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B How come escape velocity isn't imaginary?

  1. Nov 30, 2016 #1
    Going through several definitions, it appears that escape velocity is equal to the potential energy. That is:$$\frac{1}{2}m v^2=-\frac{G M m}{r}$$but if I solve for velocity, $v$, I get:$$v=\sqrt{-2\frac{G M}{r}}$$So how do I get an escape velocity that isn't imaginary?
     
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  3. Nov 30, 2016 #2

    Ibix

    User Avatar
    Science Advisor

    It's the velocity such that the kinetic energy at launch is equal to the potential energy difference between infinity and the point of launch. So$$\frac 12mv^2=0-\left(-\frac{GMm}r\right)$$
     
  4. Nov 30, 2016 #3
    No, at escape velocity the sum of kinetic and potential energy is zero.
     
  5. Nov 30, 2016 #4
    That makes sense. Thank you for the clarification.
     
  6. Nov 30, 2016 #5

    Dale

    Staff: Mentor

    And in this sum PE is always negative so KE is always positive.
     
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