Central Configurations and Action Minimizing Orbits in Kite Four-Body Problem

Abstract

In the current article, we study the kite four-body problems with the goal of identifying global regions in the mass parameter space which admits a corresponding central configuration of the four masses. We consider two different types of symmetrical configurations. In each of the two cases, the existence of a continuous family of central configurations for positive masses is shown. We address the dynamical aspect of periodic solutions in the settings considered and show that the minimizers of the classical action functional restricted to the homographic solutions are the Keplerian elliptical solutions. Finally, we provide numerical explorations via Poincaré cross-sections, to show the existence of periodic and quasiperiodic solutions within the broader dynamical context of the four-body problem.

Restricted Concave Kite Five-Body Problem

“…Tey also discussed one convex and two concave cases of central confguration in detail. Benhammouda et al [17] studied the central confguration of the kite four-body problems. Tey considered two diferent types of symmetrical confgurations.…”

“…Te whole system rotates with constant angular velocity, the centrifugal force compensates for the Newtonian attraction, and the fve bodies are in equilibrium 2 Advances in Astronomy in such a rotating system, the so-called "relative equilibria solutions." Te central confguration of this particular was derived in [17], and we will give a brief review of their results with minor improvement. If we denote by r j the position of the body with mass m j , and by r ij � ‖r j − r j ‖, the distance between the body with mass m j and the body with mass m i , then the algebraic system of equations that must be satisfed for the bodies to be in noncollinear central confguration is…”

“…| represent the area of the triangle determined by the sides ‖r i − r j ‖ and ‖r i − r k ‖. After eliminating redundant equations due to the symmetries of the problem, i.e., f 13 � f 24 ≡ 0, f 12 � f 23 , f 14 � f 34 , and m 1 � m 3 , [17], we get the following two equations:…”