Research Article  Open Access
On the Symmetric Central Configurations for the Planar 1 + 4Body Problem
Abstract
We study the planar symmetric central configurations of the 1 + 4body problem where the symmetry axis does not contain any infinitesimal masses. Under certain assumptions, we find analytically some central configurations for suitable positive masses and also get some numerical results of symmetric central configurations where infinitesimal masses are not necessarily equal.
1. Introduction
A very old problem in celestial mechanics is the study of central configurations for the nbody problem. One of the reasons why central configurations are interesting is that they allow us to construct exact solutions of the nbody problem. Central configurations also have other interesting properties in the study of the nbody problem (see [1–14] for details).
In this paper, we consider a restricted version of the problem of planar central configurations; that is, we study the limit case of one large mass and N small masses as the small masses tend to zero (planar body problem). This problem was first considered by Maxwell [15] trying to construct a model for Saturn’s rings. Hall [16] studied the planar central configuration of the body problem where the N small masses are equal. He found that when N is sufficiently large, the only possible relative equilibrium is Maxwell’s ring, that is, a regular Ngon with a central mass, and that other configurations are possible for small N. Moeckel [17] found a necessary and sufficient condition for the linear stability of relative equilibria of the body problem with N small but not necessarily equal masses. Recently, these configurations have attracted the attention of astronomers. Renner and Sicardy [18] suggested that the presence of coorbital satellites might explain, at least partly, the confinement of Neptune’s ring arcs. Cors et al. [19] proved that there are only three symmetric central configurations of the body problem with four separate identical satellites. Albouy and Fu [2] proved that all central configurations of the body problem are symmetric which settles the question in this case. Oliveira and Cabral [20] showed that, for the planar body problem where the satellites have different infinitesimal masses and two of them are diametrically opposite in a circle, the configurations are necessarily symmetric and the other satellites have the same mass. Moreover, they prove that the number of central configurations in this case is in general one, two, or three, and in the special case, where the satellites diametrically opposite have the same mass, they prove that the number of central configurations is one or two and give the exact value of the ratio of the masses that provides this bifurcation. Many other results can be found in [9, 21, 22, 23, 24, 25]. Here, we study the planar symmetric central configurations of the body problem where the symmetry axis does not contain any infinitesimal masses, and also the satellites may have different masses.
2. Preliminaries
Consider n particles of masses in subject to their mutual Newtonian gravitational interaction. In an inertial reference frame and choosing appropriate units, the equations of motion arewhereis the Newtonian potential of system (1). The position vector is often referred to as the configuration of the system.
A configuration with is called a central configuration if there exists some positive constant λ, called the Lagrangian multiplier, such that
We are interested in the planar body problem, where the big mass is equal to 1 with position . The remaining N bodies with positions , called satellites, have masses , where and is a small parameter that tends to zero. In all central configurations of the planar body problem, the satellites lie on a circle centered at the big mass ([26]); that is, the satellites are coorbital.
We exclude collisions in the definition of central configuration and take the angles between two consecutive particles as coordinates; we refer to [2, 19, 26] for details. In these coordinates, the space of configuration is the simplex:and the equations characterizing the central configurations of the planar body problem arewhere .
In the case of the foursatellite system, Equation (5) is
The function defined above plays a key role in this problem (Figure 1). The following two lemmas state some properties of f and its derivatives which will be used to prove our results. Lemma 1 can be found in [2, 20], and Lemma 2 can be proved straightforwardly.
Lemma 1. The function satisfies the following:(1)(2)(3)(4)In , there is a unique critical point of f, where , such that in and in . In , there is a unique critical point of f, where , such that in and in .
Lemma 2. , and(1) in and in (2) in and in (3) in and in
A central configuration of the planar body problem is symmetric with respect to a straight line L containing the central body; since central configuration is invariant under rotation transformation, we have the following:(1)When N is even, either where the symmetry axis L contains two satellites (Figure 2(a)), or where the symmetry axis L contains no satellites (Figure 2(b)).(2)When N is odd, where the symmetry axis L contains one satellite. In the case of the body problem, the symmetric central configuration contains or . Oliveira and Cabral [20] have completed the first case , where the symmetry axis contains and , and . We consider the symmetric central configuration of the body problem with , where the symmetry axis does not contain any satellites, and and are not necessarily equal.
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3. Main Results
For the symmetric central configuration of the body problem, we consider the case . Using the property of f that , system (6) becomeswhere .
Letthen system (10) can be rewritten as
With simple computation, we have for any .
Lemma 3. Let be a symmetric central configuration solution of system (6) with ; then for and , there can be at most one to be zero.
Proof. First, we prove that , and cannot be zero simultaneously. Since , and , . By , we have or . If , then , which is a contradiction with the fact . If , then and which is a contradiction with .Moreover, when some two of and , equal zero, with simple calculation, we obtain that they are all equal to zero. Thus, this proves Lemma 3.
By Lemma 3, we consider the following cases that one of and , equals zero while the others are not zero, and also the case that all of , , and are not zero.
implies , or .
When , we get the following proposition which has already been obtained in [20], and we omit the proof.
Proposition 1 (see [20]). Consider a symmetric configuration of the 1 + 4body problem with and . There exists exactly one central configuration (Figure 3(a)) if and only if .
