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B How is Angular Momentum conserved in orbits?

  1. Dec 13, 2016 #1
    So a light particle is orbiting a massive particle by gravity.

    We take both particles as spot particles.

    The light particle makes an eccentric orbit where maximum radius of the orbit equals 2 and minimum radius equals 1.

    I suppose the mass of the massive particle such that the speed of the light particle at the farthest radius is 1 m/s

    According Newton when you double the radius of an orbit its speed decreases by a factor of square root of two

    So when the orbit is at a distance of 1 m its speed will be equal to square root of 2=1.41 m/s

    So how is angular momentum conserved here for initial angular momentum is mvr=2 and final angular momentum=1.41?
     
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  3. Dec 13, 2016 #2

    A.T.

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    This is General Relativity not Newton.
     
  4. Dec 13, 2016 #3

    Bandersnatch

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    He means a particle that is light, not a particle of light. I.e., a test particle.

    Have you actually changed the orbit here, though? What is the difference between the speed of a particle at the periapsis of an elliptical orbit vs the speed of a particle on a circular orbit with the radius equal to the apoapsis distance?
     
  5. Dec 13, 2016 #4

    Dale

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    Can you post the formula that you are referring to here. I don't recognize this immediately.

    Please post a link or LaTeX formula, not just a jumble of characters, and any additional information that you have on this formula you referenced
     
  6. Dec 13, 2016 #5
  7. Dec 13, 2016 #6

    Nugatory

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    That's for a circular orbit (and note that "THE radius" is not defined for an elliptical orbit).
    Also, the magnitude of the angular momentum is only proportional to the product of the distance and the speed in circular orbits; for all others it's proportional to the product of the radius and the transverse component of the velocity. (It just so happens that in a circular orbit the transverse component of the velocity is the speed, so you can get by with using the speed in that special case).

    So you're trying to use the circular orbit simplifications on an elliptical orbit, and that won't work. To take on elliptical orbits you're going to have to learn and use the general expression for angular momentum: ##\vec{L}=\vec{r}\times\vec{p}## where ##\vec{p}=m\vec{v}## is the linear momentum.
     
  8. Dec 13, 2016 #7
    Thanks a lot that was very enlightening but Im not sure conservation of angular momentum would allow me to solve the problem because an eccentric orbit satellite is not an isolated system but interacts with earth.

    For example the rotation of earth is slowed down by tides and in exchange to keep angular momentum constant the moon goes higher.

    How do i know that in an eccentric orbit angular moment is not interacting with earth as it varies radius if i dot take earth as a point mass but consider it has some moment of inertia?

    Myself to solve this proble Id use potential energy and forget all about angular momentum.
     
  9. Dec 13, 2016 #8

    Nugatory

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    The mass of a satellite is enough less than the mass of the earth that we can ignore the effects on the earth, and we still get extremely accurate results. Likewise, when we solve planetary motion problems, we treat the sun as a fixed point mass, and we still get good results.
    You need both to solve the general problem of a small mass orbiting a larger fixed mass under the influence of gravity; if you ignore angular momentum you'll find yourself with more unknowns than you have equations. You'll likely encounter this solution towards the end of the first-semester of a calculus-based physics program; or you can google for "Kepler's laws derivation".
     
  10. Dec 13, 2016 #9
    i found the formula to solve this problem::


    v=squareroot of(mu(2/r-1/a))
    from:
    https://en.wikipedia.org/wiki/Elliptic_orbit

    and this is not the conservation of momentum formula

    how can it be true this formula and the momentum conservation formula at the same time if they expect different results to solve this problem?
     
  11. Dec 13, 2016 #10

    Nugatory

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    That formula is derived from conservation of energy and conservation of momentum: Start with ##F=Gm_1m_2/r^2##, use conservation of energy and conservation of momentum to calculate the orbital speed of an object in an elliptical orbit, and you can calculate that result. In fact, if you go back to the wikipedia page and follow the link directly above the formula you will see how this is done. (Or, you can get hold of a suitable textbook, something like Kleppner and Kolenkow).

    As for the resolution of the paradox that you think you've found: If you do a bit of algebra, you will find that there is a relationship between ##a## and ##r## that makes it possible for that formula to be true and for angular momentum to be conserved (the product of speed and distance is the same at the points of closest approach and greatest distance) if ##a## is chosen properly.
     
