Relative Equilibria in curved restricted 4-body problems

arXiv:1801.02126v1 [math.CA] 7 Jan 2018

RELATIVE EQUILIBRIA IN CURVED RESTRICTED 4-BODY PROBLEMS Florin Diacu1,2,3 and Sawsan Alhowaity3,4 1

Yangtze Center of Mathematics, Sichuan University, Chengdu, China 2 Yale-NUS College, National University of Singapore, Singapore 3 Department of Mathematics and Statistics, University of Victoria, Canada 4 Department of Mathematics, University of Shaqra, Saudi Arabia [email protected], [email protected] Abstract. We consider the curved 4-body problems on spheres and hyperbolic spheres. After obtaining a criterion for the existence of quadrilateral configurations on the equator of the sphere, we study two restricted 4-body problems, one in which two masses are negligible, and another in which only one mass is negligible. In the former we prove the evidence square-like relative equilibria, whereas in the latter we discuss the existence of kite-shaped relative equilibria.

1. Introduction The classical N-body problem has a long history. Isaac Newton first proposed it in 1687 in his first edition of Principia in the context of the Moon’s motion. He assumed that universal gravitation acts between celestial bodies (reduced to point masses) in direct proportion with the product of the masses and in inverse proportion with the square of the distance. The study of the N-body problem was further advanced by the Bernoulis, Lagrange, Laplace, Euler, Cauchy, Jacobi, Dirichlet, Poincaré and many others. The idea of extending the gravitational force between point masses to spaces of constant curvature occurred soon after the discovery of hyperbolic geometry. In the 1830s, independently of each other, Bolyai and Lobachevsky realized that there must be an intimate connection between the laws of physics and the geometry of the universe, [2], [29], [25]. A few years earlier, Gauss had interpreted Newton’s gravitational law as stating that the attracting force between bodies is inversely proportional with the area of the sphere of radius equal to the distance between the point masses (i.e. proportional to 1/r 2 , where r is the distance). Using this idea, Bolyai and Lobachevsky suggested that, should space be hyperbolic, the attracting force between bodies must be inversely proportional to the hyperbolic area of the corresponding hyperbolic sphere (i.e. proportional to 1/ sinh2 (|κ|1/2 r), 1

2

Florin Diacu and Sawsan Alhowaity

where r is the distance and κ < 0 the curvature of the hyperbolic space). This is equivalent to saying that, in hyperbolic space, the potential that describes the gravitational force is proportional to coth(|κ|1/2 r). The above analytic expression of the potential was first introduced by Schering, [33], [34], and then extended to elliptic space by Killing, [22–24]. But with no physical ways of checking the validity of this generalization of the gravitational force, it was unclear whether the cotangent potential had any physical meaning, the more so since Lipschitz had proposed a different extension of the law, which turned out to be short lived, [28]. The breakthrough came at the dawn of the 20th century when Liebmann made two important discoveries, [26], [27]. He showed that two basic properties of the Newtonian potential are also satisfied by the cotangent potential: (1) in the Kepler problem, which studies the motion of one body around a fixed centre, the potential is a harmonic function (i.e. a solution of the Laplace equation in the Euclidean case, but of the Laplace-Beltrami equation in the non-flat case); (2) in both the flat and the non-flat case, all bounded orbits of the Kepler problem are closed, a property discovered by Bertrand for the Newtonian law, [1]. These similarities between the flat and the curved problem convinced the scientific community that the cotangent potential was the natural way to express gravity in spaces of constant curvature. The curved N-body problem became somewhat neglected after the birth of general relativity, but was revived after the discretization of Einstein’s equation showed that an N-body problem in spaces of variable curvature is too complicated to be treated with analytical tools. In the 1990s, the Russian school of celestial mechanics considered both the curved Kepler and the curved 2-body problem, [24], [35]. After understanding that, unlike in the Euclidean case, these problems are not equivalent, the latter failing to be integrable, [35], the 2-body case was intensively studied by several researchers of this school. More recently, the work of Diacu, Santoprete, and Pérez-Chavela considered the curved N-body problem for N > 2 in a new framework, leading to many interesting results, [3–20], [32]. Other researchers developed these ideas further, [20] [30], [31], [37–40], and the problem is growing in popularity. In this short note we prove three results. The first is a criterion for the existence of quadrilateral relative equilibria on the equator of the sphere. The second shows that if two masses are negligible and the other two are equal, then square-like relative equilibria exists on spheres, but—surprisingly—not on hyperbolic spheres. The element of surprise arises from the fact that, in the general problem, squarelike equilibria exist both on the hyperbolic sphere and on the sphere (except for the case when they are on the equator), [5]. Finally we prove that if only one mass is negligible and the other three are equal, some kite-shaped relative equilibria exist on spheres, but not on hyperbolic spheres.

