Re: [amsat-bb] School days on Oscar 27 - 28 January

[w4sm@xxxxxxxxxx (Stacey Mills)]

At 03:29 PM 1/27/99 -0500, you wrote:
>
>P.S.  Who knows how fast AO27 travels?  Let me know if you know the
>answer.

AO-27's perigee = 774 km, apogee = 787 km
Assume orbit is circular at 780.5 km.
Earth's mean orbital radius = 6367 km.
therefore one orbit on AO-27 = (6367+780.5)*2*PI = 44,909 km./orbit

AO-27's mean motion = 14.278 orbit/day
24hr/14.278 = 1.68 hrs/orbit

44,909 km/orbit / 1.68 hrs/orbit = 26,731 km/hr 
--or--
26,731 km/hr * 0.621 mi/km =  16,600 mi/hr (give or take)



-- 
 __________________________________________________________________________  
  Stacey E. Mills, W4SM    WWW: 	http://www.cstone.net/~w4sm/ham1.html 
   Charlottesville, VA	   PGP key:	http://www.cstone.net/~w4sm/key.asc    	
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