**Previous message:**John Locker: "Calculating satellite movement in the arc."**In reply to:**John Locker: "Calculating satellite movement in the arc."**Next in thread:**John Locker: "Re: Calculating satellite movement in the arc."**Reply:**John Locker: "Re: Calculating satellite movement in the arc."**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]

The motion over a segment does not accumulate monotonously over a day. After 23h 56m (more or less) the satellite is at the same latitude/declination. A 1-degree inclination object is at maximum latitude 6h after ascending node pi/180*(38000+6378) km above the equatorial plane. At that time the "vertical" speed is 0. In-between the latitude varies as a sine function of orbital longitude/"time", so the "vertical" speed near the nodes is 1/6 * pi/2 = 0.26 degrees/h. The apparent motion over these six hours is 1 degree iff the satellite is close to horizon, up to (38000+6378)/38000 = 1.17 degrees iff directly overhead. At the nodes the apparent speed is 0.26 to 0.31 d/h. You can import the output from a textual prediction program like HighFly or Track16 into Excel to get a graph, or even better to a graphic program like SkyMap set for Horizontal map projection. The inclination of 24931 is 0.039, not 0.9 in my alldat.tle Your result 663 km is the average "horizontal" motion in four minutes (111 km at surface level), largely compensated by Earth's rotation (geostationary!) /Björn ----------------------------------------------------------------- To unsubscribe from SeeSat-L, send a message with 'unsubscribe' in the SUBJECT to SeeSat-L-request@satobs.org List archived at http://www.satobs.org/seesat/seesatindex.html

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