**Previous message:**Greg Roberts: "Obs 25/28 October,2005"**Next in thread:**Dinogeorge@aol.com: "Re: Fun with 2-line elsets"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]

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 ------------------------------------------------------------------------- Subscribe/Unsubscribe info, Frequently Asked Questions, SeeSat-L archive: http://www.satobs.org/seesat/seesatindex.html

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