@MarkAdler's comment led me to ask Why would a slow spiral from a C3 of zero take about 2.4 times as much ΔV as an impulsive maneuver? which resulted in this tidy and efficient @MarkAdler answer which points to another thoughtful answer about slowly spiraling out from a circular orbit to escape in the limit of very weak prograde propulsion, which (at first counterintuitively) slows you down while raising your orbit.

Below that answer is yet another easter-egg-like comment gem.

Always aligned with the velocity vector. That is the most efficient use of the thrust in order to increase the specific energy. The final γ is 31°.

Question: In this context, what is the angle γ? How is it defined?

The phenomenon is correct but 31° was a typo and should be about 39.2° instead.

  • $\begingroup$ It seems that $\gamma$ is the flight path angle of the orbit. The final value appears to have a dependance on the value of the applied constant acceleration (the value of flight path angle of the second plot is near zero as I can see). $\endgroup$
    – Julio
    Commented Jul 12, 2018 at 9:06
  • $\begingroup$ @Julio I don't know what "flight path angle" means in general, and especially for a spacecraft escaping to infinity. If there is a diagram that shows how this angle would be defined in this case, that would be great! $\endgroup$
    – uhoh
    Commented Jul 12, 2018 at 9:32

1 Answer 1


Image illustrating the flight path angle, $\gamma$, as requested by uhoh's comment



It is just the angle between the velocity vector and the tangential orbit component, $\vec{e}_{\theta}=[-\sin(\theta), \cos(\theta), 0]^T$, assuming the orbit plane is the $XY$ plane. The bounds for $\gamma(t)$$\in$$[-\pi/2, \pi/2]$, an the zeros happens at periapsis and apoapsis (or always if the orbit is circular).

I have seen people taking the flight path angle as the one subtended between $\vec{e}_r$ and $\vec{v}$ ($\beta$ in the image). However, in my opinion, it is quite confusing since it not coincide with the aircraft definition of flight path angle (using $\gamma$ does coincide).

In the Mark Adler's simulations, there appear to be two cases, one with high constant acceleration value to highlight the spiral behaviour where the final flight path angle seems to be 31º and one with a low value of the acceleration where the final path angle seems to be near 0º (at look), so it has clearly a dependance of the applied acceleration. More acceleration changes the orbit more faster to elliptic ones (and hence final flight path angle is higher, if termination does not occur at periapsis or apoapsis) and low acceleration seems to keep the orbit quasi-circular when raising (hence the final flight path angle is almost null).

  • $\begingroup$ Thanks for the excellent explanation! I'm give this some thought. $\endgroup$
    – uhoh
    Commented Jul 12, 2018 at 10:54
  • $\begingroup$ I need some help! I've just asked How to calculate the flight path angle, γ, from a state vector? $\endgroup$
    – uhoh
    Commented Jul 14, 2018 at 6:27
  • 1
    $\begingroup$ what-when-how.com/space-science-and-technology/… $\endgroup$
    – Julio
    Commented Dec 16, 2018 at 0:52
  • $\begingroup$ Thanks for that! I've added it back to the answer to credit the image as well. That's a nifty tutorial of the problem, maybe the best I've ever seen for this problem! $\endgroup$
    – uhoh
    Commented Dec 16, 2018 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.