# Orbital vocabulary confusion! How can the tangential velocity of an elliptical Kepler orbit not be tangent to the orbit?

I'm now officially confused about the usage of "tangential" when breaking down orbital velocity components. It started with edits and comments on this answer to Orbital speed is (vector) sum of tangential and normal speed?

I've (possibly/probably incorrectly) used "tangential" to refer to the velocity component perpendicular to radial in Low-thrust spiraling to escape, is the flight path angle (gamma) at C3=0 always 39 degrees? and also in How to calculate the flight path angle, γ, from a state vector?. I say possibly/probably inccorectly because the the velocity should always be tangent to the orbit. But rather than correct me, @MarkAdler's answer to the first question continues the distinction between tangential velocity and the direction of motion:

Below is the same plot for when accelerating tangentially, as opposed to in the velocity direction.

and @TomSpilker's answer to the second question does likewise:

In addition to $$\gamma$$, the angle between the tangential direction and the velocity vector, there is $$\beta$$, the angle between the radial direction and the velocity vector.

However, the diagram below from Julio@'s answer to ** suggests the component perpendicular to the radial direction might be called normal velocity.

Question: How can the tangential velocity of an elliptical Kepler orbit not be tangent to the orbit, but instead be perpendicular to the radial component? Help me Mr. Wizzard!

Source

• There seems to be indeed some confusion of vocabulary, with "tangential" used instead of "transverse". See, for example, daviddarling.info/encyclopedia/T/transversevel.html . – Litho Oct 8 '19 at 15:39
• @Litho Oh, that's cool! Would you be willing to post that exact comment as another answer? It helps me understand why the authors of the linked answers didn't correct me. – uhoh Oct 8 '19 at 15:44
• @Litho wait, I think I see what's happening. Searching that image, I found this Astronomy presentation (starting on slide 2) In this context the velocity is tangent to the sphere that contains the object being observed from the center of the sphere. For stellar motion, there is radial velocity (usually from Doppler shift) and tangential velocity (usually from proper motion). That's a different context than the velocity components of an object in a Kepler orbit being analyzed mathematically, but it seems to have "leaked" – uhoh Oct 8 '19 at 15:49
• I don't know what the correct term for $v_n$ in your diagram is, but surely the correct term for $v$ is simply "velocity". And yes, the velocity of any object moving along a continuous and differentiable trajectory is always tangential to said trajectory, by definition. – Ilmari Karonen Oct 8 '19 at 22:05

If "tangential velocity" is not tangental to the orbit, what else can it be tangential to? I studied orbital mechanics in Howard Curtis' Orbital Mechanics for Engineering Students, and while I don't claim that it holds absolute truth, it uses the same definition as the diagram you posted (but less clutter):

(Figure 2.8 from H. Curtis, Orbital Mechanics for Engineering Students)

The velocity $$v := \dot{r}$$ is the sum of the radial velocity component $$v_r$$ and the orthogonal component $$v_\perp$$ (note that all are vectors). In the book $$v_\perp$$ is named "azimuth component", but I find that term not very clarifying. By definition, $$v$$ is tangential to the orbit.

How can the tangential velocity of an elliptical Kepler orbit not be tangent to the orbit, but instead be perpendicular to the radial component?

It cannot be. The tangential velocity is tangential to the orbit. Anything else does not make sense in my (humble) opinion.

• Thank you for the concise yet complete answer! I agree with this 100%, and that my usage of "tangential" for the azimuthal component in those older posts was incorrect. (The users who posted the two answers I've linked to are both quite comfortable correcting me as the need arises, so the mystery remains why they've both gone along with said usage in their posts.) – uhoh Oct 8 '19 at 8:06
• There's some additional comments below the question you might want to check out. – uhoh Oct 8 '19 at 16:22