Relative equilibria in curved restricted 4-body problems

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2. Equations of motion We consider the motion of four bodies on 2-dimensional surfaces of constant curvature κ, namely spheres S2κ for κ > 0, the Euclidean plane R2 for κ = 0, and hyperbolic spheres H2κ for κ < 0. We will arrange these objects in R3 such that they all have a common point at which lie all the north poles of the spheres and the vertices of the hyperbolic spheres, to all of which the plane R2 is tangent. If we fix the origin of a coordinate system at this point, then we can write 1

S2κ := {(x, y, z) | κ(x2 + y 2 + z 2 ) + 2κ 2 z = 0} for κ > 0, 1

H2κ := {(x, y, z) | κ(x2 + y 2 − z 2 ) + 2|κ| 2 z = 0 z ≥ 0} for κ < 0. Consider now four point masses, mi > 0, i = 1, 2, 3, 4, whose position vectors, velocities, and accelerations are given by ri = (xi , yi , zi ), r˙ i = (x˙ i , y˙ i , z˙i ), ¨ri = (¨ xi , yï , zï ), i = 1, 2, 3, 4. Then, as shown in [9], the equations of motion take the form h i  2 κrij  m x − 1− xi j j  PN 2   x ¨ = − κ(˙ri · r˙ i )xi  i 3/2 j=1,j6=i  κr 2  ij 3  r 1−   h 4 2 ij i   κr   mj yj − 1− 2ij yi PN yï = j=1,j6=i − κ(˙ri · r˙ i )yi 3/2 κr 2  ij 3  1− rij   h 4 i  2  κrij   m z − 1− zi j j PN  2   z ¨ = − (˙ri · r˙ i )(κzi + σ|κ|1/2 ), i = 1, 2, 3, 4, i  3/2 j=1,j6=i  κr 2  1− ij r3 4

ij

where σ = 1 for κ ≥ 0, σ = −1 for κ < 0, and ( [(xi − xj )2 + (yi − yj )2 + (zi − zj )2 ]1/2 for κ ≥ 0 rij := [(xi − xj )2 + (yi − yj )2 − (zi − zj )2 ]1/2 for κ < 0

for i, j ∈ {1, 2, 3, 4}. The above system has eight constraints, namely κ(x21 + yi2 + σzi2 ) + 2|κ|1/2 zi = 0, κri · r˙ i + |κ|1/2 z˙i = 0, i = 1, 2, 3, 4. If satisfied at an initial instant, these constraints are satisfied for all time because the sets S2κ , R2 , and H2κ are invariant for the equations of motion, [5]. Notice that

4


for κ = 0 we recover the classical Newtonian equations of the 4-body problem on the Euclidean plane, namely N X mj (rj − ri ) ¨ri = , 3 r ij j=1,j6=i

where ri = (xi , yi , 0), i = 1, 2, 3, 4.

3. Relative equilibria Relative equilibria are solutions for which the relative distances remain constant during the motion. We first introduce some coordinates (ϕ, ω), which were originally used in [9] for the case N = 3, to detect relative equilibria on and near the equator of S2κ , where ϕ measures the angle from the x-axis in the xy-plane, while ω is the height on the vertical z-axis. In these new coordinates, the constraints become x2i + yi2 + ωi2 + 2κ−1/2 ωi = 0, i = 1, 2, 3, 4. With the notation, Ωi = x2i + yi2 = −κ−1/2 ωi (κ1/2 ωi + 2) ≥ 0, ωi ∈ [−2κ−1/2 , 0], i = 1, 2, 3, 4, where equality occurs when the body is at the North or the South Pole of the sphere, the (ϕ, ω)-coordinates are given by the transformations 1/2

xi = Ωi

1/2

cos ϕi , yi = Ωi

Thus the equations of motion take the form  1/2 mj Ωj sin(ϕj −ϕi ) ˙  −1/2 PN  ϕ ¨ = Ω − ϕ˙Ωi Ωi i  i i j=1,j6=i κρ2   ρ3ij (1− 4ij )3/2  i h mj  −1/2 PN  ω ¨ = Ω  i i j=1,j6=i   

ωj +ωi + ρ3ij

where

κρ2 ij 2

(ωi +κ−1/2 )

κρ2 1− 4ij

3/2

sin ϕi .

