anywhere near the Kármán line because the velocity required to generate lift is too high and the reentry heating will doom it. The domain is well-split because the intermediate range is forbidden to both kinds of vehicles for long.

As you approach the Kármán line, the terms from the equations of motion from orbital dynamics become dominant and the terms from aerodynamics become corrections to them. At lower altitudes the situation is reversed. Continuing above the Kármán line, lift becomes negligible fairly quickly. There is, in fact, not much distance between the Kármán line and the altitude at which the mean free path becomes the same as the craft's cross section, which causes aerodnyamic lift to drop to zero, leaving only impact lift.

• To me this Karman line definition works for an infinitely large flat Earth. I know no airplane that doesn't, in the long term, follow the curvature of Earth's surface. No spacecraft can travel on a straight line that doesn't follow this curvature either, it could approach it by providing one 40g constant super-burn towards an hyperbolic escape trajectory, looking like a straight line relative to the ground. Why this difference between both cases in this definition?
– user19132
Commented Apr 29, 2019 at 20:21
• @qqjkztd: The result is the Kármán line is approximately where the orbital domain takes over because orbital dynamics exceed aerodynamics. Commented Apr 29, 2019 at 20:23
• +1 for "If you can remain above the Kármán line you are a spacecraft." But Wikipedia is not a reliable source for a definition of the Karman line. That quoted section is an internet explanation for something that was never well documented historically. The definition is 100 km. "Full stop." The justification of that number is shaky. Let's not relitigate Karman! How many variants of the “Was Karman wrong?” question is enough for one user?
– uhoh
Commented Apr 30, 2019 at 0:16
• @uhoh: I can't help the fact this user will not believe that these things are more usefully defined by dominant term than some small term. Commented Apr 30, 2019 at 0:21