Can the centrifugal force still be ignored as the lift force is a small fraction of the gravitational force at the Kármán line altitude? [duplicate]

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 TuggyJan 13 at 4:42

• @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. – Conelisinspace Jan 12 at 18:33