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Kinetic Energy, What is K2/K1

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A rock is dropped from a distance RE above the surface of the Earth, and is observed to have a kinetic energy of K1, when it hits the ground. An identical rock is dropped from 2RE above the surface and has kinetic energy K2 when it hits the ground. Re is the radius of the earth. What is K2/K1.

    a) 2 b) 4/3 c) 3/2 d) 2/3 e)4
    2. Relevant equations


    3. The attempt at a solution

    K1= (-GMm)/(RE+ RE)= (−GMm)/2RE K2 = (-GMm)/(RE+ 2RE)= (−GMm)/3RE

    K2/K1 = [(−GMm)/3RE]/[(−GMm)/2RE] = [(−GMm)/3RE] * [(2RE/(−GMm)] = 2/3 = d).
    However the answer is 4/3, where did I go wrong?






     
  2. jcsd
  3. Feb 18, 2015 #2

    gneill

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    Staff: Mentor

    You want to compare the changes in PE that each rock undergoes when they fall. Where does each rock end up at then end of its fall? Not at infinity...
     
  4. Feb 18, 2015 #3
    Oh I see where I made a mistake, thank you!
     
  5. Apr 21, 2015 #4
    But how do we compare the changes in PE do we do mgRe?
     
  6. Apr 21, 2015 #5

    haruspex

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    No. Those distances are too great to be taking g as a constant. Use PEgrav=-GMm/r, as in the OP.
     
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