    Last edited: Dec 13, 2016
  12. Dec 13, 2016 #11
    Yes my apologies youre right I calculated the a for speeds 1 and 2 and radius 2 and 1 and i obtained and a=0.33

    I was wondering now if conservation of momentum couldnt be used to calculate potential energy instead of integrating to infinity.

    The particle at 1 m radius is going at 2m/s and at 2 m radius at 1 m/s.

    So the diference in cuadratic gravity potential energy between 2 and 1 m radius would be the difference between kinetic energy at both 1 and 2 radius

    i just take that instead of shooting the projectile straight from earth i shoot it in a slight curve that would keep angular momentum conservation true
     
  13. Dec 13, 2016 #12
    Does not matter how you shoot it. Conservation of momentum in the reference frame centered on Earth is a consequence of the fact that the gravitational force is radial (central) not of the way you initiate the motion.

    You could calculate the change in PE by knowing the change in KE, it's just conservation of energy.
     
  14. Dec 13, 2016 #13
    Ok thanks so lets see if I analize the problem correctly:

    By conservation of momentum when the radius of a satellite halves its speed doubles hence its kinetic energy cuadruples.

    If the kinetic energy at radius 2 is x at radius 1 would be 4x so i substract both and obtain that when the radius doubles potential energy triples.

    would this be right?
     
  15. Dec 13, 2016 #14
    Also I have another question concerning orbits and momentum:

    I throw a 1 kg ball horizontally at speed 0.01 m/s, I consider it starts and orbit interrupted by the ground.

    In the initial instant there will be an angular momentum of 1*0.01*6000 km

    but in the instant before it touches the ground the vector will be aiming almost down so its angular momentum would be something like 1*0.01*100m(the arm between the vector of the ball and the center of earth)

    So how would you explain this?
     
  16. Dec 13, 2016 #15
    PE is inverse proportional to distance between the bodies. You don't need all this.

    Your description is not correct. Neither the conclusion. You calculate the change in KE, which is equal to the change in PE. You cannot draw conclusions about what the actual value does in the way you wrote above.

    Note: This is about your post 13.
     
  17. Dec 13, 2016 #16

    Nugatory

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    You mean "conservation of angular momentum", not "conservation of momentum". They're different things (and momentum is not conserved in planetary motion because the planet is not a closed system).
    No. The potential energy of an object of mass ##m## at a distance ##r## from a massive object is ##-\mu{m}/r## (using the convention that the potential energy at infinity is zero, the sensible way of working this problem).
     
  18. Dec 13, 2016 #17

    Nugatory

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    That's not "throwing" the ball, it's "dropping" it... although you are correct that it will enter into a very eccentric elliptical orbit that will be interrupted by the ground.

    What is the transverse component of that velocity vector at that point? That's what goes into the angular momentum calculation.... Not the speed.
    And where did you get 100m from? The ball is still near as no never mind 6000 km from the center of the earth.

    I said several posts above that you have to learn and use the definition of angular momentum to take on these problems. So we have ##\vec{L}=\vec{r}\times\vec{p}##. What are ##\vec{r}## and ##\vec{p}## at the moment of release and at the moment of impact with the ground? You'll have to write them in polar coordinates.
     
  19. Dec 13, 2016 #18

    Dale

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  20. Dec 13, 2016 #19
    Thanks a lot Nugatory and Dale I think Im begining to make sense of it:

    Angular momentum conservation just refers to the transversal component of the velocity of the particle.

    Could I conclude then that to use conservation of angular momentum to obtain velocity is only valid at the perigee and apogee for those are the only places where net velocity equals the transversal component?
     
  21. Dec 13, 2016 #20

    Nugatory

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    No. You could conclude that it is especially easy to use conservation of angular momentum to obtain the velocity at one of those two points when you already know the velocity at the other.

    But if you want to understand orbital motion, it makes a lot more sense to use conservation of angular momentum and conservation of energy to work out the equations of orbital motion; you've already posted a link to a wikipedia page that has links to how to do that. And once you've done that, then you know the velocity everywhere in the orbit, not just at perigee and apogee.
     
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