J. Geod. Sci. 2017; 7:43–50
Research Article
Open Access
L.E. Sjöberg*, E.W. Grafarend, and M.S.S. Joud
The zero gravity curve and surface and radii for
geostationary and geosynchronous satellite
orbits
DOI 10.1515/jogs-2017-0005
Received January 15, 2017; accepted April 20, 2017
Abstract: A geosynchronous satellite orbits the Earth
along a constant longitude. A special case is the geostationary satellite that is located at a constant position above
the equator. The ideal position of a geostationary satellite
is at the level of zero gravity, i.e. at the geocentric radius
where the gravitational force of the Earth equals the centrifugal force. These forces must be compensated for several perturbing forces, in particular for the lunisolar tides.
Considering that the gravity field of the Earth varies not
only radially but also laterally, this study focuses on the
variations of zero gravity not only on the equator (for geostationary satellites) but also for various latitudes.
It is found that the radius of a geostationary satellite
deviates from its mean value of 42164.2 km only within
±2 m, mainly due to the spherical harmonic coefficient J22 ,
which is related with the equatorial flattening of the Earth.
Away from the equator the zero gravity surface deviates
from the ideal radius of a geosynchronous satellite, and
more so for higher latitudes. While the radius of the former surface increases towards infinity towards the poles,
the latter decreases about 520 m from the equator to the
pole. Tidal effects vary these radii within ±2.3 km.
Keywords: Geostationary satellite, geosynchronous satellite, zero gravity surface
1 Introduction
Geostationary and geosynchronous satellites are important for society, e.g. for tele- communication and naviga-
*Corresponding Author: L.E. Sjöberg: Royal Institute of Technology (KTH) Stockholm, Sweden, E-mail: [email protected]
E.W. Grafarend: Department of Geodesy and Geoinformatics
Stuttgart University Stuttgart, Germany
M.S.S. Joud: Royal Institute of Technology (KTH) Stockholm,
Sweden
tion. Both types of satellites have a period of revolution
around the Earth’s rotational axis that equals one day,
which makes these satellites to stay fixed at a specific longitude. An ideal geostationary satellite, a special case of
the geosynchronous satellite, is always in the equatorial
plane of the Earth, while the orbital plane of the latter is
inclined vs. the celestial equatorial plane, which makes
the satellite move periodically between its northern and
southern latitudes, being approximately equal to plus and
minus of the inclination angel of the orbital plane.
The orbital plane of the geostationary satellite is subject to a precession due to tidal forces of the moon and the
sun and the non-spherical shape of the Earth. This implies
that the inclination of the plane varies within ±15 degrees
away from the equator with a period of about 53 years. The
flattening of the Earth also causes a slow longitudinal drift
of the satellite towards one of the stable equilibrium points
at 75∘ .3E (over Indonesia) and104∘ .7W (in the Eastern
Pacific), “the graveyards of geostationary satellites”. Also
solar wind and radiation pressure make the satellite to
slowly drift away from the undisturbed orbit. All these perturbations require orbital manoeuvres to keep the satellite
in a requested position (Soop 1994).
The orbits of the above satellites can be determined
analytically and/or directly monitored by observations
(Soop 1994), such as satellite laser ranging and C-band
ranging (e.g. Guo et al., 2010; Yang et al. 2013), GNSS or
Doppler shift observations by LEO satellites (Knogl et al.
2011).
Here we will consider the problem that the Earth is
not at spherical homogenous ball, and the gravity field is
not radially symmetric but varies also laterally in space. In
case of the geostationary satellite, orbiting the Earth along
the equator, this means that the centrifugal force acting
on the satellite matches the gravitational force at various
radii, depending on the longitude position of the satellite.
As gravity is defined as the sum of the gravitation and the
centrifugal force per unit mass, the geostationary satellite
is located on the zero gravity surface, which is at the focus
of this study. However, as will be shown, geosynchronous
© 2017 L.E. Sjöberg et al., published by De Gruyter Open.
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44 | L.E. Sjöberg, E.W. Grafarend, and M.S.S. Joud
satellites (with exception for geostationary satellites) typically orbit the Earth below the zero gravity surface.
Although most of the orbital theory given here should
be well-known among satellite orbit engineers, it is probably not the case for geodesists. Also, “the zero gravity surface” is a new concept.
times mass of the Earth,) it follows from Eqs. (3) and (4a)
that
√︀
3
(5a)
r = r̄ = kM/ω2 ,
and by differentiating this equation one can see the change
in radius (dr) for any change in gravitation (dg) :
dr = r̄
2 Some basic equations
The energy integral for satellites
V+
kM m kM s v2
+
=
+ c,
lm
ls
2
(1)
where V is the gravitational potential at the satellite generated by the Earth’s mass, k is the gravitational constant,
M m and M s are the masses of the moon and the sun at distances l m and l s from the satellite, respectively, v is the
total velocity of the satellite and c is an integration constant, was presented by Bjerhammar (1967). Disregarding
the tidal effects one obtains an equivalent formula for the
Earth’s gravitation (g) by the expression
g=−
∂V 1 ∂v2
=
.
∂r
2 ∂r
(2)
Here r is the geocentric radius of the satellite. This formula
disregards the minor influences of solar radiation, air drag
and tidal effects of the planets.
2.1 Approximate radius of a geostationary
satellite
A geostationary satellite orbits the Earth in its equatorial
plane. In this case the satellite is moving with the same
angular velocity (ω) as the Earth rotates, and the velocity
of the satellite becomes
and Eq. (2) yields
or
v = ωr,
(3)
g = rω2 ,
(4a)
G = g − rω2 = 0,
(4b)
where G is gravity. Hence, the geostationary satellite is located at the radius of zero gravity above the Earth, which
implies that the gravitational force of the Earth is completely balanced by the centrifugal force. By approximating g by the zero degree harmonic/Keplerian term (kM/r2 )
of the Earth’s gravitation, where kM is the geocentric gravitational constant, (i.e. Newton’s gravitational constant
dg
g
.
(5b)
Bode and Grafarend (1982) estimated r̄ to 42164.26 km by
using the constants from Geodetic Reference System (GRS)
1967 (Table 1). Taking the constants from GRS1980 the radius for zero gravity becomes 42164.17 km, i.e. the estimates of the radius differ 90 m between the reference systems, mainly due to the change in kM.
3 The precise radius of a
geostationary satellite
As the gravity field of the Earth varies laterally from point
to point, Eq. (5a) cannot hold as a precise estimate, but g
must be determined by a more exact formula than above.
Here we will first improve the gravity G of Eq. (4b) by approximating it by the normal gravity field of Pizzetti (1911)
and Somagliana (1929).[Ardalan and Grafarend (2001) presented an approximate formula for the normal gravity on
the level ellipsoid accurate to 10−7 mGal. However, here we
need gravity at satellite level along the equator.] Then we
will also compute the radius along the equator for the vanishing gravity from an Earth Gravitational Model (EGM).
3.1 The zero gravity radius by normal gravity
The normal gravity field is the field generated by a level ellipsoid. As derived in Somigliana (1929; see also Heiskanen
and Moritz 1967, Sect. 2-8 and Caputo 1967) using Jacoby ellipsoidal coordinates (u, β, λ), the normal gravity vector is
the gradient of the normal potential, with the two components:
[︃
)︂]︃
′ (︂
1 kM ω2 a2 E q
sin2 β 1
+
−
− ω2 u cos2 β
𝛾u = −
w B2
2
6
B2 q0
(6a)
and
(︂ 2 2
)︂
1
ω a q
𝛾β = −
−
(6b)
+ ω2 B sin β cos β,
w
B q0
where
1
q = q(u) =
2
[︂(︂
)︂
(︂ )︂
]︂
u2
E
u
1 + 3 2 arctan
−3
u
E
E
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(6c)
The zero gravity curve and surface and radii for geostationary and geosynchronous satellite orbits |
45
Table 1: The radii for zero gravity determined for GRS67 and GRS80 for a Keplerian orbit by Eq. (5a).
