Coordinated Orbit Transfer for
Satellite Clusters
FUMIN ZHANG AND P.S. KRISHNAPRASAD
Institute for Systems Research and Department of Electrical and Computer Engineering,
University of Maryland at College Park, College Park, Maryland, USA
ABSTRACT: We propose a control law that allows a satellite formation to
achieve orbit transfer. During the transfer, the formation can be either maintained or modified to a desired formation. Based on the orbit transfer control
law proposed by Chang, Chichka, and Marsden for single satellite, we add coupling terms to the summation of Lyapunov functions for single satellites. These
terms are functions of the difference between the mean anomalies (or perigee
passage times) of formation members. The asymptotic stability of the desired
formation in desired orbits is proved.
KEYWORDS: momentum map; control Lyapunov function; formation control;
orbit transfer
INTRODUCTION
This paper is concerned with the problem of achieving a satellite formation near
a designated elliptic orbit. For orbits near the earth, one can use a space shuttle to
place satellites into specified relative positions. What we want to consider here is the
case when members of a cluster have been placed relatively far apart. They have to
use their on-board thrusters to get to the desired orbits and achieve the desired formation. A similar case is when the entire formation needs to be restructured for mission-related reasons. We want the formation to be maintained to some extent during
the transfer and be reestablished after the transfer.
Our approach is to use Lyapunov functions to design the control laws for orbit
transfer. The Lyapunov function achieves a local minimum when the correct orbit
and formation are reached. In Reference 1, a Lyapunov function is expressed as a
quadratic function of the differences of orbital elements between current orbit and
the destination orbit. However, in that work convergence of the associated control
law is not proved.
In this paper, we develop new control laws for orbit transfer of formations and
give a proof of convergence. This algorithm is based on a Lyapunov function on the
shape space of elliptic orbits in our previous work,2 the work of Chang, Chichka, and
Marsden,3 and a result of Cushman and Bates.4 However, the most significant extension is the addition of a coupling term that is a function of the difference between
Address for correspondence: P.S. Krishnaprasad, Institute for Systems Research and
Department of Electrical and Computer Engineering, University of Maryland at College Park,
College Park, MA 20742, USA.
krishna@isr.umd.edu
Ann. N.Y. Acad. Sci. 1017: 112–137 (2004). ©2004 New York Academy of Sciences.
doi: 10.1196/annals.1311.008
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
113
the osculating perigee passage times. For circular orbits, we use the ascending node
passage time instead of the perigee passage time.
In the next section, we develop formulæ used in the proofs of our theorems.
Thereafter, we give a brief summary of results published elsewhere.2,3 are given.We
then introduce the definition of periodic satellite formations, followed by our main
results and proofs about orbit transfer of periodic formations. Simulation results are
provided in the final section of our paper.
CONTROL OSCULATING ELEMENTS
If the mass of a satellite is small compared to the mass of the Earth, the Kepler
two body problem can be approximated by a one center problem as follows:
(1)
mq̇˙ = – ∇V G + u,
where q ∈ 3 is the position vector of the satellite relative to the center of the Earth,
m is the mass of the satellite, VG is the gravitational potential of the Earth, u is the
control force plus other disturbances. Without considering higher order terms, VG
takes the form
µ
V G = – m -------- .
(2)
q
Let p = mq̇ be the momentum vector of the satellite. For simplicity we assume that
all the satellites considered in this paper have unit mass.
We define the following:
l(t) = q(t) × p(t)
q(t)
A(t) = p(t) × l(t) – µ -------------q(t)
A(t)
e(t) = --------------µ
(3)
h(t) 2
a(t) = ---------------------------µ ( 1 – e(t) 2 )
1
r(t)
cos ( E(t) ) = -------- ⎛ 1 – --------⎞
e(t) ⎝
a(t)⎠
M (t) = E(t) – e(t) sin ( E(t) ),
where h(t) = l(t) and r(t) = q(t) (these formulæ can be found in textbooks on
celestial mechanics5); l is the angular momentum vector and A is called the Laplace
vector (both are conserved if u(t) ≡ 0); a is the length of the semimajor axis; e is the
eccentricity; E is the eccentric anomaly; and M is the mean anomaly. The last equation is Kepler’s equation. When e(t) = 0, the eccentric anomaly E(t) is defined to be
M(t), which is measured from the line of ascending node. For now, we assume that
e(t) ≠ 0.
Note that these formulæ are valid for all t and all the elements are differentiable
on 3 × 3 − {0}. Hence, we can take derivative on both sides of Equations (3). By
using the property that l, A, a, and e are conserved when u(t) ≡ 0, we have
114
ANNALS NEW YORK ACADEMY OF SCIENCES
∂l
∂l
l˙(t) =
q̇ +
ṗ
∂q
∂p
=
∂l
∂l
q̇ + ( – ∇V G + u )
∂q
∂p
∂l
= l˙ u = 0 + u(t)
∂p
(4)
∂l
u(t).
∂p
Note that we used the property that l is conserved when u(t) ≡ 0. Similarly,
=
∂A
u(t).
(5)
∂p
We obtain the derivatives of the orbital elements in the APPENDIX, the results are listed below
Ȧ(t) =
1 ∂A T
ė(t) = --- ⎛ ⎞ Â ⋅ u(t) ,
µ ⎝ ∂ p⎠
where  is the unit vector along the direction of A. We also have
2
2ae
∂l T
∂A T
ȧ(t) = ----------------------- ⎛ ⎞ l ⋅ u(t) + ----------------------- ⎛ ⎞ Â ⋅ u(t)
µ ( 1 – e 2 ) ⎝ ∂ p⎠
µ ( 1 – e 2 ) ⎝ ∂ p⎠
Ė =
cos ( E )
r
µ 1--- ------------------ė – -------------------------ȧ
--- +
a r e sin ( E )
a 2 e sin ( E )
∂A T
∂l T
Ṁ = n + η ⎛ ⎞ Â ⋅ u(t) + ξ ⎛ ⎞ l ⋅ u(t),
⎝ ∂ p⎠
⎝ ∂ p⎠
where n =
(6)
(7)
(8)
(9)
µ ⁄ a 3 and
2 ( 1 – e cos ( E ) ) 2
ξ(l, A, E) = – ---------------------------------------------µae ( 1 – e 2 ) sin ( E )
cos ( ( E ) – e ) 2 ( 1 – e cos ( E ) ) 2
η(l, A, E) = ------------------------------- – ---------------------------------------- .
µe sin ( E )
µ ( 1 – e 2 ) sin E
(10)
Note that ξ and η are infinite if sin(E) = 0. Physically, this means that when the satellite is passing perigee or apogee, the control can cause a sudden jump of the mean
and eccentric anomalies. To prevent this from happening in our control laws, we turn
off the control when sin(E) = 0.
ORBIT TRANSFER OF SINGLE SATELLITE
For a single satellite on an elliptic orbit, the set D of ordered pairs (l, A) is a subset
of 3 × 3 with Euclidean norm,
(11)
D = { ( l, A ) ∈ 3 × 3 A ⋅ l = 0, l ≠ 0, A < m 2 µ }.
q̇ 2 + VG be the total energy of the satellite. Let P be the set of all pairs
Let W =
(q, p) with Euclidean norm. Then, we define a set Σe by
1
--- m
2
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
Σ e = { ( q, p ) ∈ P W (q, p) < 0, l ≠ 0 }.
115
(12)
By definition of l and A, we have a mapping π : Σe → D, (q, p) (l, A).
Theorem 1: (Chang–Chichka–Marsden)3
The following hold:
1. Σe is the union of all elliptic Keplerian orbits.
2. π(Σe) = D and Σe = π−1(D).
3. The fiber π−1(l, A) is a unique (oriented) elliptic Keplerian orbit for each
(l, A) ∈ D (see also Ref. 4, page 58).
The mapping π is a continuous mapping because (l, A) are continuous with respect
to (q, p).
Corollary 2:
π−1(K) is compact for any compact set K ⊂ D (c.f., Ref. 2).
If we have n ≥ 1 satellites, let D = D1 × D2 × … × Dn and let
d = d1
D1
+ d2
D2
+ … + dn
Dn .
We define the shape space of elliptic Keplerian orbits to be the set D with the distance function d .The above theorem and corollary can be extended to ( D , d ).
To control the orbit transfer of a single satellite, one considers a Lyapunov
function3
1
V (q, p) = --- ( l – l d
2
2
+ A – Ad
2 ),
(13)
where (ld, Ad) is the pair of the angular momentum vector and Laplace vector of the
target elliptic (circular) orbit. The derivative of V along the integral curve of the system is
V̇ = ( l – l d ) ⋅ l˙ + ( A – A d ) ⋅ Ȧ
∂l
∂A
u + ( A – Ad ) ⋅ u
∂p
∂p
= [ ( l – l d ) × q + l × ( A – A d ) + ( A – A d ) × p × q ] ⋅ u.
