On the Dumb-Bell Equilibria in the Generalized Sitnikov Problem Текст научной статьи по специальности «Физика»

Russian Journal of Nonlinear Dynamics, 2022, vol. 18, no. 4, pp. 577-588. Full-texts are available at http://nd.ics.org.ru DOI: 10.20537/nd221203

NONLINEAR PHYSICS AND MECHANICS

MSC 2010: 37N05

On the Dumb-Bell Equilibria in the Generalized

Sitnikov Problem

P. S. Krasilnikov, A. R. Ismagilov

This paper discusses and analyzes the dumb-bell equilibria in a generalized Sitnikov problem. This has been done by assuming that the dumb-bell is oriented along the normal to the plane of motion of two primaries. Assuming the orbits of primaries to be circles, we apply bifurcation theory to investigate the set of equilibria for both symmetrical and asymmetrical dumb-bells.

We also investigate the linear stability of the trivial equilibrium of a symmetrical dumb-bell in the elliptic Sitnikov problem. In the case of the dumb-bell length l ^ 0.983819, an instability of the trivial equilibria for eccentricity e G (0, 1) is proved.

Keywords: Sitnikov problem, dumb-bell, equilibrium, linear stability

1. Introduction

It is known that the massless particle in the Sitnikov problem can move along the Oz axis with increasing amplitude, its motion may be chaotic, stable or unstable [1-7]. In the generalized Sitnikov problem, the barycenter of a rigid body can move along the Oz axis also. Thus, the article [8] describes the translational-rotational motion of a homogeneous rod where its barycenter moves along the normal to the plane of motion of the to primaries, whilst the rod itself rotates continuously around the normal, forming a constant angle of | with it (the manifold "gravitational propeller"). It is also shown that this manifold includes, as a special case, two types of motions of the rod. The first type is rotations with a constant angular velocity, which coincides with the angular velocity of the primaries. The second type is uneven rotations in the plane of motion of the two primaries. A manifold of motions also exists, where the

Received October 24, 2022 Accepted November 21, 2022

The research was carried out at the Moscow Aviation Institute with the financial support of the Russian Science Foundation, project no. 22-21-00560.

Pavel S. Krasilnikov krasil06@rambler.ru Albert R. Ismagilov arism8@mail.ru

Moscow Aviation Institute (National research university) Volokolamskoe sh. 4, Moscow, 125993 Russia

rod moves translationally along the normal, being directed along it. In a more a generalized setup, some translational-rotational motions of a rigid body are studied in the framework of the restricted three-body problem. A review of such studies until 1990 is contained in [9, 10]. The article [11] studies the positions of the relative equilibrium of a symmetric body when the satellite's barycenter remains at the libration point of the restricted three-body problem. A similar study was carried out for a gyrostat satellite in [12]. In [13, 14], a proof is provided of the existence and stability of special solutions for a dumb-bell satellite located at a triangular libration point of the circular restricted three-body problem. In [15, 16], some periodic modes of orbit-attitude behaviors are studied for spacecraft in the Earth-Moon system. In [17], the motion of a homogeneous rod in the field of attraction of two motionless attracting centers of equal masses, spaced from each other at a purely imaginary distance of 2ic, is investigated.

Also of note are studies concerned with the motion of a rod and a dumb-bell in a central force field. In [18], stationary orbit-attitude motions of a homogeneous rod in a central gravitational field are studied in greatest detail, and the types of motions are described: "the spoke", "the float", and "the arrow". The paper [19] considers the planar problem of the motion of an asymmetric dumb-bell in a central gravitational field, and studies the existence and stability of steady motions; the case of a symmetrical dumb-bell was considered in [20].

The aim of our paper is to investigate dumb-bell equilibria in the generalized Sitnikov problem for the case e = 0. It is assumed that the dumb-bell is oriented along the normal to the plane of motion of two primaries. To this end, we study the equations of equilibrium using the theory of bifurcations. The case e = 0 requires applying the methods of analyzing the Hill-type equation. This is carried out in the last section of the paper.

