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Homework Help: What is the radius of a geosynchronous satellite?

  1. Dec 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the radius R of the orbit of a geosynchronous satellite that circles the earth. (Note that R is measured from the center of the earth, not the surface.) You may use the following constants:

    * The universal gravitational constant G is [tex]6.67 \times 10^{-11}\;{\rm N \; m^2 / kg^2}[/tex].
    * The mass of the earth is [tex]5.98 \times 10^{24}\;{\rm kg}[/tex].
    * The radius of the earth is [tex]6.38 \times 10^{6}\;{\rm m}[/tex].


    The correct answer is [tex]4.23\times10^7\;{\rm m}[/tex], but I get a different answer.

    2. Relevant equations

    [tex]T=2\pi\sqrt{\frac{R^3}{GM}}[/tex]

    3. The attempt at a solution

    Since T is measured in seconds, and there are 86,400 seconds in a day, some simple algebra gives me the answer of 1,994,400,816 m. What am I doing wrong?
     
  2. jcsd
  3. Dec 13, 2009 #2

    ehild

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    Homework Helper

    The formula is correct, do your calculation again. Use the normal form of the numbers.

    ehild
     
  4. Dec 13, 2009 #3
    what is the "normal form" for these numbers?

    EDIT: I did the calculation a different way, and I got the right answer, but the only thing I did differently was the order that I did the algebra in. I don't know where my error lies.

    I normally did:

    [tex]\frac{86400\times\sqrt{GM}}{2\pi}=\sqrt{R^3}[/tex]

    and solve for R, but instead, this time I did:

    [tex]86400^2=4{\pi^2}\frac{R^3}{GM}[/tex]
    [tex]\frac{86400^2\times{GM}}{4{\pi^2}}=R^3[/tex]

    and it worked..... what did i do wrong the first time?

    if you square the first equation, you get the same thing......


    2nd edit, nevermind, i know what I did wrong.

    this has been a nice lesson for me in making sure I properly place parenthesis in a calculator to ensure teh correct order of operations. because [tex]x/2\pi[/tex] is not interpreted as [tex]x/(2\pi)[/tex], but instead as [tex]\frac{x}{2}\pi[/tex]
     
    Last edited: Dec 13, 2009
  5. Dec 13, 2009 #4

    ehild

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    Homework Helper

    Never forget the parentheses in the denominator!!!!:rofl:

    ehild
     
  6. Dec 13, 2009 #5

    Borek

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    Staff: Mentor

    No idea why radius of a satellite should depend on the Earth mass.
     
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