Transcript Orbital dynamics and restricted 3

Planet Formation
Topic:
Orbital dynamics and
the restricted 3-body problem
Lecture by: C.P. Dullemond
Objects of the Solar System
Gas Giants
Rocky Planets
0.300.47 AU
e = 0.2
Venus Earth Mars
0.72 AU
1 AU
1.381.67 AU
e = 0.1
Asteroids
& TNOs
DwarfPlanets
Planets
Mercury
Jupiter
Saturn
4.95-5.46 AU
e = 0.05
9.05-10.1 AU
e = 0.05
Uranus Neptune
18-20 AU
e = 0.05
30 AU
Ceres
Pluto
2.552.99 AU
e = 0.08
29.648.9 AU
e = 0.25
Asteroid Belt
(2.0-3.4 AU)
1 AU
Ice Giants
10 AU
Logarithmic distance scale
+ Haumea,
Makemake,
Eris
Kuiper Belt
(30-50 AU)
Kepler orbits, eccentricity
Kreisbahn (Mplanet << M*):
WK =
torbit =
GM *
(Kepler Frequency)
a3
2p
WK
uK = WK a =
v(t) = WK t
Planet
Sonne
v
a
GM *
a
(Kepler
Velocity)
y
(„True anomaly“)
x
X(v) = acos(v)
Y (v) = asin(v)
Kepler orbits, eccentricity
Elliptic orbit (Mplanet << M*):
WK =
torbit =
GM *
(Kepler Freq.)
a3
2p
WK
Planet
r
Apoapsis
(Aphelion)
a („Semi-major axis“)
e („eccentricity“)
M(t) = WK t
(„Mean anomaly“)
M = E - esin(E)
cos(E) - e
cos(v) =
1- ecos(E)
a(1- e 2 )
r(v) =
1+ ecos(v)
Sonne
ea
a
v
Periapsis
(Perihelion)
Fokus
y
x
E(t) („Eccentric anomaly“ , solve numerically)
v(t) („True anomaly“)
X(v) = r(v)cos(v)
Y (v) = r(v)sin(v)
Kepler orbits, eccentricity
Elliptic orbit (Mplanet << M*):
WK =
torbit =
GM *
(Kepler Freq.)
a3
2p
WK
Planet
r
Apoapsis
(Aphelion)
ea
a
v
Periapsis
(Perihelion)
Fokus
Total energy:
Utot = Ukin +Upot º
Sonne
1 2 GM
GM
v =2
r
2a
Note: Total energy depends only on a!
Angular momentum:
L = GM*a (1- e2 )
Note: eccentricity thus leads to an
angular momentum deficit.
y
x
Orbital elements
Compared to the
equatorial plane
of the solar system
(or exoplanetary
system), and
compared to a
reference direction,
you can uniquely
orient the ellipse.
Semi-major axis a,
eccentricity e and
true anomaly ν,
together with
orientation (i, ω, Ω)
are: orbital elements.
From: Wikipedia: http://en.wikipedia.org/wiki/File:Orbit1.svg
Guiding-center and epicyclic motion
If an orbit is almost circular, then we can describe this elliptic orbit
as a circular orbit with epicyclic motion superposed on it.
X(v) = r(v)cos(v)
Y (v) = r(v)sin(v)
e=0.3
ˆ = r(v)cos(v - M )
X(v)
Yˆ (v) = r(v)sin(v - M )
e=0.3
Guiding-center and epicyclic motion
When using another orbital frequency as the reference frame than
the orbital frequency of the particle, this epicyclic motion looks like:
X(v) = r(v)cos(v)
Y (v) = r(v)sin(v)
e=0.3
ˆ = r(v)cos(v -1.1M )
X(v)
Yˆ (v) = r(v)sin(v -1.1M )
e=0.3
Guiding-center and epicyclic motion
For small eccentricity
the epicycle becomes
an ellipse.
As we will see (and use)
later: these small
deviations from Kepler
can be described as
velocity disturbances
Δv.
guiding
center
epicycle
Even kids are familiar with this... ;-)
What is the „restricted 3-body problem“?
• Three bodies, one of which (M3) is a „test particle“:
M3 <<< M2 < M1
• Bodies 1 and 2 only feel each other‘s gravity and thus
perfectly follow a Kepler orbit.
• Assume that body 1 and 2 are in perfect circular orbits
• Put the center of coordinate system at center of mass
• Body 3 feels bodies 1 and 2.
• Resulting motions:
–
–
–
–
Some orbits are stable
Some orbits are unstable (body 3 gets ejected)
Some orbits are chaotic: Chaos theory!
Chaotic orbits are unpredictable on the long run.
