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 ro- tates 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 lon- gitude 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.
5.2.2 Maximum Latitude Attained
The ground track moves between two bounding latitudes, φ =+φm in the northern hemisphere and φ = −φm in the southern hemisphere. Considering the maximum positive value of Z, obtained for sin n(t − tAN) = 1, we obtain
sin φm =sini, (5.18) which implies that:
• for prograde satellites (0 i π/2), φm = i, • for retrograde satellites (π/2 i π), φm = π − i .
This value φm (φm 0) is called the maximum attained latitude.
Example 5.1. Calculate the maximal latitude attained by the Chinese satellite FY- 1A, in Sun-synchronous orbit at an altitude of h = 901 km.
Given the altitude, we determine the inclination using (4.67). In the present exam- ple, we can use the simplified formula (4.74):
−3 ◦ ◦ ∆i = (901 − 800) × 4.17 × 10 ≈ 0.4 ,i= i0 +∆i =98.6+0.4=99.0 ,
◦ which yields the maximal attained latitude φm = 180 − i =81.0 . Note that this is the latitude reached by the ground track, not by oblique sightings by instruments carried aboard. The ground track of the satellite thus remains within the bounding latitudes 81◦Nand81◦S. 5.2 Ground Track of Satellite in Circular Orbit 181
5.2.3 Equatorial Shift
The difference in longitude between two consecutive ascending nodes λ1 and λ2 is called the equatorial shift and denoted by ∆λE, i.e.,
∆λE = λ2 − λ1 .
Rough Calculation
It is often sufficient to carry out a quick calculation of the equatorial shift, whichisthendenotedby∆0λE. Indeed we may say to a first approximation that, during one revolution of period T (and we may take the Keplerian period here), the orbit of the satellite will not have moved relative to , whilst the Earth makes one complete turn every day, i.e., it rotates through 15◦ per hour, or 1◦ every 4 min relative to this same frame. In this context, we do not bother with the precession of the orbit, or the Earth’s motion relative to the Sun over the time taken for the satellite to complete one revolution. This ˙ amounts to using the relations ψ ≈−Ω˙ T and Ω˙ T ≈ 2π/JM. We have the simplified relation T [min] ∆ λ [degrees] = − , (5.19) 0 E 4 where the minus sign indicates a shift westwards. Observing that the value of one degree of longitude is 111.3 km on the equator, we can also write
∆0λE [km] = −27.82T [min] . (5.20) Using the daily orbital frequency ν (number of round trips per day), we 4 obtain ∆0λE [degrees] = −360/ν and ∆0λE [km] ≈−4 × 10 /ν.
Exact Calculation
During one revolution lasting T = Td, the orbital plane will have rotated throughanangleψ with respect to T. The exact value of the equatorial shift as given by (5.14) with t − tAN = T is therefore ˙ ∆λE = ψT = −(Ω˙ T − Ω˙ )T. (5.21) We note the following points, which follow from (5.21):
• The equatorial shift is always negative, since Ω˙ T is greater than Ω˙ .The shift is westward for a satellite below the geosynchronous orbit. • For satellites in Sun-synchronous orbit, we saw the specific value of ψ˙, according to (5.16). Over one nodal period T ,wehave
˙ 2π ∆λE = ψT = − T. (5.22) JM 182 5 Orbit and Ground Track of a Satellite
Writing the angles in degrees and the time in minutes, we obtain (5.19). For a Sun-synchronous satellite, the approximate formula is identical to the exact one. This is because the two approximations we made in the rough calculation (neglecting precession and the annual motion of the Earth) exactly balance for this type of satellite. • For satellites in geosynchronous orbit, T =2π/Ω˙ T and Ω˙ is negligible. (In any case, it is not the leading term in the perturbation treatment for this type of satellite.) In this case, we thus have
∆λE = −Ω˙ TT = −2π =0 mod(2π) . (5.23)
There is no equatorial shift for such a satellite. The projection of two consecutive ascending nodes does not move on the Earth. (If the satellite is geostationary, we cannot even speak of an ascending node.) Example 5.2. Calculate the equatorial shift for the satellite Meteor-3-07.
The characteristics of the orbit of this satellite are given in Example 4.2. For the quick calculation, we use (5.20) with T = 109.4min.Then
∆0λE = −27.82 × 109.4=−3044 km .
For the exact calculation, with the values already given, viz.,
−7 −1 Ω˙ T = 729.212 × 10 rad s , Ω˙ = −1.429 × 10−7 rad s−1 , −7 −1 Ω˙ T − Ω˙ = 730.641 × 10 rad s , and T = 6 565.28 s, we obtain
◦ ∆λE = −0.479 7 rad = −27.48 = −3 059.51 km .
