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Orbital Period

  1. Oct 28, 2008 #1
    GPS (Global Positioning System) satellites orbit at an altitude of 2.0×10^7 m.

    Find the orbital period.(hours)

    I use the equation T=(2pi/sqrt(GM))r^(3/2).

    Shouldnt that be the correct way of doing it? I know im supposed to add the altitude to the radius of the earth, G is known, M of earth is known. I just dont understand where I am going wrong with this. It should be a simple problem yet I am not able to get the correct answer. is it possible im doing something as silly as not converting something somewhere?

    If anyone could offer some help, it would be greatly appreciated.
  2. jcsd
  3. Oct 28, 2008 #2


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    What's the problem you are having?
    Lookup the values for 'G' and 'M' (of the earth)
    Add the Earth's radius to the altitude and plug into the equation.

    Make sure everythign is in the correct units m/s/kg

    The equation is normally written T = 2pi sqrt( r^3 / GM), might be simpler to use.
  4. Oct 28, 2008 #3
    I used the formula you provided and I still got the answer wrong.I cant imagine im capable of messing this up this bad.

    R=6.37*10^6 + 2*10^7

    the answer I got was =.001617844
    Rounding to two significant figures= .0016

    This is driving me insane.I know that whatever it is im doing wrong,is going to end up being something small and infinitely stupid.
  5. Oct 28, 2008 #4


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    I think a bit of finger trouble on the calculator.
    It's always worth being able to do a rough approx just using the rules of exponents for large number calculations. Remember to multiply simply add the exp, to divide subtract.

    ps. the answer isn't exact becaue I only did a few decimal places - but you see the idea

    GM = 6.7E-11 * 6E24 = 6.7*6 E(24-11) = 40 E13 m^3 s^-2
    r = 6.4E6 + 20E6 = 26E6 m
    r^2 = 26*26*26 E(6+6+6) = 17500E18 m^3

    sqrt( r^3/GM) = sqrt( 17500E18 / 40E13 ) = sqrt(437E5) = 6600
    t = 2pi*6600 = 41500 seconds =11 h
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