3body problem, restricted & planar & circular

(1)

3body problem, 3body problem,

restricted & planar restricted & planar

& circular

(2)

ω =

(

⁰⁰n

)

^v=

(

^˙x^˙y˙z

)

P

B

r

₁

r

₂

y x

¨x−2 n ˙y=∂U

∂ x

¨y +2 n ˙x=∂U

∂ y

¨z=∂U

∂z U =1

2 n²(x²+ y²)+μ₁

r₁ +μ₂

r₂ Pseudo− potential

Hill’s equations.

1

2( ˙x²+ ˙y²)=U +C

2 Jacoby constant

¨r +2( ω × v)+ ω ×( ω × r)=−∇ V

-Mass of test body=0

-Second body on circular orbit -Planar (no inclination)

Rotating

reference

frame ω=n

(3)

Normalized units:

G=1 G (m₁+m₂)=1 μ= m₂

m₁+m₂ G μ₁=G m₁=1−μ=x₂

μ₂=G m₂=μ =x₁ x₁=μ= m₂

m₁+m₂ x₂=1−x₁ a=1 n=

√

^G⁽^m¹⁺^a³^m²⁾ ⁼¹

˙x ¨x− ˙x 2 n ˙y= ˙x ∂U

∂ x

˙y ¨y + ˙y 2 n ˙x= ˙y ∂U

∂ y

˙x ¨x + ˙y ¨y = ˙x ∂U

∂x + ˙y ∂U

∂ y

∫

t₀ t₁

( ˙x ¨x+ ˙y ¨y )dt =

∫

t₀ t₁

˙x ∂U

∂ x dt +

∫

t₀ t₁

˙y ∂U

∂ y dt

⇒ 1

2 ( ˙x²+ ˙y²)∣_x^x_0,^1,_y^y¹₀ =

∫

x₀ x1

∂U

∂ x dx+

∫

y₀ y1

∂U

∂ y dy 1

2 x˙²+ ˙y²=U +C 2

(4)

Hamiltonian approach

L=T +V =1

2 ( ˙x²+ ˙y²)+U (x , y )+U_c(x , y) L=1

2 ( ˙x²+ ˙y²)+

(

^1−μ^r1

)

⁺ ^r^μ2

+1

2(x²+ y²)+x ˙y− y ˙x

The Lagrangian of the system in the rotating reference frame is:

The equations of motions are:

d dt

∂ L

∂ ˙q_i− ∂ L

∂q_i=0

The Hamiltonian is:

H =

∑

i

p_iq˙_i−L (q_i, ˙q_i, t ) where p_i=∂ L

∂ ˙q_i p_x=∂ L

∂ ˙x= ˙x− y p_y=∂ L

∂ ˙y = ˙y + x H = p_x ˙x+ p_y ˙y − L=

x˙²− ˙x y+ x ˙y + ˙y²−1

2( ˙x²+ ˙y²)−U ( x , y)−1

2(x²+y²)−x ˙y + y ˙x=

1

2 ( ˙x²+ ˙y²)−U ( x , y)−1

2 (x²+ y²)=1

2 v²− ¯U ( x , y ) The Hamiltonian is the Jacoby integral

(5)

To go back to Hill’s equations from the Hamiltonian, Hamilton equations must be used:

p ˙

_x

=− ∂ H

∂ x

˙x= ∂ H

∂ p

_x

p ˙

_x

= ∂ L

∂ x = ˙y + ∂ U

∂ x p ˙

_y

= ∂ L

∂ y =− ˙x + ∂ U

∂ y

¨x− ˙y= ˙y+ ∂ U

∂ x

¨y + ˙x= ˙x + ∂ U

∂ y

¨x−2 ˙y= ∂ U

∂ x

¨y +2 ˙x= ∂ U

∂ y

(6)

x₁= m

M x₂= m

M (1−x₁) ⇒ x₁(M +m)=m ⇒ x₁= m

M +m=μ B

S P

r₁ r₂

x

y

μ

r₁

1-μ

¨x−2 ˙y= x−(1−μ)(x +μ)

r₁³ −μ (1−μ−x)(−1) r₂³

¨y +2 ˙x= y−(1−μ) y

r₁³ −μ y

r₂³ r₁=

√

⁽^{x +μ)}²⁺ ^y²

r₂=

√

⁽^x−1+μ)²⁺ ^y²

U (x , y )=1

2 (x²+ y²)+(1−μ) r₁ +μ

r₂ W ( x , y )=−U (x , y )

Hill’s equations in

normalized units

(7)

Lagrangian equilibrium points.