2. A special case of the translational motion of a dumb-bell, its power function

Let us investigate special motions of a symmetrical dumb-bell of mass 2m1 (m1 being the mass of a point located at the end of the dumb-bell) and length l in the gravitational field of attraction of two primaries of equal mass M, which rotate around their common barycenter O in an elliptic orbit with eccentricity e. The semimajor axis of the orbits is supposed to be a. Let Oxyz be a Cartesian coordinate system where the Ox axis lies in the plane of the orbits of the primaries, the Oz axis is directed along the normal to the plane of motion of these primaries, and the Oy axis complements the coordinate system to the right. It is easy to see that the dumb-bell can translate while keeping the orientation along the Oz axis. Indeed, due to the symmetry the principal vector of the forces is directed along the Oz axis and the principal moment relative to the barycenter of the dumb-bell is zero (see Fig. 1).

The main purpose of this paper is to investigate the translational motions of the dumb-bell along the Oz axis. It is easy to see that the potential energy of the problem has the form

Here y is the gravitational constant, m1 is the mass of one of the points of the dumb-bell, z is the coordinate of the barycenter of the dumb-bell,

(2.1)

2r(t)

a- (1 -e2) 1 + e ■ cos v'

Fig. 1

v is the true anomaly. Then the equation of motion of the dumb-bell along the Oz axis takes the form

z + Ym

z +

(r2(i) + (z + i)2)3/2 (r2(t) + (*-i)2)

+

z

2\3/2

0.

(2.2)

Obviously, z = 0 is a trivial dumb-bell equilibrium. The equations linearized in the vicinity of this equilibrium have the form

u i = u2,

v,2 = -2jm (V2(i) + j

2\ -5/2

r2(t) - -

(2.3)

u

where ui = z, u2 = Z.

3. Relative equilibria of a symmetrical dumb—bell in a circular

case

Consider the case e = 0. Let us assume, without loss of generality, that the sum of the masses 2m of two primaries is equal to one, and that the constant distance a is equal to one. This can be achieved by choosing ones for distance and mass.

As follows from the above, we have trivial equilibrium z = 0 where e = 0. Note that this equality does not exhaust all equilibria of a dumb-bell, since there are bifurcations with respect to the parameter l. Indeed, let us write the bifurcation equations:

(ffl dz

0,

crll dz2

0.

(3.1)

2

2

2

After calculating, we get

Z+2

2\ 3/2

r2+{z-~) | + (z-- ) [r2+ \Z + -

2

3/2

= 0,

r2 - 2 z +

r2 + z -

5/2

+ r2 - 2 z -

r2 + z +

5/2

(3-2)

0.

Here r = It. is easy to see that the first equality has a trivial solution 2 = 0. From the second equation we find the bifurcation value I = I* = at, z = 0.

To describe the entire set of solutions to Eq. (3.2), we write it in the variables x = z + y = z — The transformed equation admits three types of real solutions x = x(s), y = y(s) given in parametric form, provided that the inequality x(s)y(s) < 0 is fulfilled:

I. x = -s, y = s II. x = x(s), y = s < 0 III. x = -x(s), y = s > 0,

3/2

, , y/2 ( -4s4 - s2 + Vl6s8 + 24s6 + 9s4 + s2 .

= -i^TI- >0' SGR-

Here s is a real parameter. Returning to the original notation, we obtain two types of solutions satisfying the condition l > 0:

I. z = 0 II. z

x (s) s , _

+ 7, I = x(s) - s, s < 0.

22

0.3 n

0.2-

0.1-

-0.1-

-0.2-

-0.3 J

Fig. 2

The bifurcation diagram, which is a set of equilibrium curves indicating stable and unstable equilibria, is shown in Fig. 2. The Poincaré coefficient ^^ is positive at black points, and negative at white points, so stable equilibrium corresponds to black points and unstable equilibrium to white points. The bifurcation diagram is of fork type. Hatching marks areas where < 0.

2

2

2

2

2

l

l

l

l

2

2

2

2

Note that at the bifurcation point z = 0, I = ^ the second and third derivatives of the potential energy are equal to zero, but the fourth is greater than zero. Therefore, the bifurcation point is Lyapunov stable.

