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Find eccentricity of orbit given position and velocity

  1. Jan 30, 2014 #1
    1. The problem statement, all variables and given/known data

    Hey, I need to find whether an asteroid is in a "bound orbit" around the Sun.
    Furthermore, describe the shape of the orbit (elliptical or circular).
    The only information I have been given is its velocity and position at a random point.

    v_o=61m/s
    r_o=13.75AU=2.06E12m

    I also need to find its orbital energy.

    2. Relevant equations
    vis-viva eqn: v^2=GM((2/r)-(1/a))
    angular momentum L=mvr
    r_p=a(1-e) position at perihelion


    3. The attempt at a solution

    First i used the vis-viva eqn with my v_o and r_o but a (semi-major axis) is an unknown.
    so a=((2/r_o)-(v_o^2)/GM)^-1
    a=1.03E12m

    Then i used momentum conservation with the observed position and the position at perihelion (where velocity is a maximum)

    m(v_o)(r_o)=m(v_p)(r_p)
    r_p=(v_o)(r_o)/(v_p) call this eqn 1

    I computed the vis-viva eqn at perihelion, and i subbed in eqn 1 and a=1.03E12m

    so i got a quadratic:
    v_p^2=GM[((2v_p)/(v_o)(r_o))-(1/a)

    (v_p)^2 - 2.11E6(v_p)+1.29E8=0

    The solutions are 61m/s or 2.11E6m/s.

    I used assumed 2.11E6m/s =v_p since it is larger.. and 61m/s is my original velocity.

    then i found r_p using eqn 1
    r_p=5.9E7m

    finally i used the eccentricity eqn e=1-(r_p/a)
    however r_p is of order 10^7 and a is of order 10^12, therefore i got e~1.
    that implies my orbit is a straight line.... so where did i go wrong?
    also, this feels like ALOT of steps so am i over thinking it??

    Oh and for orbital energy I think the equation is E=-GM/2a. is that correct?
     
  2. jcsd
  3. Jan 30, 2014 #2

    D H

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    Staff Emeritus
    Science Advisor

    That was a whole lot more work than was needed to answer those questions!

    Specific orbital energy is indeed ##-\frac {GM}{2a}## , but it's also ##\frac 1 2 v^2 - \frac{GM} r## . You don't need to find the semi major axis to find the specific orbital energy. Just compute it directly. What's the sign of this result? If it's positive, the comet is on an escape trajectory. If it's negative, it's in a bound orbit.

    Regarding the nature of the orbit, what's the circular velocity at 13.75 au? If the comet's speed is something different than this, the orbit can't be circular.
     
  4. Feb 1, 2014 #3
    Okay thank you.

    So I used E=K+U = 0.5v^2 -GM/r to find the orbital energy and I got E=-6.46E7 (joules?)

    By the way, do we just assume the mass of the asteroid is negligible in the energy eqn?

    So then the asteroid is in a bound orbit since the energy is negative, right?

    However, how would I find the circular velocity using just radius? I think the only thing I can compute for circular motion is a=v^2/r...
     
  5. Feb 1, 2014 #4
    Ohhh would I use v_circular=sqrt(GM/r) ?

    then i get v_circular=8035m/s which is much greater than v=61m/s
    So that means its an ellipse?
     
  6. Feb 1, 2014 #5

    D H

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    Staff Emeritus
    Science Advisor

    The energy is negative, so it's in a bound orbit -- either circular or elliptical. You've correctly found that velocity is not that of circular orbit at that distance, so the orbit isn't circular. That leaves one choice.

    Note that if you had been told the velocity was 8035 m/s the answer would have been "not enough information". All that this would tell you is that r=a. The velocity has to be equal to circular orbital velocity and normal to the radial displacement vector to have a circular orbit. You weren't told anything regarding the direction of the velocity vector.

    Aside: Since you were not told anything about the direction of the velocity vector, you can't solve for the eccentricity. All you do is find a lower bound on the eccentricity (the upper bound is of course one). In this case, that lower bound is rather large. This is a highly eccentric orbit.
     
  7. Feb 1, 2014 #6
    Okay, thank you so much for all your help.
     
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