Re: Fun with 2-line elsets

From: b_gimle@algonet.se
Date: Wed Nov 02 2005 - 05:14:01 EST

  • Next message: Ted Molczan: "RE: Elements of 05042A & B and 05011B"

    The quantity 'a' usually refers to the semi-major axis,
    ie the average orbital radius. Thus your constant K
    is eight times the standard G*M*(P/2/pi)^2, where
    GM = 398600.44 km^3s^-2 giving K=290784000 km^3min^-2
    
    The Epoch of TLEs is conventionally at the time of ascending
    node preceding the observation. For a circular orbit, this
    makes AOP+MA=360, more or less for an eccentric orbit.
    (AOP indicates the fraction of the period from perigee to the Epoch)
    
    The Mean Motion is 1/anomalistic period, ie perigee to perigee.
    This difference is large for LEO objects, and low inclinations.
    I believed the relation of 'a'  to the orbital period refers 
    to the nodal period, but see below!
    
    Furthermore, these quantities vary along one orbit, and
    TLEs are a mathematical model of 'mean elements'.
    The conversion to a, apogee, perigee can be found in
    the 'C' code for e.g. SeeSat, and in Fortran in the
    SpaceTrack Report. AMSAT may have conversion programs with
    more data, but there I found only the SpaceTrack Report link
    http://www.amsat.org/amsat/ftp/docs/spacetrk.pdf
    
    Many prediction programs and TLE maintenance programs
    give you this information.
    With the elset below, Orbitel gives  5147 x 709 km
    
    and Rob Matson's SkyMap shows
    Epoch day/date:          05302.05363195 [2005/10/29 01:17:14]
    Inclination:              28.3130
    Right Ascension:         161.3956
    Eccentricity:            .2384000
    Argument of perigee:     103.4509
    Mean anomaly:            256.5486
    Mean motion (revs/day):   9.67055069... and during prediction:
    period=148.91 min (1440/9.67055069=148.90569) 709.4 x 5146.5 km
    
    MStar 3 CentRk
    1 25725U 99023B   05302.05363195 0.00000080  00000-0  18269-3 0    06
    2 25725  28.3130 161.3956 2384000 103.4509 256.5486  9.67055069    03
    
    18.9 days earlier:
    1 25725U 99023B   05283.15350902 0.00000100  00000-0  22838-3 0    07
    2 25725  28.3130 211.1600 2384000  22.2028 337.7967  9.67052341    03
    so the perigee advanced 81.25 degrees --> 81.25/360/19.25 = 0.011724
    The number of nodal periods/day is thus 0.011724387+9.67055069 =
    9.682275, and the nodal period 1440/9.682275=148.725
    
    With 148.725 I get a=9298.59 km and orbit 703.7 x 5137.2 km
    With 148.906 a=9306.1    and orbit    709.4 x 5146.5
    
    Surprisingly (?) it appears that it is correct to use the anomalistic 
    period.
    
    /Björn
    
    
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