Re: MAP vectors (updated)

From: Tony Beresford (aberesford@iprimus.com.au)
Date: Sun Jul 01 2001 - 06:02:04 PDT

  • Next message: Russell: "JUN29.OBS"

    At 13:27 1/07/01 , McConahy, Ralph wrote:
    >
    >Just received the following...
    >
    >MAP (SATID 0102701; NORAD # 26859) SEPARATION PLUS 6HR. STATE
    >....
    Here is my suggestion for a TLE, based on this information.
    The argument of perigee,the eccentricity , RA of the node and
    the inclination are taken  
    directly from the Mean equator of date values. I had to calculate the
    mean motion I give the unperturbed value. The checksums are not correct.
    Other values of interest are coded
    into the data. The perigee height of the current orbit is 182Km, 
    with an apogee height of 296,000Km approx. 
    
    MAP
    1 26859U 01027A   01181.89580000  .00000000  00000-0 +00000-0 0 00112
    2 26859 28.7308  034.6199 9570422 048.5164 000.8347 00.14547093800014
    
    Assummimg the drag area
    corresponds to an average surface area, and ignoring solar panel
    glints I predict a magnitude around 16 at apogee of current orbit.
    The corresponding magnitude at the Sun Earth L2 is 19.5
    This is rather faint but not beyond the capabalities of
    those who find/follow asteroids. There are some who have only
    lx-200 30cm telescopes( and CCD's) who could make observations on
    such an object.
    Tony Beresford
    Adelaide, So. Australia
    
    
    
    >SPACECRAFT MASS 836.46 KG
    >SPACECRAFT AREA 20.438 SQUARE METERS
    >COEFFICIENT OF DRAG 2.2
    >COEFFICIENT OF SOLAR RADIATION 1.3
    >
    >
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