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Q on satellite orbits + orbital drift

  1. Jul 12, 2008 #1
    Im aware that due to the oblate nature of the Earth, the satellite's line of apside and line of nodes rotate according to the following formulas (where i = angle of inclination of orbit to Earth and K is mean motion per day)

    Variance in right ascension of ascending node ([tex]\Omega[/tex]) due to rotation of line of nodes

    d[tex]\Omega[/tex]/dt = -Kcos(i)

    Variance in argument of perigree ( [tex]\omega[/tex] )due to rotation of line of apsides

    d[tex]\omega[/tex]/dt = K(2 - 2.5sin(i)*sin(i) )

    New argument of perigee (taking into consideration rotation of line of apsides)

    [tex]\omega[/tex] = [tex]\omega[/tex]o + d[tex]\omega[/tex]/dt(t-to)

    where [tex]\omega[/tex]o is argument of perigee at epoch, to is time at epoch

    In a textbook example I am reading, the author is trying to visualise the drift of the orbits as a result of just the rotation of the line of apsides (in other words i=90 degrees and the orbit is polar).

    He starts of by assuming the situation where the perigee is exactly over the ascending node (in other words [tex]\omega[/tex] = 0 degrees) and d[tex]\omega[/tex]/dt = - K/2. One orbital period (Pa) later, he states that the perigee would appear south of the equator (in other words [tex]\omega[/tex] = -KPa/2.

    Mathematically, he is correct, but I cannot seem to visualise how the perigee would appear south of the equator. Would anyone be able to help me out here ?

    If I am not clear, could anyone let me know ?
  2. jcsd
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