= ( l – ld ) ⋅
(14)
If we let the control be
u = – [ ( l – l d ) × q + l × ( A – A d ) + ( ( A – A d ) × p ) × q ],
(15)
then V̇ ≤ 0 along the trajectory of the closed loop system. The following theorem is
proved.
Theorem 3: (Chang–Chichka–Marsden)
There exists c > 0 such that if V(q0, p0) ≤ c, by applying the control law in
Equation (15), the trajectory of the closed loop system starting at (q0, p0) will
asymptotically converge to the target orbit π−1(ld, Ad).
116
ANNALS NEW YORK ACADEMY OF SCIENCES
PERIODIC FORMATION
Suppose we have a formation consisting of m satellites. Let Oj denote the orbit of
the jth satellite. The orbit Oj and the position of the satellite on Oj can be described
either by the six orbital elements (aj, ej, Ij, Ωj, ωj, τj) or by (lj, Aj, τj) where a denotes
the length of semimajor axis, e is the eccentricity, I is the inclination, Ω is the longitude of the ascending node, ω is the argument of the perigee, and τ is the perigee passage time. If Oj is elliptic or circular, except for singular cases for which some of the
orbital elements are not well defined, Theorem 1 tells us that the two expressions are
equivalent. Among all possible formations, we are interested in formations with periodic shape changes. We introduce the following definition.
Definition 1:
A formation is periodic when aj, the length of semimajor axis of orbit Oj,
satisfies aj = a > 0 for all j = 1, 2, …, m.
This definition is valid since all the satellites in a periodic formation have the
same orbital period
2π a 3
T = ----------------.
µ
(16)
Thus, although the shape of the formation is varying, it is varying periodically.
On the other hand, given a set of orbits with identical length semimajor axis, there
are infinitely many possible periodic formations. We need to specify their relative
positions on the orbits to determine a specific formation. The difficulty is that the relative positions are complicated functions of time. However, the differences between
the perigee passage times, (τi − τj), are constants. Because the mean anomaly is
M i = n i ( t – τ i ),
(17)
where ni = 2π / T, then (Mi − Mj) are constants. By specifying the values of (τi − τj)
or (Mi − Mj) for all i and j, a periodic formation can be uniquely determined.
TRANSFER OF SATELLITE FORMATIONS
BETWEEN ELLIPTIC ORBITS
With all the tools we need developed, we are now ready to determine our control
laws. To illustrate our ideas, we show the calculations for two satellites. A control
law for more than two satellites can be developed based on the two-satellite case in
a natural way.
To set up a periodic formation of two satellites, one can control each satellite separately to transfer to its target orbit. However, this will not assure the correct values
for (τi − τj) or (Mi − Mj). To do that, additional terms involving (τi − τj) or (Mi − Mj)
should be added in the summation of the Lyapunov functions for single satellites.
This extension results in a cooperative orbit transfer of multiple satellites.
We introduce a variable ϒi, defined by
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
⎧ ⎛ ai ⎞ 3 / 2
⎪ ⎝ -----⎠ M i , if E i ∈ ( – ε, π + ε )
⎪ ad
ϒi = ⎨
⎪ ⎛ ai ⎞ 3 / 2
( 2π – M i ), if E i ∈ ( π – ε, 2π + ε )
⎪ ⎝ ---⎠
⎩ ad
117
(18)
for i = 1, 2, where ad is the common length of the semimajor axes for the destination
orbits.
Here, the trouble of using different expressions for the cases Ei ∈(− ε, π + ε) and
Ei ∈(π − ε, 2π + ε) is caused by the fact that Ei ∈S1 a circle. Two coordinate charts
are required on S1. Here we pick the charts to be
ψ 1 : ( – ε, π + ε ) → ( – ε, π + ε ) s.t. E i E i
(19)
ψ 2 : ( π – ε, 2π + ε ) → ( – ε, π + ε ) s.t. E i ( 2π – E i ).
The value of ε is chosen so that the two satellites are in the same chart. Because in a
satellite formation the angular separations between satellites are usually small, the
value of ε is small.
For Ei ∈(− ε, π + ε), we have
a 3/2
3 a
ϒ̇ i = ⎛ -----i ⎞ Ṁ i + --- -----i- ȧ i M i
⎝a ⎠
2 a3
d
d
=
∂A T
∂l T
µ
------ + ρ(l i, A i, E i) ⎛ --------i⎞ Â i ⋅ u i(t) + ζ(l i, A i, E i) ⎛ -------i-⎞ l i ⋅ u i(t),
⎝∂p ⎠
⎝∂ p ⎠
a d3
i
i
(20)
where
a 3/2
3 a
2
ζ = ⎛ -----i ⎞ ξ(l i, A i, E i) + --- -----i- ----------------------- M i
⎝a ⎠
2 a 3 µ(1 – e 2)
d
i
d
(21)
a 3/2
3 a 2a i e i
-M .
ρ = ⎛ -----i ⎞ η(l i, A i, E i) + --- -----i- ---------------------⎝a ⎠
2 a 3 µ(1 – e 2) i
d
i
d
For Ei ∈(π − ε, 2π + ε),
a 3/2
3 a
ϒ̇ i = – ⎛ -----i ⎞ Ṁ i + --- -----i- ȧ i ( 2π – M i )
⎝a ⎠
2 a3
d
d
∂A T
∂l T
µ
= – ------ + ρ(l i, A i, E i) ⎛ --------i⎞ Â i ⋅ u i(t) + ζ(l i, A i, E i) ⎛ -------i-⎞ l i ⋅ u i(t),
⎝
⎠
⎝
3
∂ pi
∂ p i⎠
ad
(22)
where
a 3/2
3 a
2
ζ = – ⎛ -----i ⎞ ξ(l i, A i, E i) + --- -----i- ----------------------- ( 2π – M i )
⎝a ⎠
3 µ(1 – e 2)
2
a
d
i
d
a 3/2
3 a 2a i e i
- ( 2π – M i ).
ρ = – ⎛ -----i ⎞ η(l i, A i, E i) + --- -----i- ---------------------⎝a ⎠
2 a 3 µ(1 – e 2)
d
i
d
(23)
118
ANNALS NEW YORK ACADEMY OF SCIENCES
Note that we have terms that explicitly contain Mi. If not handled well, these terms
cause discontinuity in our control algorithm when the satellites enter a new chart.
The reason for us to pick the particular charts (ψ1, ψ2) is to reduce the discontinuities
in the derivatives of ϒi caused by changing charts.
We design a Lyapunov function on the phase space of the two satellites. This single function has different expressions in different charts. The Lyapunov function is
⎧
ϒ 1 – ϒ 2 – φ⎞ 2
⎪ V 1 + V 2 + 4 sin ⎛ --------------------------, if E i ∈ ( – ε, π + ε )
⎝
⎠
⎪
4
V = ⎨
2
⎪
⎛ ϒ 1 – ϒ 2 + φ⎞
⎪ V 1 + V 2 + 4 sin ⎝ ----------------------------⎠ , if E i ∈ ( π – ε, 2π + ε ),
4
⎩
(24)
where
1
V 1 = --- ( l 1 – l d
1
2
1
V 2 = --- ( l 2 – l d
2
2
2
2
+ A1 – Ad
+ A2 – Ad
1
2
2)
(25)
2 ).
Here, ( l d 1 , A d 1 ) and ( l d 2 , A d 2 ) specify the orbits in a two-satellite periodic formation and φ specifies the desired (M1 − M2) on these orbits.
We calculate the derivative of V as
−φ
ϒ1 – ϒ2 +
-⎞ ( ϒ̇ 1 – ϒ̇ 2 ).
V̇ = V̇ 1 + V̇ 2 + sin ⎛ --------------------------(26)
⎝
⎠
2
The choice of − or + depends on the value of Ei as in the definition of V. From calculations performed in the single satellite case,
V̇ i = [ ( l i – l d ) × q i + l i × ( A i – A d ) + ( ( A i – A d ) × p i ) × q i ] ⋅ u i ,
i
i
i
(27)
for i = 1, 2. Thus,
−φ
ϒ1 – ϒ2 +
-⎞ l 1⎞ × q 1
V̇ = ⎛ l 1 – l d + ζ 1 sin ⎛ --------------------------⎝
⎠ ⎠
⎝
1
2
ϒ1 – ϒ2 −
+ φ⎞ ⎞
- Â 1
+ l 1 × ⎛ A 1 – A d + ρ 1 sin ⎛ --------------------------⎝
⎠ ⎠
⎝
1
2
ϒ1 – ϒ2 −
+ φ⎞ ⎞
- Â 1 × p 1⎞ × q 1 ⋅ u 1
+ ⎛ ⎛ A 1 – A d + ρ 1 sin ⎛ --------------------------⎝
⎠ ⎠
⎝⎝
⎠
1
2
−φ
ϒ1 – ϒ2 +
-⎞ l 2⎞ × q 2
+ ⎛ l 2 – l d + ζ 2 sin ⎛ --------------------------⎝
⎝
⎠ ⎠
2
2
ϒ1 – ϒ2 −
+ φ⎞ ⎞
- Â 2
+ l 2 × ⎛ A 2 – A d + ρ 2 sin ⎛ --------------------------⎝
⎠ ⎠
⎝
2
2
ϒ1 – ϒ2 −
+ φ⎞ ⎞
- Â 2 × p 2⎞ × q 2 ⋅ u 2 .
