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I Is tangential velocity in elliptical orbits stable?

  1. May 11, 2016 #1
    Hi, my name is Andreas, and I am new to this forum. I had a couple of questions regarding orbital dynamics, since I am not very familiar with the field:

    Is tangential velocity in elliptical orbits stable like in circular orbits? If not, how exactly does it change? How do you calculate orbital velocity from tangential velocity? What's the mechanism that keeps elliptical orbits from dissipating?
  2. jcsd
  3. May 11, 2016 #2
    well, i will like to know what you mean by stable velocity.
    my guess is that by stable you mean the magnitude being same at each point on its path.
    the evidence is that you call it stable in circular orbits.
    if you look at the motion of a body on elliptical path its angular velocity is changing ,so the speed will vary -keeping areal velocity constant as per Keplar's law.
    the change in velocity does not mean that the planets are dissipating its total energy-the kinetic energy may be changing but the potential energy is also changing and its a conservative system.
    the area swept by the the radius vector drawn from center of the force to the body sweeps equal area in equal time , that means the Areal velocity is constant
    mathematically it means 1/2. r^2 . d(theta)/dt = constant. so its tangential velocity will be changing - it will be slower when r is large and faster when r is smaller.
  4. May 11, 2016 #3
    That's 100% correct. I kinda struggle with the terminology, because English is not my first language.

    Oooh, thanks now I get it. I think I kinda confused the two. By the way, is there a simple derivation of this from Newton's laws?
  5. May 11, 2016 #4
    if you calculate the area swept by the radius vector say in infinitesimal time dt -then the radius vector must have turned by an angle d(theta) and have swept ds a length of an arc- one can use the relation d(theta)= ds /r
    the small triangle formed with base ds has almost same r as two sides ;
    so the area can be written as dA = 1/2 . base . altitude = 1/2. ds. r = 1/2. r.d(theta).r = 1/2. r^2,d(theta)
    the motion has happened in dt time
    Therefore dA/dt = 1/2 . r^2. d(theta)/dt
    but the angular momentum of the body is m. r^2. d(theta)/dt and is a constant of motion as no external torque is acting on the body and mass m of the body being constant - r^2. d(theta)/dt must be constant , therefore areal velocity must be constant.
  6. May 11, 2016 #5
    Thanks, that's a good answer. I am slightly confused though on the matter of orbital and tangential velocity. How do I calculate the orbital velocity if I know the tangential velocity in an ellipse?
  7. May 11, 2016 #6
    what is your definition of orbital velocity?
    if it is orbital speed an average speed of completing one rotation then its a scalar and can be calculated ;
    if you denote it by (a path length/time taken ) then the tangential velocity magnitude can give you a picture as its a curved path.
  8. May 11, 2016 #7
    Yeah, it's orbital speed. I'm sorry, as I have already said, I struggle a bit with the terminology. I want to know how to calculate it if we know the tangent velocity at a given point of the ellipse. Also how to find the "instantaneous" orbital speed. Is it just equal to the magnitude of the tangent velocity or am I making a mistake here?
  9. May 11, 2016 #8


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    A circle is to an ellipse like a square is to a rectangle. Ellipses are much, much more complicated than circles or rectangles however. There is some similarity between a pendulum and a elliptical orbit. In both cases the total energy =potential+kinetic is conserved and constant (but the ratio PE:KE is changing periodically). In our Solar System, most of the planets and moons are in near circular orbits. So, circles are good approximations. Of course, this isn't true for most comets! (which are on highly elliptical or even parabolic 'orbits'). Tangental velocity is instantaneous velocity. The path length of an ellipse does NOT result in a elementary formula like the formulas for circle or rectangles. (see elliptic integrals, non-elementary functions.) These are almost never discussed in introductory physics or math courses. While velocity is constant in a circular orbit, it is NOT in an elliptical orbit. Ellipses are non-elementary.
  10. May 11, 2016 #9
    Yeah, but the question is, how do you find the orbital speed if you know either the periapsis or the apoapsis and the tangential velocity (and the mass of each body, of course)? I guess this probably requires first finding the shape of the ellipse and the path length, and then the time it takes, right? Am I missing something here?
  11. May 11, 2016 #10


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    I don't understand the difference you make between orbital and tangential velocity. For me, these are synonymous.
  12. May 11, 2016 #11


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    If you switch the word Orbital to Angular (and this may be what the OP means) the difference is that the velocity along the tangent is not the same as the radius times the angular velocity. That's just Geometry. It could be as simple as that.
  13. May 12, 2016 #12
    Yeah, that's probably it. I know, I've probably confused all of you very much with my poor terminology, so sorry about that.
  14. May 12, 2016 #13


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    I think this could be the nub of the question, the rest being to do with definitions. There is no loss mechanism in an orbit that's well out of the drag of an atmosphere so the total Energy must remain constant. At any Point, the PE plus the KE will be the same as at any other point. In an elliptical orbit, the satellite moves from a position of minimal GPE when it is going fastest (Max KE). It then climbs away, gaining GPE and losing KE, until it is at its perigee, when its GPE is highest (i.e. furthest out of the potential well of the planet). At apogee and perigee, the tangential velocity is the same as its angular velocity times radius. Keppler's Laws were found by straight observation (and mind bending) and with no idea of the Inverse Square Law that Newton later showed to be operating and producing an elliptical orbit.
  15. May 12, 2016 #14
    Didn't Newton base his inverse square law on Kepler's laws? Am I missing something here? I probably am...
  16. May 12, 2016 #15


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    Yes, of course. But the ISL was how Newton explained a 'physical reason how Kepler's Laws are followed. Kepler was just describing the observed pattern of things and not the mechanics of it. At the time, it was thought that there had to be a 'motor' driving it all around (just as you always need something to keep an earthbound mechanism going). Isaac out them all straight!! Respect.
  17. May 12, 2016 #16
    What a lovely discussion. Usually there is someone asking about something. But, because the person wants information, as opposed to already having it, the question asker often uses an improper term or makes a cognitive error about something considered obvious to those who know the subject a bit better. What usually follows is a smarmy diatribe from someone who clearly knows something about the answer but is more interested in pointing out what's wrong with the question (and the "what's wrong" part usually reveals almost exactly what the person wants to know anyway.)

    At that point most questioners just back down -- and this repeated tone is what keeps me away from these forums. I just hate seeing it.

    And so ***my hearty congratulations*** Drvrm for the intelligent, caring, and human way you have addressed this question. Let it be a standard!
  18. May 12, 2016 #17
    I registered just to make that point -- never felt compelled to register for these forums in the several years I've been visiting. Thank you drvrm whomever you are. I think I'd like you even if you smelled bad.
  19. May 13, 2016 #18
    Thank you everyone for the great replies! I have one more question now: how do I determine 1) the shape of an ellipse and 2) the time one revolution takes if I know the tangential velocity at a given point in the orbit and the gravitational acceleration?
  20. May 13, 2016 #19


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    I think you should find what you need in this link - or the one it links to, for the orbital equation.
  21. May 14, 2016 #20
    Oh OK, thank you!
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