The lift force for an airplane is: $$ F_L = 0.5 \rho v^2 S C_L $$

According to this talk page the lift coefficient for a supersonic airplane is: $$ C_L = \frac{4\alpha}{\sqrt{M^2-1}} $$ where $\alpha$ is the angle of attack in radians and $M$ is the Mach number.
(According to one of the editors, instead of 4$\alpha$ the numerator could be 4$\sin(\alpha)$, with $\alpha$ in degrees.)

To calculate the different forces acting on a supersonic Kármán plane we can take the North American X-15 as an example.

With 4$\alpha$ = 2 and $M$ = 25 the lift coefficient would become: $C_L$ = 0.08 .

With $\rho$ = 5,6 x 10$^-$$^7$, $v$ = 7.5 km/sec and $S$ = 18.6 : $F_L$(X-15) = 23.4

The gravitational force is:
$$F_G = \frac{GM_Em}{(R+h)^2} $$

With $h$ = 100 and $m$ = 7000 : $F_G$(X-15) = 66,667 and thus $F_L$(X-15) < 0.04 % of $F_G$(X-15)

Because this example shows that the lift force at the Kármán line altitude is only a small fraction of the gravitational force how can the question "What would a "Kármán plane" look like, a bird or a plane ? " still ignore then the acceleration downwards to the centre of the Earth ?


marked as duplicate by JCRM, uhoh, Mark Omo, Dr Sheldon, Nathan Tuggy Jan 13 at 4:42

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.

  • $\begingroup$ 1) Your final question: Where is gravity ignored in the linked question? It's the central part of the equation. 2) Please don't just take arbitrary approximations that may be valid for common plane designs at common altitudes and supersonic speeds and use them for calculations without knowing if they are still valid at hypersonic speeds and air densities more than 1000x lower. You need to do a full hydrodynamic simulation to get any reasonable numbers so far from any common environment. $\endgroup$ – asdfex Jan 12 at 18:00
  • $\begingroup$ @asdfex I don't say that gravity is ignored, the huge difference between the lift force and the gravitational force is ignored there, they are set equal. $\endgroup$ – Conelisinspace Jan 12 at 18:26
  • $\begingroup$ @asdfex I don't have the means to do (hydro-) ? dynamic simulations, so all we can do are rough calculations that give an indication that the $C_L$ will become small and as a consequence the lift force will too. $\endgroup$ – Conelisinspace Jan 12 at 18:33
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    $\begingroup$ @uhoh: Consensus is complicated, but the case for not double-counting votes on similar answers does not need to be voted on. $\endgroup$ – Nathan Tuggy Jan 13 at 5:06
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    $\begingroup$ He's been using diaeresis for a while, @uhoh, but since the other contributor used them. $\endgroup$ – JCRM Jan 13 at 8:58