Orbit Determination with the two-body Integrals. III

0.65 MB·English

by  Giovanni F. Gronchi

Orbit Determination with the
two-body Integrals. III
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2 G. F. Gronchi∗, G. Bau`†, S. Maro`‡

n Dipartimento di Matematica, Universit`a di Pisa,
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J Largo B. Pontecorvo, 5, Pisa, Italy

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January 16, 2015

]
h
p
-
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t Abstract
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m
We present the results of our investigation on the use of the two-
[ body integrals to compute preliminary orbits by linking too short arcs

1 of observations of celestial bodies. This work introduces a significant
v improvement with respect to the previous papers on the same subject
0
[3], [4]. Here we find a univariate polynomial equation of degree 9
0
6 in the radial distance ρ of the orbit at the mean epoch of one of
3 the two arcs. This is obtained by a combination of the algebraic
0
integrals of the two-body problem. Moreover, the elimination step,
.
1
whichin[3],[4]wasdonebyresultanttheorycoupledwiththediscrete
0
5 Fourier transform, is here obtained by elementary calculations. We
1 also show some numerical tests to illustrate the performance of the
:
v new algorithm.
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1 Introduction
a
This paper is related to the research started in [3], [4], where some first
integralsofthetwo-bodyproblemwereusedtowritepolynomialequationsfor
the linkage problem with optical observations of asteroids and space debris.
∗[email protected]
†[email protected]
‡[email protected]
1
In [3] the authors consider polynomial equations with total degree 48, that
are consequences of the conservation of angular momentum and energy, and
they propose two methods to search for all the solutions using algebraic
elimination theory. The same equations were first introduced in [7], but their
algebraic character was not fully exploited at that time. In [4] the authors
introduce for the same purpose new polynomial equations, with total degree
20, using the angular momentum integral and a projection of Laplace-Lenz
vector along a suitable direction. Both in [3] and in [4] the number of the
considered equations is equal to the number of unknowns.
Here we improve significantly the previous results by writing an overdeter-
mined polynomial system (more equations than unknowns) which is proved
to be generically consistent, i.e. the set of solutions in the complex field is
not empty. For generic values of the data, by variable elimination, we obtain
a system of two univariate polynomials of degree 10 with a greatest common
divisor of degree 9.
We also discuss the case where the oblateness of the Earth is relevant, so that
we add the perturbation of the J term to the Keplerian potential. In [2]
2
and [4] this problem was faced by an iterative scheme, writing at each step
polynomialequationswiththesamealgebraicstructureasintheunperturbed
case (without J effect). In this paper, the overdetermined system that can
2
be written following the steps of the unperturbed case has the same algebraic
structure but is generically inconsistent. However, we can use the iterative
scheme mentioned above by neglecting one polynomial of the system.
This paper is organized as follows. After recalling the linkage problem
and some preliminaries on the two-body integrals (Sections 2, 3), in Section 4
we discuss the polynomial equations that can be written for this problem,
including the overdetermined system (9) which is the object of this work. In
Section 5 we show how the elimination steps can be carried out, and we prove
the consistency of system (9). In Section 6 we discuss the spurious solutions
of (9), and illustrate two methods to discard them. Section 7 is devoted to
the linkage problem with the J effect. Some numerical tests are presented
2
in Section 8. In the Appendix we discuss a simple way to filter the pairs of
attributables to be linked with this algorithm.
2
2 Linkage of too short arcs
We consider objects moving in a central force field. Let us fix an inertial
reference frame, with the origin at the center of attraction O, which is the
center of the Sun (Earth) in the asteroid (space debris) case. Assume the
position q and velocity q˙ of the observer are known functions of time. We
describe the position of the observed body as the sum
r = q+ρeρ,
with ρ the topocentric distance and eρ the line of sight unit vector. We
choose spherical coordinates (α,δ,ρ) ∈ [−π,π)×(−π/2,π/2)×R+, so that
eρ = (cosδcosα,cosδsinα,sinδ).