+ ⎛ ⎛ A 2 – A d + ρ 2 sin ⎛ --------------------------⎝
⎠ ⎠
⎝⎝
⎠
2
2
To obtain V̇ ≤ 0 , we let
(28)
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
119
ϒ1 – ϒ2 −
+ φ⎞ ⎞
- l1 × q1
u 1 = – sin2 ( E 1 ) ⎛ l 1 – l d + ζ 1 sin ⎛ --------------------------⎝
⎠ ⎠
⎝
1
2
−φ
ϒ1 – ϒ2 +
-⎞ Â 1⎞
+ l 1 × ⎛ A 1 – A d + ρ 1 sin ⎛ --------------------------⎝
⎠ ⎠
⎝
1
2
−φ
ϒ1 – ϒ2 +
-⎞ Â 1⎞ × p 1⎞ × q 1
+ ⎛ ⎛ A 1 – A d + ρ 1 sin ⎛ --------------------------⎝
⎠ ⎠
⎝⎝
⎠
1
2
ϒ1 – ϒ2 −
+ φ⎞ ⎞
- l2 × q2
u 2 = – sin2 ( E 2 ) ⎛ l 2 – l d + ζ 2 sin ⎛ --------------------------⎝
⎠ ⎠
⎝
2
2
(29)
ϒ1 – ϒ2 −
+ φ⎞ ⎞
- Â 2
+ l 2 × ⎛ A 2 – A d + ρ 2 sin ⎛ --------------------------⎝
⎠ ⎠
⎝
2
2
−φ
ϒ1 – ϒ2 +
-⎞ Â 2⎞ × p 2⎞ × q 2 .
+ ⎛ ⎛ A 2 – A d + ρ 2 sin ⎛ --------------------------⎝
⎠ ⎠
⎝⎝
⎠
2
2
Note that the factors sin2(Ei) cancel the term sin(Ei) in the denominators of ζi and ρi.
This results in a continuous control law that will be 0 when Ei = 0, π.
Another issue is that when ei = 0, Ei is not well defined. Thus, we should not allow
A d i = 0. Furthermore, we should choose our initial value of function V such that the
subset of the space Σ e1 × Σ e2 , where Ai = 0, will not be reached by the controlled
dynamics.
Let z = (q1, p1, q2, p2). We now proceed to find the initial condition
z 0 = ( q 1(0), p 1(0), q 2(0), p 2(0) )
for z, such that the set
S M = { z V (z) ≤ V (z 0) }
(30)
is a compact subset of Σ e1 × Σ e2 − {z | A1 = 0 or A2 = 0}. This is a necessary step
because we want to apply LaSalle’s invariance principle to prove our main result.
Lemma 4:
Let
c < min { c 1, c 2 } ,
(31)
⎧1
⎫
1
1
c i = min ⎨ --- A d 2, --- l d 2, --- ( µ – A d ) 2 ⎬
i
i
i
2
2
2
⎩
⎭
(32)
where
for i = 1, 2. Then the set
S M = { z V ( z) ≤ c }
(33)
is a compact subset of Σ e1 × Σ e2 − {z | A1 = 0 or A2 = 0}.
Proof:
The first observation is that the set
S 1 = { ( q 1, p 1 ) V (q 1, p 1) ≤ c *, c * < c 1 }
(34)
120
ANNALS NEW YORK ACADEMY OF SCIENCES
is a subset of Σ e1 , that is, S1 ∩ Σ e1 = S1. In fact, c1 is the supremum of V1(q1, p1) on
the set Σ e1 − {(q1, p1) | A1 = 0}. To see this, we solve a constrained maximization
problem as follows:
⎧1
sup ⎨ --- l 1 – l d
1
⎩2
2
+ A1 – Ad
2⎫
1
⎬
⎭
(35)
under the constraints
A 1 ⋅ l 1 = 0, l 1 ≠ 0 , A 1 ≠ 0, and A 1 < µ.
(36)
First, we need to calculate the supremum of the unconstrained maximization problem. It is easy to see that this value is infinite. Then, we need to calculate the minimum value subject to l1 = 0. The result is 1--2- l d 2 , achieved when A = Ad. Similarly,
the minimum subject to A1 = 0 is --12- A d 2 . We should also calculate the minimum
value subject to A = µ. By applying the Lagrange multiplier method, we found
this value to be 1--2- ( µ – A d ) 2 . Thus,
⎧1
⎫
1
1
c 1 = min ⎨ --- A d 2, --- l d 2, --- ( µ – A d ) 2 ⎬
1
1
1
2
2
2
⎩
⎭
(37)
is the supremum of V1 on the set Σ e1 − {(q1, p1) | A1 = 0}. Thus, we have proved that
S1 ⊂ Σ e1 − {(q1, p1) | A1 = 0}.
Another observation is that π(S1) is a compact subset of D1. Thus, by Corollary 2,
S1 is a compact subset of Σ e1 .
We can make the same arguments for the case when i = 2 to prove that S2 is a compact subset of Σ e2 − {(q2, p2) | A2 = 0}. Hence, by letting c < min{c1, c2}, it is true that
S M ⊂ S 1 × S 2 ⊂ Σ e × Σ e – { z A 1 = 0 or A 2 = 0 }.
1
2
(38)
Thus, SM is a compact subset of Σ e1 × Σ e2 − {z | A1 = 0 or A2 = 0}.
We can now apply LaSalle’s invariance principle to show that the trajectory of the
closed loop system, starting within SM, converges to the maximal invariant subset of
SM, where u(t) = 0 is satisfied for all t.
Proposition 5:
With V, c, and ui given, as in (24), (31), and (29), the trajectory starting from
the point (q10, p10, q20, p20) that satisfies
V (q 10, p 10, q 20, p 20) ≤ c
(39)
will converge to the set for which
li = ld
i
Ai = Ad
i
M1 – M2 = φ
are satisfied for i = 1, 2.
Proof:
To calculate the invariant set, let u1 = 0. When sin(E1) = 0, we obtain
(40)
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
121
ϒ1 – ϒ2 −
+ φ⎞ ⎞
⎛ l – l + ζ sin ⎛ --------------------------- l1 × q1
1
d
1
⎝
⎠ ⎠
⎝
1
2
−φ
ϒ1 – ϒ2 +
-⎞ Â 1⎞
+ l 1 × ⎛ A 1 – A d + ρ 1 sin ⎛ --------------------------⎝
⎠ ⎠
⎝
1
2
(41)
−φ
ϒ1 – ϒ2 +
-⎞ Â 1⎞ × p 1⎞ × q 1
+ ⎛ ⎛ A 1 – A d + ρ 1 sin ⎛ --------------------------⎝
⎠ ⎠
⎝⎝
⎠
1
2
= 0.
By taking inner products on both sides with q1(t) we obtain
ϒ1 – ϒ2 −
+ φ⎞ ⎞ ⎞
⎛ l × ⎛ A – A + ρ sin ⎛ --------------------------- Â 1 ⋅ q 1 = 0.
d1
1
⎝
⎠ ⎠⎠
⎝1 ⎝ 1
2
(42)
This is equivalent to
−φ
ϒ1 – ϒ2 +
-⎞ Â 1⎞ = 0.
( l 1 × q 1(t) ) ⋅ ⎛ A 1 – A d + ρ 1 sin ⎛ --------------------------⎝
⎠ ⎠
⎝
1
2
(43)
Let
−φ
ϒ1 – ϒ2 +
-⎞ Â 1 ,
B = A 1 – A d + ρ 1 sin ⎛ --------------------------(44)
⎝
⎠
1
2
then Equation (43) means that B is perpendicular to the vector l1 × q1(t). We see that
the vector B should stay in the plane spanned by l1 and q1(t), as shown in FIGURE 1.
However, from the assumption that u1(t) = 0, we know that A1 and l1 are constant
vectors. The time varying vector B will sweep a line segment passing the fixed point
(A1 − A d 1 ). The direction of this line segment is aligned with  1 . Thus, vector B
will be the intersection of the (l1, q1) plane and this line segment. Because q1(t) is
sweeping the orbital plane, the (l1, q1) plane is identical at t and t + kT1, where k is
an integer and T1 is the period of the first satellite. Since the line segment is not
changed, the intersection points in these cases must be identical. Thus, we must have
(45)
B(t) = B(t + kT 1).
