The Optimal Orbital Control and Constellation of Dive and Ascent Earth
Observing Satellites
Tianshuang Fu and Fumiaki Imado (Shinshu University)
Abstract: We proposed the use of a Dive and Ascent earth observing Satellite (DAS) to observe the
particular area which is in the emergency situation or suffered by a natural disaster. In this paper, we
selected the forces of the thrusters fixed to the satellite body three axes as control variables, and obtained
the optimal thrust patterns by using the Steepest Descent Method. Some of new ideas and study results
are presented. We also proposed the ideas of constellations about the DAS here.
降下・上昇地球観測衛星の最適軌道制御およびコンステレーション
付
天爽，今度 史昭（信州大学）
摘要： われわれは降下・上昇地球観測衛星を用いて有事状態の地域と自然災害の被災地等を迅
速的に観測することを提案してきた．本稿において，衛星機体軸に固定されたスラスタの推力を
制御変数とし，最急降下法を用いて最適な推力パターンを求めた．ここで新しいアイデアおよび
研究結果を示し，また DAS のコンステレーションを提案する．
1. INTRODUCTION
Some natural disasters occur on the earth every
year, and some areas are still in the emergency situation
even now. In order to take countermeasures against
disasters and emergencies, we need to obtain the detailed
information of these particular areas quickly. If we obtain
the information by using a conventional earth observing
satellite, it is very difficult to get what we need, because
the orbit of the satellite is fixed. Therefore we propose
the use of DAS to perform the observing mission.
This paper is structured as follows: We derive the
mathematical model of the system and the Steepest
Descent Method in section 2. We considered the zonal
second coefficient J2 and treat the atmospheric drag as
the disturbance force. In section 3, we show the optimal
controls for various angle differences of the argument of
latitude (  ) between the DAS and the observing point.
In section 4, we show the results of extending the time of
the DAS to reach the observing point. In section 5, we
introduce some constellations about the DAS. Finally we
summarize the study of this paper in section 6.
2. MATHEMATICAL MODEL
In this paper, we treat the earth as an ellipsoid of
revolution. We show its parameters in table 1.
Table 1 The parameters of the earth
Angular velocity
  7292.115108 rad  s 1
e
Equatorial radius
Re  6378 .137 km
Gravity constant
e  398600.4km3 / s 2
Zonal second coefficient
J 2  1082.627 106
  1 / 298 .257
Flattening rate
We use the geocentric equatorial inertial coordinate
system  , the orbit reference coordinate system  R , and
the orbit coordinate system  O (for details see reference
[10]).
Z
ωe
YR
ZR
Xo
Zo
Y
Equatorial plane
i
Yo
η
Ω
X
Vernal　equinox
XR
Fig.1 Coordinate systems and symbols
The transformation matrix A from  to  O is
given by Eq. (2.1).
 S C  C Ci S 

A
 Si S
 C C  S Ci S 

C S i 

S i C
 Ci 
 C S   S Ci C  S S i 
(2.1)
where the symbols “C” and “S” show the abbreviation of
“Cos” and “Sin”. The symbols  ,  and i show the
argument of latitude, the right ascension of the ascending
node and the inclination of the DAS’s orbit, respectively.
The transformation matrix B from  O to  is the inverse
of A.
We employed the Steepest Descent Method to
 S S   C Ci C
solve the optimal control problems. The system
equations of the problems are shown as follows,
f1  x  V x


f 2  y  V y


f 3  z  V z

2
2

 f  x    e 1  3Re J  5 z  1 x  ( f  d ) / m


4
2
x
x

r 3  2r 2  r 2 

(2.2)

