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Acceleration Due to Gravity

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data
    At what altitude above the earth's surface is the acceleration due to gravity equal to g/3?



    2. Relevant equations

    ag=G Me/(Re+H)^2

    3. The attempt at a solution

    I solved for H and got H=√3 Re-Re
    using this i got 17*10^6
    but it is saying this is not correct
     
  2. jcsd
  3. Apr 13, 2012 #2

    gneill

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    Staff: Mentor

    Can you show the details of your calculation? What value did you use for Re?
     
  4. Apr 13, 2012 #3

    D H

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    Staff Emeritus
    Science Advisor

    17*10^6 what? Distance isn't just a number. It has units. What are the units of this number 17*10^6, what units is the answer supposed to be in, and what value did you use for the radius of the Earth?
     
  5. Apr 13, 2012 #4
    for re i used 6.37*10^6m
    the answer is also sopossed to be in meters

    i started with ag=g Me/(Re+H)^2
    then plugged in 1/3*9.8=G Me/(Re+H)^2
    I reduced that to ag=1/(Re+H)^2=1/3(1/Re^2)
    from there i got
    (Re+H)^2=3Re^2
    Re+H=sqrt(3)*Re
    H=sqrt(3)Re-Re
     
  6. Apr 13, 2012 #5

    gneill

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    Staff: Mentor

    Your derivation looks okay, so you must be having finger problems with the calculator :smile:
     
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