System
GRS 1967
GRS 1980
kM [m3 s−2 ]
3986030 × 108
3986005 × 108
ω [s−1 ]
7292115.1467 × 10−11
7292115 × 10−11
r̄ [km]
42164.26
42164.17
3.2 The zero gravity radius by an EGM
q0 = q(b)
u2
q (u) = 3 1 + 2
E
(︂
′
and
)︂ (︂
√︁
w=
u
E
1 − arctan
E
u
(6d)
)︂
−1
(6e)
u2 + E2 sin2 β
.
(6f)
B
Here B = u2 + E2 , and a and E = ae are the semi-major
axis and linear eccentricity of the level ellipsoid. Also e is
the eccentricity and u, β and λ(= 0) are the ellipsoidalharmonic coordinates with u along the semi-minor axis
of the reference ellipsoid and β being the reduced latitude. (The longitude component λ of normal gravity always vanishes, as the level ellipsoid is rotationally symmetric around the z-axis.)
As the magnitude of the gravity vector is
√︁
(7)
𝛾 = 𝛾u2 + 𝛾β2 .
√
and on the equator β = 𝛾β = 0, it follows that
(︂
)︂
1 kM ω2 a2 E q′ (u)
2
−
−ω u .
𝛾 = |𝛾u | =
w B2
6q0
B2
V=
(︂ )︂n+1
n0
∑︁
R
V n (θ, λ),
r
where
V n (θ, λ) =
(9a)
where
kM a2 Eq′ (u)
−
.
(9b)
6q0
ω2
Here the term C is first assumed as a constant. In that case
Cardano’s method shows that only one real root exists, and
it becomes:
√︃√︂
√︃
√︂
2
6
3 C
3
C
E
C2 E6 C
+
+
−
+
− .
(10)
u=
2
4
27
4
27 2
C=
′
However, the term q (u) in C is slightly dependent on u,
and therefore the primary solution of Eq. (10) needs to be
iterated.
Once u is determined, the zero gravity radius above
the equator follows from
√︀
r = B = u2 + E2 .
(11)
Using the parameters of GRS 1967/1980 one obtains the radius 42164.79/42164.70 km.
n
kM ∑︁
A nm Y nm (θ, λ),
R
(12b)
m=−n
the gravitation experienced by the satellite becomes
n0
∑︁
n=0,n≠1
(8)
(12a)
n=0,n≠1
g=
Hence, equating 𝛾 to zero, zero gravity is given by the (approximate) depressed cubic formula
u3 + uE2 − C = 0,
There will be a small variation of the radius of zero gravity
along the equator, mainly due to the tesseral spherical harmonics of degree and order 2 [i.e. A22 and A2−2 in Eq. (12b)]
. As these coefficients are of the order of ±2 10−6 , one can
expect deviations of the radius of the order of a few metres
[when considering Eqs. (5b), (12b) and (13)]. Starting from
the spherical harmonic expansion of the Earth’s gravitational potential truncated at some degree n0 :
n+1
R
(︂ )︂n+2
R
V n (θ, λ).
r
(13)
Here (θ, λ) is the horizontal position in co-latitude and longitude, respectively, R is a reference radius (e.g. sea level
radius) for the EGM with coefficients A nm .
Inserting Eq. (13) into Eq. (4b) one obtains for a geostationary satellite with θ = π/2:
√︂
(︂
)︂
∆ ∆2
3 kM
r=
... ,
(14a)
(1 + ∆) =r̄ 1 + −
3
9
ω2
where
∆=
n0 (︂ )︂n ∑︁
n
∑︁
R
n=2
r
A nm Y nm (π/2, λ).