FIGURE 1. The relationship between l1, q1, A1 − A d 1 , and B (see Proposition 5).
122
ANNALS NEW YORK ACADEMY OF SCIENCES
Without lost of generality, suppose at time t, E1(t), E2(t) ∈ (− ε, π + ε). Then, (45)
requires that
ϒ 1(t) – ϒ 2(t) – φ⎞
ρ 1(E 1(t)) sin ⎛ -------------------------------------⎝
⎠
2
(46)
(
t
+
kT
)
–
ϒ
(
t
+
kT
)
–
φ
ϒ
1
1
2
1
= ρ 1(E 1(t + kT 1)) sin ⎛ -----------------------------------------------------------------------⎞ .
⎝
⎠
2
Let k = 1 in Equation (46), because ρ1(E1(t)) = ρ1(E1(t + T1)), the first observation
we make is that the two satellites must have the same period. In fact, suppose that at
time t0 Equation (46) is satisfied. Then, at time t0 + T1, since E1(t0) = E1(t0 + T1),
ϒ1(t0) = ϒ1(t0 + T1), and all the angles (anomalies) are in the interval [0,2π), we must
have ϒ2(t0) = ϒ2(t0 + T1). However,
a 3/2
(47)
ϒ 2(t 0) – ϒ 2(t 0 + T 1) = – ⎛ ----2-⎞ ( M 2 ( t 0 + T 1 ) – M 2(t 0) ) .
⎝a ⎠
d
Then ϒ2(t0) = ϒ2(t0 + T1) is satisfied only if M2(t0 + T1) = M2(t0). Thus, we have
T1 = k1T2, where k1 is a positive integer. Remember, we can apply the same argument to the second satellite to obtain T2 = k1T1. Thus, we must have T1 = T2. Hence,
on the invariant set, we have proved that a1 = a2.
On the other hand, for a specific time t, we know that there exists t* ∈ [0, T1) such
that
π + f 1(t) = f 1(t + t *),
(48)
where f1 is the true anomaly of the first satellite. The value of t* depends on t. The
plane spanned by (l1, q1) at time t is also identical to the plane spanned by (l1, q1) at
time t + t*. Thus, we must have
B(t) = B(t + t *),
(49)
which requires that
ϒ 1(t) – ϒ 2(t) – φ⎞
ρ 1(E 1(t)) sin ⎛ -------------------------------------⎝
⎠
2
ϒ 1(t + t *) – ϒ 2(t + t *) + φ⎞
- .
= ρ 1(E 1(t + t *)) sin ⎛ -----------------------------------------------------------⎝
⎠
2
(50)
Furthermore, a1 = a2 implies that M1(t) − M2(t) = M1(t + t*) − M2(t + t*), and one can
verify that
ϒ 1(t) – ϒ 2(t) – φ⎞
ϒ 1(t + t *) – ϒ 2(t + t *) + φ⎞
- = – sin ⎛ ------------------------------------------------------------ .
sin ⎛ -------------------------------------⎝
⎠
⎝
⎠
2
2
For (50) to be satisfied, one possibility is that
ϒ 1(t) – ϒ 2(t) – φ⎞
- = 0.
sin ⎛ -------------------------------------⎝
⎠
2
(51)
(52)
Another possibility is that
ρ 1(E 1(t)) = – ρ 1(E 1(t + t *)) .
(53)
By definition of ρ1, one can verify that (53) can only be satisfied when t takes value
from a set of measure 0. Thus, for (50) to be satisfied, (52) must be true. Because of
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
123
(52), the time varying parts in Equation (41) vanish. We can make the same argument as in the proof for the single satellite case3 to show that
l1 = ld
1
(54)
A1 = Ad .
1
We can apply similar arguments for the second satellite. Thus, we have
li = ld
i
Ai = Ad
(55)
i
ϒ1 – ϒ2 = ±φ
for i = 1, 2. By definition of ϒ1 and ϒ2 in Equation (18), we have
a 1⎞ 3 / 2
a 3/2
⎛ ---M 1 – ⎛ ----2-⎞ M 2 = φ.
⎝a ⎠
⎝a ⎠
d
(56)
d
However, we already know a1 = a2 = ad, so we conclude that
M 1 – M 2 = φ.
(57)
This proposition can be easily extended to m satellites where m > 2. Let the mth
satellite be the leader of the formation. Let
m
− φi 2
ϒ1 – ϒm +
⎞ ⎞,
V = ∑ ⎛ V i + 4 sin ⎛ -----------------------------(58)
⎝
⎠ ⎠
⎝
4
i
where i = 1, 2, 3, …, m, φm = 0, and
1
(59)
V i = --- l i – l d 2 + A i – A d 2 .
i
i
2
It is easy to see that the control for the ith satellite with i ≤ m − 1 is the same as the
control for the (i, m) pair of satellites. The initial value of V should be less than the
minimum value of ci for i = 1, 2, …, m − 1. Then the solution will converge to the set
where
li = ld
i
Ai = Ad
i
(60)
M i – M n = φi
are satisfied for i = 1, 2, …, m.
TRANSFER OF SATELLITE FORMATIONS TO CIRCULAR ORBITS
Formations on circular orbits are of great importance.6 If the destination of at
least one of the satellites in a formation is circular, there is difficulty in determining
the mean anomaly. Since the Laplace vector A vanishes, the perigee point cannot be
determined. In this case, the position of a satellite on the circular orbit is usually
described by using the argument of latitude, defined as the angle θ between the line
124
ANNALS NEW YORK ACADEMY OF SCIENCES
of ascending node and the position vector q. It is easy to see that on an elliptic orbit,
this angle is
θ = ω+ f,
(61)
where ω is the argument of perigee and f is the true anomaly. On a circular orbit, we
have
θ = n ( t – τ ),
(62)
where τ is the time when the satellite passes the ascending node for the first time. As
can be seen, θ serves as the mean anomaly for circular orbit.
Although θ and θ̇ can be easily determined, as discussed in the APPENDIX, they
are not suitable for our Lyapunov function based control laws because θ is not affine
in time on an elliptic orbit. Thus, we use the angle Mθ, which is the mean anomaly
measured from the line of node. Mθ can be determined from θ, but if the orbit is close
to circular, we can use θ as an approximation for Mθ. Our analysis in the APPENDIX
shows that the derivative of Mθ is
r ω2
cos ( I ) ∂l T
- ------------------ ⎛ ⎞ n̂ ⋅ u
Ṁ θ = n – ----------------------⎝ ⎠
a 2 1 – e 2 h sin ( I ) ∂ p
r sin ( E ) – r ω sin ( E ω )⎞ 1 ⎛ ∂A⎞ T
--– ⎛ sin ( E ) – sin ( E ω ) + --------------------------------------------------Â ⋅ u
⎝
⎠ µ ⎝ ∂ p⎠
a( 1 – e2 )
(63)
r 2 – r ω2 ⎛ ⎛ ∂A⎞ T
+ -----------------------------Â ⋅ u⎞ .
⎝⎝ ⎠ ⊥ ⎠
µa 2 e 1 – e 2 ∂ p
Here, ω is the argument of perigee, E and Eω are the eccentric anomalies corresponding to θ − ω and ω, rω denotes the distance between the ascending node and the
gravitational center. We let n̂ = ( l x2 + l y2 )−1/2[− ly, lx, 0]T, where n̂ is the unit vector
pointing to the ascending node. Let  ⊥ be the unit vector perpendicular to  and l
forming an acute angle with n̂ .
We now study the case when we want to set up a formation in which all the members are on circular orbits with the same radius. According to the previous discussion
under PERIODIC FORMATION, this formation is a periodic formation. Let τi be the time
when the ith satellite passes the line of ascending node for the first time. Similar to
the perigee passage time in our elliptic orbit case, τi is a constant of the free motion
for satellite i. Hence, by specifying the differences τi − τj or θi − θj between satellites,
a periodic formation on circular orbits can be uniquely determined.
Similar to the elliptic orbit case, we introduce a variable ψi, which is defined as
⎧ ⎛ ai ⎞ 3 / 2
⎪ ⎝ -----⎠ M θi , if θ i ∈ ( – ε, π + ε )
⎪ ad
ψi = ⎨
⎪ ⎛ ai ⎞ 3 / 2
( 2π – M θ ), if θ i ∈ ( π – ε, 2π + ε )
⎪ ⎝ ---⎠
i
⎩ ad
(64)
for i = 1, 2, where ad is the common length of the semimajor axes (or radius) for the
destination orbits. Here, the reasons for using two charts are the same as in the case
of elliptic orbits.