The equatorial shift of the satellite Meteor-3-07 is thus 3 059.5 km westward (see also Fig. 5.3 and Table 5.1).
5.2.4 Apparent Inclination
Definition and Calculation of Apparent Inclination
The apparent inclination is the angle between the ground track and the equa- tor. This angle i differs from the angle i representing the inclination of the satellite, which is the inclination of the orbital plane of the satellite with respect to the equatorial plane. This happens because i is measured in , whereas i is measured in T. To calculate i ,weconsiderin T the tangent plane to the sphere of radius R at the point on the Earth’s surface corresponding to the ascending node, using the orthogonal unit vectors eλ and eφ already defined. This ascending node is denoted by N in and N0 in T.Attimet = tAN,thethreepoints 5.2 Ground Track of Satellite in Circular Orbit 183
Meteor-3-07 Altitude = 1194.6 km a = 7572.703 km Orbit Inclination = 82.56 ° Terrestrial frame Period = 109.42 min * rev/day =13.16 >>>> Time span shown: 109.42 min = 0.08 day Equat. orbital shift = 3059.5 km ( 27.5 °)
Projection: Orthographic Map centre: 12.0 ° N; 122.0 ° W Asc. node: -133.95 ° [07:43 UTC] Property: none Aspect: Oblique T.:Azimuthal Graticule: 10° [ -90.0 / +78.0 / -148.0 ] Gr.Mod.: GEM-T2
Meteor-3-07 Altitude = 1194.6 km a = 7572.703 km Ground track Inclination = 82.56 ° Terrestrial frame Period = 109.42 min * rev/day =13.16 >>>> Time span shown: 109.42 min = 0.08 day Equat. orbital shift = 3059.5 km ( 27.5 °)
Projection: Plate-carrée Map centre: 0.0 ° ; 0.0 ° Asc. node: -133.95 ° [07:43 UTC] Property: none Aspect: Direct App. inclin. = 86.93 ° T.:Cylindrical Graticule: 10° [ +0.0 / +0.0 / +0.0 ] Gr.Mod.: GEM-T2
Figure 5.3. Orbit and ground track of the satellite Meteor-3-07 over one revo- lution. The distance between the two successive ascending nodes is the equatorial shift 184 5 Orbit and Ground Track of a Satellite
Figure 5.4. Diagrams for the description of apparent inclination. (a) Inclination i and apparent inclination i . The subsatellite point S0 and the point on the ground track corresponding to the ascending node have been represented in T.(b)Com- ponents of the satellite velocity in the frame T. The velocity vector v and its projection vxy onto the equatorial plane have been represented in the frame mov- ing with the Earth
S0 (the subsatellite point), N and N0 coincide. An infinitesimal time dt later, N and S0 have moved away from N0, as shown in Fig. 5.4a. In the frame T(N0, eλ, eφ), the components of the vectors N 0N and NS0 are, setting dt =1, −(Ω˙ − Ω˙ ) n cos i N N = T , NS = . 0 0 0 n sin i
We thus deduce the components of N 0S : n cos i − (Ω˙ − Ω˙ ) N S = T , 0 n sin i and the apparent inclination is given by n sin i tan i = . (5.24) n cos i − (Ω˙ T − Ω˙ )
In terms of the daily recurrence frequency κ defined by (4.32), we may write 5.2 Ground Track of Satellite in Circular Orbit 185
sin i tan i = . (5.25) cos i − (1/κ)
Note that we always have i i. For a Sun-synchronous satellite, we may replace κ by ν, since according to (4.33), these two daily frequencies are equal.
Example 5.3. Calculate the angle between the ground track of the TRMM satellite and the equator when the satellite crosses the ascending node.
For TRMM (i =34.98◦), we calculated the nodal precession rate in Example 4.1. We obtain
sin 34.98 tan i = =0.7604 , cos 34.98 − 0.0649
i =37.25◦ ,i − i =2.27◦ .
Example 5.4. Calculate the apparent inclination for the ground track of a geosyn- chronous satellite.
For a geosynchronous satellite, we have Ω˙ T/n =1andthetermΩ˙ is negligible. Equation (5.24) becomes „ « sin i cos(i/2) π i tan i = = − =tan + , cos i − 1 sin(i/2) 2 2
◦ i ◦ i i =90 + ,i− i =90 − . 2 2 When i is very small, e.g., i =1◦,wehavei =90.5◦: the ground track is not a point but a small line segment almost perpendicular to the equator, between latitudes 1◦Nand1◦S, which transforms into a figure of 8 (lemniscate) when i increases, growing larger with i. The first operational geosynchronous satellite, Syncom-2, had inclination 32.8◦. Its ground track made an angle of 106.4◦ with the equator, or an angle of 16.4◦ with the nodal meridian, as can be seen from the upper part of Fig. 5.10.
