Can the ISS achieve any lift by using it's solar panels as 'wings' against the sparse atmosphere? My understand is they started moving the solar panels into a 'low drag' orientation when on the shadowed side of Earth to save on the not insignificant drag the panels created. But if you tilted them even slightly from horizontal would you get a slight lift?
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7$\begingroup$ Ironically, to move up you must thrust forward which actually slows you down! Just pushing radially out will not help, and there's no simple, easy way to get forward thrust from drag with solar panels. The force you mention would indeed be called lift, since it has a component perpendicular to the direction of motion, but the effect would not be to raise the altitude of the orbit. $\endgroup$– uhohJul 13, 2017 at 0:59
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1$\begingroup$ Would it create the least drag then? If it's between the panels being flat (lowest drag), open (highest drag), or a very slight angle to create a lift but not enough of a lift to increase altitude but enough to cause the least loss of altitude of the 3 scenarios. Wouldn't that be the optimum position on the shadowed side of the Earth $\endgroup$– Brooks NelsonJul 13, 2017 at 1:28
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$\begingroup$ One question at a time is best. Better wait for someone to post a real answer, and then ask a follow-up question as a question/ $\endgroup$– uhohJul 13, 2017 at 1:39
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$\begingroup$ The solar "panels" are not rigid. They consist of cells on a flexible film. The ends are held apart by very thin structures not designed for any lift forces. $\endgroup$– HobbesJul 13, 2017 at 5:37
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3$\begingroup$ No, the ISS is above the Karman line. Above the Karman line, you would need to travel faster then orbital velocity to generate lift. the ISS orbits at orbital velocity, not faster. Furthermore, the panels are not wings, so the lift Coefficient (C_L) would be miniscule... $\endgroup$– PolygnomeJul 13, 2017 at 7:49
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5 Answers
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I can see your reasoning, however it's not going to work. Aerodynamically the byproduct of lift is drag, meaning when you create lift you also create drag as well. An efficient wing creates less drag per unit of lift, but the drag is still there. A wing works because the pressure of the air above it lower than the pressure below, pushing the wing in the direction of the lower pressure (ok, that's a simplification), to do this with any efficiency wings must be a certain shape to create that pressure and to allow smooth airflow. Solar panels are not wings and will not have smooth airflow or a shape designed to produce lift (just the opposite), so if you tried to use the ISS solar panels to create aerodynamic lift you would create infinitesimal lift for the drag produced.
In any case lift is not what you want, even if it was for free.
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GdD correctly answered the question you asked but I see a misunderstanding that needs correction:
Lift is of no value to a spacecraft. Spacecraft stay aloft because the outward "force" from their speed matches the downward pull of gravity. In the time it takes the spacecraft to fall a foot it's moved forward enough that it's a foot farther away from the Earth. The two movements cancel, the spacecraft stays up. (Note that a spacecraft in a non-circular orbit does not have such a perfect balance and there's an eternal cycle of trading speed for height and the trading the height back for speed. They stay up for the same reason, though.)
Suppose you push that spacecraft up, what happens? It of course heads up--but as it heads up it loses speed climbing away from Earth's gravity. Soon this loss of speed overcomes the shove you gave it, it ceases heading out and heads back in again. Now it speeds up, eventually that overcomes the inwards velocity and it heads out again.
Note that the end effect of pushing the spacecraft up is to change it from a circular orbit to an elliptical one. The higher you pushed it the lower it goes on the other side of the orbit--and the atmosphere isn't linear. You have increased the drag on the spacecraft. Your attempt to push it up (even if it didn't cost you speed due to drag) has actually accelerated the orbital decline, if you push hard enough you'll actually cause it to re-enter.
In practice the only way to counteract orbital decay is to push the craft forward.
Anyone who is serious about learning about such things should play Kerbal Space Program. While the orbital mechanics aren't perfect they're good enough to give a good grounding in how things will actually behave when you move them. And, yes, I have seen the effect of burning outward sending you into the planet many a time, although I've always corrected the orbit before impact. A burn straight out is the fastest, easiest (but not most fuel efficient!) way to get to something following you not too far behind in your orbit.
