#### 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.