Dynamics of the solar system

Planets: Wanderer Through the Sky

Ecliptic

The zodiac

Geometry of the Solar System Nearly all planets rotate synchronously around the Sun

Spin distribution of the planets

The school of Athens (Vatikan) Raffael 1510 Sokrates Platon + Aristoteles Zarathustra R Pythagoras Ptolemäus Euklid

The World of Ptolemäus Platon and Aristoteles Aristoteles Earth: center of the Universe Everything else rotates in circles around Earth

The Epicycles of Ptolemäus

The Heliocentric Concept (Nikolaus Kopernikus) (1473-1543) Which world model is correct?

The Problem with Mercury (Nikolaus Kopernikus) (1473-1543) Why a circle?

Johannes Kepler Tycho Brahe (1546-1601) No circle! (1571-1630)

Kepler s Laws: 1. Planets move in ellipses with the Sun in one of their focal points. Eccentricity: Ellipticity: Perihelion distance: Aphelion distance:

Elliptical orbits: Orbits with eccentricity e < 1 are bound ellipses a p a b

Jupiter: Eccentricity: 0.05 Perihelion: 4.95 AU Aphelion : 5.46 AU Eccentricity distribution of extrasolar planets

Pluto: The 5 moons of Pluto Pluto: discovered 1840 on a 248 yrs orbit Uranus: a = 30.11AU e = 0.0090

3:2 Resonance

2. The position vector covers equal areas in equal times 3. The square of the orbital period is proportional to the cube of the length of the major axis.

Mercury 0.387 0.241 0.206 1.002 Venus 0.723 0.615 0.007 1.001 Earth 1.000 1.000 0.017 1.000 Mars 1.524 1.881 0.093 1.000 Jupiter 5.203 11.86 0.048 0.999 Saturn 9.537 29.42 0.054 0.998 Uranus 19.19 83.75 0.047 0.993 Neptun 30.07 163.7 0.009 0.986 Pluto 39.48 248.0 0.249 0.999 But: Kepler s laws cannot explain the distances of the planets.

Orbital radii in AU 0.4 0.7 1 1.5 5.2 9.5 19.2 30. 40

The Titius-Bode Law: Planet Titius-Bode Observed Distance (AU) Distance c A similar law works for many moons The Moons of Uranus

Sir Isaac Newton (1642 1727)

m Planet Sun This 2-body problem consists of 3 differential equations of second order. Its solution requires 6 constants of integration which are the initial conditions: The solution determines the velocity and location of each planet as function of time relative to the Sun.

The Orbital Plane: This is true for all radial forces!!! y Axisymmetry: h_z conserved, no orbital plane r Planet Triaxial: no h conservation no orbital plane angular momentum vector The planet moves in one plane. Its orbit is then completely specified by its radius r und its position angle Sun Circle x period angular velocity

The second Keplerian Law e = 0

Circular orbit: (e = 0 and r = a): 3rd Keplerian Law:

The second Keplerian Law e = 0.5

The second Keplerian Law e = 0.97

(Keplers third law) + e = 0 No dependence on e or h Closed ellipses Only true if f (r)! GM (r) = GM 0 r 2 r 2 e = 0.5 e = 0.97 + +

Chaos starts as soon as N > 2 Here: a 3 body encounter

Ionisation: No energy conservation for each particle!!!!! Exchange: b

The Chaotic Moon The Moon s orbit relativ to the Sun is almost circular The Moon s orbit shows complex perturbations as a result of the gravitational interaction with the Sun (3-body problem) This made it difficult to use the Moon in order to determine the geographical length. Even famous mathematicans and astronomers like Newton or Laplace were not able to predict the Moon s orbit with high accuracy.

Tycho Brahe wanted to determine the Moon s orbit with high accuracy. For that he determined precisely the time of linar eclipses and developed a theory. With his new theory and a position of the Moon on December, 28th, 1590 he predicted that the next lunar eclipse will start on Decémber, 30th 1590 at 6:24pm. The eclipse however began while Brahe was still having dinner and was almost over when Brahe finally started his measurements at 6.05 pm. Brahe then stopped investigations of the Moon for some time.

The moon has a stabilising effect on the orbit of the Earth Numerical simulation of the solar flux onto Earth at 67 North Past with Moon Future without Moon

Chaos in the solar system An error of 15m in the position of the Earth will in 100 million years have grown to an error of 150 million km = 1 AU. It is impossible to determine the precise position of the Earth over several Gyrs Mercury

Henry Poincare (1854-1912) Poincare showed that the 3-body problem in general has no analytical solution and can only be solved approximately. Lagrange-system He also showed that the solutions can be rather complex and often chaotic. However analytical solutions exist.

Chenciner Montgomery system

The large Moons of Saturn (Cassini-Huygens) chaotic rotation (vulcanism)

The dance of the moons Janus and Epimetheus: Almost on the same orbit meet every 4 years

e = 0.97

(Keplers third law) e = 0 + Given T and a we can determine M+m e = 0.5 e = 0.97 + +

Searching for extrasolar planets

Dynamics of binaries

The spectroscopic method (G. Marcy und P. Butler, San Francisco) System from above (unlikely) System from the side (more likely)

The solar system The spectroscopic method m/s 20 10 0-10 -20 0 10 20 30 Years In 1995 Marcy & Butler had their spectrograph optimized enough to detect velocity shifts of order a few m/s..