So, the equilibrium z = 0 is stable for I ^ ^ and unstable for I > Additional

equilibria z = 1 / 0 existing for I > ^ are Lyapunov stable. They consist of the pairs located symmetrically with respect to the origin. Note that the stability (instability) of equilibrium is considered with respect to perturbations that preserve the conditions y = x = 0. Using the energy integral

m1

f! 2

+ n(z) = C,

we draw phase portraits of oscillations for different values of I. Figure 3 shows the phase portrait of oscillations for the case I < The case I > ^ is shown in Fig. 4.

Fig. 3

Fig. 4

4. Translational motions of an asymmetric dumb—bell, equilibria

Let us also study the partial motions of an asymmetric dumb-bell, when two point masses m1 and m2 are located at the ends of a rod of length l, while m1 = m2. As in the symmetrical case, translational motions of the dumb-bell along the Oz axis are possible. The potential energy of the system has the form

n = -

2^mm1

2jmm2

l + +

Here m1 is the mass of the upper point and m2 is the mass of the lower point. Let us put A The equilibrium equation of an asymmetric dumb-bell is of the type

2 +

1 ( 4 + (*"2

3/2

+ A 2 -

1 (

4 + [Z+2

3/2

0.

_ 1112.

m.

(4.1)

This equation is invariant under the change of variables A ^ A , z ^ —z. Therefore, we can consider the case 0 < A < 1, l > 0 where z G (—x>, +rc>). Figures 5 and 6 show the equilibrium curves for different values of A.

Fig. 5

Fig. 6

It follows from Figs. 5 and 6 that the equilibrium curves have the form of oblique parabolas if z > 0. If z < 0, then the set of equilibria is a falling curve passing through zero. As A ^ 0, the upper parabola goes to infinity and the lower curve tends asymptotically to the straight line z = — This straight line corresponds to the position of the dumb-bell when its upper mass ml is at the origin, regardless of the length l, and the lower mass is m2 = 0 (the case of the classical Sitnikov problem).

Note that the vertices of parabolas are bifurcation points. Taken together, they form a continuous curve, which satisfies Eqs. (4.1) and the condition = 0:

3 (, + §)2

+

1

/

-A

3 (z - I)2

1

\

n n3/2

4 + ^-2) ) y

Calculations show that, the curve consisting of bifurcations points (I*, z) at. different. A is as follows (Fig. 7). Typical phase portraits of oscillations of the geometric center of an asymmetric dumb-bell are shown in Figs. 8 and 9.

5. Linear stability of a symmetrical dumb—bell in an elliptic problem

Let us describe some results on the linear stability of the trivial equilibrium z = 0 in the elliptic case. We have

2r = a(1 — e cos E),

Fig. 7. Curve of bifurcation parameters l*,z

Fig. 8

Fig. 9

where E is an eccentric anomaly. We assume, without loss of generality, that y = 1, a = 1, 2m = 1. The equation describing the linear oscillations of the barycenter of the dumb-bell takes the type of the Hill equation:

z + 8g(t, e, l) • z = 0, 0 <e< 1, l> 0, g(t, e, l) = ((1 - ecos E(t))2 + l2)-5/2 • ((1 - ecos E(t))2 - 2l2)

(5.1)

The dependence of E on time t is expressed by the Kepler equation

E — e sin E = n(t — t ).

As a consequence, we get

È = -----. (5.2)

(1 - e cos E) v 7

Here n = = 1 is the mean motion.

To study the stability of a trivial solution to Eq. (5.1), we use Lyapunov's theorems [21] (see also [22]). The following theorem gives sufficient conditions for instability.

Theorem 1. If the function 8g(t, e, l) is such that it can only take negative or zero values (not being identically zero), then the roots of the characteristic equation corresponding to this equation will always be real, one of them will be greater than one and the other less than one.

The condition for the function 8g(t, e, l) to be negative has the form (1 — ecos E)2 < 2l2. This implies the dumb-bell instability condition:

1 + e).

In the particular case e = 0, we have the previously obtained result: the equilibrium z = 0 is unstable for I > Obviously, the condition I > \pi gives instability for all e e [0, 1]. The following theorem yields sufficient conditions for stability.

