5 Orbit and Ground Track of a Satellite

5 Orbit and Ground Track of a Satellite 5.1 Position of the Satellite on its Orbit Let (O; x, y, z) be the Galilean reference frame already defined. The satellite S is in an elliptical orbit around the centre of attraction O. The orbital plane P makes a constant angle i with the equatorial plane E. However, although this plane P is considered as fixed relative to in the Keplerian motion, in a real (perturbed) motion, it will in fact rotate about the polar axis. This is precessional motion,1 occurring with angular speed Ω˙ , as calculated in the last two chapters. A schematic representation of this motion is given in Fig. 5.1. We shall describe the position of S in using the Euler angles. 5.1.1 Position of the Satellite The three Euler angles ψ, θ and χ were introduced in Sect. 2.3.2 to specify the orbit and its perigee in space. In the present case, we wish to specify S. We obtain the correspondence between the Euler angles and the orbital elements using Fig. 2.1: ψ = Ω, (5.1) θ = i, (5.2) χ = ω + v. (5.3) Although they are fixed for the Keplerian orbit, the angles Ω, ω and M − nt vary in time for a real orbit. The inclination i remains constant, however. The distance from S to the centre of attraction O is given by (1.41), expressed in terms of the true anomaly v : a(1 − e2) r = . (5.4) 1+e cos v 1 The word ‘precession’, meaning ‘the action of preceding’, was coined by Coper- nicus around 1530 (præcessio in Latin) to speak about the precession of the equinoxes, i.e., the retrograde motion of the equinoctial points. This term was then taken up in mechanics to describe the corresponding Euler angle. In the motion of the satellite orbital plane, the word ‘precession’ clearly refers to a motion that may actually be prograde, as well as retrograde. 176 5 Orbit and Ground Track of a Satellite Figure 5.1. Precessional motion of the orbit in the frame . The orbital plane rotates about the polar axis, maintaining a fixed inclination relative to the equatorial plane (xOy). Its projection onto the equatorial plane can be used to measure Ω, the longitude of the ascending node, whose variation is given by Ω˙ . If the satellite has a prograde orbit (as here, where the ascending node has been indicated by a small black circle, the descending node by a small white circle and the latest ground trackbyadash-dotted curve), the precessional motion is retrograde, i.e., Ω<˙ 0 Since this distance is specified, the position of S is determined by composing the following three rotations, shown schematically in Fig. 5.2 and described below: • Precessional motion in E, taking the straight line Ox onto the straight line ON (= Ox1): =⇒ [P1] : Rotation through angle (Ox, Ox1)=ψ about Oz . • Rotation of E onto P about the line of nodes: =⇒ [P2] : Rotation through angle (Oz1, Oz2)=θ about Ox1 . • Motion in P which takes the straight line ON (= Ox1 = Ox2)ontothe straight line OS (or OX): =⇒ [P3] : Rotation through angle (Ox2, OX)=χ about Oz2 = OZ . It can be shown that any rotation of a solid can be decomposed into three elementary rotations about suitably chosen axes. In the case of the Euler angles, this decomposition is one-to-one with the following domains: 5.1 Position of the Satellite on its Orbit 177 Figure 5.2. The three rotations taking a given point on a sphere to another ar- bitrary point, using the three Euler angles. Black circles indicate the three axes of rotation: Oz = Oz1 for [P1], Ox1 = Ox2 for [P2], and Oz2 = OZ for [P3] ψ ∈ [0, 2π) ,θ∈ [0,π) χ ∈ [0, 2π) . The axes and angles of rotation are summarised here: P1 (Ox, Oy, Oz) −→ (Ox1, Oy1, Oz1 = Oz) , P2 (Ox1, Oy1, Oz1) −→ (Ox2 = Ox1, Oy2, Oz2) , P3 (Ox2, Oy2, Oz2) −→ (OX, OY , OZ = Oz2) . We then have the three rotation matrices: ⎛ ⎞ cos ψ − sin ψ 0 ⎝ ⎠ P1 = sin ψ cos ψ 0 , (5.5) 001 ⎛ ⎞ 10 0 ⎝ ⎠ P2 = 0cosθ − sin θ , (5.6) 0sinθ cos θ ⎛ ⎞ cos χ − sin χ 0 ⎝ ⎠ P3 = sin χ cos χ 0 . (5.7) 001 178 5 Orbit and Ground Track of a Satellite The matrix product of these three matrices gives the matrix P calculated below. We consider without loss of generality that N is on the axis Ox at the time origin. Its coordinates are thus (r, 0, 0). The coordinates of S(X, Y, Z) are obtained from those of N(x, y, z) via application of P : ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ X x r ⎝ Y ⎠ = P ⎝ y ⎠ = P ⎝ 0 ⎠ . Z z 0 We see that only the first column of the matrix P will be required for this calculation. We shall therefore calculate the matrix product P = P1P2P3 and write it in the form ⎛ ⎞ cos ψ cos χ − sin ψ sin χ cos θP12 P13 ⎝ ⎠ P = sin ψ cos χ +cosψ sin χ cos θP22 P23 , (5.8) sin χ sin θP32 P33 which gives ⎛ ⎞ ⎛ ⎞ X cos ψ cos χ − sin ψ sin χ cos θ ⎝ Y ⎠ = r ⎝ sin ψ cos χ +cosψ sin χ cos θ ⎠ . (5.9) Z sin χ sin θ Using the orbital parameters given by (5.1)–(5.4), we obtain ⎛ ⎞ ⎛ ⎞ X cos Ω cos(ω + v) − sin Ω sin(ω + v)cosi a(1 − e2) ⎝ Y ⎠ = ⎝ sin Ω cos(ω + v)+cosΩ sin(ω + v)cosi ⎠ . (5.10) 1+e cos v Z sin(ω + v)sini Consider a spherical coordinate system in the Galilean frame . The plane of reference is the equatorial plane xOy of the Earth, Oz is the polar axis and the position of Ox is fixed in space. The point S can be specified in by its spherical coordinates, the longitude λ and the latitude φ, measured with the usual convention following from the right-handed trigonometric system. The longitude of Ox (position of N at the time origin) is denoted by λ0. Hence, ⎛ ⎞ ⎛ ⎞ X cos φ cos(λ − λ0) ⎝ ⎠ ⎝ ⎠ Y = r cos φ sin(λ − λ0) . (5.11) Z sin φ We thus obtain the position of S(λ, φ) as a function of time and the other orbital parameters via X, Y, Z : Z X φ =arcsin ,λ= λ + arccos , (5.12) r 0 r cos φ λ − λ0 from the sign of Y, λ− λ0 ∈ (−π, +π] . (5.13) If |φ| = π/2, λ is not determined (and its determination would be pointless). 5.2 Ground Track of Satellite in Circular Orbit 179 5.1.2 Equation for the Ground Track In many circumstances, one needs to know the position of the satellite relative to the Earth. One must therefore represent S in the frame T,whoseaxes in the equatorial frame rotate with the Earth. The transformation from this frame to the Galilean frame is obtained by a simple rotation about the polar axis Oz with angular speed (−Ω˙ T), since T rotates in with angular speed Ω˙ T. Bear in mind that these calculations are carried out in the Galilean frame , whilst the results may be expressed in the frame of our choice. Recalling the above definition of λ0, the equations of motion of S are the same in T as in , provided that we replace the value of ψ in (5.1) by ψ = λ0 +(Ω˙ − Ω˙ T)(t − tAN) , (5.14) where the time origin, the crossing time at the ascending node N, is written t = tAN. The satellite ground track is defined as the intersection of the straight line segment OS with the Earth’s surface. Its equation is thus obtained by replacing r by R in the above equations. (For this application, we may treat the Earth as a sphere of radius R.) 5.2 Ground Track of Satellite in Circular Orbit Near-circular orbits, which may be considered as circular in a first approxi- mation, constitute a very important and frequently encountered case. Let us now study some notions developed specifically for these orbits, such as the equatorial shift or the apparent inclination. The velocity of the satellite will be calculated in Sect. 5.5. 5.2.1 Equation for Satellite Ground Track When the orbit is circular, the motion is uniform with angular frequency n, the mean motion. Using the notation introduced above, the value of χ in (5.3) can be replaced by χ = n(t − tAN) . (5.15) We thus obtain the equation for the ground track, with (5.14), (5.2) and (5.15) substituted into (5.9) and (5.11), where r has been changed to R. In this case, the ground track of the satellite is determined by two quan- tities relating to the ascending node taken as origin, namely, its longitude λ0 and the crossing time tAN, which constitute the initial conditions of the uniform motion. 180 5 Orbit and Ground Track of a Satellite Sun-Synchronous Satellites For Sun-synchronous satellites, the angle ψ takes a specific value since Ω˙ = Ω˙ S. We have seen that the two angular frequencies characterising the Earth’s (annual and daily) motion are related by (4.24). Hence, according to (5.14), 2π ψ˙ = Ω˙ S − Ω˙ T = − . (5.16) JM Using the daily orbital frequency as given by (4.25), we obtain, for Sun- synchronous satellites, the very simple relation n ψ˙ = − . (5.17) ν We shall see the very important consequences of this relation in the following chapters, in particular, when studying the crossing time of the satellite and the question of recurrent orbits.

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