A typical choice for α,δ is right ascension and declination. The velocity
vector is
r˙ = q˙ +ρ˙eρ +ρ(α˙ cosδeα +δ˙eδ), ρ˙,α˙,δ˙ ∈ R,ρ ∈ R+,
˙
where ρ˙, ρα˙ cosδ, ρδ are the components of the velocity, relative to the
observer, in the (positively oriented) orthonormal basis {eρ,eα,eδ}, with
∂eρ ∂eρ
eα = (cosδ)−1 , eδ = .
∂α ∂δ
Let (t ,α ,δ ) with i = 1...m, m ≥ 2, be a short arc of optical obser-
i i i
vations of a moving body, made from the same station. If m ≥ 3, we can
compute α, δ, α˙, δ˙, α¨, δ¨ at the mean time t¯ = 1 (cid:80)m t by a quadratic
m i=1 i
fit. From these quantities we can try to compute a preliminary orbit. When
the second derivatives are not reliable due to errors in the observations (or
not available, if m = 2) we speak of a too short arc (TSA) and, to compute
a preliminary orbit, we have to add information coming from other arcs of
observations. This is a typical identification problem, see [6].
In any case, it is possible to compute an attributable
A = (α,δ,α˙,δ˙) ∈ [−π,π)×(−π/2,π/2)×R2,
representing the angular position and velocity of the body at epoch t¯ (see
[5], [3]). The radial distance and velocity ρ,ρ˙ are completely undetermined
and are the missing quantities to define an orbit for the body.
In this paper we deal with the linkage problem, that is to join together two
TSAs of observations to form an orbit fitting all the data.
3
3 First integrals of Kepler’s motion
We consider the first integrals of the equation of Kepler’s problem
µ
¨r = − r
|r|3
as functions of the unknowns ρ, ρ˙. The angular momentum is the polynomial
vector
c(ρ,ρ˙) = r×r˙ = Dρ˙ +Eρ2 +Fρ+G,
with
D = q×eρ, E = eρ ×e⊥, F = q×e⊥ +eρ ×q˙, G = q×q˙,
where we have set
e⊥ = α˙ cosδeα +δ˙eδ.
The expression of the energy is
1 µ
E(ρ,ρ˙) = |r˙|2 − , (1)
2 |r|
where
|r| = (ρ2 +|q|2 +2ρq·eρ)1/2, (2)
|r˙|2 = ρ˙2 +|e⊥|2ρ2 +2q˙ ·eρρ˙ +2q˙ ·e⊥ρ+|q˙|2. (3)
The Laplace-Lenz vector L is given by
r (cid:16) µ (cid:17)
µL(ρ,ρ˙) = r˙ ×c−µ = |r˙|2 − r−(r˙ ·r)r˙, (4)
|r| |r|
with |r|, |r˙|2 as in (2), (3), and
r˙ ·r = ρρ˙ +q·eρρ˙ +(q˙ ·eρ +q·e⊥)ρ+q˙ ·q.
Moreover, the following relations hold for all ρ,ρ˙:
c·L = 0, µ2|L|2 = µ2 +2E|c|2. (5)
Expressions (1), (4) are algebraic, but not polynomial, in ρ,ρ˙. However, we
can introduce a new variable u ∈ R, together with the relation |r|2u2 = µ2
and we obtain
1
E = |r˙|2 −u, µL = (|r˙|2 −u)r−(r˙ ·r)r˙,
2
4
that are polynomials in ρ,ρ˙,u.
For later reference we also introduce the quantity
1
K = |r˙|2r−(r˙ ·r)r˙. (6)
2
4 Polynomial equations for the linkage
We use the notation above, with index 1 or 2 referring to the epoch. Let
˙
A = (α ,δ ,α˙ ,δ ), j = 1,2
j j j j j
be two attributables at epochs t¯. We consider the polynomial system
j
c = c , L = L , E = E , u2|r |2 = µ2, u2|r |2 = µ2, (7)
1 2 1 2 1 2 1 1 2 2
in the 6 unknowns
(ρ ,ρ ,ρ˙ ,ρ˙ ,u ,u ).
