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Why is horizontal escape velocity less ?

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data

    an asteriod with mass 6.6x10^17kg & radius 46km has escape velocity 44ms, but if the projectile is launched horizontally then the escape velocity is only 31ms, show that this is correct


    2. Relevant equations

    Vesc = SQRT (2GM/R)
    G = 6.67x10^-11


    3. The attempt at a solution

    the only thing i can think of is that the rotational speed of the asteriod is helping to reduce the escape velocity, but im not sure how to work it out ?
     
  2. jcsd
  3. Nov 22, 2009 #2

    tiny-tim

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    Hi victoriafello! :smile:
    (hmm … 31 is approx √(1/2) times 44)

    I think that's rubbish …

    escape velocity (ignoring rotation) is the same for any direction, since it's the solution to 1/2 mv2 - GM/r2 = 0 + 0.

    And although, yes, the rotation of the asteroid can make a difference, you're not given any information about it. :confused:

    Is this part of a longer question?
     
  4. Nov 24, 2009 #3
    Hi

    No its not part of a longer question thats it, the only thing thats maybe wasnt clear is that the 31ms is to get the stone into orbit, does that make any difference ?

    im really stuck with this one so any ideas would be great
     
  5. Nov 24, 2009 #4

    tiny-tim

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    D'uh! :rolleyes: yes of course it does!

    "escape velocity" means that at large r the speed is zero.

    "into circular orbit" means that at large r the speed is proportional to … ? :smile:

    But I'm still not understanding the question …

    is it a circular orbit? or just a very eccentric elliptical one, always returning to just miss the asteroid?

    and does the vertical launch also go into orbit, or does it just keep going straight up? :confused:
     
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