(14b)
m=−n
Eq. (14a) needs to be iterated by starting, e.g., from r = r̄,
and the convergence should be rapid as for R being sea
level radius R/r̄ = 6371/42157≈ 0.127 and |A nm | < 2 × 10−3
Numerical results are given below.
3.3 Numerical results.
Equations (14a) and (14b) were used with EGM GOCO05c
(Pail, et al. 2016) complete to degree 80 to determine the
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46 | L.E. Sjöberg, E.W. Grafarend, and M.S.S. Joud
variation of zero gravity and its radius along the equator.
The mean value (ḡ ) of gravitation along the equator was
determined to 0.2242179 m/s2 by Eq. (13), corresponding
to the mean radius r̄ = 42164.2 km (i.e. the same as for the
Keplerian orbit). The variations along the equator of the
gravitation (dg = g − ḡ) at the mean radius and the radius
(dr = r − r̄) are shown in Figs. 1 and 2, respectively. The
first figure shows that dg varies between −9.81 µGal and
+9.69 µGal with minima at longitudes (75∘ .1) and (254∘ .7)
and maxima at longitudes (161∘ .0) and (349∘ .7). The maxima (being unstable points) correspond to radii with deviations dr of 1.9 m and 1.6 m, while the minima (stable
points) are attained at deviations of −2.0 and −1.5 m. (See
also Table 2.)
Figure 2: The variation of the radius (r) of zero gravity along the
equator. The vertical axis shows dr = r − ¯r̄ in metre, where ¯r̄ =
42164.17 km is the average radius along the equator. The horizontal
axis shows the longitude.
dinates that can be expressed:
r ≈ r̄ + δa − r̄e cos [ω (t − t P )]
δ ≈ i sin [ϖ + ω (t − t P )]
(15)
(︀
)︀
λ ≈Ω + ϖ − α0 + ω (t0 − t P ) − 1.5 δa/r̄ ω (t − t P )
+ 2e sin ω (t − t P ) ,
Figure 1: The vertical axis shows the difference of zero gravity from
its mean value along the equator. λ-axis is longitude.
where Ω, i and ϖ are the right ascension of the ascending
node and the inclination of the orbital plane as well as the
argument of the perigee vs. the ascending node. t P is the
time for perigee passage, t0 is an arbitrarily selected reference time and the declination approximates the latitude.
3.4 The lateral motions of the satellite
4 The zero gravity surface
A perfect geostationary satellite is constantly positioned
over the equator at a specific longitude and geocentric radius. However, in reality, mainly due to perturbing forces,
this will not be the case, but there will always be small motions away from the perfect position. Small periodic motions may be acceptable, but at some point secular motions must be corrected by on-board thrusters.
However, also for the Kepler motion (with no perturbing forces), in practice the orbit is always inclined and eccentric, implying that the satellite moves away from the
stable geostationary position. In case of small eccentricity and inclination, as outlined in the Appendix based
on Soop (1994), the geostationary satellite will experience
small sinusoidal liberations (periodic motions) of its coor-
Equation (4a) or (4b) yields the radius of the zero gravity
surface only along the equator. For any other latitude zero
gravity is given by the equation
G = g − rω2 cos2 φ = 0,
(16)
where φ is the latitude.
Relying on the approximate formula for gravitation
g = kM/r2 , one obtains approximately the radius as
√︁
3
r̄ (φ) = kM/ (ω cos φ)2 .
(17)
Considering the more precise gravity given by Eq. (13), one
obtains also the improved radius:
√︃
(︂
)︂
kM
∆ ∆2
r= 3 2
1
+
∆
=
r̄
1
+
−
.
.
.
,
(18)
(
)
3
9
ω cos2 φ
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The zero gravity curve and surface and radii for geostationary and geosynchronous satellite orbits |
47
Table 2: The radius and Earth’s gravity acceleration at the extreme points.