Our calculations in the APPENDIX show that the derivative of ψ is
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
ψ̇ =
∂l T
∂A T
µ
------ + ⎛ ⎞ ( λn̂ + κl ) ⋅ u + ⎛ ⎞ ( σ Â + Â ⊥ ) ⋅ u
⎝ ∂ p⎠
a d3 ⎝ ∂ p⎠
125
(65)
where for θ ∈ (− ε, π + ε),
2
r ω2
cos ( I )
a 3 / 2 r ω cos ( I )
a 3/2
- ------------------ ≈ – ⎛ -----⎞ ----- -------------------λ = – ⎛ -----⎞ ----------------------⎝a ⎠
⎝a ⎠
a h sin2 ( I )
d
d
a 2 1 – e 2 h sin ( I )
3
κ = -------------------------- ψ
µa ( 1 – e 2 )
r sin ( E ) – r ω sin ( E ω )⎞
3e
1 a 3/2
σ = ----------------------- ψ – --- ⎛ -----⎞ ⎛ sin ( E ) – sin ( E ω ) + --------------------------------------------------⎝
⎠
⎝
⎠ (66)
2
µ
a
µ(1 – e )
a( 1 – e2 )
d
r sin ( E ) – r ω sin ( E ω )⎞
1 a 3/2
≈ – --- ⎛ -----⎞ ⎛ sin ( E ) – sin ( E ω ) + --------------------------------------------------⎠
µ⎝a ⎠ ⎝
a( 1 – e2 )
d
2
2
a 3 / 2 2 ( cos ( E ) – cos ( E ω ) )
a 3 / 2 r – rω
-,
- ≈ – ⎛ -----⎞ --------------------------------------------------- = ⎛ -----⎞ -----------------------------⎝a ⎠
⎝a ⎠
µ
2
2
d
d
µa e 1 – e
and for θ ∈ (π − ε, 2π + ε),
2
2
rω
cos ( I )
a 3 / 2 r ω cos ( I )
a 3/2
- ------------------ ≈ ⎛ -----⎞ ----- -----------------λ = ⎛ -----⎞ ----------------------⎝a ⎠
⎝
⎠ a 2 h sin ( I )
h
sin
(
I
)
a
d
d
a2 1 – e2
3
κ = -------------------------- ψ
µa ( 1 – e 2 )
r sin ( E ) – r ω sin ( E ω )⎞
3e
1 a 3/2
σ = ----------------------- ψ + --- ⎛ -----⎞ ⎛ sin ( E ) – sin ( E ω ) + --------------------------------------------------⎝
⎠
⎝
⎠ (67)
2
µ
a
µ(1 – e )
a( 1 – e2 )
d
r sin ( E ) – r ω sin ( E ω )⎞
1 a 3/2
≈ --- ⎛ -----⎞ ⎛ sin ( E ) – sin ( E ω ) + --------------------------------------------------⎠
µ⎝a ⎠ ⎝
a( 1 – e2 )
d
2
2
a 3 / 2 2 ( cos ( E ) – cos ( E ω ) )
a 3 / 2 r – rω
-.
- ≈ ⎛ -----⎞ --------------------------------------------------- = – ⎛ -----⎞ -----------------------------⎝a ⎠
⎝a ⎠
µ
d
d
µa 2 e 1 – e 2
We use the symbol ≈ when the eccentricity of the orbits are very small. When the
orbit is circular, E and Eω are not well defined, but we have Mθ = θ and
cos ( I ) ∂l T
θ̇ = n – ------------------ ⎛ ⎞ n̂ ⋅ u.
h sin ( I ) ⎝ ∂ p⎠
(68)
Thus, when e = 0
a 3 / 2 cos ( I )
λ = −
+ ⎛⎝ -----⎞⎠ -----------------h sin ( I )
a
d
3
κ = -------------------------- ψ
µa ( 1 – e 2 )
σ = 0
= 0.
(69)
126
ANNALS NEW YORK ACADEMY OF SCIENCES
This result agrees with the limits of (66) and (67) as e → 0.
Mimicking what was done in the elliptic orbit case, we design a Lyapunov function on the phase space of two satellites as follows:
⎧
ψ 1 – ψ 2 – φ⎞ 2
⎪ V 1 + V 2 + 4 sin ⎛ ---------------------------, if θ i ∈ ( – ε, π + ε )
⎝
⎠
⎪
4
V = ⎨
2
⎪
⎛ ψ 1 – ψ 2 + φ⎞
⎪ V 1 + V 2 + 4 sin ⎝ -----------------------------⎠ , if θ i ∈ ( π – ε, 2π + ε )
4
⎩
(70)
where
1
V 1 = --- ( l 1 – l d 2 + A 1 2 )
1
2
(71)
1
V 2 = --- ( l 2 – l d 2 + A 2 2 ).
2
2
The derivative of this function has similar form to Equation (28). The control law u
such that V̇ ≤ 0 can be chosen as
ψ1 – ψ2 −
+ φ⎞
- ( λ 1 n̂ 1 + κ 1 l 1 )⎞ × q 1
u 1 = – ⎛ l 1 – l d + sin ⎛ ---------------------------⎝
⎠
⎝
⎠
1
2
ψ1 – ψ2 −
+ φ⎞
- ( σ 1 Â 1 + 1 Â 1⊥ )⎞
+ l 1 × ⎛ A 1 + sin ⎛ ---------------------------⎝
⎠
⎝
⎠
2
−φ
ψ1 – ψ2 +
-⎞ ( σ 1 Â 1 + 1 Â 1⊥ )⎞ × p 1⎞ × q 1
+ ⎛ ⎛ A 1 + sin ⎛ ---------------------------⎝
⎠
⎝⎝
⎠
⎠
2
ψ1 – ψ2 −
+ φ⎞
- ( λ 2 n̂ 2 + κ 2 l 2 )⎞ × q 2
u 2 = – ⎛ l 2 – l d + sin ⎛ ---------------------------⎝
⎠
⎝
⎠
2
2
(72)
ψ1 – ψ2 −
+ φ⎞
- ( σ 2 Â 2 + 2 Â 2⊥ )⎞
+ l 2 × ⎛ A 2 – sin ⎛ ---------------------------⎝
⎠
⎝
⎠
2
−φ
ψ1 – ψ2 +
-⎞ ( σ 2 Â 2 + 2 Â 2⊥ )⎞ × p 2⎞ × q 2 .
+ ⎛ ⎛ A 2 – sin ⎛ ---------------------------⎝
⎠
⎝⎝
⎠
⎠
2
Based on Lemma 4, the following lemma is obvious.
Lemma 6:
Let
c < min { c 1, c 2 }
(73)
⎧1
1 ⎫
c i = min ⎨ --- l d 2, --- µ 2 ⎬
i
2 ⎭
2
⎩
(74)
S M = { z V ( z) ≤ c }
(75)
where
for i = 1, 2. Then the set
is a compact subset of Σ e1 × Σ e2 .
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
127
Hence we prove that, starting within the set SM, the controlled dynamics will converge to the maximal invariant set where formation relationships are obtained.
Proposition 7:
With V, c, and ui given as in (70), (73), and (72), the trajectory starting from
the point (q10, p10, q20, p20) that satisfies
V (q 10, p 10, q 20, p 20) ≤ c
(76)
will converge to the set for which
li = ld
i
Ai = 0
(77)
θ1 – θ2 = φ
are satisfied for i = 1, 2.
Proof:
Let u1 = 0, and take the inner product with q1 on both sides of the equation to obtain
(78)
( l 1 × B 1 ) ⋅ q 1 = 0,
where
−φ
ψ1 – ψ2 +
-⎞ ( σ 1 Â 1 + 1 Â 1⊥ ) .
B 1 = A 1 + sin ⎛ ---------------------------⎝
⎠
2
At the time t0 when E1(t0) = E ω1 , since M θ1 = 0 and r1 = r ω1 , we have
1(t 0) = 0 and σ 1(t 0) = 0,
(79)
(80)
hence
B 1(t 0) = A 1 .
(81)
( q1 × l1 ) ⋅ B1 = 0
(82)
On the other hand,
tells us that B1 lies in the plane spanned by l1 and q1. Thus, when q1(t0) is not aligned
with A1, B1(t0) must vanish. This implies that
(83)
A 1 = 0.
However, it is also possible that q1(t0) is aligned with A1. In this case we have
(84)
E ω = 0 and ω 1 = 0.
1
Therefore, for all time t,
a 3 / 2 r 12 – r ω2 1
1(t) = ± ⎛ ----1-⎞ --------------------------------⎝a ⎠
d
µa 2 e 1 – e 2
1 1
1
3e 1
r1
⎛
⎞
1 a 3/2
- ψ 1 – --- ⎛ ----1-⎞ sin ( E 1 ) ⎜ 1 + -----------------------σ 1(t) = ----------------------⎟ .
