INCLINATION OF THE GROUND TRACK OF A SATELLITE VERSUS LATITUDE
revosatjanvierinclination_of_iridium
revosatjanvieriridium_pente_au_sol_versus_latitude
The inclination of the trajectory of a satellite in the sky is sometimes quite different from that we expect, considering (i) inclination of the orbit. Two cases.
I)Quasi circular satellite. by example Iridium (i)=86.4 °(on january 2008)
II)Elliptic satellite example Molniya 11474 (i)=62.70,Aureole-2roc,USA 129 (On february 2008),
I)Quasi circular.
We know the relations between latitude (lat) and longitude (lo) on an orbit inclined (i). TAN(i)=TAN(lat)/SIN(Lo)
then TAN(lat)=TAN(i)*SIN(lo) and d(lat)/d(lo)=cos(lat)^2*tg(i)*cos(lo)
But cos(lo)^2=(1-Tan(lat)^2/Tan(i)^2)
1)So d(lat)/d(lo)=cos(lat)^2*tan(i)*(1-tan(lat)^2/tan(i)^2)^.5 or cos(lat)^2*(tan(i)^2-tan(lat)^2)^.5
If latitude =0° then d(lat)/d(lo)=1*Tan(i)*1=Tan(i) the ground inclination is i
If latitude =(i) then d(lat)/d(lo)=0 the ground inclination is 0 see here after the curves but these results don’t take into account the speed rotation of Earth.
In fact the ground inclination is the sum of the movement of the satellite Vsat and the spin of the Earth Vspin.
2)Vsat Vs is 40000kms/T T is period (hour).(precession is negligible)
3)Spin Earth.SE The equatorial rotation of Earth is 40000 kms in 24 h. At latitude lat the spin is Vequatorial*cos(lat).
4)Ground speed GS. We use GS^2=VS^2+SE^2-2*SE*VS*cos(i)
5)To get di, the delta angle to add to (i) to have the true inclination, we use
sin(di)/SE=sin(i)/GS
6)Results. See curves. We see that the equatorial ground inclination of iridium is 90°.The inclination 86.4° of the orbit of iridium has specially been chosen by Motorola to be 90° at the equatorial ground track with Earth spin. We observe that the inclination decreases regularly from equator to pole.( di) is always minor compared to (i). It will be quite different with elliptic orbit where (di) may be quite superior to (i).
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