λ [degree]
dg [µGal]
⧸︀
g [m s2 ]
dr [m]
r [km]
Minima
75.05
254.75
−9.8
−7.1
0.22421781 0.22421784
−1.973
−1.544
42164.347
42164.347
Maxima
160.95
349.75
9.7
7.2
0.22421801 0.22421798
1.953
1.560
42164.351
42164.351
with ∆ as in Eq. (14b) but for varying co-latitudes.
From Eqs. (17) and (18) one can see that the radius increases by the factor (cos φ)−2/3 vs. the radius of the geostationary satellite when moving away from the equator.
This is the ideal radius that matches Eq. (16), and it is illustrated in Figs. 3 and 4. One can see that the radius increases dramatically from that of the geostationary satellite at the equator to about 135 000 km at 80∘ latitude (and
towards infinity at the poles).
Figure 3: The figure shows the mean geocentric radius (r̄) along
parallel circles of the zero gravity surface between ±80∘ latitudes.
It can be seen from Fig. 4 (bottom) that the variation
of the residual radius dr = r − r̄ (φ) at different latitudes
decreases with latitude (due to the increasing distance to
the centre of the Earth). For instance, at the equator dr is
within ±2 m, while at latitude 80∘ the variation has decreased to within ±0.02 m, but the extrema occur for the
same longitudes (see Table 2).
Figure 4: The vertical axes are (top) the radius of the zero gravity
surface in km and (bottom) the change of the radius dr = r − r̄ (φ) in
m of the surface about its means r̄ (φ) over parallels. The horizontal
axis shows the longitude in degrees.
5 The ideal radius of a
geosynchronous satellite
A geosynchronous satellite orbits the Earth along a constant longitude, implying that its period and angular velocity is the same as for the Earth’s rotation around its axis
(with angular velocity ω). Generally, the orbital plane is
inclined (with the inclination angel i) with respect to the
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48 | L.E. Sjöberg, E.W. Grafarend, and M.S.S. Joud
equatorial plane, implying that the satellite moves periodically along a specific meridian between the latitudes −i
and +i. However, as also in this case the centrifugal force
should match the gravitational force, Eqs. (4a) and (4b)
hold, and the radius can be determined by Eqs. (14a) and
(14b), but now the fixed co-latitude is replaced by the variable co-latitude of the geosynchronous satellite.
Figure 5: The mean radius (r̄) of a geosynchronous satellite along
parallel circles at latitude ϕ.
In Fig. 5 the mean radius as a function of latitude
is illustrated. On the equator, it is clear that the geosynchronous satellite orbits at the same radius as a geostationary satellite (Eq. 14a) on the zero gravity surface
(Eq. 18). When migrating from the equator towards the
poles, the satellite orbit is closer to the Earth’s centre than
at the equator. For instance, the mean radius for the satellite at 60∘ latitude is about 400 m less than that at the
equator. In Fig. 6 the radii about their latitudinal means
r̄ (φ) are shown. In Fig. 6(top) one can see that the radius for a geosynchronous satellite decreases about 500 m
from the equator to latitude 80∘ , while the radius of the
zero gravity surface, shown in Figs. 3 and 4(top), increases
about 92800 km. Comparing Figs. 4(bottom) and 6(bottom) one notices that away from the equator the radius
variation around its latitude mean value is slightly smaller
for the zero gravity surface than for a geosynchronous
satellite.
Figure 6: The vertical axes show (top) the radius r (φ) in km +
42160 km and (bottom) the change of the radius dr = r − r̄ (φ) in
m along parallel circles of a geosynchronous satellite. The horizontal axis shows the longitude in degrees.
does not cause the largest perturbation of the satellite orbit, but it is due to the lunisolar tide. Considering various
perturbations he found that the radius for geostationarity
is 1.7 km higher than that for the Keplerian orbit.
