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Angular Momentum and Energy from parametric orbit equations

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Given the parametric equations for a satellite in orbit around a spherical mass find angular momentum L in terms of ε, a, k, m, where k=GMm.

    Also, find the energy E in the same terms.

    Lastly, I can only use the equations provided and "fundamental definitions."

    2. Relevant equations

    Equations provided:

    y=a[itex]\sqrt{1-ε^2}[/itex]sinψ

    x=a(cosψ-ε)

    t=([itex]\frac{T}{2π}[/itex])(ψ-εsinψ)

    [itex]\frac{T^2}{a^3}[/itex]=[itex]\frac{4π^2}{GM}[/itex]

    What I think is necessary and otherwise fundamental:
    L=rxp

    Not yet sure about the equations for energy that would be considered fundamental...

    3. The attempt at a solution

    I believe the first part is relatively straight forward even if it wouldn't otherwise be the most elegant way to go about the problem.

    L=rxp

    I can find r using a2+b2=c2:

    r2=x2+y2

    r=[itex]\sqrt{x^2+y^2}[/itex]

    r=[itex]\sqrt{a^2(cosψ-ε)^2+a^2(1-ε^2)sin^2ψ}[/itex]

    v=[itex]\frac{2π}{t}[/itex] (edit: need correction here; this is average velocity... not sure how to do this without using theta; is the nasty relationship between theta and ψ fundamental?)

    I'm given t so:

    v=T(ψ-εsinψ)

    p=mv

    p=mT(ψ-εsinψ)

    I can solve for T from the given equation that contains it:

    T=sqrt([itex]\frac{4π^2}{GMa^3}[/itex])

    p=msqrt([itex]\frac{4π^2}{GMa^3}[/itex])(ψ-εsinψ)

    Now I would just have to solve for ψ and replace it and get rid of the GM factor using k. Assuming I do my algebra and final cross product correctly does that seem like a sound solution to the angular momentum?

    With regard to the energy portion I am a little weary about how to proceed given the requirement to use only fundamental definitions. I have come across a number of formulas for E of elliptical orbits that utilize kinetic energy and effective potential. Could it be as simple as:

    E=[itex]\frac{1}{2}[/itex]mv2+[itex]\frac{GMm}{r}[/itex] and plug in the values for v and r that I already solved for?

    Thank you for your help!
     
    Last edited: Nov 26, 2013
  2. jcsd
  3. Nov 27, 2013 #2

    gneill

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    Staff: Mentor

    You can find the angular momentum given any pair of radius and velocity vectors. Since you're given the equations for x(ψ) and y(ψ) and t(ψ), you're in a position to choose an angle ψ and determine the position and velocity (hint: differentiation, chain rule). Pick an angle that makes your life easy :wink:
     
  4. Nov 27, 2013 #3
    Thank you for the advice. That seems like a simpler way to go about the problem. I think I understand what you are getting at for the most part but I'm getting stuck:

    If ψ=0:

    y=0
    x=a(1-ε)=rψ=0
    t=0

    The problem is that if I choose a value for ψ the results will all be constants so differentiation with respect to t yields 0. I couldn't, for example, find v(t)=x'(t) because that would equal zero and the velocity is not zero. ψ=π results in a similar predicament.

    Alternatively, I could differentiate x(t) under the assumption that I would laser be using ψ=0 but find v(t) before killing the time dependent variable. I'm honestly just not sure how to differentiate a function that contains variables that are dependent on the variable I am integrating with respect to. At least not without an explicit definition of that dependent variable with respect to t, for example. Is that what you mean by chain rule? Can I treat ψ as if it were t for the sake of differentiation?
     
  5. Nov 27, 2013 #4

    gneill

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    Staff: Mentor

    Differentiate first! You want to have position and velocity as functions of ψ.

    The power of the chain rule is that you can find the rates of change of a variable with respect to another variable that changes with respect to yet another variable. It's important to "get" the chain rule and what it's about. Also note that there's nothing special about "t" as a variable when it comes to the mathematics of differentiation; A function f(q) can be differentiated with respect to the variable q. A function f(s) can be differentiated with respect to the variable s. And so on.

    As an example: Suppose I have an expression v(x) for a velocity of some object with respect to position x. I'm also told that time is related to position by the function t(x). The chain rule says that I can write the acceleration dv/dt as:
    $$\frac{dv}{dt} = \frac{dv}{dx}\cdot \frac{dx}{dt}$$

    Use the same concept in your problem to find the x and y velocities.
     
  6. Dec 1, 2013 #5
    Very clear, thank you. I thought that is what you were getting at but I rarely use the chain rule in this fashion. I should definitely seek out some practice doing this to drive the concept home.

    [itex]\frac{dx}{dt}[/itex]=[itex]\frac{dx}{dψ}[/itex]*[itex]\frac{dψ}{dt}[/itex]

    [itex]\frac{dy}{dt}[/itex]=[itex]\frac{dy}{dψ}[/itex]*[itex]\frac{dψ}{dt}[/itex]

    As far as [itex]\frac{dψ}{dt}[/itex] goes can I just find [itex]\frac{dt}{dψ}[/itex] and take the inverse? I'm not sure of another way to find it.