1 2 1 2 1 2
System (7) is defined by the vector of parameters
(A ,A ,q ,q ,q˙ ,q˙ ),
1 2 1 2 1 2
and we shall discuss properties which hold for generic values of them. More-
over, (7) is composed by 9 equations with 6 unknowns. However, due to
relations (5), 2 equations can be considered as consequences of the others, so
that we are left with a system of 7 equations with 6 unknowns.
Assume the attributables A ,A refer to the same observed body, the
1 2
two-body dynamics is perfectly respected, and there are no observing er-
rors. Then the set of solutions of (7) is not empty. Taking into account
the observational errors, and the fact that the two-body motion is only an
approximation, system (7) turns out to be generically inconsistent, i.e. it has
no solution in C.
We search for a polynomial system, consequence of (7), which is generi-
cally consistent, with a finite number of solutions in C, and which leads by
elimination to a univariate polynomial equation of the lowest degree possible.
Introducing relations u2|r |2 = µ2 (j = 1,2) for the auxiliary variables u ,
j j 1
u correspondstothesquaringoperations, usedin[3], [4]tobringtheselected
2
algebraic system in the variables (ρ ,ρ ,ρ˙ ,ρ˙ ) into a polynomial form. Since
1 2 1 2
these operations are responsible of the high total degree of the resulting
5
polynomial systems (48 and 20 respectively), in writing the new equations
we try to cancel the dependence on both u , u by algebraic manipulations
1 2
of the conservation laws. First we shall consider the intermediate system
c = c , µ(L −L ) = (E −E )r , u2|r |2 = µ2, (8)
1 2 1 2 1 2 2 1 1
where u does not appear, which is still inconsistent; then we shall take into
2
account the system
c = c , (K −K )×(r −r ) = 0, (9)
1 2 1 2 1 2
where also u does not appear, whose consistency is proven in the next sec-
1
tion.
5 Elimination of variables
Inthissectionweshowthatgenericallysystem(9)isconsistent. Inparticular,
by elimination of variables, we shall end up with two univariate polynomials
of degree 10 in the range ρ , whose greatest common divisor generically has
2
degree 9. A similar procedure can be carried out by eliminating all the
variables but ρ .
1
5.1 Angular momentum equations
The conservation of angular momentum gives us 3 polynomial equations that
are linear in ρ˙ ,ρ˙ , and quadratic in ρ ,ρ . Therefore, it is natural to use
1 2 1 2
these equations to eliminate the radial velocities, as done in [3], [4]. These
equations can be written as
D ρ˙ −D ρ˙ = J(ρ ,ρ ), (10)
1 1 2 2 1 2
with J a vector whose components are quadratic polynomials in ρ , ρ . Fol-
1 2
lowing[3]weproject(10)ontothevectorsD ×(D ×D )andD ×(D ×D )
2 1 2 1 1 2
and obtain ρ˙ , ρ˙ as quadratic polynomials in ρ , ρ . With these expressions
1 2 1 2
of ρ˙ , ρ˙ we have
1 2
∆ ×(D ×D ) = 0, (11)
c 1 2
with ∆ = c −c , whatever the values of ρ ,ρ . The projection of (10) onto
c 1 2 1 2
D ×D allows us to eliminate the variables ρ˙ ,ρ˙ and yields
1 2 1 2
q(ρ ,ρ ) = q ρ2 +q ρ +q ρ2 +q ρ +q , (12)
1 2 2,0 1 1,0 1 0,2 2 0,1 2 0,0
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where the coefficients q depend only on the attributables and on the posi-
i,j
tion and velocity of the observer at epochs t¯, t¯.
1 2
In the following we shall consider the quantities introduced in Section 3 as
function of ρ , ρ only, by the elimination of ρ˙ , ρ˙ just recalled.
