Greg Roberts wrote: > (1) Cannot find a match for this in my element database updated last > night. Looks like the inclination is about 62.5 degrees and RA > Node about 53 degrees. Height of satellite when seen was > about 1500 kilometres so the standard magnitude works out a > t0 around +6.8 Almost anything could be fit to such a short arc. I could fit a typical U.S. Molniya orbit: 1 70000U 05221.76000488 .00000000 00000-0 00000-0 0 03 2 70000 63.4000 59.5279 7185867 270.0000 15.4711 2.00600000 07 Residuals are small, but could be better, and certain combinations of a higher inclination and lower argument of perigee fit much better: 1 70000U 05221.76000488 .00000000 00000-0 00000-0 0 03 2 70000 71.5881 67.2699 7236776 269.0000 15.3216 2.00600000 09 1 70000U 05221.76000488 .00000000 00000-0 00000-0 0 03 2 70000 69.5147 65.2837 7080209 260.0000 16.2775 2.00600000 02 1 70000U 05221.76000488 .00000000 00000-0 00000-0 0 03 2 70000 67.4216 63.0807 6952299 250.0000 17.5294 2.00600000 04 1 70000U 05221.76000488 .00000000 00000-0 00000-0 0 03 2 70000 66.4314 61.9588 6907254 245.0000 18.2127 2.00600000 03 The above four yield about the same path at about the same time, tonight. Unfortunately, this fits very well too, and the pass would occur far later: 1 70000U 05221.76000488 .00000000 00000-0 00000-0 0 03 2 70000 66.3786 61.8079 6714664 245.0000 19.9629 2.20000000 01 For all I know, the orbit could be that of a NOSS. If I find the time, I will try to find whether or not one could fit. Ted Molczan ------------------------------------------------------------------------- Subscribe/Unsubscribe info, Frequently Asked Questions, SeeSat-L archive: http://www.satobs.org/seesat/seesatindex.html
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