RE: Rosetta fly by: about magnitudes

From: Ted Molczan (molczan@rogers.com)
Date: Wed Mar 02 2005 - 13:02:59 EST

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    Bram Dorreman wrote:
    
    > Who can tell me the formula to calculate the stellar 
    > magnitude from a self defined absolute magnitude (phase 90 
    > and distance 1000 km), range and sun-target-observer angle?
    > 
    > Another question: what might be a reasonable value for 
    > Rosetta's standard magnitude?
    
    I have derived the following formula, for use with data provided by the
    ephemeris:
    
    mag =  32.17 + 5 * log10("delta") + 0.01 * ("S-T-O" - 90), +/- 2
    
    where:  "delta" = Target apparent range relative to observer. Units: AU
    
            "S-T-O" = Sun-Target-Observer angle; target's apparent PHASE ANGLE as
             seen at observer's location at print time. Units: DEGREES
    
    The above formula is based on my estimated standard magnitude of 6.3 +/- 2 mag
    (1000 km, phase angle 90 deg), based upon the ESA's description: "a large
    aluminium box with dimensions 2.8 x 2.1 x 2.0 metres"
    
    http://esamultimedia.esa.int/docs/science/media/rosetta2004.pdf
    
    For example, here is the relevant Horizon's output for 2005 Mar 04 21:49 UTC,
    for Bram's site coordinates:
    
    "delta" = 0.0000534570 A.U.
    "S-T-O" = 88.0465 deg
    
    Using my formula, I estimate mag = +10.8 +/- 2, in rough agreement with ESA's
    estimate of about mag +8 or +9:
    
    http://www.esa.int/esaCP/SEMMTBYEM4E_index_0.html
    
    Ted Molczan
    
    
    
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