⎝
⎠
µ ad
µ ( 1 – e 12 )
a 1 ( 1 – e 12 )⎠
⎝
Thus,
(85)
128
ANNALS NEW YORK ACADEMY OF SCIENCES
ψ1 – ψ2 −
+ φ⎞
- ( σ 1 Â 1 + 1 Â 1⊥ ),
B 1(t) = A 1 + sin ⎛ ---------------------------⎝
⎠
2
(86)
and B1(t) has to be aligned with q1(t) for all time t. If we let T1 be the orbital period
of the first satellite and T2 be the orbital period of the second satellite, then after one
period T1 the first satellite will return to the same position, which implies that
ψ 1(t + T 1) – ψ 2(t + T 1) −
+ φ⎞
- ( σ 1 ( t + T 1 ) Â 1 + 1 ( t + T 1 ) Â 1⊥ )
A 1 + sin ⎛ ----------------------------------------------------------------⎝
⎠
2
(87)
ψ 1(t) – ψ 2(t) −
+ φ⎞
⎛
= A 1 + sin ---------------------------------------- ( σ 1 ( t ) Â 1 + 1 ( t ) Â 1⊥ ).
⎝
⎠
2
Because M θ1(t) = M θ1(t + T 1) and E1(t) = E1(t + T1), we know σ1(t + T1) = σ1(t)
and 1(t + T1) = 1(t). Thus, the above equation results in
a 3/2
a 3/2
⎛ ⎛ ----1-⎞ M θ (t) – ⎛ ----2-⎞ M θ (t) – φ⎞
⎝a ⎠
1
2
⎜ ⎝ a d⎠
⎟
d
-⎟
sin ⎜ -----------------------------------------------------------------------------------2
⎜
⎟
⎝
⎠
(88)
3/2
3/2
a
a
1
2
⎛ ⎛ -----⎞ M θ (t) – ⎛ -----⎞ M θ (t + T 1) – φ⎞
⎝a ⎠
1
2
⎜ ⎝ a d⎠
⎟
d
-⎟ .
= sin ⎜ ------------------------------------------------------------------------------------------------2
⎜
⎟
⎝
⎠
Therefore, we must have
M θ (t) = M θ (t + T 1).
2
2
(89)
Hence, T1/T2 is a positive integer. On the other hand, the above arguments can be
applied to the second satellite. If A2 ≠ 0, T2/T1 must be a positive integer. Hence,
T2 = T1, which implies that a1 = a2.
Still assuming a1(t0) is aligned with A1, we know that by a property of elliptic orbits,
after half period, q1(t0 + T1/2) is aligned with A1 again, which means that
−φ
ψ 1(t 0 + T 1 ⁄ 2) – ψ 2(t 0 + T 1 ⁄ 2) +
-⎞ 1(t 0 + T 1 ⁄ 2) = 0.
sin ⎛ -----------------------------------------------------------------------------------(90)
⎝
⎠
2
According to the definition of ψ, one can verify that
ψ 1(t) – ψ 2(t) −
ψ 1(t + T 1 ⁄ 2) – ψ 2(t + T 1 ⁄ 2) −
+ φ⎞
+ φ⎞
- = – sin ⎛ --------------------------------------- .
sin ⎛ ------------------------------------------------------------------------------⎝
⎠
⎝
⎠
2
2
Therefore, the first possibility is that
1(t 0 + T 1 ⁄ 2) = – 1(t 0) = 0.
(91)
(92)
This further implies that r1(t0 + T1/2) = r1(t0). Because at time t0 the satellite is at
perigee and at time t0 + T1/2 the satellite is at apogee, we conclude that the orbit is
circular; that is, A1 = 0.
The second possibility for Equation (90) to hold is that
−φ
ψ 1(t 0) – ψ 2(t 0) +
⎞ = 0.
(93)
sin ⎛ --------------------------------------------⎝
⎠
2
However, because T1 = T2, we must have
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
129
ψ 1(t) – ψ 2(t) = ψ 1(t 0) – ψ 2(t 0),
(94)
ψ 1(t) – ψ 2(t) −
+ φ⎞
- = 0.
sin ⎛ --------------------------------------⎝
⎠
2
(95)
which implies that
One can make the same argument as in the proof of the single satellite case3 to show
that
l1 = ld
1
(96)
A 1 = 0.
To this point, we have proved that our algorithm will cause either A1 or A2 to vanish.
Without loss of generality, we proceed by assuming A1 = 0. Then, from u1 = 0, we get
−
1 – ψ 2 + φ⎞
⎛ l – l + sin ⎛ ψ
---------------------------- ( λ 1 n̂ 1 + κ 1 l 1 )⎞ × q 1 = 0.
(97)
⎝
⎠
⎝ 1 d1
⎠
2
Let
ψ1 – ψ2 −
+ φ⎞
- ( λ 1 n̂ 1 + κ 1 l 1 ).
D 1 = l 1 – l d + sin ⎛ ---------------------------⎝
⎠
1
2
(98)
D1 must be aligned with q1 and also lies in the plane passing (l1 − l d 1 ) spanned by
l1 and n1. Because the plane is fixed and q1 is periodic in time, at time t and time
(t + T1), the intersection of q1 and the plane must be the same. This tells us that T1/T2
is a positive integer. Now, if A2 = 0, the above arguments can be repeated so that we
conclude T2/T1 is also a positive integer. If A2 ≠ 0, we can repeat the arguments following Equation (84) to conclude that T2/T1 is also a positive integer. Therefore, we
conclude that T1 = T2 and a1 = a2.
According to a property of circular orbits, at times t and (t + T1/2), the intersection
of q1 and the plane spanned by l1 and n1 passing l1–ld1 should be the same, that is,
D 1(t) = D 1(t + T 1 ⁄ 2).
(99)
Thus,
−φ
ψ 1 ( t ) – ψ 2( t ) +
-⎞ ( λ 1(t)n̂ 1 + κ 1(t)l 1 )
sin ⎛ --------------------------------------⎝
⎠
2
ψ 1(t + T 1 ⁄ 2) – ψ 2(t + T 1 ⁄ 2) −
+ φ⎞
- ( λ 1(t + T 1 ⁄ 2)n̂ 1 + κ 1(t + T 1 ⁄ 2)l 1 ).
= sin ⎛ ------------------------------------------------------------------------------⎝
⎠
2
Because of (91), Equation (100) is satisfied only if
ψ 1(t) – ψ 2(t) −
+ φ⎞
- = 0
sin ⎛ --------------------------------------⎝
⎠
2
(100)
(101)
or
λ 1(t)n̂ 1 + κ 1(t)l 1 = – λ 1(t + T 1 ⁄ 2)n̂ 1 – κ 1(t + T 1 ⁄ 2)l 1 .
Equation (102) implies that for all time t,
κ(t) = – κ(t + T 1 ⁄ 2),
(102)
(103)
which further implies that
π
M θ (t) = --- .
1
2
(104)
130
ANNALS NEW YORK ACADEMY OF SCIENCES
This is impossible. Thus, the only possibility is that
ψ 1(t) – ψ 2(t) −
+ φ⎞
- = 0
sin ⎛ --------------------------------------(105)
⎝
⎠
2
Because of (105), the time varying parts in the equation u1 = 0 vanish. We can make
the same argument as in the proof for the single satellite case3 to show that
l1 = ld
1
(106)
A 1 = 0.
Now let u2(t) = 0 and substitute Equation (105). We conclude that the second satellite will achieve
l2 = ld
2
(107)
A 2 = 0.
Thus, we have
li = ld
i
Ai = 0
(108)
ψ 1 – ψ 2 = ±φ
for i = 1, 2. By definition of ψ1 and ψ2 in (64), we have
a 1⎞ 3 / 2
a 3/2
⎛ ---M θ – ⎛ ----2-⎞ M θ = φ.
⎝a ⎠
⎝
1
2
a ⎠
d
(109)
d
However, we already know a1 = a2 = ad and that the orbits are circular; thus, we conclude that
θ 1 – θ 2 = φ.
(110)
We mention that this algorithm is not limited to set up formations on circular
orbits. With slight modification to the proof one can show the convergence result
holds for elliptic orbits.
SIMULATION RESULTS
To verify our algorithms, a series of simulations were carried out using Matlab.
First, we show a controlled transfer of two satellites from orbit [a, e, i, ω, Ω] =
[20, 0.1, π/4, π/2, 0] with initial separation of mean anomaly π/90 to the orbit
[a, e, i, ω, Ω] = [20, 0.05, π/3, π/2, 0] with final separation of mean anomaly π/18 ≈
0.175. Only relative motion between the satellites is plotted in an inertial frame centered at one of the satellites. FIGURE 2 displays the desired relative motion between
the satellites. FIGURE 3 displays the relative motion between the satellites using the
proposed control algorithm. The final orbits for the two satellites are:
[a1, e1, I1, ω1, Ω1] = [25.10, 0.0495, 1.0472, 1.6231, 0]
[a2, e2, I2, ω2, Ω2] = [25.12, 0.0507, 1.0472, 1.6231, 0]
and the final separation of mean anomaly is 0.168. As can be seen, the desired orbit
and separation are achieved with small error. We noticed that by adjusting the
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
131
FIGURE 2. The desired final relative motion of two satellites (length unit, one tenth
of Earth radius).
