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Orbital speed definition in wiki does not state clearly - it is just tangential speed component or square root of squares of normal and tangential speed (full speed vector).

When we say that moon orbital speed is 1 km per sec, we don't know its tangential speed (speed along its trajectory), right? We need some complicated math to calculate it (because of normal speed component inside that 1 km/sec orbital speed)?

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"For any object moving through space, the velocity vector is tangent to the trajectory."(Citing https://en.m.wikipedia.org/wiki/Orbital_state_vectors).

Therefore, it's both of these components, there is no normal component1 (relative to the central body)... the orbital speed represents the intensity of the orbital velocity which is always tangent to the ellipse (or parabola/hyperbola) the body in orbit describes.

The Moon's orbital speed being 1 km/s means it moves along an ellipse around the Earth at a speed of 1 km/s (this speed is roughly constant, so that ellipse can sometimes be approximated to a circle).

1However you can separate orbital motion into two components normal to each other (the most popular such separation is to the radial component and one normal to it, but this is unnecessary for circular orbits where there is no radial velocity, only radial acceleration which causes the curvature of the trajectory), and in that case the orbital speed will indeed be the square root of sum of the squares of those two.

$$v^2 = v_{||}^2 + v_{\perp}^2$$

But in two-body Kepler orbits there is no velocity component normal to the plane of the orbit, which is the plane containing the $\mathbf{v_{||}}$ and $\mathbf{v_\perp}$, which is also the plane containing $\mathbf{r}$ and $\mathbf{v}$.

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    $\begingroup$ I added a bit to your answer, feel free to modify further roll back. $\endgroup$
    – uhoh
    Oct 7, 2019 at 0:26
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    $\begingroup$ @Tosic I am now officially confused! Orbital vocabulary emergency! How can the tangential velocity of an elliptical Kepler orbit not be tangent to the orbit? $\endgroup$
    – uhoh
    Oct 8, 2019 at 4:19
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    $\begingroup$ @uhoh I got confused as well so I made the answer more general ... I am glad your question was answered and things were cleared up, I think the way I edited it now is the most "comfortable", but please do correct me if I am wrong. $\endgroup$
    – Tosic
    Oct 8, 2019 at 8:40
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    $\begingroup$ You are right, it is not a straight line, and there is a radial force (the gravitational force, which is the centripetal force of this rotational movement), but that does not mean there is a radial component of velocity in a circular orbit (it means there is normal acceleration), the curvature is caused because after time dt, you add the vector an*dt (an is the normal acceleration) to the velocity and thus change its orientation (not its intensity)... or at least that's what I was taught at school :-) Note: there is no "speed component", speed is the intesity of velocity, which is a vector. $\endgroup$
    – Tosic
    Oct 8, 2019 at 8:48
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    $\begingroup$ @CodeComplete The vector diagram in the answer to the question which uhoh linked should make the situation clear. $\endgroup$
    – PM 2Ring
    Oct 9, 2019 at 5:35

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