Chin. Phys. B Vol. 24, No. 4 (2015) 048402
Orbit equation (4) contains three independent parameters:
h, e, and f , and it describes the conic sections in the orbital
plane. In the actual calculation, the angular momentum h and
true anomaly f are frequently replaced by the semi-major axis
a and mean anomaly M, respectively. Describing the orienta-
tion of an orbit in three dimensions requires three additional
parameters, called the Euler angles, which are illustrated in
Fig. 1.
First, we locate the intersection of the orbital plane with
the equatorial (XY ) plane. That line is called the node line.
The point on the node line where the orbit passes above the
equatorial plane from below it, is called the ascending node.
The node line vector 𝑁 extends outward from the origin
through the ascending node. At the other end of the node
line, where the orbit dives below the equatorial plane, is the
descending node. The angle between the positive X axis and
the node line is the first Euler angle Ω, the right ascension of
the ascending node. The right ascension is a positive number
lying between 0
◦
and 360
◦
.
The dihedral angle between the orbital plane and the
equatorial plane is the inclination i, measured according to the
right-hand rule, that is, counterclockwise around the node line
vector from the equator to the orbit. The inclination is also
the angle between the positive Z axis and the normal plane of
the orbit. The inclination is a positive number between 0
◦
and
180
◦
.
The third Euler angle ω, the argument of perigee, is the
angle between the node line vector 𝑁 and the eccentricity vec-
tor 𝑒, measured in the plane of the orbit. The argument of
perigee is a positive value between 0
◦
and 360
◦
.
The orbit of the space target can be determined by the
six orbital elements (
a, M, i, ω, e, Ω
), where the mean
anomaly can be expressed as
M = E − e sin E, (5)
with E being the eccentric anomaly, and the relationship be-
tween the eccentric anomaly and the true anomaly is
tan( f /2) =
p
(1 + e)/(1 − e)tan(E/2). (6)
Considering the two-body motion, the coordinates of the
target’s center of mass in the geocentric observation coordi-
nate system xyz can be expressed as
x
S
, y
S
, z
S
=
r, 0, 0
, (7)
where r is the distance from target center to geocentre, and it
can be obtained from Eq. (4).
Assuming that χ
t
0
= (
a, M
t
0
, i, ω, e, Ω
) is the target’s
orbital element at the initial observation time, ignoring the in-
fluences of precession and nutation on the celestial pole and
equinox, the coordinate system xyz can be rotated into coor-
dinate system XYZ by a sequence of three rotations. The first
step is to rotate an angle f
t
0
+ ω around the z axis clockwise.
The second step is to rotate an angle i around the x axis clock-
wise. Finally, we clockwise-rotate around the z axis through
an angle Ω . Here, f
t
0
is the true anomaly of the initial obser-
vation time; it can be obtained by the initial orbital elements
through Eqs. (5) and (6).
Except the mean anomaly, the remaining orbital elements
are all constants. The mean anomaly can be obtained by
M
t
0
= n(t
0
−t
epoch
) + M
epoch
, (8)
where t
epoch
is the epoch time of initial orbital elements and t
0
is the epoch time of the initial observation time.
At any observation time t, to transfer the coordinate sys-
tem xyz into XY Z, the only difference relative to initial ob-
servation time t
0
is that the rotation angle around the z axis
is f
t
+ ω rather than f
t
0
+ ω, where f
t
is the true anomaly of
observation time t.
Therefore, the transformation matrix xyz into XY Z at the
observation time t can be written as
𝑅
x−X
= 𝑅
z
(−Ω)𝑅
x
(−i)𝑅
z
(− f
t
− ω), (9)
where 𝑅
x
(•) and 𝑅
z
(•) are the rotation matrices around the x
axis and z axis, respectively.
Without considering the polar motion, the difference be-
tween the geocentric equatorial coordinate system XY Z and
the geocentric Earth-fixed coordinate system X
0
Y
0
Z
0
is the
Earth’s rotation angle (i.e., Greenwich hour angle S
G
). Then
the transformation matrix XY Z into X
0
Y
0
Z
0
can be expressed
as
𝑅
X−X
0
= 𝑅
z
(S
G
t
0
+ n
E
t), (10)
where S
G
t
0
is the Greenwich hour angle of the initial observa-
tion time t
0
, n
E
is the Earth’s rotation angular velocity, and its
period is 86164.098903691 s.
The Greenwich hour angle can be calculated from the fol-
lowing equation:
S
G
= 18
h
.6973746 + 879000
h
.0513367t
J
+0
s
.093104t
2
J
− 6
s
.2 × 10
−6
t
3
J
, (11)
where t
J
can be expressed as
t
J
=
JD(t) − JD(J2000.0)
36525.0
, (12)
with JD(t) being the Julian day at the observation time t, and
JD(J2000.0) the Julian day corresponding to epoch J2000.0.
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