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

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

Is not the equation above, meaning that gravitational force minus the lift force equals the force caused by the acceleration towards the centre of the Earth, the right one for an airplane flying and maintaining the altitude of the Kármán line ?
Or is the following equation, derived from this question that ignores the acceleration towards the centre of the Earth, the right one ? $$\frac{GM_Em}{(R+h)^2} - \frac{\rho v^2 S C_L}{2} = 0$$

If that equation would be right the North American X-15 would have to have a $C_L>200$ !

Lift force X-15: 0.5 x 5.6 x 10$^-$$^7$ x 56.25 x 10$^6$ x 18.6 x 0.08 = 23.4

Gravitational force: $$\frac{4. 10^7.10^7 .7000}{ 42.10^6.10^6} = 66.7 . 10^3 $$

  • $\begingroup$ Let me give a question as response: Imagine the plane not flying straight ahead (be it at constant altitude or with constant direction), but flying in a large circle. Besides the force turning the plane (acting perpendicular to all the other forces) there should be no difference, agree? Where do you see the centrifugal force in this case? $\endgroup$
    – asdfex
    Commented Dec 26, 2018 at 15:22
  • $\begingroup$ @asdfex Interesting thought experiment ! I guess the centrifugal force's direction stays lying in the plane of the circle, so when the circle gets larger and larger it will follow the curvature of the Earth more and more, so with reference to the Earth it will turn from horizontal to vertical. $\endgroup$
    – Cornelis
    Commented Dec 26, 2018 at 16:44
  • $\begingroup$ where did $v^2$ and $C_L$ disappear to on your fourth line? $\endgroup$
    – user20636
    Commented Dec 27, 2018 at 10:23

2 Answers 2


Just looking at your equations, I can say the first one seems the be closer to the truth. At Karman's line(which seems to be the limit between 'atmosphere' and space), you have rho(density of air) closer to 0. Infinitezimally close to 0. so the second term of the equation would be 0. If you consider this in your second equation, you would get that the gravitaional force at karmans line would be 0, which is false. However, you kinda fail to account for something, the air-induced drag. If you have aerodynamic lift, you have air drag. This is the reason why you cannot have an orbiting body in Earth's atmosphere, the air drag would simply slow it down too fast. Also, if you want to keep level flight at karmans line you have another small problem, you alpha would need to be close to 0, alpha being the Angle of attack. This means that CL would be also close to 0


I think I solved it... I'm not sure but I think I'm on the right path.

You were very close to the right solution, and you were very correct the linked equation is useless in describing actual trajectory of a "Karman Plane" - it only works for flat Earth :) Karman's Line is an abstraction that has very little physical meaning in the real world - it's derived from a lousy hybrid of two equations:

  • the aviation equation for level flight: $mg = {\rho v^2 S C_l \over 2}$ (weight = lift; $W=L$); assumptions: flat Earth, constant gravity

  • the circular orbit equation: $v^2 = {GM\over r}$ or 'unabridged' ${G M_e m \over r^2 } = {m v^2 \over r}$ (centripetal acceleration = gravitational acceleration, $W=ma$); assumptions: vacuum. (also, $r = R+h$)

The hybrid extends the level flight equation by the gravity drop-off with altitude, but doesn't account for Earth being round, a simplification one can't afford in orbital mechanics.

Your version correctly expands the orbital equation by giving it an aerodynamic component: $W-L=ma$, where

  • Lift $L = {\rho v^2 S C_l \over 2}$; (with $\rho$ being a function of height; proportional to pressure approximated as $P_0 e^{- {h \over H}}$)

  • Weight $W = {G M_e m \over (R+h)^2 }$

  • centripetal acceleration $a = {m v^2 \over R+h}$

Now, why your equation won't help:

If we take standard lift, it will never vanish; its density component is dropping off exponentially with altitude, and as orbital speed also drops with altitude, the velocity component will drop with it, so it's a huge $x e^x$ drop-off rate, but never zero. This makes it impossible to find the altitude where the airplane "doesn't produce lift". Angle the solar panels of ISS right and they will still produce some millinewtons of lift.

But what we can do is to take negative lift coefficient, try to prevent the spaceplane from being ejected into a higher orbit, and find an altitude where we'll fail that. Fly the airplane "belly-up" and see how high can it go at speed slightly exceeding the orbital speed for given altitude before it gets ejected into an elliptical orbit as its dwindling lift fails to prevent that.

So let's switch the sign: $W+L=ma$

The velocity is slightly larger than necessary. Not infinitesimally but of order of just a couple m/s.

$v_k^2 = {GM_e \over r}+\epsilon$

The equation will take form:

${G M_e m \over r^2 }+{\rho v_k^2 S C_l \over 2}= {m v_k^2 \over r}$

Let's substitute the $v_k^2$

${G M_e m \over r^2 }= ({m \over r} - {\rho S C_l \over 2} )v_k^2$

${G M_e m \over r }= ({m } - {\rho r S C_l \over 2} )v_k^2$

${G M_e m \over r }= ({m } - {\rho r S C_l \over 2} )( {GM_e \over r}+\epsilon)$

Simplify it a bit, solve for r

$m = (m - {\rho r S C_l \over 2} )( 1+{ \epsilon r \over GM_e})$

$0 = - {\rho r S C_l \over 2} - {\rho \epsilon r^2 S C_l \over 2GM_e} + { m \epsilon r \over GM_e}$

$0 = - {GM_e \rho r S C_l } - {\rho \epsilon r^2 S C_l } + { 2m \epsilon r}$

$0 = (2m \epsilon - {GM_e \rho S C_l })r - {\rho \epsilon r^2 S C_l }$

$r = {2m \epsilon - {GM_e \rho S C_l } \over {\rho \epsilon S C_l } }$

and reorder into aerodynamic and gravitational parts.

$r = {2m \over \rho S C_l} - {GM_e \over \epsilon } $

And so, the equation to solve would be

$R+h = {2m \over \rho(h) S C_l} - {GM_e \over \epsilon } $

Normally, the $GM_e \over \epsilon$ part should be very large - but even for quite small $\epsilon$, $\rho(h)$ diminishing exponentially should create growth of the aerodynamic part rapid enough to give a good solution.

  • 1
    $\begingroup$ @uhoh: "In practice, supporting full weight wouldn't be necessary to maintain altitude because the curvature of the Earth adds centrifugal lift as the airplane reaches orbital speed. However, the Kármán line definition ignores this effect because orbital velocity is implicitly sufficient to maintain any altitude regardless of atmospheric density." - I'd say this is a fault in the definition. $\endgroup$
    – SF.
    Commented Dec 27, 2018 at 1:44
  • $\begingroup$ but solving that equation is complicated by $C_L$ not being constant $\endgroup$
    – user20636
    Commented Dec 27, 2018 at 11:43
  • $\begingroup$ @JCRM: You set $C_L$ and $S$ to sane values of an airplane that 'might attempt this', at least that's how Karman went about it. $\endgroup$
    – SF.
    Commented Dec 27, 2018 at 20:36
  • $\begingroup$ The velocity change in your $\epsilon$ needs to be more than 6 km/sec for a positive $R+h$ $\endgroup$
    – Cornelis
    Commented Jan 5, 2019 at 16:17

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