 3Re2  z 2 

e
 f 5  y   1 
J 2  5  1 y  ( f y  d y ) / m

r 3  2r 2  r 2 

 e  3Re2  z 2 

 f 6  z   3 1  2 J 2  5 2  3  z  ( f z  d z ) / m
r 
r
 r



f 7  m  k ( f xo  f yo  f zo )

where k  1 /( g 0 I sp ) , g 0  9.80665m / s 2 ,and I sp is the
fuel specific impulse. The state variables of this system
are
x  ( x , y , z , V x , V y , V z , m) T
(2.3)
where x, y and z are the components of the inertial
coordinates, Vx ,V y and Vz are the components of
velocity vector, m is the satellite’s mass. Three thrust
components in the direction of three axes of  O are
selected as control variables
u  ( f xo , f yo , f zo )T
(2.4)
If the satellite body attitude is controlled to direct along
with these three axes, then f xo , f yo and f zo are considered
as thrust forces of thrusters fixed to the satellite body
three axes. The following constrains are also imposed on
control variables.
 f max  f xo (t )  f max

(2.5)
 f max  f yo (t )  f max
 f
 max  f zo (t )  f max
The control force components f x , f y and f z in Eq. (2.2)
are expressed in relation to control variables
 f x   S C  C Ci S   S i S   C C  S Ci S   f xo 
  
 
 f y    S S   C Ci C S i C  C S   S Ci C  f yo 
f  
 f zo 
C S i
Ci
 S S i
 z 
 
(2.6)
The d x , d y and d z in Eq. (2.2) are disturbance force
components. In this study, the DAS dives into very low
altitude, and the atmospheric drag is far greater than the
other disturbance forces, therefore we only considered
the atmospheric drag here. Its components are given by
1

