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Can anyone explain the need for elipses?

  1. Apr 23, 2009 #1

    NWH

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    I was wondering, why is it that the planets orbit in elipses? I find it strange that it's path would be eliptical, is it the result of some kind of force of nature keeping it fixed in it's orbit? Or is that simply the way it is? I can't help but feel there's something deeper to it...
     
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  3. Apr 23, 2009 #2

    DaveC426913

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    The ellipse is the natural path of an object trading off two opposing phenomena (gravitational attraction versus inertia/momentum). What is not very natural is a circular orbit. (A circular orbit is merely a normal elliptical orbit with zero eccentricity.)
     
  4. Apr 23, 2009 #3
    My understanding is that Kepler empirically determined that the orbits of planets are ellipses based on Tycho Brahe's observational data. Later, Newton postulated his famous general mechanical force laws and the gravity force law that says gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance. From this, he mathematically proved that this generates stable elliptical orbits for a two body system.
     
  5. Apr 23, 2009 #4

    NWH

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    I see, thanks. So, what if the orbit was, for sake of this example, circular? What would be predicted to happen now that Kepler's ideas had been proven mathamatically? How would that change the scenario? Would the force of gravity start to take over where it otherwise wouldn't have? Or could it actually be stable? The way I'm thinking right now, is that the inertia/momentum is what's keeping that orbit stable, would I be right in saying that?

    Sorry for all the questions, I appreciate your help...
     
  6. Apr 24, 2009 #5

    Nabeshin

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    Circles are just special cases of ellipses where the eccentricity is zero. The only reason they aren't observed is because the positioning of an astronomical body has to be so perfect that it almost never occurs.

    Circles actually represent the lowest eccentricity case for an orbit (e=0), and since eccentricity is always positive, the value can only increase, degrading into an elliptical orbit (0<e<1). Once e=1, the orbit is parabolic (a very special case. One that is again, not usually observed in nature for similar reasons to that of the circle). And if 1<e<infinity, the orbit is hyperbolic. The reason they're all ellipses is because neither parabolas nor hyperbolas are stable, so there's no time to observe these kind of orbits because they quickly leave the host star.
     
  7. Apr 24, 2009 #6

    DaveC426913

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    This does not merely apply to orbits.

    Have you ever seen one of these coin funnels in malls?
    http://us.st12.yimg.com/us.st.yimg.com/I/igumballs_2049_5927318
    You drop a coin in on its edge and watch it roll round and round until it finally drops out the drain in the middle/bottom.

    If it could discount friction, the coin would follow an elliptical path, speeding up as it neared the vortex of the funnel, then slowing down as its speed carried it past the vortex to outer the perimeter again.

    OK, well actually, it is still the exact same thing (gravity trading off versus inertia) but I want you to see that this is a natural and basic consequence of the two forces involved.

    A Youtube video of coins (trying) to follow an elliptical path.
     
    Last edited by a moderator: Sep 25, 2014
  8. Apr 24, 2009 #7

    NWH

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    Yeah, I had been using that analogy to think about this. But wouldn't the eliptical orbit require some kind of escape from the gravitational field to be eliptical? Or are gravitational fields not circular (in this instance, for example)? I'm having a hard time understanding where the energy comes from to throw it outward like that, as the coin always falls to the center and never orbits the machine. Would the mere absence of friction make it go round forever?
     
  9. Apr 24, 2009 #8

    DaveC426913

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    Yes. Without friction, the elliptical path of both coin and planet is stable.

    Think about throwing a ball in the air. (Let's ignore air friction.)
    It starts off with high speed moving upward.
    As it rises, it gains potential energy but loses kinetic energy.
    As it reaches its zenith, it has max potential energy and zero kinetic energy. Note though, that is has exactly as much energy as it had when it left your hand.
    As it begins to fall, it trades off potential energy for kinetic energy.
    Once it reaches the ground, it has max kinetic energy and zero potential energy, yet still has exactly as much energy as when it left your hand.

    If you manage to somehow have your ball not hit the Earth (say you replace the Earth with a black hole of identical mass but which is only one mile in diameter), the ball will fall all the way to centre, picking up kinetic energy as it goes. It will loop around and come back, shooting past you to its max height again.

    And voila! the ball is in orbit!

    More generally, everytime someone throws a baseball, they are putting the ball into orbit around the Earth's centre - barring air friction and impact with the Earth's surface.
     
  10. Apr 24, 2009 #9

    NWH

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    I see, great explination. I think I get it now, thanks...
     
  11. Apr 24, 2009 #10

    malawi_glenn

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    just solve the motion of a body in a spherical symmetric gravity potential, you will get the ellipic motion. One of these solutions is the circular one (which is a special case of elliptic, as mentioned)
     
  12. Apr 24, 2009 #11

    DaveC426913

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    I question if this answer is in the spirit of the OP's knowledge level.
     
  13. Apr 24, 2009 #12

    NWH

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    The chances of me calculating anything right now are little to none...
     
  14. Apr 24, 2009 #13

    Chronos

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    GR plays a role in this as well, bear in mind how GR solved the precession problem with Mercury.
     
  15. Apr 25, 2009 #14

    malawi_glenn

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    OP's answer in post #4 is full of physical quantities that is introduced in courses in Newtonian Mechanics.

    One can not prove by intuitive arguments that the paths are elliptical, one must show that angle covered over one period is 2 pi radians.
     
  16. Apr 25, 2009 #15

    atyy

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    Would it be too much of a cheat to invoke angular momentum conservation?
     
  17. Apr 25, 2009 #16

    malawi_glenn

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    Why would angular momentum conservation show that orbits must be elliptic? It just tells that something is conserved, not that the orbit is closed. As far as I know, Angular momentum is conserved if you have secular motion as well (general relativity).
     
  18. Apr 25, 2009 #17

    Nabeshin

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    If I remember correctly, conservation of angular momentum makes the proof of conic section orbits easier. So, in a sense, I guess you could consider it "cheating" because Newton couldn't invoke such methods.
     
  19. Apr 25, 2009 #18

    malawi_glenn

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    Then angular momentum is not conserved in GR since then the orbits are not closed, if we turn the argument.

    If you invoke Angular Momentum conservation, you must prove it as well.
     
  20. Apr 25, 2009 #19

    Nabeshin

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    Well the proof is pretty trivial. Call some value r x v angular momentum, and it's fairly straightforward to show from the gravitation equation that d/dt (r x v) = 0, so angular momentum is conserved. Then if you use this value it makes the derivation flow pretty nicely. Not that much extra work.
     
  21. Apr 25, 2009 #20

    malawi_glenn

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    Of course, everything that is spherical symmetric angular momentum must be conserved, i meant that angular momentum is conserved in GR as well (as far as I know, but orbits are not ellipses)

    I looked up how to 'derive' ellipses from angular momentum conservation, but that derivation is as long as the one I meant so...
     
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