¨x=0 ˙x=0

¨y=0 ˙y=0 ∇ U =−∇ W =0

∇ U =r −(1−μ) r₁

r₁³ −μr₂ r₂³ r=(1−μ)r₁+μr₂

r=

(

^xy

)

^r¹⁼

(

^x+μ^y

)

^r²⁼

(

^x−1+μ^y

)

(1−μ)r₁

(

¹⁻¹^r1

3

)

^+μ^r²

(

¹⁻¹^r2

3

)

⁼⁰

Triangular Lagrangian points

L

₄^and

L

₅^{: r}₁^{= r}₂⁼¹

Solving equation along x-axis

L

₁

, L

₂

, L

₃

:

x=−1−μ+7

12 μ +O (μ³) L₃ x=1−μ±

(

^μ3

)

¹³⁺^{O (μ}

2

3) L_1, L₂

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L₄⇒

(

¹²

^√

²³^−μ

)

^⋅^a

L₅⇒

(

⁻¹² ^−μ

^√

²³

)

^⋅^a

Around these points move Trojan asteroids.

L₁⇒

(

^1−μ+⁰

⁽

^μ³

⁾

¹³

)

^⋅^a ^L²^⇒

(

^1−μ−⁰

⁽

^μ³

⁾

¹³

)

^⋅^a

These

lagrangian points define Hill’s sphere

L₃⇒

(

⁻¹⁻⁰⁵¹² ^μ

)

^⋅^a

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0 velocity curves: They cannot be crossed by a body!! They define the regions were stable motion is

possible

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Different curves for different value of the Jacoby

constant C = 2U

(11)

Mars, Jupiter, Uranus and Neptune have

Trojan asteroids.

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Hill’s lobe: where the gravity of the body dominates respect to the central (more massive) body.

r

_Hill

= 1− μ

₂

+ ( ^μ 3

²

)

1/ 3

r

_Hill

= ( ^m

¹

^m ⁺

²

^m

²

^⋅ ¹ ³ )

^{1 /3}

^⋅ ^a

r

_Hill^Jupiter

=0.355 au r

_Hill^Earth

=0.01 au r

_Roche

= R

_M

( ² ^ρ ^ρ

^Mm

)

(1/3 )

R

_M

primary body

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Cataclismic variables.

(14)

(15)

From Hsieh & Haghighipour, 2016, Icarus, Volume 277, p. 19-38

Most comets have T < 3 while most asteroids

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P67: the comet

explored by ROSETTA mission

The comet is made of two pieces (lobes) which

independently formed in the outer Kuiper Belt.

(17)

Formed in the Kuiper Belt, due to

repeated close encounters with Neptune

it was injected in a elliptic orbit crossing

the inner regions of the SS.

(18)

Close

images of the comet surface.

(19)

A peculiar Plutino triple system: Lempo.

Hiisi and Praha.

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Trojan motion: effects of mass growth and migration of the planet. Trojan capture.

¨ϕ+ 27

4 μn²_Pϕ=0

With n_P mean motion of the planet and φ the difference between between the

longitude of the planet and that of the body.

This equation can be re-written in Hamiltonian form:

H=1

2 a_P² ϕ˙²+1

2 ω a_Pϕ² Where the libration frequency is given by

ω²=27

4 μn_P² μ= m_P M_s+m_P

To test what it happens when we change either the semi-major axis or the mass of the planet we

introduce an adiabatic invariant.

J =

∫

period

p dq Assuming that in a period the

relevant properties of the planet do not change i.e. a_P, A and ω are

costant over a period T. Since:

It is the hamiltonian of a harmonic oscillator, the solution is

ϕ (t )= A

2 cos(ω t+α) Where A is the libration

amplitude (the ½ is by definition of amplitude)

p=a_P ˙ϕ q=a_P ϕ

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the integral becomes J =

∫

0 2 π

a_P ˙ϕ a_Pd ϕ=

∫

0 T

a_P² A²

4 ω²sin²( ωt +α)dt J =a_P² A²

4 ω²

∫

0 2 π ω

sin²(ωt +α)dt Recalling that

∫

^sin²⁽^{ax )dx=}₂^x⁻_{4 a}¹ sin(2 a x )

we get J =a_P² A²

4 ω²2 π

2 ω =π

4 A²a_P² ω=

π

4 A²a²_P

√

⁽ ²⁷⁴ ^m^P^m⁺^P^M^s⁾

√

⁽^G(m^a^P⁺^P³ ^M^s⁾⁾⁼

π

4 A²

√

⁽²⁷⁴ ^G)m^{1 /2}^P ^a^{1 / 2}^P

The conservation of J in presence of adiabatic variations of both a_P and m_P leads to:

A_f

A_i =

(

^a^a^{P , f}^{P ,i}

)

⁽^{1/ 4 )}

(

^m^m^{P , f}^{P ,i}

)

⁽^{1/ 4 )}