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Effect of Instantaneous Velocity Change on Orbit

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    A small particle of mass m is on a circular orbit of radius R around
    a much larger mass M. Suppose we suddenly increase the speed at which the mass m is moving
    by a factor (that is, v[itex]_{final}[/itex] = α * v[itex]_{initial}[/itex], with α > 1). Compute the major axis, minor axis,
    pericentre distance, and apocentre distance for the new orbit; express your answers in terms
    of R and α alone

    2. Relevant equations

    Vis-Viva Equation:

    (αv)[itex]_{initial}[/itex][itex]^{2}[/itex] = GM [ [itex]\frac{2}{R}[/itex] - [itex]\frac{1}{a}[/itex] ]

    Speed of circular orbit:

    v[itex]_{initial}[/itex] = [itex]\sqrt{\frac{GM}{R}}[/itex]

    Pericentre distance:

    a(1 - e)

    Apocentre distance:

    a(1 + e)

    Semi-minor axis:

    [itex]b^{2}=a^{2}(1-e^{2})[/itex]

    3. The attempt at a solution

    By inserting the initial orbital speed into the vis-viva equation I was able to find the semi-major axis as required:

    a = [itex]\frac{R}{2-α^{2}}[/itex]

    The problem I'm having now is that I can't find the semi-minor axis without the eccentricity of the new elliptical orbit, or the distance between the two foci, and I can't find a way to eliminate them.
     
  2. jcsd
  3. Jan 25, 2012 #2
    have you drawn a diagram and labeled everything?
     
  4. Jan 25, 2012 #3
    Yeah, but it's hasn't really helped any more than trying to manipulate formulas.

    I have an idea rolling around that the since the velocity increased while the particle was at a Radius of R, then R will be its pericenter, which would allow me to find the eccentricity, but I'm not sure if that's right, or how to justify it if it is.
     
  5. Jan 25, 2012 #4
    Well at the point it obtains an instantaneous velocity increase it moves off tangent to the circle it was already traveling in, faster than it was going so the inward pull of the gravity will not pull it back to the original path. This means it's not going to get any closer to the larger mass so it's not going to go any faster so it has to be the position closest to the large mass at that point. So go from there.
     
  6. Jan 25, 2012 #5

    gneill

    User Avatar

    Staff: Mentor

    It's a good idea, and a correct one. The particle starts in a circular orbit of radius R and you're adding to its velocity (KE) in a direction tangent to the circle. There are only two places in an elliptical orbit where the velocity is tangent to the radius vector, and that's at periapsis and apoapsis, and since closed orbits repeat it must be one or the other. Since the new velocity is greater than that required for circular motion, the radius must be increasing from that point as time increases and so it's at periapsis.

    If it turns out that the new orbit is not closed (what values of [itex] \alpha [/itex] will cause this?) then indeed, periapsis is the only choice.
     
  7. Jan 25, 2012 #6
    Thanks for the help guys.
     
  8. Jan 25, 2012 #7
    Glad you got it.
     
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