    [itex]\frac{dx}{dψ}[/itex]=-asin(ψ)

    [itex]\frac{dy}{dψ}[/itex]=a[itex]\sqrt{1-ε^2}[/itex]cos(ψ)

    Sorry for the delay. It has been a crazy week!
     
  7. Dec 1, 2013 #6

    gneill

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    Staff: Mentor

    Yes, yes you can :smile:

    And those derivatives look fine, too.

    Understood. Been there. Done that.
     
  8. Dec 1, 2013 #7
    Wow, that was a fast response. Thanks!

    Okay, here is [itex]\frac{dψ}{dt}[/itex]:

    [itex]\frac{dt}{dψ}[/itex]=[itex]\frac{T-εTcos(ψ)}{2π}[/itex]

    [itex]\frac{dψ}{dt}[/itex]=[itex]\frac{2π}{T-εTcos(ψ)}[/itex]

    [itex]\frac{dx}{dt}[/itex]=-asin(ψ)*[itex]\frac{2π}{T-εTcos(ψ)}[/itex]

    vx=[itex]\frac{dx}{dt}[/itex]=[itex]\frac{-asin(ψ)(T-εTcos(ψ))}{2π}[/itex]

    vy=[itex]\frac{dy}{dt}[/itex]=a[itex]\sqrt{1-ε^2}[/itex]cos(ψ)*[itex]\frac{2π}{T-εTcos(ψ)}[/itex]

    If ψ=π/2

    vx=[itex]\frac{-aT}{2π}[/itex]

    vx=0

    v=vx=[itex]\frac{-aT}{2π}[/itex], where ψ=π/2

    Also,

    x=-aε

    y=a[itex]\sqrt{1-ε^2}[/itex]

    r=[itex]\sqrt{x^2+y^2}[/itex]=a

    L=[itex]\frac{-a^2mT}{2π}[/itex]

    Does that look about right?
     
  9. Dec 1, 2013 #8

    gneill

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    Staff: Mentor

    Check your expression for ##v_x##. It looks like you've inverted second term from the previous line (how did the ##2\pi## end up in the denominator?).

    Also, You might make a better choice for the angle; the one you've chosen does not make the radius and velocity vectors perpendicular, so you'd need to do a full cross product to find the angular momentum. Consider either perigee or apogee...
     
  10. Dec 1, 2013 #9
    Whoops, you're absolutely right. I was using the wrong function for [itex]\frac{dψ}{dt}[/itex]

    vx=[itex]\frac{dx}{dt}[/itex]=[itex]\frac{-2πasin(ψ)}{T-εTcos(ψ)}[/itex]

    vy=[itex]\frac{dy}{dt}[/itex]=[itex]\frac{2πasqrt(1-ε^2)cos(ψ)}{T-εTcos(ψ)}[/itex]

    When ψ=0,

    vx=0

    v=vy=[itex]\frac{2πasqrt(1-ε^2)}{T(1-ε)}[/itex]

    r=x=a(1-ε)

    L=rmv=[itex]\frac{2πa^2msqrt(1-ε^2)}{T}[/itex]

    edit: Of course I now need to get rid of the T term. I'll do that now. I hope the rest looks alright.
     
  11. Dec 1, 2013 #10

    gneill

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    Staff: Mentor

    Yes, that looks much better :approve:

    By the way, you can use \sqrt{} to create a square root in LaTex :wink:
     
  12. Dec 1, 2013 #11
    Ohh, I always wondered about that. Heck yeah, thanks! How about division within a square root?

    k=GMm => GM=k/m

    T=[itex]\frac{\sqrt{4π^2a^3}}{\sqrt{GM}}[/itex]

    T=[itex]\frac{\sqrt{4π^2a^3m}}{\sqrt{k}}[/itex]

    L=a2m([itex]\frac{\sqrt{1-ε^2}}{\sqrt{a^3m/k}}[/itex])

    As far as E goes:

    E=[itex]\frac{1}{2}[/itex]mv2+[itex]\frac{GMm}{r}[/itex]

    E=[itex]\frac{1}{2}[/itex]mv2+[itex]\frac{k}{r}[/itex]

    Energy will be conserved, also, so I could use the r and v vectors when ψ=0. Assuming this formulation is correct.
     
  13. Dec 1, 2013 #12

    gneill

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    Staff: Mentor

    You can put any expression you wish inside the {} of the \sqrt, including \frac{}{} expressions...

    Gravitational potential energy usually has its zero reference at infinity, and so takes on negative values for all r. Fix up the sign of the potential energy and you should be good to go.

    I'm a bit surprised that they chose to define k = GMm. Usually it's defined to be k = GM, or more accurately, k = G(M + m). When m << M then it becomes G ≈GM. This would make it consistent with the "standard" definitions used in orbital mechanics.
     
  14. Dec 1, 2013 #13
    Oh crap, it's the little obvious things I always seem to overlook. Good catch, thanks. That always confuses me a bit though because total energy should be the sum of the kinetic and potential but because potential is represented negatively it's the difference?

    With regard to k, you're right. My prof likes to change things up. I like it because it helps keep us from relying purely on first-level pattern matching. It is clearly defined this way, though. k=GMm.
     
  15. Dec 1, 2013 #14

    gneill

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    Staff: Mentor

    It's still the sum... it just happens that one term is negative :wink:

    Well, so be it then :smile:
     
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