Dynamics of Kuiper belt objects Yeh, Lun-Wen

Outline Motivations Motivations Sun-Neptune-Eris-Particle system Sun-Neptune-Eris-Particle system Numerical method Numerical method Results Results Next Step Next Step

Motivations Planet migration model is one most popular model for the formation of Kuiper belt. (Malhotra 1993,1995; Hahn & Malhotra1999, 2005; Gomes 2003, 2004; Levison & Morbidelli 2003; Tsiganis et al., 2005) Planet migration model is one most popular model for the formation of Kuiper belt. (Malhotra 1993,1995; Hahn & Malhotra1999, 2005; Gomes 2003, 2004; Levison & Morbidelli 2003; Tsiganis et al., 2005) Sun + 4 giant planets + numerous particles. Sun + 4 giant planets + numerous particles. Because of the computing source the interaction between particles was usually neglected. Because of the computing source the interaction between particles was usually neglected. Considering the influence of larger planetesimals in the model is a way to approach the complete model. Considering the influence of larger planetesimals in the model is a way to approach the complete model.

At present there are three dwarf planets by the definition of IAU. (Ceres: D~940km, Pluto: D~2300km, Eris: D~ At present there are three dwarf planets by the definition of IAU. (Ceres: D~940km, Pluto: D~2300km, Eris: D~ 2400km) 2400km) Some papers have considered the effect of Pluto on the Some papers have considered the effect of Pluto on the Plutinos. (Yu & Tremaine 1999; Nesvorny & Roig 2000) Plutinos. (Yu & Tremaine 1999; Nesvorny & Roig 2000) We would like to know the gravitational influence of Eris on the Kuiper belt objects. We would like to know the gravitational influence of Eris on the Kuiper belt objects.

a=67 AU, e=0.44, i=44 0. a=67 AU, e=0.44, i=44 0. T=557 years. T=557 years. mass= M E, diameter=2400 km. mass= M E, diameter=2400 km.

Although we don’t know where Eris was formed and how long Eris has stayed in the present orbit, but Eris should have some influence on the Kuiper belt. Although we don’t know where Eris was formed and how long Eris has stayed in the present orbit, but Eris should have some influence on the Kuiper belt. The coagulation model (ex: Kenyon et al. 2007) implies that Eris-size objects were formed in a more massive disk (~30M E ) with low e and i (e, i~10 -3 ) planetesimals. The coagulation model (ex: Kenyon et al. 2007) implies that Eris-size objects were formed in a more massive disk (~30M E ) with low e and i (e, i~10 -3 ) planetesimals. Under the planet migration model (ex: Hahn & Malhotra 2005), the scattered disk objects originated from low e, low i and smaller a orbit due to Neptune’s migration. The migration timescale of Neptune is about 10 Myr. Therefore Eris-size objects may leave disk between 10 Myr to 4.5 Gyr. Under the planet migration model (ex: Hahn & Malhotra 2005), the scattered disk objects originated from low e, low i and smaller a orbit due to Neptune’s migration. The migration timescale of Neptune is about 10 Myr. Therefore Eris-size objects may leave disk between 10 Myr to 4.5 Gyr.

The mass of Eris to the total mass of KBOs is about 0.1 (0.0028/0.03; Soter 2007) hence Eris may still collide and scatter with many other KBOs. The mass of Eris to the total mass of KBOs is about 0.1 (0.0028/0.03; Soter 2007) hence Eris may still collide and scatter with many other KBOs.

Hahn & Malhotra 2005

The main belt is not eccentricity enough. The main belt is not eccentricity enough. Absence of higher order resonance. (ex: 5:2) Absence of higher order resonance. (ex: 5:2) High inclination KBOs are deficient. High inclination KBOs are deficient. Simulation: # of (3:2+2:1) / # of MB~2. Simulation: # of (3:2+2:1) / # of MB~2. Observation: # of (3:2+2:1) / # of MB~0.06. Observation: # of (3:2+2:1) / # of MB~0.06.

Hahn & Malhotra 2005

We consider the Sun-Neptune-Eris-particle system in the late stage of planet migration model (the last several hundreds Myrs) and assume Eris at present orbit. We consider the Sun-Neptune-Eris-particle system in the late stage of planet migration model (the last several hundreds Myrs) and assume Eris at present orbit.