Ω˙2 − (κωi + κ1/2 )( 4Ωii + ϕ˙ i 2 Ωi + ω˙i2 ),

Ω˙ i = −2κ−1/2 ω˙ i (κ1/2 ωi + 1) 1/2

1/2

ρ2ij = Ωi + Ωj − 2Ωi Ωj cos(ϕi − ϕj ) + (ωi − ωj )2 , i, j = 1, 2, 3, 4, i 6= j.


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4. Relative equilibria on the equator If we restrict the motion of the four bodies to the equator of S2κ , then ωi = −κ−1/2 , ω˙ i = 0, Ωi = κ−1 , i = 1, 2, 3, 4, and the equations of motion take the form 3/2

ϕï = κ

4 X mj sin(ϕj − ϕi ) , i = 1, 2, 3, 4. 3 | sin(ϕ j − ϕi )| j=1,j6=i

For the relative equilibria, the angular velocity is the same constant for all masses, so we denote this velocity by α 6= 0 and take ϕ1 = αt + a1 , ϕ2 = αt + a2 , ϕ3 = αt + a3 , ϕ4 = αt + a4 , where a1 , a2 , a3 , a4 are real constants, so ϕï = 0, i = 1, 2, 3, 4. Using the notation s1 :=

κ3/2 sin(ϕ1 − ϕ2 ) κ3/2 sin(ϕ2 − ϕ3 ) κ3/2 sin(ϕ3 − ϕ1 ) , s := , s := , 2 3 | sin(ϕ1 − ϕ2 )|3 | sin(ϕ2 − ϕ3 )|3 | sin(ϕ3 − ϕ1 )|3

s4 :=

κ3/2 sin(ϕ4 − ϕ1 ) κ3/2 sin(ϕ2 − ϕ4 ) κ3/2 sin(ϕ3 − ϕ4 ) , s := , s := , 5 6 | sin(ϕ4 − ϕ1 )|3 | sin(ϕ2 − ϕ4 )|3 | sin(ϕ3 − ϕ4 )|3

we obtain from the equations of motion that  −m2 s1 + m3 s3 + m4 s4 = 0    m s − m s − m s = 0 1 1 3 2 4 5  −m1 s3 + m2 s2 − m4 s6 = 0    −m1 s4 + m2 s5 + m3 s6 = 0.

To have other solutions of the masses than m1 = m2 = m3 = m4 = 0, the determinant of the above system must vanish, which is equivalent to s1 s6 + s3 s5 = s2 s4 . We have thus proved the following result. Theorem 1. A necessary condition that the quadrilateral inscribed in the equator of S2κ , with the four masses m1 , m2 , m3 , m4 > 0 at its vertices, forms a relative equilibrium is that s1 s6 + s3 s5 = s2 s4 .

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5. Equivalent equations of motion Let us now introduce some equivalent equations of motion that are suitable for the kind of solutions we are seeking. First, rewriting the above constraints as κ(x2i + yi2 ) + (|κ|1/2 zi + 1)2 = 1, and solving explicitly for zi , we obtain q −1/2 zi = |κ| [ 1 − κ(x2i + yi2 ) − 1].

The idea here is to eliminate the four equations involving z1 , z2 , z3 , z4 , but they still appear in the terms rij2 in the form σ(zi − zj )2 as σ(zi − zj )2 = hp

κ(x2i + yi2 − x2j − yj2)2 i2 . q 2 2 2 2 1 − κ(xi + yi ) + 1 − κ(xj + yj )

The case of physical interest is when κ is not far from zero, so the above expression exist even for κ > 0 under this assumption. Then the equations of motion become  h i κρ2  mj xj − 1− 2ij xi  P    − κ(x˙i 2 + y˙i 2 + κBi )xi xï = N 3/2  j=1,j6=i κρ2  ij  3 ρij 1− h 4 i 2 κρij  m y − 1− yi  j j PN 2    y ¨ = − κ(x˙i 2 + y˙i 2 + κBi )yi , i 3/2 j=1,j6=i  κρ2  ij  3 1−

4

ρij

where

κ(Ai − Aj )2 p , ρ2ij = (xi − xj )2 + (yi − yj )2 + √ ( 1 − κAi + 1 − κAj )2 Ai = x2i + yi2 ,

Bi =

(xi x˙i + yi y˙i )2 , 1 − κ(x2i + yi2 )

i = 1, 2, 3, 4.