1 2 1 2
5.2 Bivariate equations for the linkage
By subtracting E r to both members of the Laplace-Lenz equation, and
2 2
using the conservation of energy, we obtain
µL −E r = µL −E r . (13)
1 1 2 2 2 2
Equation (13) can be written
1
∆ +( |r˙ |2 −u )∆ = 0, (14)
K 1 1 r
2
where we have set
∆ = K −K , ∆ = r −r ,
K 1 2 r 1 2
with K as in (6). Note that the variable u does not appear in (14).
2
We can also eliminate u by cross product with ∆ :
1 r
∆ ×∆ = 0. (15)
K r
For brevity we set ξ = ∆ ×∆ , and we note that
K r
1
ξ = (|r˙ |2 −|r˙ |2)r ×r −(r˙ ·r )r˙ ×∆ +(r˙ ·r )r˙ ×∆ . (16)
2 1 1 2 1 1 1 r 2 2 2 r
2
Remark 1. By developing the expressions of r ×r , r˙ ×∆ , r˙ ×∆ as
1 2 1 r 2 r
polynomials in ρ ,ρ we obtain that the monomials in ξ with the highest total
1 2
degree, which is 6, are all multiplied by eρ ×eρ.
1 2
In the following section we shall prove that for generic values of the data the
system
q = 0, ξ = 0 (17)
is consistent, that is the set of its roots in C is not empty.
Note that, if q = 0, the vector ξ is parallel to the common value c = c of
1 2
the angular momentum.
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5.3 Consistency of the equations
The proof of the consistency relies on some geometrical considerations. In
particular it is relevant to check whether the angular momentum vector is
orthogonal to the line of sight. We introduce the quantities
c = c ·eρ, i,j = 1,2.
ij i j
More explicitly we have
c = q ×eρ ·e⊥ρ +q ×eρ ·q˙ ,
11 1 1 1 1 1 1 1
c = eρ ×e⊥ ·eρρ2 +q ×eρ ·eρρ˙ (ρ ,ρ )+
12 1 1 2 1 1 1 2 1 1 2
+ (eρ ×q˙ +q ×e⊥)·eρρ +q ×q˙ ·eρ,
1 1 1 1 2 1 1 1 2
and similar expressions for c , c . In particular, equations c = 0 and
22 21 11
c = 0 represent straight lines in the plane ρ ρ , while c = 0 and c = 0
22 1 2 12 21
give conic sections, see Figure 1.
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4
3
P2 C
2
1
ρ2 0
−1
−2
−3 P1
−4
−5
−2 0 2 4 6 8 10
ρ1
Figure 1: In the plane ρ ρ , for a test case, we draw the curves q = 0 (black),
1 2
c = 0 and c = 0 (light gray), and the straight lines c = 0, c = 0
12 21 11 22
(dashed), and ρ = ρ(cid:48),ρ = ρ(cid:48) (dotted).
1 1 2 2
8
Consider the point C = (ρ(cid:48)(cid:48),ρ(cid:48)(cid:48)) defined by c = c = 0, so that
1 2 11 22
q ×q˙ ·eρ q ×q˙ ·eρ
ρ(cid:48)(cid:48) = 1 1 1, ρ(cid:48)(cid:48) = 2 2 2. (18)
1 eρ ×e⊥ ·q 2 eρ ×e⊥ ·q
1 1 1 2 2 2
In Lemma 1 we shall prove that C lies on the conic q = 0 and is the only
point where both angular momenta c , c vanish. The straight line ρ = ρ(cid:48)(cid:48)
1 2 2 2
generically meets q = 0 in another point P = (ρ(cid:48),ρ(cid:48)(cid:48)), where the angular
2 1 2
momenta do not vanish. Similarly, the straight line ρ = ρ(cid:48)(cid:48) generically meets
1 1
q = 0 in another point P = (ρ(cid:48)(cid:48),ρ(cid:48)), where the angular momenta are not
1 1 2
zero, see Figure 1. For q = 0, the vector r ×r gives the direction of c = c ,
1 2 1 2
therefore from the equations r ×r ·eρ = 0, j = 1,2 we obtain
1 2 j
q ×q ·eρ q ×q ·eρ
ρ(cid:48) = 1 2 2, ρ(cid:48) = 1 2 1. (19)
1 eρ ×eρ ·q 2 eρ ×eρ ·q
1 2 2 1 2 1
Note that
q = −E ·D ×D = −(eρ ×e⊥ ·q )(eρ ×eρ ·q ),
2,0 1 1 2 1 1 1 1 2 2
so that q (cid:54)= 0 implies that both ρ(cid:48) and ρ(cid:48)(cid:48) are well defined.
2,0 1 1
In a similar way we obtain
q = E ·D ×D = (eρ ×e⊥ ·q )(eρ ×eρ ·q ),
0,2 2 1 2 2 2 2 1 2 1
so that q (cid:54)= 0 implies that both ρ(cid:48) and ρ(cid:48)(cid:48) are well defined.
0,2 2 2
Generically we have
q ,q (cid:54)= 0. (20)
2,0 0,2
Lemma 1. If (20) holds, then the point C = (ρ(cid:48)(cid:48),ρ(cid:48)(cid:48)) given by c = c = 0
1 2 11 22
satisfies q(ρ(cid:48)(cid:48),ρ(cid:48)(cid:48)) = 0. Moreover, in C we have c = c = 0 and C is the
1 2 1 2
unique point in the plane ρ ρ where both angular momenta vanish.
1 2
Proof. Using relation r ·D = 0 we obtain
j j
c ×D = −(r˙ ·D )r , j = 1,2.
j j j j j
Moreover, condition (20) yields eρ ×q (cid:54)= 0, so that r (cid:54)= 0.
j j j
From relations
r˙ ·D = −c ·eρ
j j j j
9
we have
c ×D = 0 if and only if c = 0 (21)
j j jj
for j = 1,2. Therefore c = c = 0 implies
11 22
∆ ·D ×D = 0,
c 1 2
that together with (11) gives
c = c . (22)
1 2
Finally, relations (21), (22) imply c = 0, j = 1,2. The uniqueness immedi-
j
ately follows from the definition of C.
(cid:3)
Lemma 2. In the point C = (ρ(cid:48)(cid:48),ρ(cid:48)(cid:48)) generically we have ξ (cid:54)= 0.
1 2
Proof. By Lemma 1, if (20) holds, we have c = c = 0 in C, so that
1 2
(cid:16) r r (cid:17)
2 1
µ(L −L )−(E −E )r = µ − −(E −E )r
1 2 1 2 2 1 2 2
|r | |r |
2 1
∆ 1
= −µ r + (|r˙ |2 −|r˙ |2)r .
2 1 2
|r | 2
1
Therefore we have
1
∆ ×∆ = (|r˙ |2 −|r˙ |2)r ×r . (23)
K r 1 2 1 2
2
We show that the right-hand side of (23) does not vanish in C. In fact, by
projecting r ×r onto q ,q we obtain
1 2 1 2
q ·r ×r = ρ(cid:48)(cid:48)[ρ(cid:48)(cid:48)(eρ ×eρ ·q )−q ×q ·eρ],
1 1 2 1 2 1 2 1 1 2 1
q ·r ×r = ρ(cid:48)(cid:48)[ρ(cid:48)(cid:48)(eρ ×eρ ·q )−q ×q ·eρ],
2 1 2 2 1 1 2 2 1 2 2
and the expressions in the brackets vanish only if
ρ(cid:48) = ρ(cid:48)(cid:48), ρ(cid:48) = ρ(cid:48)(cid:48).
1 1 2 2
Moreover, ρ(cid:48)(cid:48) = ρ(cid:48)(cid:48) = 0 occurs only if q × q˙ · eρ = 0, j = 1,2. Thus
1 2 j j j
r ×r generically does not vanish. Using c = 0 projected e.g. onto e⊥, we
1 2 1 1
can prove that in C the quantity |r˙ |2 depends only on the data A ,q ,q˙
1 1 1 1
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