Sun-Neptune-Eris-Particle system S N E P

η ξ y x nt S N E P

For particles:

For Eris:

For particles:

Jacobi constant For Eris:

We We used Runge-Kutta-Fehlberg method to solve the differential equations. This method combines 4th and 5th order Runge-Kutta method to choose a suitable step size for each step. We did some modification in the step size determination in order to have a more significant criterion to choose the step size. Numerical method

yiyi y i+1 t y y = f(t,y) hihi RK4 RK5

yiyi y i+1 t y y = f(t,y) hihi RK4 qh i ∆ □ X R ≡ | ∆ | / (| X |+ | □ |) < TOL

The initial conditions: The initial conditions: : a=30.07AU (1); e=0.0; θ=0.0 Neptune: a=30.07AU (1); e=0.0; θ=0.0 : a=67.73AU (2.252); e=0.44; ω=0.0; θ=0.0 Eris: a=67.73AU (2.252); e=0.44; ω=0.0; θ=0.0 : 50 test particles in 3:2 mean motion resonance: a=39.4 ± 0.3 AU (1.31 ± 0.01); e=0.0,0.1,0.2,0.3,0.4; a=39.4 ± 0.3 AU (1.31 ± 0.01); e=0.0,0.1,0.2,0.3,0.4; randomly assign ω,θ. randomly assign ω,θ. : 50 test particles in 2:1 mean motion resonance: a=47.7 ± 0.3 AU (1.59 ± 0.01); e=0.0,0.1,0.2,0.3,0.4; a=47.7 ± 0.3 AU (1.59 ± 0.01); e=0.0,0.1,0.2,0.3,0.4; randomly assign ω,θ. randomly assign ω,θ. : 50 test particles in main belt: a=40.3 – 46.8 AU (1.34 – 1.56); e=0.0,0.1; a=40.3 – 46.8 AU (1.34 – 1.56); e=0.0,0.1; randomly assign ω,θ. randomly assign ω,θ.

t 0 =0, t f =10 6 T N (T N =165 yr) t 0 =0, t f =10 6 T N (T N =165 yr) ∆t out =10 2 T N ∆t out =10 2 T N ∆t max =10 -1 T N, ∆t min = T N ∆t max =10 -1 T N, ∆t min = T N TOL= TOL= The criterions for stopping the program: The criterions for stopping the program: when r 1000AU when r 1000AU when the particle collides with Neptune when the particle collides with Neptune when the particle collides with Eris when the particle collides with Eris when t=t f when t=t f

Results

a=1.30, e=0.0, curlypi=2.18, theta=6.23

3to2: T S =438 yr 2to1: T S =845 yr

remaining total # / initial total # = 26/50

remaining total # / initial total # = 42/50

remaining total # / initial total # = 48/50

There two reasons cause the number of surviving particles of 3 :2 smaller than that of 2:1. One is that the synodic period of 3:2 is smaller than that of 2:1. Another is that the q=30 AU line in the a-e diagram crosses more region in 3:2 than that in 2:1 mean motion resonance. There two reasons cause the number of surviving particles of 3 :2 smaller than that of 2:1. One is that the synodic period of 3:2 is smaller than that of 2:1. Another is that the q=30 AU line in the a-e diagram crosses more region in 3:2 than that in 2:1 mean motion resonance.

remaining total # / initial total # = 116/150

remaining total # / initial total # = 144/150

# in 3to2: 13 # in MB: 57 # in 2to1: 21 (3to2+2to1)/MB = 0.6

Construct three dimensional model. Construct three dimensional model. Run more particles to enhance the statistical significance. Run more particles to enhance the statistical significance. From estimation of Hahn 2005 there are ~ 10 5 KBOs in the Kuiper belt. How many particles I will use depends on the computing sources. From estimation of Hahn 2005 there are ~ 10 5 KBOs in the Kuiper belt. How many particles I will use depends on the computing sources. Consider more Eris-size objects. How many? How to choose initial conditions? I am still thinking…….. Consider more Eris-size objects. How many? How to choose initial conditions? I am still thinking…….. Next Step Thank you ^^….. Any comment or suggestion?