It is obvious that for κ = 0 we recover the classical Newtonian equations of motion of the planar 4-body problem.


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6. The case of two negligible masses We now consider the case when two out of the four given masses are negligible, m3 = m4 = 0. Then the equations of motion become  h i κρ2  12 x x − 1− m  2 2 1 2    x ¨ = − κ(x˙1 2 + y˙1 2 + κB1 )x1 1 3/2  2  κρ  1− 12 ρ3 h 2 2 12 i κρ  m2 y2 − 1− 212 y1     y¨1 = − κ(x˙1 2 + y˙1 2 + κB1 )y1 3/2  2  κρ  1− 12 ρ312 4

      x¨ =   2 

      x¨3 =         y¨3 =   

      x¨4 =         y¨4 =   

     y¨2 =   

h

h

h

h

κρ2 1− 413

3/2

m1 y1 − 1−

κρ2 1− 413

h

m1 x1 −

h

κρ2 1− 414

κρ2 14 2

3/2

where ρ2ij = ρ2ji , i 6= j,

i

3/2

ρ312

+

h

κρ2 1− 423

m2 y2 − 1−

κρ2 1− 232

+

h

+

h

m4 x4 −

ρ314

y4

ρ314

− κ(x˙2 2 + y˙2 2 + κB2 )y2

y2

+

x4

i

i

ρ312

h

y3

i

− κ(x˙2 2 + y˙2 2 + κB2 )x2

x2

m2 x2 − 1−

x3

ρ313

3/2

m1 y1 − 1−

κρ2 1− 412

i

κρ2 13 2

3/2

κρ2 1− 212

ρ313

κρ2 1− 214

κρ2 1− 414

κρ2 13 2

i

κρ2 12 2

3/2

κρ2 1− 412

m1 y1 −

m1 x1 − 1−

m1 x1 − 1−

− κ(x˙3 2 + y˙3 2 + κB3 )x3

23 i

κρ2 32 2

− κ(x˙3 2 + y˙3 2 + κB3 )y3

i

− κ(x˙4 2 + y˙4 2 + κB4 )x4

42 i

− κ(x˙4 2 + y˙4 2 + κB4 )y4 ,

y3

ρ332

κρ2 1− 442

3/2 κρ2 42 2

3/2

x3

ρ3

κρ2 1− 242

m4 y4 − 1−

3/2

3/2

κρ2 1− 442

i

κρ2 32 2

x2

ρ3

y2

ρ342

κ(x2i + yi2 − x2j − yj2 )2 q . ρ2ij = (xi − xj )2 + (yi − yj )2 + p [ 1 − κ(x2i + yi2 ) + 1 − κ(x2j + yj2 )]2

We can now show that when m1 = m2 =: m > 0 and m3 = m4 = 0, square-like relative equilibria, i.e. equilateral equiangular quadrilaterals, always exist on S2κ , but not on H2κ .

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m3 m4

m1

m2

Figure 1. The case of 2 equal masses and 2 negligible masses. Theorem 2. In the curved 4-body problem, assume that m1 = m2 =: m > 0 and m3 = m4 = 0. Then, in S2κ , there is a circle of radius r < κ−1/2 , parallel with the xy-plane, such that a square configuration inscribed in this circle, with m1 , m2 at the opposite ends of one diagonal and m3 , m4 at the opposite ends of the other diagonal, forms a relative equilibrium. But in H2κ , there is no such solution. Proof. We must check the existence of a solution of the form q = (q1 , q2 , q3 , q4 ) ∈ S2κ , qi = (xi , yi ), i = 1, 2, 3, 4. x1 = r cos αt,

y1 = r sin αt,

x2 = −r cos αt, x3 = r cos(αt + π/2) = −r sin αt, x4 = −r cos(αt + π/2) = r sin αt,

