In this video the person resolves the momentum vector into two components, tangential and radial. But isn't the velocity at every point on the orbit tangential?
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$\begingroup$ Does the statement "the velocity at every point on the orbit must be tangential" apply to elliptical orbits? $\endgroup$ – Arturs C. Oct 26 '18 at 6:41
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1$\begingroup$ Possible duplicate: physics.stackexchange.com/q/349811/2451 $\endgroup$ – Qmechanic♦ Oct 26 '18 at 6:55
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$\begingroup$ @ArtursC. Yes,it does! $\endgroup$ – ayc Oct 26 '18 at 7:42
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1$\begingroup$ Please include the explanation you're asking about in your question. $\endgroup$ – Dmitry Grigoryev Oct 26 '18 at 10:58
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1$\begingroup$ @DmitryGrigoryev , I had actually explained what I had to ask but it seems like my question was edited for reasons unknown. $\endgroup$ – Mike Victor Oct 26 '18 at 12:10
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I think it's just a misunderstanding!
But the velocity at every point on the orbit must be tangential right?
Yes,it is and that's why the actual momentum vector is tangential to the ellipse
this person resolves the momentum vector into two components , tangential and radial
And yes he did.But,you should notice that he called one radial and the other perpendicular i.e the resolution is done according to the line joining the object's(planet's) location and sun and the object has radial velocity because radial velocity is defined as the component of the object's velocity that points in the direction of the radius connecting the object and the point.
And if you look at what you are calling as tangential velocity you would notice that this component, i.e perpendicular to line joining planet and sun, isn't tangential to the ellipse.It's just perpendicular to the line joining the planet and the ellipse.
Conclusion: The planet always has velocity tangential to the ellipse and the velocity perpendicular to the line joining planet and object isn't tangential to the ellipse at all instants.
Note:
Although in your question you particularly ask about momentum I just used the term velocity rather than momentum because I think it is easier to understand this way.
If you need momentum at any instant just multiply total velocity with mass(p=mv)
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$\begingroup$ To put it another way the decomposition into "tangential" and "radial" parts is in polar coordinates rather than relative the ellipse defined by the orbit. As such the "tangential" part might be better called "azimuthal". $\endgroup$ – dmckee♦ Oct 26 '18 at 15:58
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$\begingroup$ @dmckee.Could you be a little bit more specific. $\endgroup$ – ayc Oct 26 '18 at 16:29
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$\begingroup$ The overall velocity vector can always be equal to its component perpendicular to the line between the two objects, but this happens only when the orbit is perfectly circular. $\endgroup$ – zwol Oct 26 '18 at 16:39
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2$\begingroup$ He is working in a cylindrical coordinate system. It is obvious what the "radial direction" means, but "tangential" is a poor name for the component perpendicular to the radial direction, though the alternatives aren't much better. In cylindrical polar coordinates in 3D the directions are often called axial, radial, and circumferential rather than tangential. The tangent to the ellipse is not the same as the "tangential direction," except at the points where the radial coordinate is a maximum or a minimum. $\endgroup$ – alephzero Oct 26 '18 at 17:04
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