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Elastic gravitational collision

  1. Mar 24, 2016 #1
    • Member advised to use the formatting template for questions posted to the homework forums
    Hi all.
    Our lecturer gave us an exercise the other day regarding an elastic gravitational collision between a planet and a satellite where the satellite slingshots using the gravitational field of the planet. The question asks to show that ##v_{f} - v_{i} = 2v_{0}## where ##v_{f}## is the final velocity of the satellite, ##v_{i}## is it's initial velocity and v_{0} is the orbital speed of the planet. The hint is to translate this problem into the COM reference frame, which I did do, and got:

    ##p'_{i} = m(v_{i}- v_{0})##

    and

    ##p'_{f} = m(v_f + v_0)##

    . I was told by someone that I can't do this since the planet isn't an inertial frame, I thought it would be ok since the acceleration of M is extremely small,. But I'm still not sure since if the equations for the momentum of the system from the lab's reference frame are written out and then translated into the planet's reference then there's an ##M\delta v## term in the second equation which is absurd since the planet can't be moving in it's own ref frame. Also this implies that ##v_{f} + v_{0}## and ##v_{i} - v_{0}## are equal whilst they're in opposite directions. I hope someone can set me on the right track .
     
  2. jcsd
  3. Mar 24, 2016 #2

    ehild

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    You can consider the planet as of infinite mass. The energy is conserved in gravitational interaction , so the speed of the satellite with respect to the planet remains the same, while the velocity changes sign. Initially the satellite moves toward the planet, finally it goes away from it. See:https://en.wikipedia.org/wiki/Gravity_assist

    200px-Gravitational_slingshot.svg.png
     
  4. Mar 24, 2016 #3
    Right. I had realized that the conservation of kinetic energy meant that the magnitude of the two velocities would be the same, but I just found it hard to convince myself that the change in momentum of the satellite can be ignored in the calculations, but if it has 'infinite' mass then that provides a sort of momentum sink right?
     
  5. Mar 24, 2016 #4

    ehild

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    It is the change of velocity of the planet that is ignored because of its "infinite" mass.
     
  6. Mar 24, 2016 #5

    haruspex

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    No. When one mass is very much larger than the other, we can ignore the KE change of the larger. If the mass ratio is M >> 1, momentum conservation says the the ratio of velocities is M (favouring the smaller mass) so the ratio of energies is also M, favouring the smaller mass.
    The change in momentum of the planet was not ignored in ehild's analysis; indeed, it must be equal and opposite to the satellite's change in momentum.
     
  7. Mar 25, 2016 #6

    ehild

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    To increase the speed of the satellite by slingshot, the satellite and the planet have to travel in opposite directions in the lab's frame of reference. If the orbital speed of the planet is U0 and the initial velocity of the satellite is vi along the positive x axis, the velocity of the planet is -U0.
    In case of elastic collision between two bodies with masses m1 and m2, initial velocities V1i and V2i, the equations for momentum and energy are
    m1V1i+m2V2i=m1V1f+m2V2f
    m1V1i2+2V2i2+m1V1f2+2V2f2
    Rearrange the equations to collect the '1' terms at one side and the '2' terms at the other side:
    m1(V1i-V1f)=-m2(V2i-V2f) (1)
    m1(V1i2-V1f2)=-m2(V2i2-V2f2)
    Divide the second equation by the first one. You get
    (V1i+V1f)=(V2i+V2f), (2) so
    V1f=-V1i+(V2i+V2f).
    In this problem, '1' means the satellite, and '2' means the planet. V2i=-u0 as they travel at opposite directions.
    m1/m2<<1, you can see from equation (1) that V2i≈V2f = - u0.
    The satellite travels with higher speed in the opposite direction after the slingshot.
     
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