FIGURE 3. The relative motion achieved by our control law for elliptic orbits (length
unit, one tenth of Earth radius).
132
ANNALS NEW YORK ACADEMY OF SCIENCES
weight of the terms in our Lyapunov functions, we can control the distribution of
errors among the orbital elements of final orbits.
Similar experiments were performed to set up a two satellite formation in circular
orbits. Initially, the two satellites were on an elliptic orbit with [a, e, i, ω, Ω] =
[20, 0.1, π/4, π/2, 0] and the separation of argument of latitude equal to π/90. Their
destination orbit was circular with [a, e, i, ω, Ω] = [25, 0, π/3, π/4] and separation π/18.
Using our algorithm, the final orbits obtained are:
[a1, e1, I1, Ω1] = [25.082, 0.0023, 1.0472, 0.7854]
[a2, e2, I2, Ω2] = [25.078, 0.0022, 1.0472, 0.7854]
and the final separation equal to 0.174. Again, the formation is achieved with small
error.
SUMMARY AND FUTURE DIRECTIONS
In this paper we have proposed control algorithms that can be used to set up periodic formations of satellites on elliptic and circular orbits. The shape space formed
by the angular momentum vectors and Laplace vectors is appropriate to describe satellite formations. The control laws we propose are based on a Lyapunov function on
this shape space and proved to be convergent. We have not considered the effect of
perturbations, such as the J2 effect. This is currently being investigated.
ACKNOWLEDGMENTS
This research was supported in part by the National Aeronautics and Space
Administration under NASA-GSFC Grant NAG5-10819, by the Air Force Office of
Scientific Research under AFOSR Grant F49620-01-0415, by the Army Research
Office under ODDR&E MURI97 Program Grant DAAG55-97-1-0114 to the Center
for Dynamics and Control of Smart Structures (through Harvard University), and
under ODDR&E MURI01 Program Grant DAAD19-01-1-0465 to the Center for
Communicating Networked Control Systems (through Boston University).
REFERENCES
1. SCHAUB, H., S.R. V ADALI, J.L. JUNKINS & K.T. ALFRIEND. 2000. Spacecraft formation
flying control using mean orbit elements. AAS J. Astronaut. Sci. 48(1): 69–87.
2. ZHANG, F. & P.S. K RISHNAPRASAD. 2002. Formation dynamics under a class of control
laws. In Proc. 2002 American Control Conference, Anchorage, Alaska. 1678–1685.
3. CHANG, D.E., D. C HICHKA & J.E. MARSDEN. 2002. Lyapunov-based transfer between
elliptic Keplerian orbits. Discrete Contin. Dynam. Syst. B 2: 57–67.
4. CUSHMAN R.H. & L.M. B ATES. 1997. Global Aspects of Classical Integrable Systems.
Birkhauser Verlag, Switzerland.
5. DANBY, J.M.A. 1962. Fundamentals of Celestial Mechanics. Macmillan, New York.
6. KRISHNAPRASAD, P.S. 2000. Relative equilibria and stability of rings of satellites. In
Proc. 39th IEEE Conference on Decision and Control, Sydney, Australia. 1285–1288.
7. BURNS, J.A. 1976. Elementary derivation of the perturbation equations of celestial
mechanics. Am. J. Phys. 44(10): 944–949.
8. VINTI, J.P. 1998. Orbital and Celestial Mechanics. AIAA. Reston, Virginia.
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
133
APPENDIX
In this appendix we derive the perturbation equations for the orbital elements
used in our control algorithm. The control forces are continuous functions of time.
Hence, the derivatives are taken along the controlled dynamics of a satellite.
Starting with the eccentricity, we have
A
∂A
1 ∂A T
A ⋅ Ȧ
ė(t) = ------------- = ------------- ⋅ u(t) = --- ⎛ ⎞ Â ⋅ u(t),
µ A ∂p
µ ⎝ ∂ p⎠
µ A
(111)
where  is the unit vector along the direction of A. We also have
1 ∂l T
l⋅l
ḣ(t) = ------- = --- ⎛ ⎞ l ⋅ u(t)
h ⎝ ∂ p⎠
l
(112)
and
1
ȧ(t) = ----------------------- ( 2hḣ + 2µaeė )
µ( 1 – e2 )
(113)
∂l T
∂A T
2
2ae
= ----------------------- ⎛ ⎞ l ⋅ u(t) + ----------------------- ⎛ ⎞ Â ⋅ u(t).
µ ( 1 – e 2 ) ⎝ ∂ p⎠
µ ( 1 – e 2 ) ⎝ ∂ p⎠
We now derive the formulæ for Ṁ . Since r2 = q q, we have rṙ = q p. Hence
h2 = l ⋅ l = ( q × p ) ⋅ ( q × p )
2 1
= ( q ⋅ q ) ( p ⋅ p ) – ( q ⋅ p ) ( q ⋅ p ) = r 2 µ ⎛ --- – ---⎞ – r 2 ṙ 2 .
⎝ r a⎠
(114)
On the other hand
h 2 = µa ( 1 – e 2 ).
(115)
µa ( 1 – e 2 )
2 1
ṙ 2 = – -------------------------- + µ ⎛ --- – ---⎞ .
⎝ r a⎠
r
(116)
Thus,
After applying r = a(1 − e cos(E)), on simplification we obtain
e sin ( E )
ṙ = µa -------------------.
r
Taking derivatives on both sides of r = a(1 − e cos(E)),
ṙ = ȧ ( 1 – e cos ( E ) ) + a ( ė cos ( E ) – e sin ( E )Ė ) .
(117)
(118)
Thus,
r
acos ( E )
ṙ
Ė = ----------------------- – -------------------------ȧ + ----------------------- ė
ae sin ( E )
ae sin ( E ) a 2 e sin ( E )
=
cos ( E )
r
µ 1--- ------------------ė – -------------------------ȧ.
--- +
a r e sin ( E )
a 2 e sin ( E )
(119)
Taking derivatives on both sides of M = E − e sin(E),
Ṁ = Ė – ė sin ( E ) – e cos ( E )Ė = ( 1 – e cos ( E ) )Ė – ė sin ( E ).
Combining this equation with Equations (119), (113), and (111), we obtain
(120)
134
ANNALS NEW YORK ACADEMY OF SCIENCES
r ( 1 – e cos ( E ) )
µ1
cos ( E ) ( 1 – e cos ( E ) )
Ṁ = ( 1 – e cos ( E ) ) --- --- + ⎛ ---------------------------------------------------- – sin ( E )⎞ ė – -------------------------------------ȧ
⎠
ar ⎝
e sin ( E )
a 2 e sin ( E )
=
( 1 – e cos ( E ) ) 2
µ cos ( E ) – e
----- + -------------------------- ė – ------------------------------------- ȧ
e sin ( E )
ae sin ( E )
a3
(121)
∂l T
∂A T
= n + η ⎛ ⎞ Â ⋅ u(t) + ξ ⎛ ⎞ l ⋅ u(t),
⎝ ∂ o⎠
⎝ ∂ p⎠
where n =
µ ⁄ a 3 and
2 ( 1 – e cos ( E ) ) 2
ξ(l, A, E) = – ---------------------------------------------µae ( 1 – e 2 ) sin ( E )
cos ( E ) – e 2 ( 1 – e cos ( E ) ) 2
η(l, A, E) = -------------------------- – ---------------------------------------- .
µe sin ( E ) µ ( 1 – e 2 ) sin ( E )
(122)
The inclination I is defined as the angle between the z-axis and the angular
momentum vector l. It can be determined from
lz
(123)
cos ( I ) = ---.
h
Taking derivatives on both sides, we obtain
l z l x l˙x + l z l y l˙y – ( l x2 + l y2 )l˙z
– sin ( I ) I˙ = – ------------------------------------------------------------h3
(124)
sin ( I )
˙
= – --------------- [ cos ( I ) sin ( Ω ), – cos ( I ) cos ( Ω ), – sin ( I ) ]l ,
h
because
–l y
–l y
-.
= ----------------cos ( Ω ) = ------------------(125)
h sin ( I )
l x2 + l y2
Thus,
1
∂l
(126)
I˙ = --- [ cos ( I ) sin ( Ω ), – cos ( I ) cos ( Ω ), – sin ( I ) ] ⎛ ⎞ u.
