This question already has an answer here:

The equation for an airplane in orbit with the air providing lift would be:

$$\frac{GM_Em}{(R+h)^2} - \frac{\rho(h) v^2 S C_L}{2} = \frac{mv^2}{R+h} $$

$GM_E$ is Earth's standard gravitational parameter,
$R$ is Earth's radius and $h$ the altitude of the airplane above the surface,
$\rho(h)$ is the air density at altitude $h$ and $S$ is the airplane's wing area,
$C_L$ is the airplane's lift coëfficiënt.

From Wikipedia, about the definition of the Kármán line:

For an airplane flying higher and higher, the increasingly thin air provides less and less lift, requiring increasingly higher speed to create enough lift to hold the airplane up. It eventually reaches an altitude where it must fly so fast to generate lift that it reaches orbital velocity.

But does the airplane ever reach orbital velocity ?

Rearranging the equation above:

$$\rho(h) v^2 S C_L = \frac{2m}{R+h}(\frac{GM_E}{R+h} - v^2) $$ Orbital velocity $v_0$ can be gotten from the vis-viva equation:

$$v_0^2= \frac{GM_E}{R+h} $$ Substituting the $v_0^2$ will give: $$ \rho(h) v^2 S C_L = \frac{2m}{R+h}(v_0^2 - v^2) $$ $$ R+h = \frac{2m}{\rho(h) S C_L}(\frac{v_0^2}{v^2} - 1) $$

When all the variables and constants are positive, $v$ must be less than $v_0$.


marked as duplicate by JCRM, uhoh, David Hammen, Magic Octopus Urn, Mark Omo Jan 5 at 22:37

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.

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    $\begingroup$ questions about how airplanes fly should be asked in Aviation Stack Exchange $\endgroup$ – uhoh Jan 5 at 12:19
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    $\begingroup$ nope, you're arguing about wikipedia again. But if it's just me that thinks that you've got nothing to worry about. $\endgroup$ – JCRM Jan 5 at 16:27
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    $\begingroup$ You have some good questions just move past this subject. $\endgroup$ – Muze Jan 5 at 21:39
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    $\begingroup$ Karman line varies depending on the density of the air and changes altitude and should be called Karman area. It is an imperfect line because of that. $\endgroup$ – Muze Jan 5 at 21:41
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    $\begingroup$ @Conelisinspace yes here is some more support and stop wasting all your reputation. $\endgroup$ – Muze Jan 6 at 18:00

No, if your aeroplane follows that formula, always flying at $C_{L_{max}}$ it will never reach orbital velocity.

But as we've observed in MANY other version of this question, the aeroplane never needs to generate sufficient lift to support itself solely by lift, as centrifugal force contributes.

The Kármán line is the altitude where the speed necessary to aerodynamically support the airplane's full weight equals orbital velocity.

(my emphasis, from Wikipedia)

However, a vehicle in orbit at that altitude is likely to wish to minimise its drag, which it would likely do by changing its AoA to reduce its $C_L$ to zero, (which would likely also minimise its cross section), and so it would need to be going at orbital velocity to maintain its altitude

  • $\begingroup$ Thank you for this answer. Because it is the only one that answers my specific question, I would say it is not a duplicate. $\endgroup$ – Conelisinspace Jan 6 at 16:24
  • $\begingroup$ In an answer to another question it was calculated that already far below the Kármán altitude the lift force would become less than the centrifugal force. So at the Kármán altitude the so called "full weight" would be far less than half of that. $\endgroup$ – Conelisinspace Jan 6 at 16:33
  • $\begingroup$ If you are referring to this answer, it makes an error. That said, The 100km value, and the concept it is derived from are different (each vehicle design would have its own Karman line) $\endgroup$ – JCRM Jan 6 at 17:05
  • $\begingroup$ I would really like to see that error. $\endgroup$ – Conelisinspace Jan 6 at 18:04
  • $\begingroup$ that A is constant @Conelisinspace $\endgroup$ – JCRM Jan 6 at 19:09

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