<p>"For any object moving through space, the velocity vector is tangent to the trajectory."(Citing <a href="https://en.m.wikipedia.org/wiki/Orbital_state_vectors" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Orbital_state_vectors</a>).</p>

<p>Therefore, it's both of these components, there is no normal component<sup>1</sup> (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.</p>

<p>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).</p>

<p><sup>1</sup>However 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. </p>

<p><span class="math-container">$$v^2 = v_{||}^2 + v_{\perp}^2$$</span></p>

<p>But in two-body Kepler orbits there is no velocity component <em>normal</em> to the plane of the orbit, which is the plane containing the <span class="math-container">$\mathbf{v_{||}}$</span> and <span class="math-container">$\mathbf{v_\perp}$</span>, which is also the plane containing <span class="math-container">$\mathbf{r}$</span> and <span class="math-container">$\mathbf{v}$</span>.</p>

CC BY-SA 4.0
2019-10-07T00:26:40.943Z
I added a bit to your answer, feel free to modify further roll back.
uhoh
CC BY-SA 4.0
2019-10-07T09:55:25.163Z
Perpendicular component is perpendicular to the tangent to trajectory (not to the two-body orbit plane itself). So still not clear - "orbital speed" in terms of wiki definition and moon's 1 km/s includes that perpendicular component or not.
Code Complete
CC BY-SA 4.0
2019-10-07T19:31:19.937Z
What @uhoh probably meant is that you can "separate" the velocity into two vectors that are normal to one another. The Moon's orbital speed being 1 km/s means the tangent speed is 1 km/s and the "normal speed" is 0 so it could move in a circle while affected by gravity.
Tosic
CC BY-SA 4.0
2019-10-08T04:19:25.877Z
@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?](https://space.stackexchange.com/q/39237/12102)
uhoh
CC BY-SA 4.0
2019-10-08T08:05:59.420Z
@Tosic Moon's orbit is not a straight line (relative to moon-earth center of mass, or relative to earth or relative to whatever) => normal (perpendicular) speed component cannot be zero! Normal speed component arises because of the curvature of the trajectory (trajectory's being different from a straight line)!  Even ideal circle orbit has non-zero normal speed!
Code Complete
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2019-10-08T08:40:01.060Z
@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.
Tosic
CC BY-SA 4.0
2019-10-08T08:48:42.547Z
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.
Tosic
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2019-10-09T05:35:34.973Z
@CodeComplete The vector diagram in the answer to the question which uhoh linked should make the situation clear.
PM 2Ring
CC BY-SA 4.0
2019-10-09T17:52:03.327Z
if there is a normal force component (perpendicular to the trajectory), it means that there is respective acceleration component (it is a vector directed exactly as that force component) and it means that there is respective speed component (it is a vector directed exactly as that acceleration component). So normal speed component should exist!
Code Complete
CC BY-SA 4.0
2019-10-09T18:59:00.157Z
I disagree. There is acceleration in some direction, we agree on that (and that is because of II Newton's law). Now, that means that after every infinitely small amount of time the velocity (not speed, this is crucial) changes by a*dt where a is a vector and dt is a scalar whose value is that of the small change. Try drawing this and adding these two vectors. The fact dt is infinitely small will mean the speed (the intensity of velocity) will not change, but its direction will (by an infinitely small amount), and that applied infinitely many times will create a circle.
Tosic
CC BY-SA 4.0
2019-10-06T15:07:01.657Z
2019-10-08T08:38:31.250Z
2019-10-08T08:38:31.250Z
Tosic
<p><a href="https://en.wikipedia.org/wiki/Orbital_speed" rel="nofollow noreferrer">Orbital speed</a> 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).</p>

<p>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)?</p>

<orbit><the-moon><orbital-mechanics><orbital-elements>
Orbital speed is (vector) sum of tangential and normal speed?
Code Complete
# Orbital speed is (vector) sum of tangential and normal speed?

(Asked by: username_0 on 2019-10-06)

[Orbital speed](https://en.wikipedia.org/wiki/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)?

**Tags:** orbit, the-moon, orbital-mechanics, orbital-elements


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## 1 answers

### Answer by username_1 on 2019-10-06. Score: 2

"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 $\mathbf{v}$.

> **Comments:**
> **username_2**: I added a bit to your answer, feel free to modify further roll back.  
> **username_0**: Perpendicular component is perpendicular to the tangent to trajectory (not to the two-body orbit plane itself). So still not clear - "orbital speed" in terms of wiki definition and moon's 1 km/s includes that perpendicular component or not.  
> **username_1**: What @username_2 probably meant is that you can "separate" the velocity into two vectors that are normal to one another. The Moon's orbital speed being 1 km/s means the tangent speed is 1 km/s and the "normal speed" is 0 so it could move in a circle while affected by gravity.  
> **username_2**: @username_1 I am now *officially confused!* [Orbital vocabulary emergency! How can the tangential velocity of an elliptical Kepler orbit not be tangent to the orbit?](https://space.stackexchange.com/q/39237/12102)  
> **username_0**: @username_1 Moon's orbit is not a straight line (relative to moon-earth center of mass, or relative to earth or relative to whatever) => normal (perpendicular) speed component cannot be zero! Normal speed component arises because of the curvature of the trajectory (trajectory's being different from a straight line)!  Even ideal circle orbit has non-zero normal speed!  
> **username_1**: @username_2 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.  
> **username_1**: 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.  
> **username_3**: @CodeComplete The vector diagram in the answer to the question which username_2 linked should make the situation clear.  
> **username_0**: if there is a normal force component (perpendicular to the trajectory), it means that there is respective acceleration component (it is a vector directed exactly as that force component) and it means that there is respective speed component (it is a vector directed exactly as that acceleration component). So normal speed component should exist!  
> **username_1**: I disagree. There is acceleration in some direction, we agree on that (and that is because of II Newton's law). Now, that means that after every infinitely small amount of time the velocity (not speed, this is crucial) changes by a*dt where a is a vector and dt is a scalar whose value is that of the small change. Try drawing this and adding these two vectors. The fact dt is infinitely small will mean the speed (the intensity of velocity) will not change, but its direction will (by an infinitely small amount), and that applied infinitely many times will create a circle.  

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