The lunisolar perturbations of geostationary and
geosynchronous satellite orbits are both periodical and
secular. The acceleration vector of a satellite (being the
negative of the gravitational acceleration) due to each of
the celestial bodies (with mass M i ) is given by
(︃
)︃
′
r
r
dr̈ = −dg = kM i
−
,
(19)
r3 (︀r′ )︀3
′
6 Effects of tidal perturbations
Capderou (2004, p.159) emphasized that at the radius of
a geosynchronous satellite the Earth’s flattening term J2
where r and r are the vectors from the celestial body to the
satellite and Earth’s center, respectively. Inserting dg into
Eq. (5b) one obtains the change of radius due to the tidal
effect. Our numerical calculations show that the lunisolar
tide vary the satellite radius between −2.26 and 1.80 km.
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The zero gravity curve and surface and radii for geostationary and geosynchronous satellite orbits |
7 Concluding remarks
While the radius of a geostationary satellite (42164.2 km)
varies only within ±2 m along the equator, the ideal radius
of a geosynchronous satellite, mainly due to the flattening of the Earth, decreases about 522 m from the equator
to the pole. However, these figures do not include the lunisolar tidal effect, which alter the radius between −1.80
and +2.26 km.
Along the equator the radius of the zero gravity surface agrees with the radii of geostationary satellites. Away
from the equator geosynchronous satellites fly below the
zero gravity surface, whose radius approaches infinity at
the poles. The zero gravity surface is the ideal surface for
conducting zero-gravity experiments and engineering production.
Acknowledgement: This paper was mainly prepared
while the first author was a visiting scientist at Stuttgart
University, Institute of Geodesy. He is grateful to the
Alexander-von-Humboldt foundation for providing the
financial support.
References
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the international gravity formula accurate to sub-nanoGal level. J.
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the Division of Geodesy Division, Royal Institute of Technology,
Stockholm, Sweden
Bode A and Grafarend E W, 1982. The telluroid mapping based on normal gravity potential including the centrifugal term. Boll. Geod.
Szie. Aff. XLI(1):21-56
Capderou M, 2004. Satellites - orbits and missions, Springer Co.
Caputo M, 1967. The Gravity Field of the Earth, Academic Press , New
York - London
Guo R, Hu Xiao G, Tang B, Huang Y, Liu L, Cheng L and Feng
H E (2010) Precise orbit determination for geostationary satellites with multiple tracking techniques, Astronomy, 55 (8): 687692,doi: 10.1007/s11434-010-0074-x
Heiskanen W A and Moritz H, 1967. Physical Geodesy, W.H. Freeman
and Co., San Francisco
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Symposium ELMAR-2011, 14-16 September 2011, Zadar, Croatia,
pp. 325-328
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Model GOCO05c. GFZ Data Services. http://doi.org/10.5880/
icgem.2016.003
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dall’ ellipsoide, Atti Accad. Sci., Torino, 46, 331.
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Somagliana C, 1929. Teoria generale del campo gravitazionale dell’
ellisoide di rotazione. Mem. Soc. Astron. Ital., Vol. IV.
Soop E M, 1994. Handbook of geostationary orbits. Springer Co.
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A Appendix. The satellite position
in a near geostationary Keplerian
orbit
The satellite position is usually expressed in a “space fixed
system, which is close to an inertial system. The position
vector vs. the Earth’s gravity center can be written
⎞
⎛
⎞
⎛
X
cos δ cos α
⎜
⎟
⎜
⎟
(A1)
ρ̄ = ⎝ Y ⎠ = r ⎝ cos δ sin α ⎠ ,
Z
sin δ
where δ is the declination of the satellite and α is the right
ascension (corresponding to latitude and longitude in the
Earth fixed system).
