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Determine Orbital Period Of Satellite

  1. Jan 24, 2013 #1
    1. The problem statement, all variables and given/known data
    An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.32 m/s^2. Determine the orbital period of the satellite [in minutes].


    2. Relevant equations

    [itex]g= G \frac{M_E}{r^2}[/itex] Solving for [itex]r[/itex], and G is a constant

    [itex]T^2=K_sr^3[/itex] Solving for T, and [itex]K_s[/itex] is a constant.


    3. The attempt at a solution

    Solving the first equation for [itex]r[/itex], I obtain the value of [itex]6.30440 \cdot 10^{13}~m[/itex]; and substituting this value into the second equation, I get an orbital period of [itex]1.24 \cdot 10^{21} min[/itex] (this value has been converted from seconds to minutes). However, this isn't the correct. What did I do incorrectly?
     
  2. jcsd
  3. Jan 24, 2013 #2

    SteamKing

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    What value of Ks are you using?
     
  4. Jan 24, 2013 #3
    That would be the one they provided in the textbook, 2.97 x 10^-19.
     
  5. Jan 24, 2013 #4

    haruspex

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    Did you stop to think how far that is? Looks like you forgot to sqrt.
     
  6. Jan 24, 2013 #5
    I re-worked the problem, but it is still incorrect.

    I recalculated the distance at which the satellite is suspended above the earth:

    [itex]r= \sqrt{\frac{G(5.97 \cdot 10^{24})}{6.32}} = 7940022.399~m[/itex]

    Using this, I evaluated the period equation: [itex]T= \sqrt{K_s(7940022.399)^3} = 12.2~s[/itex]

    None of this is correct. What am I doing wrong? I am not sure if these are the proper equations.
     
  7. Jan 24, 2013 #6

    Dick

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    You probably don't want to use the Kepler equation. The acceleration due to gravity should be the same as the centripetal acceleration v^2/r. Can you figure out a way to use that?
     
  8. Jan 24, 2013 #7
    You don't even need to know the mass of the earth to do this. All you need to know is that g varies inversely as the square of the distance from the center of the earth and that the radius of the earth is about 6450 km. So

    [tex](9.8)(6450)^2=6.32r^2[/tex]

    What does this give you for r?
     
  9. Jan 24, 2013 #8

    haruspex

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    Yes, simply adjusting from g by the ratio of squares of the radii is much the simplest, but in terms of where the OP went wrong it was in using Ks, which incorporates the mass of the sun, instead of (what I assume may be called) Ke.
     
  10. Jan 24, 2013 #9
    As I'm sure you're aware, in the next step, he didn't need to know the mass of the earth either. He could just simply have used:
    [tex]\omega^2r=6.32[/tex]
    with [itex]\omega=\frac{2\pi}{T}[/itex]

    Chet
     
  11. Jan 25, 2013 #10
    All right, I tried it a different way, according to the suggestion of Dick.

    I still used the same method to find the radius at which the satellite is revolving.
    That being: [itex]r=7940022.399~m[/itex].

    Then, [itex]6.32=\frac{v^2}{7940022.399} \implies v = 7083.85~m/s[/itex]

    Furthermore, [itex]v=r\omega \implies \omega = 8.92 \cdot 10^{-4}~rad/s[/itex]

    Using the relationship Chestermiller alluded to, [itex]\omega = \frac{2\pi}{T} \implies T=0.001401476~s[/itex]

    However, after all is said and done, this is the incorrect answer.
     
  12. Jan 25, 2013 #11
    You messed up the algebra in the very last step. Otherwise, OK.
     
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