By Guochang Xu
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So setting ϑ as zero, the equation of motion is then UN CO μ r¨ − rϑ˙ 2 = − 2 , r rϑ¨ + 2˙rϑ˙ = 0. Fig. 19), it turns out to be d(r2 ϑ˙ ) = 0. 20) RO O Because rϑ˙ is the tangential velocity, r2 ϑ˙ is the two times of the area velocity of the radius of the satellite. 20) and comparing it with the discussion in Sect. 21) r2 ϑ˙ = h. 21), one gets and dr dϑ d dr = = dt dϑ dt dϑ TE dϑ = hu2 dt DP h/2 is the area velocity of the radius of the satellite. 19), the equation has to be transformed into a differential equation of r with respect to variable f .
4 will be used for describing the orbit of the satellite. The vector form of the equation of motion can be rewritten through the three components x, y and z(r = (x, y, z)) as G. Xu, Orbits, c Springer-Verlag Berlin Heidelberg 2008 25 26 3 Keplerian Orbits μ x, r3 μ y¨ = − 3 y, r μ z¨ = − 3 z. 5) or in vector form: r × r¨ = 0. 8) lead to EC TE d(y˙z − zy) ˙ = 0, dt d(zx˙ − x˙z) = 0, dt d(xy˙ − yx) ˙ = 0, dt ˙ d(r × r) = 0. 10) UN where A, B, C are integration constants; they form the integration constant vector h.
42) q˙ = ⎝ 1 − e2 cos E ⎠ 1 − e cos E 1 − e2 0 0 RR The second part of above equation can be derived from the relation between E and f . The state vector of the satellite in the orbital coordinate system can be rotated to the ECSF coordinate system by three successive rotations. First, a clockwise rotation around the 3rd-axis from the perigee to the node is given by (see Fig. 4) CO R3 (−ω ). Next, a clockwise rotation around the 1st-axis with the angle of inclination i is given by R1 (−i). UN Finally, a clockwise rotation around the 3rd-axis from the node to the vernal equinox is given by R3 (−Ω).