By renumbering these satellites from the third one counterclockwisely, we can find that the case with is equivalent to the case with . Thus, we just consider the case that and get the following theorem.
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Theorem 1. Consider a symmetric configuration of the 1 + 4body problem with and . There exists exactly one central configuration (Figure 3(b)) with if and only if with satisfy , where is the unique root of in .
When , we get the following theorem.
Theorem 2. Consider a symmetric configuration of the 1 + 4body problem with and . There exists no central configuration for any with .
Also by renumbering these satellites from the third one counterclockwisely, we can find that the case with is equivalent to the case with . Thus, we just consider the case that and get the following theorem.
Theorem 3. Consider a symmetric configuration of the 1 + 4body problem with and . There exists exactly one central configuration (Figure 4) with if and only if with satisfy and , where is the unique root of in .
When , and , with numerical evidence, we get the following conjecture.
Conjecture 1. Consider a symmetric configuration of the 1 + 4body problem with , , and in order that the four satellites are in a central configuration; follow , or , and . Furthermore, for suitable with , there exist two central configurations for any (in Table 1) and one symmetric central configuration for any (in Tables 2 and 3).



4. The Proof of Theorems 1–3
4.1. The Proof of Theorem 1
When , , and , system (10) reduces to
By the first and the fourth equation of (13), we have , and by the second and the third equation of (13), we get the same conclusion. This means is a necessary condition for the existence of the central configuration under these assumptions. Thus, Equation (13) is equivalent to
System (14) yieldsand the sign of must be the same for all .
While , let , then and , where . Substituting these assumptions into (15), we have
For , and . So
We are going to prove that monotonically decreases in . With simple computation,
We divide the proof into two cases:
Case 1. (). When , , , , and . So
Case 2. (). When ,
Sothat is,
Considering the function , it is easily computed thatfor , . Then, . So
Hence, by (22) and (24), we again get
Cases 1 and 2, and (17) give that monotonically decreases in . It is easy to see that as and as . So there is exactly one root of in . By (13), we get . Also by choosing suitable satisfying , we can get exactly one central configuration satisfying (13). Thus, Theorem 1 is proved.
4.2. The Proof of Theorem 2
When , , system (10) becomes
With simple calculation, system (26) is equivalent to
For and , we get and where . The second equation of (27) becomes
We consider four subcases according to in , , or , respectively, or according to equal to or π. Assuming that , we have , and then from the plot of f, we have and . Now suppose that , then ; similarly, we have and . When , we have , and this implies and . Finally, when or π, . The above analysis shows that (28) does not hold. Theorem 2 is proved.
4.3. The Proof of Theorem 3
When and , system (10) becomes
From the second and the third equation, we obtain and . This means is a necessary condition for the existence of the central configuration under these assumptions. Then, system (29) is equivalent to
implies that , or .
When , where and for . The third equation of (30) means that and have the same sign. When and , we have . When and , we have . Then, we consider two subcases according to in or . Hence, we must solve
Clearly, in because in .
It is easy to see that as , as , and as . So must have exactly one root in . Then, by the third equation of system (30), we get .
When , where . If , then , which is a contradiction with the fact that and must have the same sign. If , then , and again, we have the same contradiction.
Now consider , then where . From the plot of f, we have and . This means the second equation of (30) does not hold.
The above analysis shows that and must be satisfied for the existence of the central configuration. Also by choosing suitable satisfying and , we can get exactly one central configuration satisfying (29). Thus, Theorem 3 is proved.
5. The Numerical Evidence for Conjecture 1
Now consider , and . System (10) can be written as the following homogeneous linear system of equations:where , .
In order to get the nonzero mass vector, the determinant of the matrix A vanishes, that is,and the determinant of the matrix A is the square of the lefthand side of the equation above.
Since the matrix A is antisymmetric, from a classical theorem of linear algebra, the rank of A is even. Under the above constraint, it is easy to get that the rank of matrix A is 2. This means we can perform row operations on the matrix A to create a new matrix that produces an equivalent system. The following are some notations of three elementary matrix row operations. means interchanging the ith and jth rows. means replacing the ith row with a times itself. means replacing the jth row with the sum of ith and jth rows. For , and , performing the row operations and on matrix A, we get the new matrix and denote it by . And then performing the row operations and on matrix , we get matrix . Finally, performing the row operations and on matrix , we get the following matrix:
Thus, under these assumptions and constraint (33), system (32) reduces to
By (33), and cannot be zero; otherwise, or which is a contradiction with the above assumptions in this section. Then, Equation (35) implies that follow , or , and .
When and , because of constraint (33), Equation (35) becomes
When and , because of constraint (33), Equation (35) reduces to
We only consider the definite positive values of , so the signs of and must be the same. When , we have and ; then, and . Thus, . In the following, we just consider .
For , we get . Let
Obviously, the signs of and in , , and are the same; thus, the mass ratio is guaranteed to be positive.
For , we just consider the region below the straight line (Figure 5(a)). The sign of is determined in Figure 5(b). The regions , , and can be seen in Figure 5(c), and the curve is plotted in Figure 5(d).
(a)
(b)
(c)
(d)