The space fixed vector can also be determined from the
satellite coordinates (x, y, z) in the orbital plane, where
⎞
⎛
⎞
⎛
x
cos f
⎜
⎟
⎜
⎟
(A2)
⎝ y ⎠ = r ⎝ sin f ⎠ ,
z
0
where f is the true anomaly of the satellite. The space fixed
position vector is related to the plane coordinate vector by
the three 3-D rotations
(︁
)︁T
ρ̄ = R3 (−Ω)R1 (−i) R3 (−ϖ) x y z
,
(A3)
where R k (ψ) implies a rotation around the k-th axis by
the angle ψ, the argument of the perigee vs. the ascending
node, and the arguments Ω, i and ω are the right ascension of the ascending node and the inclination of the orbital plane and the argument of the perigee, respectively.
The result of the transformation is
⎛
⎞
cos Ω cos (ϖ + f ) − sin Ω sin (ϖ + f ) cos i
⎜
⎟
ρ̄ = r ⎝ sin Ω cos (ϖ + f ) + cos Ω sin (ϖ + f ) cos i ⎠ .
sin (ϖ + f ) sin i
(A4)
By comparing the Z-coordinates from Eqs. (A2) and (A4)
one obtains
sin δ = sin (ϖ + f ) sin i,
(A5)
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which implies that δ ≤ i for constant ϖ and i. As the declination (in the space fixed system) is close to the latitude
(in the Earth fixed system), it holds also approximately that
the latitude is limited to within ±i when the satellite moves
in its plane.
The last two equations show that the declination (and
thereby the latitude) as well as the radius will have a sinusoidal motion in the slightly inclined satellite orbit.
The satellite longitude (λ) is the difference between its
right ascension and that at Greenwhich (α0 ). Considering
also Eq. (A9) one obtains:
λ ≈ f + Ω + ϖ − α0 .
Slightly inclined satellite orbit
The satellite orbit is always somewhat elliptic, and, considering no perturbing forces, the relation between the
geocentric radius and the true anomaly is given by Kepler’s
first law:
r=
a(1 − e2 )
,
1 + e cos f
(A6)
With reference to Soop (1994, Sect. 2.3), assuming that the
eccentricity is small, we may linearize this equation to
r = r̄ + δa − r̄e cos (f ) ,
(A7)
where e is the eccentricity and a is the semi-major of the
orbital ellipse.
Assuming also that the inclination of the orbital plane
is small, one may use the approximations cos i ≈ 1 and
sin i ≈ i . Then, by comparing Eqs. (A1) and (A4) one obtains also
X/r = cos δ cos α ≈ cos (Ω + ϖ + f )
(A8a)
Y /r = cos δ sin α ≈ sin (Ω + ϖ + f ) ,
(A8b)
By also inserting Eq. (A11) one arrives at
λ ≈ Ω + ϖ − α0 + ω (t − t P ) ,
ḟ = ω
[︀(︀
)︀]︀
1 − 1.5δa/r̄ + 2e sin ω (t − t P ) ,
(︀
)︀
λ ≈Ω + ϖ − α0 + ω (t0 − t P ) − 1.5 δa/r̄ ω (t − t P )
+ 2e sin ω (t − t P ) ,
where t0 is an arbitrary selected reference epoch.
(A9)
In this case one also obtains from (A5) that
(A10)
Inserting the simple approximation
f ≈ ω(t − t P )
(A11)
into Eqs. (A7) the satellite radius is given as a function of
time (t), with t P being the time of perigee passage, as
r ≈ r̄ + δa − r̄e cos [ω (t − t P )] .
(A12)
Similarly Eq. (A10) becomes
δ ≈ i sin [ϖ + ω (t − t P )] .
(A16)
yielding
implying that
δ ≈ sin (ϖ + f ) i.
(A15)
but, as pointed out by Soop (1994), this is only a zero-order
approximation, and the first order-approximation needs
an improved temporal change of the true anomaly by
and
f ≈ α − Ω − ϖ.
(A14)
(A13)
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Download Date | 6/15/17 2:48 PM
(A17)