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Escape velocity of a satellite

  1. Apr 3, 2014 #1
    The escape velocity of a satellite circularly orbiting a large body comes from conservation of energy. Are there any modifications that must be made for the escape velocity of an elliptical orbit?

    Thanks in advance!
     
  2. jcsd
  3. Apr 3, 2014 #2

    Bandersnatch

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    When you look at the formula for escape velocity, you can see that it only depends on the distance from the massive body and that body's mass.
    In other words, it doesn't matter what you do around that mass, just where you are.
     
  4. Apr 3, 2014 #3
    If I trying to find the optimal point for escape in an elliptical orbit, can I just differentiate the standard escape velocity eqn with respect to r? I know it must be either the apogee or the perigee...is there a better approach?
     
  5. Apr 3, 2014 #4

    D H

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    Escape velocity is a speed, not a velocity. The easiest way to attain that velocity is for a vehicle to thrust in the direction it is already going. The minimum delta V needed to escape at some point is the difference between the current velocity at that point and the escape velocity at that point. It is this value you want to minimize, not the escape velocity.
     
  6. Apr 3, 2014 #5
    So I tried doing precisely that, I minimized a 'thrust' but the expression I arrive at doesn't look like the perigee.... ImageUploadedByPhysics Forums1396542421.448070.jpg
     
  7. Apr 3, 2014 #6
    That depends on your definition of "optimal". For rockets the optimal point is characterized by the highest velocity:

    http://en.wikipedia.org/wiki/Oberth_effect
     
  8. Apr 3, 2014 #7

    D H

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    You used the wrong expression for orbital velocity.

    A good place to start is the vis-viva equation. Unfortunately, you'll get no joy here if you differentiate the resulting delta V with respect to r. The problem is there's nothing that constrains r in either the vis-viva equation or the escape velocity equation.

    You'll need something such as the orbital equation of motion, ##r=\frac{a (1-e^2)}{1-e\cos\theta}##. Now you should get something useful. In particular, you should find that extrema occur at apofocus and perifocus. So now it's just a matter of determining which is best, which is worst.
     
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