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No, but to some sense, the opposite can be done. If you want to lower the orbit of a satellite for some reason, you set the panels to be in the "drag" position, where they are perpendicular to the direction of motion as much as you can. If you want to keep the drag minimized, as the ISS does, you move the solar panels such that they are aligned with the direction of velocity, to lower your drag. Usually satellites only will do this when in eclipse mode, as they can't receive sunlight anyways during that period of time.
answered Jul 13, 2017 at 12:18
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Short Answer: No.
Long Answer:
Once you pass the Karman-line air becomes so thin, that individual particles are not collisionally well-coupled to each other any more.
One can also express this as the particle mean-free path $\lambda_{mfp}$ being greater than the mean particle distance $d$.
Below the Karman-line you have approximately $\lambda_{mfp} \sim d$, which keeps all particles well-coupled to each other. This then allows to describe the gas as a fluid, which is where the notion of lift comes from. Because a fluid can obey Bernoulli's principle, lift can be generated.
However above the Karman-line the Bernoulli principle cannot act, because the gas there is not a fluid, or $\lambda_{mfp} > d$. This means lift in space doesn't exist.
Drag, however is a more universal phenomenon, related to how entropy loves to eat up slowly all structure in the universe. This means that random collisions with particles will always nibble away your momentum, no matter how dense the surrounding medium is.
answered Jul 14, 2017 at 7:58
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$\begingroup$ Very good explanation that lift does not exist in space, but drag does. Another explanation: lift is generated by a air pressure difference between top and bottom of a wing, but in space there is no pressure difference possible, the pressure is zero below and above the wing. $\endgroup$– UweJul 14, 2017 at 8:22
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2$\begingroup$ This is totally wrong. You don't need Bernoulli's principle or pressure differences or continuum fluid flow to generate lift. A simple flat plate tilted into the flow will produce lift at all Knudsen numbers. The resulting lift and drag can be calculated using simple momentum conservation, as mass is deflected down. Lift in "space", really the exoatmosphere, does exist. $\endgroup$ Jul 15, 2017 at 5:28
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$\begingroup$ @Good point, however orbit degradation speaks a different language. Averaged over all particle encounter one always looses momentum at $Kn<1$. $\endgroup$ Jul 15, 2017 at 10:08
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But if you tilted them even slightly from horizontal would you get a slight lift?
Lift is ultimate caused by directing the flow of oncoming air downward. However, even if a spacecraft does just that, it's not going to accomplish much.
Assuming Earth acts as a point mass, the specific energy (energy divided by mass) of a spacecraft orbiting the Earth is $$E = \frac{v^2}2 - \frac{\mu_E}{r}$$ where $v$ is the magnitude of the spacecraft's velocity, $\mu_E$ is the Earth's gravitational parameter, and $r$ is the distance between the center of the Earth and the spacecraft's center of mass. Differentiating with respect to time results in $$\dot E = \vec v \cdot \dot {\vec v} + \frac{\mu_E}{r^2}\dot r$$ Assuming the vehicle is subject only to gravitational and drag forces results in $\dot{\vec v} = \vec a_{\text{grav}} + \vec a_{\text{drag}}$, which in turn results in $$\dot E = \vec v \cdot \vec a_{\text{drag}}$$ While the component of acceleration normal to the velocity vector can change the orbital eccentricity or the orbital plane, it has zero effect on the orbital energy. Only the component of velocity that is parallel or anti-parellel to the velocity vector changes the orbital energy, and in the case of drag, that component is always anti-parallel (directed against the velocity vector).
While tilting the arrays with respect to horizontal can increase lift, this has no effect on orbital energy because lift is orthogonal to the velocity vector. On the other hand, tilting the arrays with respect to horizontal markedly increases the cross section to drag. This in turn increases the component of drag acceleration that is anti-parallel to the velocity vector.
answered Jul 15, 2017 at 12:05