y2 = −r sin αt, y3 = r sin(αt + π/2) = r cos αt, y4 = −r sin(αt + π/2) = −r cos αt,

where

x2i + yi2 = r 2 , ρ2 = ρ213 = ρ214 = ρ223 = ρ224 = 2r 2 , ρ212 = ρ234 = 4r 2 . Substituting these expressions into the system, the first four equations lead us to m , α2 = 3 4r (1 − κr 2 )3/2

whereas the last four equations yield 2

α =

2m(1 −

ρ3 (1 −

κρ2 ) 2

κρ2 3/2 ) (1 4

− κr 2 )

.


9

m3

m2

m1

m4 Figure 2. The case of two equal masses and two negligible masses.

So, to have a solution, the equation 2

2m(1 − κρ2 ) m = 2 4r 3(1 − κr 2 )3/2 ρ3 (1 − κρ4 )3/2 (1 − κr 2 ) must be satisfied. This equation is equivalent to

8r 3 (1

1 1 = √ 2 3/2 3 − κr ) 2 2r (1 −

κr 2 3/2 ) 2

,

which leads to 3κr 2 = 2. Obviously, in the case of H2κ , we have κ < 0, so this equation has no solutions. For S2κ , it leads to p r = 2/3κ−1/2 .

Since r < κ−1/2 , such a solution always exists in S2κ .

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7. The case of one negligible mass Let m1 , m2 , m3 > 0 and assume that m4 = 0. Then the equations of motion take the form  h h i i κρ2 κρ2  m3 x3 − 1− 231 x1 m2 x2 − 1− 212 x1     + − κ(x˙1 2 + y˙1 2 + κB1 )x1 x¨1 = 3/2 3/2  2 2  κρ κρ  1− 12 1− 31 ρ3 ρ3 h 4 2 12 i h 4 2 31 i κρ κρ  m2 y2 − 1− 212 y1 m3 y3 − 1− 231 y1     y¨1 = + − κ(x˙1 2 + y˙1 2 + κB1 )y1 3/2 3/2  2 2  κρ κρ  3 3 12 31 1−

      x¨2 =   

h

m1 x1 − 1−

h

κρ2 1− 412

1−

ρ12

4

κρ2 12 2

3/2

i

i

m3 x3 − 1−

x2

ρ312

h

ρ31

4

+

h

κρ2 1− 432

κρ2 32 2

3/2

i

ρ3

x2

− κ(x˙2 2 + y˙2 2 + κB2 )x2

32 i

κρ2 κρ2  m1 y1 − 1− 212 y2 m3 y3 − 1− 232 y2     y¨2 = + − κ(x˙2 2 + y˙2 2 + κB2 )y2 3/2 3/2  2 2  κρ κρ  1− 412 1− 432 ρ312 ρ332  h h i i κρ2 κρ2  m1 x1 − 1− 213 x3 m2 x2 − 1− 232 x3     x¨ = + − κ(x˙3 2 + y˙3 2 + κB3 )x3 3/2 3/2  2  3 κρ2 κρ  ρ3 ρ3 1− 13 1− 32 h 4 2 13 i h 4 2 32 i κρ κρ  m1 y1 − 1− 213 y3 m2 y2 − 1− 232 y3     y¨3 = + − κ(x˙3 2 + y˙3 2 + κB3 )y3 3/2 3/2  2 2  κρ κρ  1− 413 1− 432 ρ313 ρ332 i i i h h h  κρ2 κρ2 κρ2 14 42 43  x4 x4 x4 m1 x1 − 1− 2 m2 x2 − 1− 2 m3 x3 − 1− 2    x ¨ = + + 4  3/2 3/2 3/2  κρ2 κρ2 κρ2   1− 414 1− 442 1− 443 ρ314 ρ342 ρ343     −κ(x˙4 2h+ y ˙4 2 + κB h i i i4 4 )x h 2 2 2 κρ κρ κρ  m2 y2 − 1− 242 y4 m3 y3 − 1− 243 y4 m1 y1 − 1− 214 y4     + + y ¨ = 3/2 4 3/2 3/2   κρ2 κρ2 κρ2  3 3 14 42 43 ρ14 ρ42 ρ343 1− 4 1− 4 1− 4     −κ(x˙4 2 + y˙4 2 + κB4 )y4 . We will next show that if the non-negligible masses are equal, then there exist some kite-shaped relative equilibria.