Equations of motion for test particle
Remember from chapter „Turbulence“ Section „Magnetorotational
instability“ the equation of motion of a test particle in a rotating frame:
x - 2Wy = 3
GM
x + fx
3
r0
y + 2Wx = f y
(from chapter
„Turbulence“)
Now we do the same (though now we put x=0 at the center of
mass of the entire system), we drop the fx and fy forces but now
we include the forces of both the star and the planet.
¶Feff
x - 2Wy = ¶x
¶Feff
y + 2Wx = ¶y
Coriolis forces
Gravity and
centrifugal forces
With the „effective potential“ given by:
Feff
GM1 GM 2 1 2 2
=- W r
r - r1 r - r2 2
Exercise: re-derive these equations.
Effective potential, Lagrange points
r1
r2
r1 M 2
=
r2 M1
L4
Effective potential in the
co-rotating frame:
L3
L1
L5
Gravitational
potential
centrifugal
kinetic
energy
Example: M2/M1=0.1
L2
Effective potential, Lagrange points
r1
r2
r1 M 2
=
r2 M1
L4
Effective potential in the
co-rotating frame:
L3
L1
L5
Gravitational
potential
centrifugal
kinetic
energy
Example: M2/M1=0.01
L2
Jacobi‘s Integral
The full 3-D set of equations is:
¶Feff
x - 2Wy = ¶x
¶Feff
y + 2Wx = ¶y
¶F eff
z=¶z
now multiply by:
x
y
z
and add them all up:
¶Feff
¶Feff
¶Feff
xx + yy + zz = xyz
¶x
¶y
¶z
Jacobi‘s Integral
¶Feff
¶Feff
¶Feff
xx + yy + zz = xyz
¶x
¶y
¶z
This can be integrated once, to obtain:
1
2
2
2
2
x
+
y
+
z
(
) = -Feff + C
Traditionally the constant C is written as -½CJ:
x 2 + y2 + z 2 = -2Feff -CJ
CJ is called Jacobi‘s constant or Jacobi‘s Integral of motion.
For the restricted 3-body problem it is the only integral of motion,
i.e. there exist no closed-form solutions.
Zero-velocity curves / surfaces
CJ = -2Feff - ( x + y + z
2
2
2
)
>0
<0
Jacobi‘s constant is some kind of energy, sometimes called
„Integral of relative energy“. It is the rotational equivalent of minus
twice the total (potential + kinetic) energy of a test particle in a nonrotating system: -2Utot = -2F - ( X 2 +Y 2 + Z 2 )
Since the kinetic term <0, we know that a given particle
on a given orbit (with a given constant CJ), can only reach
points (x,y,z) where the effective potential obeys:
1
Feff (x, y, z) < - CJ
2
Remember:
Feff < 0
Zero-velocity curves / surfaces
1
(allowed region in x,y,z)
Feff (x, y, z) < - CJ
2
The boundaries of this region are called the „zero velocity curves“ (in
2-D) or „zero velocity surfaces“ (in 3-D). They are the potential lines.
For CJ<0 no such restrictions exist (all points are allowed). More
precisely: for CJ<min(-2Φeff) no such restriction exist. But for
CJ>min(-2Φeff) there exists regions in (x,y,z) which are inaccessible
for the particle. If CJ is sufficiently large, these inaccessible regions
can even completely surround the star or planet system or both:
= Not allowed region
Hill sphere (=Roche lobe)
Not exactly a sphere, but approximately.
It is the largest zero-velocity surface
surrounding only the planet.
It is the sphere of influence of the planet.
æ M planet ö
rHill = ç
÷ aplanet
è 3M * ø
1/3
Meaning of Hill sphere
• The Hill sphere plays a key role in planet formation:
– To add mass to a planet, we must put the mass into the Hill
sphere of the growing planet, because only then it that mass
gravitationally bound to it.
– Any circumplanetary disk or moons must be inside the
planet‘s Hill radius
– We will see later that the ratio of the Hill radius to the
protoplanetary disk‘s thickness plays a key role in planet
migration.
– Any object that is larger than its own Hill sphere will be
sheared apart by the tidal forces of the star. This will lead us
to the definition of the „Roche density“ as the minimal
density an object needs to remain gravitationally coherent
and survive tidal forces. See next chapter.
Potential field lines as approximate orbits
If x + y + z << Feff then one can approximately write:
2
2
2
1
Feff » - CJ = constant
2
This means that approximately the test particle moves along
the potential field lines. To be more precise: the guiding center
will do this; the test particle will epicycle around this guiding
center. You can find orbits without epicycles, in which case the
test particle indeed moves approximately along the field lines.
It turns out to be also a fairly good approximation if the condition
x 2 + y2 + z 2 << Feff does not hold.
This leads to a special set of orbits.
Kepler- and horseshoe orbits
Kepler orbit
around star only
Kepler orbit
around planet only
Kepler orbit
around both
Horseshoe orbit
Trojans of Jupiter
L4 and L5 are stable Lagrange points, while L1, L2 and L3 are not.