2012
DOI: 10.1088/0951-7715/25/2/343
|View full text |Cite
|
Sign up to set email alerts
|

Four-body co-circular central configurations

Abstract: We classify the set of central configurations lying on a common circle in the Newtonian four-body problem. Using mutual distances as coordinates, we show that the set of four-body co-circular central configurations with positive masses is a two-dimensional surface, a graph over two of the exterior side-lengths. Two symmetric families, the kite and isosceles trapezoid, are investigated extensively. We also prove that a co-circular central configuration requires a specific ordering of the masses and find explici… Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
1
1
1

Citation Types

2
62
0

Year Published

2013
2013
2024
2024

Publication Types

Select...
10

Relationship

0
10

Authors

Journals

citations
Cited by 49 publications
(64 citation statements)
references
References 25 publications
(93 reference statements)
2
62
0
Order By: Relevance
“…We also believe that similar results can be obtained taking the same structure with two equal co-circular central configurations (see Cors and Roberts 2012), instead of two equilateral triangles.…”
Section: Spatial Central Configurations For the 7-body Problemsupporting
confidence: 74%
“…We also believe that similar results can be obtained taking the same structure with two equal co-circular central configurations (see Cors and Roberts 2012), instead of two equilateral triangles.…”
Section: Spatial Central Configurations For the 7-body Problemsupporting
confidence: 74%
“…The characterization of the convex central configurations with an axis of symmetry and the concave central configurations of the 4-body problem when the masses satisfy that m 1 = m 2 ̸ = m 3 = m 4 is done inÁlvarez and Llibre [6]. Results on the co-circular 4-body problem can be found in [14].…”
mentioning
confidence: 99%
“…Thus, the angular velocity is (cf. equations (11)), α 2 (θ) = − ∂U ∂θ 1 / sin θ 1 cos θ 1 = n j=2 1 sin 3 d ij sin θ 1 cos θ j − sin θ j cos θ 1 cos(j 2π n − 2π n ) sin θ 1 cos θ 1 = n j=2 1 − cos(j 2π n − 2π n ) sin 3…”
Section: Remark 12mentioning
confidence: 99%