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Universal gravitational constant and satellite

  1. Jun 30, 2008 #1
    A satellite is designed to orbit earth at an altitude above it's surface that will place it in a gravitational field with a strength of 4.5N/kg

    a) calculate the distance above the surface of at which the satellite must orbit

    g= GMp/r^2

    r^2 = (6.67*10^-11)(5.98*10^24)/4.5 N/kg

    r = sqrt(8.86*10^15) = 94147166 Meter


    the answer i got was 94 million meters now thats wrong. can some please help me with the calculations? thanks
     
    Last edited: Jun 30, 2008
  2. jcsd
  3. Jun 30, 2008 #2

    D H

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    Show your work, please. You made a mistake along the way but it's a bit hard to see where when you don't show the calculations.
     
  4. Jun 30, 2008 #3
    i'm not sure how i should re-write the formula. because the answer should be a couple of

    hundred km from the earths surface
     
  5. Jun 30, 2008 #4

    D H

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    Show how you got the answer that you got: Show how you used the equation and what your intermediate steps were.
     
  6. Jun 30, 2008 #5
    i re-wrote it again. thats for the distance
     
  7. Jun 30, 2008 #6

    D H

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    You made a mistake with your exponents. What is [tex]10^{-11}\cdot10^{24}\;[/tex]?
     
  8. Jun 30, 2008 #7
    it's in the formula universal gravitational formula
     
  9. Jun 30, 2008 #8
     
  10. Jun 30, 2008 #9

    D H

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    See post #6.
     
  11. Jun 30, 2008 #10
  12. Jun 30, 2008 #11

    D H

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    Bingo.
     
  13. Jun 30, 2008 #12
    i still think the answer is wrong, over 9000 Km from the earths surface?
     
  14. Jun 30, 2008 #13

    D H

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    You almost have the right answer. The distance that you calculated is not the distance above the surface of the Earth.
     
  15. Jun 30, 2008 #14
    can you give me some hints on how to rearange the formula please? thanks

    or am i suppose to subtract the answer from the radius of the earth? thanks again
     
  16. Jun 30, 2008 #15

    D H

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    Think of it this way: You are standing on the surface of the Earth. Using the distance from the surface as r in [itex]a=GM/r^2[/itex] would mean you would have an enormous gravitational acceleration, and heaven forbid if you lay down inches from the surface. That r is the distance from the center of the Earth.
     
  17. Jun 30, 2008 #16
    okay. show me were i went wrong with the calculation please
     
    Last edited: Jun 30, 2008
  18. Jun 30, 2008 #17

    D H

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    I already did: "That r is the distance from the center of the Earth."
     
  19. Jun 30, 2008 #18
    so r=9.0*10^6 - 6.38*10^6m ? because that answer got me 2.6*10^6 which is still too

    much i see that you said that r is the distance from the earth. but it still doesn't tell me how

    i'm calculating it incorrectly
     
  20. Jun 30, 2008 #19

    D H

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    Actually, that is too little. You very first answer was off by a factor of exactly 10. You had eight significant digits in your initial answer (too many), somewhere along the line you dropped to only one significant digit (which is too few).

    What makes you think you answer is wrong? Do you know the right answer?
     
  21. Jun 30, 2008 #20
    off by a factor of 10? does that mean the answer is 900km? and if it is, how did i make such

    an error (asking myself). but anyway based on the calculations the answer comes out to be

    9000km unless i missed something again. do you see what i did wrong in my math?
     
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