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Gravity on the height of 12740 km ?

  1. Feb 8, 2015 #1
    1. The problem statement, all variables and given/known data
    So the problem I have is calculating the gravity acceleration on the height of 12740km or 2 times the radial distance of Earth. The problem is relatively simple and I think that I have it right but the result doesn't match with the book.

    ## g_0=9,81 \frac {m}{s^2} ##
    ##R=6370km ##
    ## h=R=6370km ##
    2. Relevant equations
    ## g = g_0 \frac{R^2}{(R+h)^2} ##

    3. The attempt at a solution
    Well given that I have everything except ## g ## I could have just put everything in and be done with it. I went with canceling out the symbols and got ## g = \frac {g_0}{2} ## This resulted in 4.905. When calculating everything the result was 2.4525. The result specified in the book is 1,09 m/s^2. I'm really confused :)
     
  2. jcsd
  3. Feb 8, 2015 #2

    gneill

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    Staff: Mentor

    A height is a distance above the Earth's surface. You're given a height of 2R, so the radial distance is 3R...
     
  4. Feb 8, 2015 #3
    I must have explained it badly. The height given is 1R

    EDIT: The title is wrong, sorry..
     
  5. Feb 8, 2015 #4

    gneill

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    Staff: Mentor

    That would make the book's answer incorrect...
     
  6. Feb 8, 2015 #5
    But what about the double answer I got ? One is 4.905 the other 2.4525 ? Kinda weird that the the other is 2x smaller than the first :P
     
  7. Feb 8, 2015 #6

    gneill

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    Staff: Mentor

    Can you show your algebraic steps for your attempt by cancellation?
     
  8. Feb 8, 2015 #7
    Never mind, after another try at the calculation I got that 2.4525 is correct, thanks for the help :)
     
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