NEW STACKED CENTRAL CONFIGURATIONS FOR THE PLANAR 6-BODY PROBLEM. Gokul Bhusal, Hamas Tahir, Dr. Xie. 8/03/2017 Summer Research 2017

Transcription

1 NEW STACKED CENTRAL CONFIGURATIONS FOR THE PLANAR 6-BODY PROBLEM. Gokul Bhusal, Hamas Tahir, Dr. Xie. 8/03/2017 Summer Research 2017

2 Outline of the presentation: Celestial mechanics, Central configuration, Two bodies, Three bodies, Stacked central configuration, Six body Problem, Special Cases, Future Plans.

3 Celestial mechanics: Celestial mechanics is the branch of astronomy that deals with the motions of celestial objects. Historically, celestial mechanics applies principles of physics (classical mechanics) to astronomical objects, such as stars and planets, to produce ephemeris data. As an astronomical field of study, celestial mechanics includes the sub-fields of Orbital mechanics (astrodynamics), which deals with the orbit of an artificial satellite, and Lunar theory, which deals with the orbit of the Moon. Newtonian N- body problem is a significant part of celestial mechanics.

4 What is Central configuration? A central configuration is a special arrangement of point masses interacting by Newton's law of gravitation with the following property: 1.Gravitational acceleration vector produced on each mass by all the others should point toward the center of mass, 2. proportional to the distance to the center of mass.

5 Significance of central configuration: Central configurations is one of the oldest problems and now it is one of the problems of the twentieth century. Solving this problem has been motivated by the desire to understand the motions of the Sun, Moon, planets and the visible stars. The idea is also important that we send satellites in space, we need to consider that where do they collide or expand and how do they form the central configuration.

6 The Basic C.C equations: The Equation for central configuration is: where rij= qi qj is the Euclidean distance between particles i and j (the gravitational constant is normalized to G= 1). c is center of mass.

7 Three Body Problem: The classical Newtonian three-body gravitational problem occurs in Nature exclusively in an astronomical context and was the subject of many investigations by the best minds of the 18th and 19th centuries. The first and simplest periodic exact solution to the three-body problem is the motion on collinear ellipses found by Euler (1767)

8 Three Body Problem: The first nontrivial examples of central configurations were discovered by Euler in 1767, who studied the case N=3, d=1, that is, three bodies on a line (Euler (1767)). When two masses are equal, one can get a central configuration by putting an arbitrary mass at their midpoint (a centered 2-gon). For three unequal masses it is not obvious that any central configurations exist. But Euler showed that, in fact, there will be exactly one equivalence class of collinear central configurations for each possible ordering of the masses along the line..

9 Thee body problem: Lagrange found next example in the planar three-body problem N=3,d=2. Remarkably, an equilateral triangle is a central configuration, not only for equal masses, but for any three masses m1,m2,m3. Moreover, it is the only noncollinear central configuration for the three-body problem (Lagrange (1772)).

10 Stacked Central Configuration: A stacked central configuration in the n body problem is one that has a proper subset of the n bodies forming a central configuration. Hampton (2005) provides a new family of planar central configurations for the five-body problem with an interesting property: two bodies can be removed and the remaining three bodies are already in a central configuration. Such configurations are called stacked central configurations. We want to see whether the stacked Central configuration exist or not in a six body configuration.

11 Six Body Problem: In our case the triangle 456 is fixed and the triangle 123 is moved up and down to study the possible set of configurations. m1 := (!"#,b), m2:= (#"!, b), m3:= (0,a), % % m4:= (0,0), m5:= (1, 3), m6:= (-1, 3)

12 For the planar central configurations instead of working with Newtonian Equation, we shall use the Dziobek equations (see Hagihara 1970, p. 241) for 1 i < j n, where Rij = 1/rij^3 and ijk = (ri rj ) (ri rk). As usual denotes the cross product of two vectors. For example: For a three body problem, f12 = m3 R13 R = 0 For the case of Six bodies, we will have a set of fifteen Equations.: f12, f13, f14, f15, f16, f23, f24, f25, f26, f34, f35, f36, f45, f46, f56

13 THEOREM 1: Assume that the point r4,r5 and r6(with positive masses m4,m5 and m6) are at the vertex of an equilateral triangle whose sides have length 2. And the points r1, r2 and r3(with positive masses m1, m2 and m3) is also a vertex of an equilateral triangle with no fixed size. The triangle 123 can expand and moves up and down with different value of a and b. Then the mass m1 must be equal to m2 and mass m5 must be equal to m6. Proof: The equation f12 = m3 R13 R23 Δ123 + m4 R14 R24 Δ124 + m5(r15 R25) Δ125 + m6 R16 R26 Δ126 = 0 f12=0+0+ (m5 m6)(r15 R16) Δ 125 = 0 Thus m5 = m6 In the same way the equation f56 = 0 goes over to (m1-m2)(r15-r16) Δ 156 = 0: Therefore, m1 = m2:

14 The terms in f34 = 0, cancel out with each other and the other twelve equations are as follows: F14

15 The six equations we are left are::

16 We use equation f13 = 0, to get, the value of m4 in term of m5. We use equation f45 = 0, to get the value of m1 in term of m3. We use equation f36 = 0, to get the value of m3 in term of m5.

17 The other three equations were named g11.g12 and g13 and plot in the graph:

18 The common Graph of g11, g12 and g15:

19 Special Cases: The centroid of Fixed triangle Δ456 is (0, D % ). When Both Triangle have same centroid: When both the triangle have same centroid, we get the relation: a = 2 3-2b

20 Theorem 2: Assume that the centroid of both triangle 123 and 456 have same centroid, and a>b then mass m1 must be equal to m3. Proof: F f45=m1{(r14-r15) Δ,451+(R24- R25) Δ 452}+m3(R34-R35) Δ 453=0 For the case of same centroid, (R24-R25) is always zero. And, (R14-R15) Δ,451=(R34-R35) Δ 453 So, m1=m3

21 Theorem 3: Assume that the centroid of both triangle 123 and 456 have same centroid, and a>b then mass m4 must be equal to m5. F f23=m5{(r25-r35)δ,235+(r26- R36) Δ 236}+m4(R24-R34) Δ 234=0 (R25-R35) is always equals to zero. So, m4=m5

22 Range of a Less than 1.15 The Results: C.C exist or not Greater than 4.0

23 For the Case of Hexagon: Hexagon is a special case on this family of central configuration. In hexagon all masses are equal. And central configuration exist. All the six equation is zero in the case of hexagon

24 Special case: Theorem 4: Assume that the triangle 456 is fixed in x-y plane, The triangle 123 is moving up and down along y-axis in the range of: 0<b< 3. The mass m1 and m2 move along the line r46 and r45 respectively. The lines r12 and r56 are parallel so that r12= r24. Proof: f14 = m2 R12 R m3 R13 R m5 R15 R m6 R16 R f14=0+(+ve)+(+ve)+0 >0 So C.C doesn t exist.

25 Future plans: Try to Find all possible Central configuration. Type all the results using LaTex.. Try to learn all the graph.

26 Acknowledgement: MF82dWRKQnJJQlU Thank you Dr. Xie for your time, support, and advice. Supported by the Wright W. and Annie Rea Cross Mathematics, Summer Undergraduate Research Scholarship.

27 Thank you so much for your attention:

28 Thank you so much for your attention: