Re: Calculating satellite movement in the arc.

From: Björn Gimle (b_gimle@algonet.se)
Date: Thu Oct 23 2003 - 05:45:32 EDT

  • Next message: John Locker: "Re: Calculating satellite movement in the arc."

    The motion over a segment does not accumulate monotonously over a day.
    After 23h 56m (more or less) the satellite is at the same
    latitude/declination.
    A 1-degree inclination object is at maximum latitude 6h after ascending
    node pi/180*(38000+6378) km above the equatorial plane. At that time the
    "vertical" speed is 0. In-between the latitude varies as a sine function
    of orbital longitude/"time", so the "vertical" speed near the nodes is
    1/6 * pi/2 = 0.26 degrees/h.
    The apparent motion over these six hours is 1 degree iff the satellite is
    close to horizon, up to (38000+6378)/38000 = 1.17 degrees iff directly
    overhead. At the nodes the apparent speed is 0.26 to 0.31 d/h.
    
    You can import the output from a textual prediction program like HighFly
    or Track16 into Excel to get a graph, or even better to a graphic program
    like SkyMap set for Horizontal map projection.
    
    The inclination of 24931 is 0.039, not 0.9 in my alldat.tle
    
    Your result 663 km is the average "horizontal" motion in four minutes
    (111 km at surface level), largely compensated by Earth's rotation
    (geostationary!)
    
    /Björn
    
    
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