Calculating the Inclination from the Apparent Inclination
We have obtained i as a function of i (and a). It is sometimes useful to do the opposite calculation, obtaining i as a function of i (and a). Using (4.1) and expressing Ω/n˙ by means of J2 and i, (5.24) can be replaced by the following equation:
A cos i − B = C sin i, (5.26) 186 5 Orbit and Ground Track of a Satellite
where 3 R 2 Ω˙ 1 A =1− J ,B= T ,C= . (5.27) 2 2 a n tan i
The aim now is to solve this for i. To begin with, we take n = n(a, i)=n0 in B so that this term does not depend on i. In a second step, we substitute into B the value of n obtained using the previous value of i.Themethod converges very quickly. Equation (5.26) transforms to a second order equation in tan(i/2). The solution, which is unique since the angle i lies in the interval [0,π], is given by √ i −C + C2 + A2 − B2 tan = . (5.28) 2 A + B We note that we have A ≈ 1, B ≈ 1/ν and hence B Calculating the Inclination from the Satellite Velocity The position and velocity of a satellite can be found either from the orbit bulletin or from remote-sensing data from the instruments aboard. The po- sition and velocity are given relative to the Earth. Let vX ,vY ,vZ be the velocity components in T. At the ascending node, the angle between the velocity vector, and hence the satellite trajectory, and the equatorial plane is the apparent inclination. Using Fig. 5.4b, we obtain the relation between i and the velocity at the node (using the absolute value for the velocity components if the nodes cannot be distinguished): vZ tan i = . (5.29) 2 2 vX + vY From the value for a and the value for i obtained in this way, we find i using (5.28). Example 5.5. Calculate the inclination of the satellite Meteor-3-07 using the com- ponents of the velocity vector. The values in Table 5.1 were obtained by interpolation of the raw remote-sensing data collected during the first revolution in which the instrument ScaRaB was 5.2 Ground Track of Satellite in Circular Orbit 187 Table 5.1. Raw remote-sensing data from the ScaRaB instrument relating to the first operational revolution of Meteor-3-07. The table shows the altitude h (in km), −1 the velocity components vX ,vY ,vZ (in km s ), the latitude φ and the longitude λ (in degrees), and the crossing time (in UT) at the ascending (A) and descending (D) nodes Time Node Latitude Longitude Altitude Velocity components 1994 02 24 φλ hvX vY vZ 07:43:29 A 0.00 −133.95 1211.658 0.280 −0.261 7.182 08:38:10 D 0.00 32.33 1188.318 −0.202 0.331 −7.204 09:32:54 A 0.00 −161.44 1211.657 −0.128 −0.361 7.182 recording. Nodes (1) and (3) were ascending (vZ > 0), whilst node (2) was de- scending (vZ < 0). We calculate i from (5.29): 7.182 7.204 7.182 tan i = =18.763 , tan i = =18.578 , tan i = =18.751 . 0.383 0.388 0.383 We then deduce i in each case and take the average between the ascending and descending nodes: i =86.933◦ . We calculate the semi-major axis a from the nodal period, since we know the time elapsed between two consecutive transits at the ascending node (see Examples 4.1 and 4.2). With T = 109 min 25 s = 109.42 min, we obtain a = 7 572.7km.We observe that the altitudes given in Table 5.1 vary by 12 km on either side of the average h = 1200 km depending on the type of node. This is explained by the fact that the orbit is slightly eccentric and the argument of the perigee is not ±90◦. With (5.28), the values of i and a give A =0.998 85 ,B=7.619 × 10−2 ,C=5.358 × 10−2 , cos i =0.129 47 . Finally we obtain the value of i, viz., i =82.561◦ . We thus find the inclination i =82.56◦ of the satellite Meteor-3-07, which is pre- cisely the value communicated by the Russian Space Agency. 5.2.5 Angle Between the Ground Track and a Meridian We calculate the angle between the satellite ground track and a meridian for an arbitrary point on the ground track. The calculation of the angle between the ground track and a line of latitude gives a generalisation of the apparent inclination. However, in practice, it is more useful to known the angle between the ground track and the north–south direction. In the frame , the satellite orbit cuts the meridian at an angle j. Refer- ring to Fig. 6.4, P is the subsatellite point (with latitude φ), N is the point 188 5 Orbit and Ground Track of a Satellite on the ground track corresponding to the ascending node (the dihedral angle at N gives the inclination i), and PQ is the meridian through P ,whereQ is on the equator. The dihedral angle at P is the angle j that we wish to determine. Using the relation (ST V) with PQN for CAB,weobtain cos i sin j = . (5.30) cos φ