 d x  2  ( h) AC D ( h)VV x

1
(2.7)
d y   ( h) AC D ( h)VV y
2

 d  1  ( h) AC ( h)VV
D
z
 z 2
where A is the reference area of the satellite,  (h) is the
atmospheric density, which is a function of the altitude,
V is the velocity of DAS, C D (h) is the atmospheric drag
coefficient, which is a function of the altitude too.
In the Steepest Descent Method some constraint
conditions
(2.8)
ψ[x(t ),t f ]
are imposed, and the terminal time t f is determined from
the following stopping condition.
[x(t f ),t f ]  0
(2.9)
These equations will be explained in the next section. An
optimal observing problem is defined as, finding the
optimal control histories under the above conditions to
maximize the performance index,
(2.10)
J  [x(t f ), t f ]  mt f
which means to minimize the fuel consumption.
3. OPTIMAL CONTROL FOR VARIOUS 
BETWEEN THE DAS AND THE OBSERVING
POINTS
Our previous studies show that the control of
inclination change requires a lot of fuel, and the satellite
loses its normal observing function, therefore we only
perform coplanar orbit transfers to observe an area. As
the earth is rotating, when we use a DAS to observe a
particular area, we must consider the next three
conditions: The distance between the area and the
observing point that is above the area; the angle
difference of the argument of latitude (  ) between the
DAS and the observing point; the time of the DAS to
reach the observing point. Table 2 shows the parameters
we have employed as the initial conditions of the
calculations.
Table 2 The parameters of satellite and initial orbit
Satellite’s parameters
Initial mass
350kg
Fuel mass
175
Shape
1×1×1(m3)
Thrusts
(±500),(±18),(±500)N
313s
I sp
Initial orbital parameters
Altitude
400km
Inclination
90°
Tangential velocity
7.66856km/s
Period
5553.63s
Initial position
(2792.867,4792.867,0)km
In order to obtain detailed information about the
area, we need to descend the DAS to low altitude to
reduce the observing distance. In this section, we set the
observing distance for 130km, because the lowest
altitude of many other dive and ascent satellites is 130km,
and the atmospheric drag is not so large. Therefore the
constraint conditions are
h(t f )  h f  0
ψ[x(t ), t f ]  
(3.1)
V zotf  0
1000
2000
3000
Time(s)
4000
5000
Fig. 3.1 Histories of thrust forces
8000
Normal orbit
4000
2000 3000
Time(s)
Parameter's magnitude(km)
6800
6750
6700
6650
6600
6550
6500
6450
4000
Velocity(km/s)
5000
a
r
p
0
1000
2000 3000
Time(s)
350
348
346
344
342
340
338
336
334
332
330
4000
m
Vzo
0
0
6000
1000
5000
Fig. 3.4 Histories of a, r and p
Mass(kg)
Thrust(N)
therefore we the history is abbreviated. The histories of
the orbital parameters are shown in figure 3.4, where a is
the semi major axis, r is the distance from the DAS to the
center of the earth, and p is the semi-latus rectum.
fzo
V
Fig. 3.3 Histories of altitude and velocity
pattern of figure 3.1. As the value of f yo is always 0,
fxo
7.95
7.9
7.85
7.8
7.75
7.7
7.65
7.6
7.55
h
0
Although we need to observe an area with arbitrary  ,
we conducted the next typical 5 cases, that is, the  are
π/2, 5π/8, 3π/4, 7π/8 and π. After finishing the observing
mission, we need to control the DAS to fly back to its
normal orbit, so the final stopping conditions of  are
3π/2, 9π/8, 7π/4, 15π/8 and 2π, and final altitude of all
cases are 400km.As the paper space is limited we only
show the results about the  are π/2 and π.
First, we show the optimal calculation results
about  is π. We obtained the optimal thrust forces’
500
400
300
200
100
0
-100
-200
-300
-400
-500
400
350
300
250
200
150
100
50
0
1000
2000 3000
Time(s)
4000
0.2
0.15
0.1
0.05
0
-0.05
-0.1
-0.15
-0.2
Velocity(km/s)
Vzo is the terminal Zo component of velocity of orbit
coordinate system. These constraints make the observing
point as the perigee or apogee of the controlled orbit. As
the DAS must flies  to the observing point, so the
stopping condition is
(3.2)
[x(t f ), t f ]   (t f ) 0    0
Atitude(km)
where h f is the terminal altitude of the calculation, and
5000
Fig. 3.5 Histories of mass and V zo
After the one cycle of controlled flying, the
remained mass of the DAS is 331.831kg. Therefore the
mass of the fuel consumption is 18.169kg, which is
5.191% of the initial mass of the satellite. If we perform
the similar control once more, the loss of the fuel is
5.191% (17.225kg) of remained mass too. That is we can
use all of the fuel to perform the same observing control
13 times (174.97kg<175kg).
We show the calculation results about  is π/2
in Fig. 3.6 and Fig. 3.7.
Z(km)
2000
0
Controlled orbit
Thrust(N )
-2000
-4000
-6000
-8000
-5000
X(km)
0
5000
-5000
0
Y(km)
5000
500
400
300
200
100
0
-100
-200
-300
-400
-500
fxo
fzo
0
1000
2000
Tim e(s)
3000
Fig. 3.2 Controlled orbit and normal orbit
Fig. 3.6: Histories of thrust forces
4000
7.8
V
7.7
200
7.6
100
7.5
0
7.4
0
1000
2000
Tim e(s)
3000
4000
Fuel consumption(kg)
Fig. 3.7 Histories of altitude and velocity
We summarize all the calculation results in table 3
and Fig. 3.8. By the results we know that the mass of fuel
consumption increase when the  becomes smaller. In
the case of  is π/2, the mass of fuel consumption is
22.444% of the initial mass, which means that we can
only perform the same observing control 2 times.
Table 3 The  and fuel consumption
Fuel consumption

