Edit: Although this question seems similar to the question "Is it correct to apply the vis-viva equation to an airplane that flies in a straight line", it is different because there the airplane flies in a straight line and in this question it is in orbit.

The vis-viva equation models the motion of an orbiting body and it applies when the only force acting on the body is it's own weight.

So is it correct to use this equation to an airplane that flies at the Kármán line with an orbital velocity that generates a strong lift force ?

I ask this question because i'm puzzled by this question from @uhoh about the Kármán plane where the vis-viva equation is used for an airplane.


closed as unclear what you're asking by uhoh, Steve Linton, JCRM, Mark Omo, Rory Alsop Dec 21 '18 at 10:03

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  • $\begingroup$ Without a specific example this question is too vague to have an answer. "Apply" and "correct" are open to way too much interpretation for this question to have a conclusive answer. If you want to know if X is correct, you have to show X. $\endgroup$ – uhoh Dec 19 '18 at 12:24
  • $\begingroup$ @uhoh i've added a link to your question about the Kármán plane. $\endgroup$ – Conelisinspace Dec 19 '18 at 13:20
  • $\begingroup$ You did that in your last question, and that was answered there. I still don't think "applied" is unambiguous. Different people can imagine applying an equation to have different meanings and functions. This is math and physics equations, and that kind of question should be clear and unambiguous. To me this looks like you are playing games with words, or trying to use a question to make a point, rather than to ask a genuine question. $\endgroup$ – uhoh Dec 19 '18 at 13:33
  • $\begingroup$ @uhoh So i changed to "use" instead of " apply" . $\endgroup$ – Conelisinspace Dec 19 '18 at 13:41
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Per definition the maximal lift generatable at the Kármán line is

$$L = m \cdot g$$

so if your body is generating significant lift at sea level it will probably still generate enough lift at that altitude to divert from a "vis-viva orbit" after a short while. The bigger problem however is drag. To quote Wikipedia:

Due to atmospheric drag, the lowest altitude at which an object in a circular orbit can complete at least one full revolution without propulsion is approximately 150 km (93 mi), whereas an object can maintain an elliptical orbit with perigee as low as about 130 km (81 mi) without propulsion.

So simplifying the orbit to have no other forces, as the vis-viva equation states, specifically to be a frictionless one would be a gross misstatement.

In general during deorbit of a spacecraft the altitude around the Kármán line is considered part of the reentry.


(Image Credit: Comparison of lifting reentry and ballistic reentry by A. Rathan Babu, P. Vijay Kumar, B. Praveen, and R. Suresh Kumar)

As you can see in the graphic:

  1. There is a measurable deceleration for a reentering spacecraft at that altitude
  2. The friction depends on the shape, size and orientation of the object

Spy satellites or earth observation satellites orbiting at similarly low altitude usually have a propulsion system firing permanently or quite frequently to keep the orbit and are built to be very aerodynamic to minimize friction.


There is enough friction at that altitude to make the vis-viva equation not applicable for even short term orbit calculation.

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    $\begingroup$ So you have ruled out mis-application. Is that the only application? $\endgroup$ – uhoh Dec 19 '18 at 12:59
  • $\begingroup$ Well, you could use it to compare to an orbit without additional forces, maybe even calculate the force from the acceleration, since you know the deviation from a frictionless orbit. $\endgroup$ – Hans Dec 19 '18 at 13:27

I'm not sure where the confusion comes from. Vis-viva is derived from a total energy $\epsilon$ that knows only kinetic and potential contribution (see the derivation)

$$ \epsilon = \frac{1}{2}mv^2 - \frac{GM}{r^2} .$$

Once the corresponding Euler-Lagrange equation is computed and transformed into the co-orbiting system, it is clear that those two energies correspond to centrifugal and gravitational force. No other force. If you'd try to apply any other force to the system, then vis-viva is not applicable to calculate $r$ or $v$. In other words vis-viva is wrong in that case.
You would either have to know a potential for the new force you're applying, in order to derive equations of motion, or if you already know the equqtions of motion then solve them and re-derive a modified vis-viva. This however might not be possible if your force is dissipative.

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    $\begingroup$ There is no confusion. The definition of the Karman line is an artificially contrived scenario where a craft experiencing both lift and drag is nonetheless traveling at "orbital velocity" but not in orbit, because it also uses propulsion. It's an awkward, artificial definition, but to reconstruct it you need an expression for "orbital velocity" and that's where vis-viva is used. OP needs to understand that anyone can come up with their own definition. It was "space lawyers" who took this idea and ran with it. $\endgroup$ – uhoh Dec 19 '18 at 15:02

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