Theorem 2. If the function 8g(t, e, l) in Hill's equation (5.1) takes only positive or zero values, with the inequality

t

1 2

0

t

T J g(t, e, (5.3)

being satisfied, then the roots of the characteristic equation are complex quantities equal in absolute value to one.

Here T = 2n is the period of the function g(t, e, l). It is easy to see that the condition for g(t, e, l) to be positive has the form

1 + e). (5.4)

Inequality (5.3) can be represented as

2n

f ((1 - e cos E)2 - 2l2) (1 - e cos E) 1 ,

2vr / ^---J—-----dE < -. 5.5

J ((1 - ecos E)2 +12)5/2 2

Analysis shows that conditions (5.4) and (5.5) are compatible only in a small neighborhood of the point ^e = 0, I = ^ j corresponding to a stable equilibrium in a circular problem, namely,

in a neighborhood (e, I) e [0, 0.01] x 0.7, ^ .

The cases I < — e), ^(1 — e) <1 < -^(1 + e) require special investigations. Further, we will consider Eqs. (5.1) and (5.2) together as a system, and solve it numerically to determine the fundamental matrix X(t) of Eq. (5.1). Then we will construct the monodromy matrix X(T), where T = 2n. The trace a(e, l) of the monodromy matrix X(2n) will determine the stability of the trivial equilibrium z = 0. If \a(e, l)| < 2, the solution z = 0 is stable. For the

case \a(e, l)\ =2, we need to check the simplicity of the elementary divisors of the monodromy matrix. If \a(e, l)\ > 2, the solution 2 = 0 is unstable.

For l e {0, 0.01, 0.184276695, 0.358553390, 0.532830085, 0.707106781, 0.707206781, 0.7658054249, 0.8244040687, 0.8830027125, 0.9416013562, 1.} and e e (0., 0.9999] we have calculated a(e, l) with an accuracy of e = 10_9. In these calculations, the step in e is 0.009999. The resulting curves are shown in Figs. 10-16. Figure 10 corresponds to the classical Sitnikov problem. From the graph in Fig. 11 one can see the instability interval at e > 0.9. Further, with an increase in the length l we see an increase in the width of the instability interval. The value

of the dumb-bell's length l, which gives instability for all e e (0, 1), is l = 0.983819. This result significantly refines the result obtained above by using the Lyapunov theorem.

2i

1-

a{e,l) 0

-1-

—2J

0.1 0.2 0.3 0.4 0.5 0.6 0.7

e

0.8 0.<i

I—¿ = 0.18431 Fig. 12

a(e,l) 0

-1-

—2J

I—¿ = 0.35861 Fig. 13

2~i

1-

a{e,l) 0-

0.1 01.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

e

-1-

-2J

I—¿ = 0.53281 Fig. 14

2-1

1-

a(e,l) 0

-1-

—2J

XI 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

e

"Mi-

Fig. 15

i).

.1 0.2

0.3 0.4

0.5 0.6

e

0.7 0.8 0.9

/ = 0.7072 / = 0.8830

/ = 0.7658 / = 0.9416

/ = 0.8244 ■1 = 1.

2i

1-

a(e,Z) 0

-1-

—2J

0.91 0.93 0.95 0.97 0.99 e

• / = 0.9777 ° / = 0.9787 • / = 0.9797

- / = 0.9807 x / = 0.9817 ♦ / = 0.9827

■ / = 0.9837

Fig. 16

Fig. 17

6. Conclusion

We have studied the presence of equilibria in the generalized Sitnikov problem, depending on the different lengths of the dumb-bell in the symmetric and asymmetric cases, and revealed the existence of a bifurcation value l*. We have investigated bifurcation diagrams for the symmetric and asymmetric cases in the circular problem. Phase portraits have been constructed for l < l* and l > l* in the case e = 0.

In the elliptic case, sufficient conditions for the linear instability and stability of a symmetrical dumb-bell have been obtained by using Lyapunov's theorems. By numerical calculations, we obtained stability intervals for eccentricities e e (0, 1) and various dumb-bell lengths l. Then

we found the smallest value of the dumb-bell length, l = 0.983819 at which an instability of the trivial equilibrium takes place for Ve G (0, 1).

Conflict of interest

The authors declare that they have no conflict of interest.

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