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Escape velocity and tangential impulse

  1. Jun 16, 2014 #1
    Here is the problem.

    A rocket of mass m is in a circular orbit around the Earth at a distance R from the center.
    (a) What tangential impulse, mΔv, must be given to the body so that it just escapes to infinity?

    My attempt:
    I set the problem in terms of energy, E=1/2m((dr/dt)^2+r^2(dø/dt)^2) V(r) and using conservation of angular momentum I get

    dr/dt=[2mMG/r-(l/mr)^2]^1/2,
    where l=mr^2dø/dt.

    Am I missing something?

    Thank you.
     
  2. jcsd
  3. Jun 16, 2014 #2

    Janus

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    Do you know how to find the orbital velocity for the object?
    Do you know how to find the escape velocity for the object?
    What's the difference?
     
  4. Jun 16, 2014 #3

    BvU

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    Hello Chris, welcome to PF.
    Just to help me understand what you are doing: when you
    what is it you are going to solve ? V = 0 ?

    Second, is there a way to justify
    when you whack this rocket with a big mΔv ?
     
  5. Jun 17, 2014 #4
    I am just setting the energy equation up. I generally find it easier to work with energy than with forces. Regarding the angular momentum, I meant that L is conserved until the impulse is delivered.
     
  6. Jun 17, 2014 #5
    The orbital velocity is jusr rdø/dt, while the escape velocity, for a stationary object shot vertically in a gravitational field, would be v=(MG/r)^1/2, where M is the mass of the planet or whatever the object is stationary on.
     
  7. Jun 17, 2014 #6

    Janus

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    First off, your escape velocity equation is incorrect (and it doesn't have to be the vertical direction)

    Secondly, what is orbital velocity in the terms of M, G and r?
     
  8. Jun 17, 2014 #7
    The orbital velocity should be v=(MG/r)^1/2 which I get from mv^2/r=mMG/r^2, right?
     
  9. Jun 17, 2014 #8

    Janus

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    Correct. Now note that this is the answer you gave for escape velocity.

    So what is the correct expression for escape velocity?
     
  10. Jun 18, 2014 #9
    So the energy is E=-mMG/2R and the escape velocity should be twice the orbital?
     
  11. Jun 18, 2014 #10

    Janus

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    You're dancing around the answer.

    Your energy equation is correct, but in it's present form doesn't help much as it does not have velocity in it.

    The other way to express the energy is as the sum of kinetic and potential energies.

    Once you have this, all you need to know is that the total energy of an object at escape velocity is 0.

    With this information, you can solve for escape velocity.
     
  12. Jun 20, 2014 #11
    Now I get v=(2MG/R)1/2.
     
  13. Jun 20, 2014 #12

    Janus

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    Good! Now compare this to the expression you got for the orbital velocity and you have your required delta v.
     
  14. Jun 20, 2014 #13
    Got it. Thank you for your infinite patience.
     
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