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Final speed of an asteroid using escape speed

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

    The escape speed from a very small asteroid is only 32 m/s. If you throw a rock away from the asteroid at a speed of 44 m/s, what will be its final speed?

    2. Relevant equations

    Ki + Ui = 1/2mv^2 + (-GMm/R) = 0 (for v<<c)

    3. The attempt at a solution

    I am unsure of how to do this problem because I don't have the masses or the radius. Any help would be greatly appreciated!
     
  2. jcsd
  3. Feb 22, 2009 #2

    Delphi51

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    Isn't it just 44 - 32?
     
  4. Feb 22, 2009 #3
    It should be SQRT((44^2)-(32^2))
     
  5. Feb 22, 2009 #4
    Thank you! However, I would like to know the steps on how you got to the answer if possible...
     
  6. Feb 22, 2009 #5

    Delphi51

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    The definition of escape speed is that it is the v that provides sufficient KE to take the object infinitely far away from the planet, where its velocity will then be zero.
    So your object will have E = 1/2*m*44^2 - 1/2*m*32^2 energy leftover.
    Put this back in E = 1/2*m*v^2 to see what the speed due to the leftover energy is.
    No doubt guitarman has it right, but yes, you most definitely want to know why!

    Thanks to guitarman for catching my mistake!
     
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