# Spiraling out from circular orbit to escape via low thrust, what is γ (gamma)?

@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.

• 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). – Julio Jul 12 '18 at 9:06
• @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! – uhoh Jul 12 '18 at 9:32

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).