(kg)
78.555
π/2(90￮)
51.678
5π/8(112.5￮)
36.215
3π/4(135￮)
￮
25.67
7π/8(157.5 )
￮
18.169
π(180 )
90
80
70
500
400
300
200
100
0
-100
-200
-300
-400
-500
fxo
fzo
0
1000
2000
3000
Time(s)
4000
5000
Fig. 4.1 Histories of thrust forces
700
600
500
400
300
200
100
0
8
7.6
V
7.4
7.2
7
0
60
50
40
30
7.8
h
Velocity(km/s)
300
h
Thrust(N)
7.9
400
the observing point in case 4, we show the results in Fig.
4.1~4.3.
Table 4 The cases of extending the time to reach to the
observing point
Cases
Ascending
Starting time of
control time (s)
Descending control(s)
Case1
50
500
Case2
50
1000
Case3
100
500
Case4
100
1000
Altitude(km)
8
V elocity(km /s)
A ltitude(km )
500
1000
2000 3000
Time(s)
4000
5000
Fig. 4.2 Histories of altitude and velocity
20
10
0
8000
90
112.5
135
The Δη(degree)
157.5
180
Fig. 3.8 The  vs. the fuel consumption
6000
Normal orbit
4000
4. EXTEND THE TIME OF THE DAS TO
REACH THE OBSERVING POINT
In section 3, we can make the fuel consumption
minimum to observe the particular area when the  is π,
and the time of the DAS to reach to the observing point
is 2697.406s. As the rotation of the earth is too slow,
sometimes it needs more than 2697.406s for the
particular area to appear on the orbital plane. Therefore,
we may be required to extend the time of the DAS to
reach the observing point so that it can observe the area.
For this purpose, we must control the DAS ascending to
higher altitude at first, and then descending to the
observing point. In this paper we calculated the next 4
cases, which obtained the start conditions of the optimal
calculations by simulations. The other conditions of the
calculations are the same as the  is π in the former
section. In these cases, we get the longest time to reach
Z(km)
2000
0
Controlled orbit
-2000
-4000
-6000
-8000
-5000
X(km)
0
5000
-5000
0
5000
Y(km)
Fig. 4.3 Controlled orbit and normal orbit
We summarized all the calculation results in table
5 and in figure 4.4, where case 0 shows the results of
section 3. We extended the time up to 2853.569s.
Therefore between 2697.406s and 2853.569s, we can
observe anywhere which appears on the orbital plane in
detail for the  is π. However, the mass of fuel
consumption is 132.345kg, which is 37.813% of the
Case
Ascending
control
time (s)
Starting time
of
Descending
control (s)
Fuel
consumption
(kg)
0
50
50
100
100
0
500
1000
500
1000
18.169
55.328
89.636
80.887
132.345
Time(s)
Case0
Case1
Case2
Case3
Case4
Time of
reaching to
the
observing
point (s)
2697.406
2755.242
2800.259
2785.165
2853.569
Controlled orbit
6000
4000
2000
0
Normal orbit
-2000
-4000
-6000
-8000
-5000
2880
2850
2820
2790
2760
2730
2700
2670
X(km)
0
5000
-5000
5000
0
Y(km)
0
30
60
90
Mass(kg)
120
150
Fig. 4.4 Fuel consumption vs. the time of the DAS to
reach to the observing point
We also can control the DAS to higher altitude and
make it observe the particular area at the apogee of the
controlled orbit. We check the case for apogee altitude is
1400km. Therefore, the terminal constrain h f for the
observing control is 1400km, while for the final staying
control the h f is 400km. The optimal calculation results
are shown in Fig. 4.5~4.7.
Thrust (N)
8000
Z(km)
initial mass. Though we use all of the fuel, we can not
perform the observing control more than 1 time.
Table 5 optimal calculation results
Fig. 4.7 Controlled orbit and normal orbit
In this case, the time of the DAS to reach the
observing point is 3069.996s. It means that we extended
the observing time for 372.59s. To perform this control
we only need 66.739kg fuel, which is 19.068% of the
initial mass, so we can use all of the fuel to perform the
same control 3 times (164.466kg). Obviously we can use
this method to extend the observing time, though the
resolution of the sensors is inferior to that of at the low
altitude.
5. THE CONSTELLATIONS OF THE DAS
To cover more area, we need the use of the
observing sensor’s pointing function. Now we consider
the pointing angle for ±44 degrees here. Obviously the
observable range β is -3.416 ￮ ~3.416 ￮ at 400km, and
11.968￮~11.968￮at 1400km (see Fig. 5.1).
P1
500
400
300
200
100
0
-100
-200
-300
-400
-500
h
fxo
fzo
D1 or D2
  44