⎝ ∂ p⎠
h
The argument of the ascending node Ω is defined as the angle between the x- axis
and the ascending node. It can be determined from
l
tan ( Ω ) = – ---x- .
(127)
ly
By taking derivatives on both sides of (127), we obtain
l y l˙x – l x l˙y
1
1
∂l
--------------------Ω̇ = – ---------------------- = – ---- [ l y, – l x, 0 ] u.
2
2
2
∂p
cos ( Ω )
ly
ly
(128)
Let n̂ = ( l x2 − l y2 )−1/2 [− ly, lx, 0]T, then
h sin ( I ) cos2 ( Ω ) ∂l T
1
∂l T
Ω̇ = --------------------------------------- ⎛ ⎞ n̂ ⋅ u = ------------------ ⎛ ⎞ n̂ ⋅ u.
⎝
⎝
⎠
2
h sin ( I ) ∂ p⎠
∂p
ly
Note that n̂ is the unit vector pointing to the ascending node.
(129)
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
135
The argument of the perigee ω is defined as the angle between the vector n̂ and
 . It can be determined from
cos ( ω ) = n̂ ⋅ Â.
(130)
˙
1
ω̇ = – ---------------- ( n̂˙ ⋅ Â + Â ⋅ n̂ ).
sin ( ω )
(131)
˙
1
 = ------ ( Id –   T ) Ȧ
µe
(132)
Taking derivatives on both sides,
Because
and
1
n̂˙ = ------------------ ( Id – n̂n̂ T )ṅ,
h sin ( I )
where Id is the 3 × 3 identity matrix, one can verify that
n̂˙ ⋅ Â = Ω̇ cos ( I ) sin ( ω )
(133)
(134)
and
˙
1
1
(135)
 ⋅ n̂ = ------ ( n̂ – cos ( ω )  ) ⋅ Ȧ = ------ sin ( ω )  ⊥ ⋅ Ȧ
µe
µe
where  ⊥ is the unit vector perpendicular to  and l forming an acute angle with
n̂ . The derivative of ω is
cos ( I ) ∂l T
1 ∂A T
(136)
ω̇ = – ----------------------- ⎛ ⎞ n̂ ⋅ u – ------ ⎛ ⎞ Â ⊥ ⋅ u .
µe ⎝ ∂ p⎠
h 2 sin2 ( I ) ⎝ ∂ p⎠
The longitude parameter θ is used instead of the true anomaly when the orbit is
circular. To determine θ, we have
(137)
q ⋅ n̂ = q x cos ( Ω ) + q y sin ( Ω ) = r cos ( θ ) .
Hence,
x
y
cos ( θ ) = -- cos ( Ω ) + -- sin ( Ω ) .
r
r
(138)
On the other hand,
l
(139)
– q × n̂ = r sin ( θ ) --- .
h
This gives the following set of equations:
l
z
sin ( θ ) ---x- = - sin ( Ω )
h
r
l
z
(140)
sin ( θ ) ---y- = – - cos ( Ω )
h
r
l
y
x
sin ( θ ) ---z = – -- sin ( Ω ) + -- cos ( Ω ) .
h
r
r
Thus, θ can be uniquely determined by using Equation (138) and any one of (140).
We now calculate the derivative of θ, which contains two parts. The first part is
due to the orbital motion of the satellite and the second part is due to the perturbation
of the line of ascending node; that is,
136
ANNALS NEW YORK ACADEMY OF SCIENCES
h cos ( I ) ∂l T
(141)
θ̇ = ----- – ------------------ ⎛ ⎞ n̂ ⋅ u.
r 2 h sin ( I ) ⎝ ∂ p⎠
We now derive Equation (63) for Ṁ θ . We start from the cosine relationship
between the true anomaly f and the eccentric anomaly E
e + cos ( f )
cos ( E ) = -----------------------------.
1 + e cos ( f )
Taking derivatives on both sides yields
(142)
( e 2 – 1 ) sin ( f )
sin2 ( f )
– sin ( E ) Ė = ------------------------------------- f˙ + ------------------------------------- ė.
2
( 1 + e cos ( f ) ) 2
( 1 + e cos ( f ) )
(143)
1 – e 2 sin ( f )
sin ( E ) = ----------------------------------- ,
1 + e cos ( f )
(144)
sin ( E )
sin ( E )
1 – e2
1 – e cos ( E )
Ė = ----------------------------- f˙ – ----------------ė = ----------------------------- f˙ – ----------------ė .
2
1 + e cos ( f )
2
1–e
1 – e2
1–e
(145)
Because
we have
We know that
f = θ – ω.
(146)
The true anomaly of the ascending node is − ω. Let Eω be the eccentric anomaly of
the ascending node. Its derivative is
1 – e cos ( E ω )
sin ( E ω )
- ė.
- ( – ω̇ ) – ------------------Ė ω = -------------------------------(147)
1 – e2
1 – e2
Let Mθ be the mean anomaly measured from the line of ascending node. By using
Kepler’s equation, it is easy to see that
M θ = E – e sin ( E ) – ( E ω – e sin ( E ω ) ) .
(148)
Thus, the derivative is
Ṁ θ = ( 1 – e cos ( E ) )Ė – ( 1 – e cos ( E ω ) )Ė ω – ė ( sin ( E ) – sin ( E ω ) )
( 1 – e cos ( E ) ) 2
= ------------------------------------- θ̇ – ė ( sin ( E ) – sin ( E ω ) )
1 – e2
( 1 – e cos ( E ) ) sin ( E ) ( 1 – e cos ( E ω ) ) sin ( E ω )⎞
– ė ⎛ --------------------------------------------------- – --------------------------------------------------------⎝
⎠
1 – e2
1 – e2
(149)
( 1 – e cos ( E ) ) 2 – ( 1 – e cos ( E ω ) ) 2
– -----------------------------------------------------------------------------------ω̇.
1 – e2
We know that
r
( 1 – e cos ( E ) ) = --- .
a
Therefore,
(150)
ZHANG & KRISHNAPRASAD: COORDINATED ORBIT TRANSFER
r 2 – r ω2
( 1 – e cos ( E ) ) 2 – ( 1 – e cos ( E ω ) ) 2
-.
= --------------------------------------------------------------------------------------------------------a2 1 – e2
1 – e2
Because of Equation (136),
r 2 – r ω2
r 2 – r ω2 ⎛ ⎛ ∂A⎞ T
-----------------------ω̇ = – -----------------------------Â ⋅ u⎞
⎝ ⎝ ∂ p⎠ ⊥ ⎠
2
2
2
2
a 1–e
µa e 1 – e
r 2 – r ω2 cos ( I ) ⎛ ∂l ⎞ T
- -----------------– ----------------------n̂ ⋅ u.
⎝ ⎠
a 2 1 – e 2 h sin ( I ) ∂ p
137
(151)
(152)
Since we already know that
h
θ̇ = ----- – Ω̇ cos ( I )
r2
1 ∂A T
ė = --- ⎛ ⎞ Â ⋅ u(t),
µ ⎝ ∂ p⎠
substitute the derivatives in (149) from (152) and (153), to obtain
(153)
cos ( I ) ∂l T
r2
Ṁ θ = n – ------------------------ ------------------ ⎛ ⎞ n̂ ⋅ u
⎝ ⎠
a 2 1 – e 2 h sin ( I ) ∂ p
r sin ( E ) – r ω sin ( E ω )⎞ 1 ⎛ ∂A⎞ T
--– ⎛ sin ( E ) – sin ( E ω ) + --------------------------------------------------Â ⋅ u
⎝
⎠ µ ⎝ ∂ p⎠
a( 1 – e2 )
r 2 – r ω2 ⎛ ⎛ ∂A⎞ T
– -----------------------------Â ⋅ u⎞
⎝⎝ ⎠ ⊥ ⎠
µa 2 e 1 – e 2 ∂ p
(154)
r 2 – r ω2 cos ( I ) ⎛ ∂l ⎞ T
- -----------------+ ----------------------n̂ ⋅ u,
⎝ ⎠
a 2 1 – e 2 h sin ( I ) ∂ p
which simplifies to
2
rω
cos ( I ) ∂l T
- ------------------ ⎛ ⎞ n̂ ⋅ u
Ṁ θ = n – ----------------------⎝ ⎠
a 2 1 – e 2 h sin ( I ) ∂ p
r sin ( E ) – r ω sin ( E ω )⎞ 1 ⎛ ∂A⎞ T
--– ⎛ sin ( E ) – sin ( E ω ) + --------------------------------------------------Â ⋅ u
⎝
⎠ µ ⎝ ∂ p⎠
a( 1 – e2 )
r 2 – r ω2 ⎛ ⎛ ∂A⎞ T
+ -----------------------------Â ⋅ u⎞ .
⎝⎝ ⎠ ⊥ ⎠
µa 2 e 1 – e 2 ∂ p
(155)