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This question already has an answer here:

Edit: This question is no duplicate because here the dropping of the atmospheric density together with the horizontal line, play an important role.
Furthermore none of the answers and question mentioned above adressed the escape velocity.

The question "What would a "Kármán plane look like, a bird, or a plane ?" uses the Kármán line definition to estimate the wing loading of such a plane.

When an airplane would follow the straight line in that definition it can be calculated that within a minute that plane would have to go beyond escape velocity because when going higher the atmospheric density drops dramatically.

To calculate where on the straight line the escape velocity will happen, we will have to know the atmospheric densities at different altitudes.
This question shows us that at 105 km altitude the density has dropped more than half the density at 100 km, and because the speed of the plane is proportional to the square root of the inverse of that density, that speed will increase with $\sqrt{2}$.
So for an orbital velocity at 100 km, this will mean the escape velocity will be reached at 105 km.
To calculate what the distance on the straight line between the 100 km altitude point and the 105 km altitude point is we can apply a right-angled triangle with one side between those points and the two other sides going to the center of the Earth.
The length of those sides are 6478 km and 6483 km respectively so the cosine of the angle at the center of the Earth is thus 6478 divided by 6483 giving an angle there of about 2$^0$.
The sine of that angle gives us that the distance on the straight line between the 100 km altitude point and the 105 km altitude point is about 226 km.

With an orbital velocity of 7 km/sec this means that in about half a minute the Kármán plane will reach the altitude where it has to have escape velocity to maintain a straight line.

As soon as the Kármán plane leaves the altitude of the Kármán line definition then, because of the straight line, it simply cannot maintain the orbital velocity determined by that altitude !

So should not a Kármán plane just follow the curvature of the Earth ?

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marked as duplicate by JCRM, uhoh, Dr Sheldon, Steve Linton, Magic Octopus Urn Dec 26 '18 at 4:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ why does it matter what it does when it is no longer at the karman line travelling at orbital velocity? It's an abstract construct. $\endgroup$ – JCRM Dec 25 '18 at 11:36
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    $\begingroup$ The ad hoc and informal way that this was first done used an instantaneous evaluation of forces and involved no propagation of motion into the future. OP continues to play games with questions by pretending the Karman line is something that it's been well established that it isn't. I recommend the OP proposes a new site called orbit golf, similar to code golf in area51 and stop playing games here. $\endgroup$ – uhoh Dec 25 '18 at 12:13
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    $\begingroup$ @Conelisinspace Again that is wrong. The calculation was instantaneous. There was no motion, no trajectory. It was evaluated at a single moment in time. Just like I've already stated above $\endgroup$ – uhoh Dec 25 '18 at 15:56
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    $\begingroup$ Your calculation is mistaken. You have failed to factor in the centrifugal force of the plane as it accelerates, which would reduce the speed requirement to generate sufficient lift to continue flying in a straight line. $\endgroup$ – JCRM Dec 25 '18 at 21:30
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No, it doesn't reach escape velocity.

For simplicity, I'm assuming the hypothetical object has a propulsion system that counteracts the drag - otherwise a steady state is impossible

As it rises, the Coriolis force decelerates the notional vehicle. The vis-viva equation can be used to determine the magnitude of this effect.

As it rises, the atmosphere thins further. Atmospheric models can be used to determine the magnitude of this effect.

Both of these effects diminish the lift. A detailed understanding of the parameters for the coefficient of lift of the imaginary vehicle will be needed to determine the variance.

Eventually, the thought experiment would reach an equilibrium altitude where gravity is equal to the sum of both the lift and the centrifugal force, and would orbit at the higher altitude, for as long as it was able to counteract drag. However, all the time the abstract object is above the Karman line then centrifugal force exceeds the lift. (To clarify: Yes, a Karman plane would follow the curvature of the Earth, however it would do it at a higher altitude than the Karman line. Or at the Karman line, but at a lower velocity. Or at the Karman line, at orbital velocity, but at an AoA that produces no lift)

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  • $\begingroup$ The definition oif the Kármán line includes that the plane travels in a straight line. To stay at that line iit has to travel faster and faster because of the dramatically decreasing atmospheric density. $\endgroup$ – Conelisinspace Dec 25 '18 at 12:45
  • $\begingroup$ @Conelisinspace No it does not. You are just referring to a random sentence in a Wikipedia article that's under development. If you want to argue for or against the sentence, the article has a talk page. $\endgroup$ – uhoh Dec 25 '18 at 14:57
  • $\begingroup$ @Conelisinspace, it says "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" Travelling in this straight line, however, moves it above the Karman line so the plane is no longer able, using lift and centrifugal force alone, to travel in a straight line. $\endgroup$ – JCRM Dec 25 '18 at 17:41
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    $\begingroup$ Hang on, @Conelisinspace. You've not been confusing "a straight line" as being the Karman line? $\endgroup$ – JCRM Dec 25 '18 at 18:32
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    $\begingroup$ @uhoh and @JCRM. I started a discussion on Meta about why this question was closed by you. space.meta.stackexchange.com/questions/1157/… $\endgroup$ – Conelisinspace Jan 3 at 18:48

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