Theorem 3. Consider the curved 4-body problem with masses m1 = m2 = m3 := m > 0 and m4 = 0. Then, in S2κ , there exists at least one kite-shaped relative equilibrium for which the equal masses lie at the vertices of an equilateral triangle, whereas the negligible mass is at the intersection of the extension of one height of


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m2

m3

m1 m4

Figure 3. A kite configuration of 3 equal masses and one negligible mass. the triangle with the circle on which all the bodies move. In H2κ , however, there are no such kite-shaped relative equilibria. Proof. We will check a solution of the form

where

x1 = r cos αt, y1 = r sin αt, 2π 2π x2 = r cos αt + , y2 = r sin αt + 3 3 4π 4π x3 = r cos αt + , y3 = r sin αt + , 3 3 π π , y4 = r sin αt − , x4 = r cos αt − 3 3 ρ212 = ρ213 = ρ223 = 3r 2 , ρ243 = ρ241 = r 2 , ρ224 = 4r 2 .

Substituting these expressions into the above system, we are led to the conclusion that the following two equations must be satisfied, m α2 = √ , 2 3 3r (1 − 3κr )3/2 4 α2 =

4r 3(1

m m + . 2 2 3/2 3 − κr ) r (1 − κr4 )3/2

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Comparing these equations we obtain the condition for the existence of the kiteshaped relative equilibria, √

1 1 1 = + . 2 2 3/2 κr 3κr 2 3/2 4(1 − κr ) 3(1 − 4 ) (1 − 4 )3/2

Straightforward computations show that r is a solution of this equation if it is a root of the polynomial P (r) = a24 r 24 + a22 r 22 + a20 r 20 + a18 r 18 + a16 r 16 + a14 r 14 + a12 r 12 + a10 r 10 + a8 r 8 + a6 r 6 + a4 r 4 + a2 r 2 + a0 , 6697290145 12 2884257825 11 18063189465 10 κ , a22 = − κ , a20 = κ , 16777216 524288 524288 21267471735 8 4241985935 9 κ , a16 = κ , a18 = − 32768 65536 737853351 6 41995431 5 584429805 7 κ , a12 = κ , a10 = − κ, a14 = − 1024 1024 64 1530101 3 109080063 4 κ , a6 = − κ, a8 = 256 8 446217 2 κ , a2 = −9318κ, a0 = 649 a4 = 8 that belongs to the interval r ∈ (0, κ−1/2 ) for S2κ , but needs only to be positive for H2κ . To find out if we have such a root, we make the substitution x = r 2 , and obtain the polynomial a24 =

Q(x) = a24 x12 + a22 x11 + a20 x10 + a18 x9 + a16 x8 + a14 x7 + a12 x6 + a10 x5 + a8 x4 + a6 x3 + a4 x2 + a2 x + a0 . By Descartes’s rule of signs the number of positive roots depends on the number of changes of sign of the coefficients, which in turn depends on the sign of κ. So let us discuss the two cases separately. In S2κ , i.e. for κ > 0, there are twelve changes of sign, so Q can have twelve, ten, eight, six, four, two, or zero positive roots, so this does not guarantee the −1 existence of a positive root. However, we can notice that Q( κ 2 ) = −2.4959 < 0 and Q(0) = 649 > 0, so a root must exist for x ∈ (0, κ−1/2 ), i.e. for r ∈ (0, κ−1), a remark that proves the existence of at least one kite-shaped relative equilibrium. In H2κ , i.e. for κ < 0, we seek a positive root of Q. But for κ < 0, there is no sign change in Q(x), so the polynomial has no positive roots. Therefore there are no kite solutions in H2κ . This remark completes the proof.


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Acknowledgment. Florin Diacu did most of the work on this paper while visiting the Yangtze Center of Mathematics at Sichuan University as Distinguished Foreign Professor in April-May 2017. He was also supported in part by a grant from the Yale-NUS College at the National University of Singapore and an NSERC of Canada Discovery Grant. Sawsan Alhowaity was funded by a scholarship from the University of Shaqra, Saudia Arabia, towards the completion of her doctoral degree at the University of Victoria in Canada.

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