Re  6378 .137 km
0
1000
2000
3000
4000
Time (s)
5000
6000
Fig. 4.5 Histories of thrust forces
1600
1400
7.6
1000
800
h
600
V
7.4
7.2
400
Velocity (km/s)
7.8
1200
Altitude (km)
Fig. 5.1 The observable range
8
Nominal orbit
ωe
Controlled orbit
h2=1400km D1
  
h1=400km
D2
Equatorial plane
7
200
0
6.8
0
1000
2000
3000 4000
Time (s)
5000
6000
Observing point P1
Observing point P2
Fig. 4.6 Histories of altitude and velocity
Fig. 5.2 The constellation with 2 DASes
There are two reasons that we need to set 2 DASes
on one orbital plane. First, each DAS just needs to cover
half of the orbital plane. Second, if one DAS had any
trouble, the other one can perform the observing mission
too. Fig. 5.2 shows the simplest constellation, which has
one polar orbit with 2 DASes. They are symmetrical with
respect to the center of the earth.
As the circumference of the equatorial plane is the
longest plane on the earth. If we can observe any points
on the equatorial plane, which means we can observe the
points on the other planes too. Now we consider about
observing at the point P1 (see Fig. 5.2). We can start the
observing at the time t0=0s, when the satellite D1 is just
passing P1, which altitude is 400km and observable
range  0 is 0 ￮ ~3.416 ￮ , while the rotating angle of the
earth  0 is 0 ￮ . Obviously,  0 >  0 . Then at the time
t1=3069.996s, we can control the satellite D2 to reach the
observing point P1, which altitude is 1400km, and obtain
the observable range 1 for 0￮~15.434￮ (3.416￮+11.968￮).
While in 3069.996s, the rotating angle of the earth  1 is
12.827 ￮ , then 1 >  1 . Therefore at any time the
observable range is larger than the rotating angle of the
earth. Consequently, we can observe any points on the
equatorial plane by D1 in 24 hours. That is we can cover
all the earth by this constellation in 12 hours.
Hence, we can use 4 DASes on 2 orbital planes,
they are orthogonal to each other. To avoid the
interference at the polar point, we can use the
sun-synchronous orbits. This constellation can cover all
the earth in 6 hours. Therefore we can cover all over the
earth in 2 hours with 12 DASes on 6 sun-synchronous
orbits, which is shown in Fig. 5.3.
Equatorial plane
Fig. 5.3 The constellation with 12 DASes
6. CONCLUSIONS AND FUTURE WORK
In this paper, we propose to observe the particular
area which is in the emergency situation or suffered by a
natural disaster with the DAS. We obtained the optimal
orbital controls by using the Steepest Descent Method.
First, we performed the optimal calculations for
various  between the DAS and the observing point.
We can observe the area with the minimum fuel
when  is π. Observing the area with small 
requires more fuel. Therefore it is a trade-off of
observing an area with a proper  and saving the fuel.
Next, we tried to extend the time of the DAS to reach the
observing point. We could extend the time up to 372.59s.
Extending the time longer requires more fuel. Therefore
it is also a trade-off of extending the time and saving the
fuel. The more useful method to extend the time is that
we control the DAS to higher altitude and make it
observe the particular area at the apogee. We proposed
some constellations of DASes. Our final purpose is
building an optimal constellation with DASes (optimal
orbit, optimal number of DAS), which can observe
anywhere on the earth in a short time with minimum fuel
consumption.
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