The hit by the Swiss On September, 8th 1995 the Swiss Mayor and Queloz (Genf) found the first planet that orbits an solar-type star. The planet around 51 Peg: a hot Jupiter 60 m/s -60 Days 0 3

Planeten Transits The Kepler mission

More than 1000 extrasolar planets detected so far

Small and low-mass planets are numerous

The habitable zone

Karl Jansky (1905-1950)

66 ESO

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0.175 0.15 )**+$ +,))$ )**-$ Keplerian Orbit 0.125 0.1 Dec '' 0.075./$ 01&'(2&$ 0.05 M = 4.3 ± 0.35!10 6 M! 0.025 0 +,,+$ 034$.5$$ 6274&8$$!"#$ %&'($ 0.05 0.025 0 0.025 0.05 0.075 R.A. ''!"#$%&'(&)(*'+(,--,.(,--/.(0#&1(&)(*'+(,--/.(,--2.(,--3.(04''&55&6(&)(*'+(,--7*.8.(0&61&'(&)(*'+(,-9-(

The Bondi Radius

Sgr A* : Eine bubble of 100 Mio. degree hot gas

Elliptical orbits: a p a b Major axis: Orbits with eccentricity 0 < e < 1 are bound ellipses The result of Newton s law of gravity is more general.

Two free parameters: p and e y = k! x 2 Circle Ellipse Parabola Hyperbola Eccentricity e : 0 < 1 1 > 1 Perihelion : a p/(1+e) = a(1-e) p/2 p/(1+e) Aphelion : a p/(1-e) = a(1+e) Unbound particles can get very close to gravitating object!

The Orbital Energy: d dt! " # 1 2!" r 2 What does this mean, physically? $ % & =!! r " ' r "! = -G M + m " " d( r ' r! = - r 3 dr " ' dr " dt = - d( dt The gravitational potential: (conservation of energy)

At Perihelion: v p! r p With: and No dependence on h!! Ellipse: Parabola: Hyperbola: The geometry is determined just by E. (see also cosmology)!!

Friedman equation Total energy: Elliptical universe (k > 0): E < 0 Parabolic universe (k = 0): E = 0 Hyperbolic universe (k < 0): E > 0 80

Halley s comet (74-78 period) 2024

Circular orbit: (e = 0 and r = a): 3rd Keplerian Law: ( virial theorem )

Circular orbit: (e = 0 and r = a): 3rd Keplerian Law: The virial theorem applies to all bound stellar systems in equilibrium.

The Virial Theorem Total energy: E = E kin + E pot 2! E kin = "E pot with (negative specific heat) Reason: If orbital energy is extracted the kinetic energy increases and the satellite moves onto an orbit with smaller radius: v a Note: For elliptical orbits the virial theorem is valid only when averaging over one orbit.

E pot =! GM r!e = 1 2!E pot

What if we would live in higher dimensional space? The inverse square law of gravity appears to be a property of 3-dimensional space. But why is space 3-dimensional? It is reasonable to generalize this law to n-dimensional space as: F(r)! r!(n!1) Stable orbits are only possible for n <= 3

What if we would live in higher dimensional space? Newton s law in spherical symmetry:!! r =!F(r) + L2 r 3 F(r)! r!(n!1) Circular orbit: F(r c ) = v 2 rot r c = L2 r c 3 Small radial perturbation x, keeping L constant: r = r c + x!! r c +!!x =!F(r c )! x df + L2 3 dr rc r c! 3x L2 r c 4 ( )!!!x + x# df + 3 F r " dr rc r r c $ & = 0 % harmonic oscillator

What if we would live in higher dimensional space? ( )!!!x + x# df + 3 F r " dr rc r r c $ & = 0 % harmonic oscillator Solution: x(t) = x 0 exp( i!t )! 2 = df + 3F(r) dr rc r r c Stable solution:! 2 > 0 Suppose: F(r) = k / r n!1 df dr + 3F r =! k ( n!1) + 3 k r n!2 r = n!2 k ( 3! (n!1)) > 0 r n!2 n < 4

The dynamics of particle systems

Why stars do not collide

Voyager 1 (1977) Sun: 800,000 km/h 62,000 km/h

Proxima Centauri 4 light years 74,000 yrs for Voyager I

5 cm?

5 cm 5000 km

Galaxies are collisionless system. Galaxy-galaxy interactions however occur frequently During the collision of a galaxy, the stars do however not collide with each other. Text

Gravitation is a Long-Range Force Stars are test particles within the potential of all the other stars

The gravitational potential Newton: m M For N point masses we have:

The gravitational potential test particle Newton: m M For N point masses we have: The gravitational acceleration is independent of the mass of the test particle.

At a given point the acceleration is: Is it possible to find a distribution of masses for any force field Or is the gravitational field peculiar?

This gravitational acceleration can be defined as gradient of a scalar (potential) We choose the zero point such that: Note: The potential is the work, exerted by the gravitational field onto a unit mass that falls from infinity to radius r! r W =!E = m! fd r! r = "m #$ #r! dr! & = m $ % " $ r! & %! % ( ( )) = "m ' $ ( r! )

The time independent gravitational potential is conservative The required work W for a point mass m is independent of its detailed orbit. W=m 2 $ 1 2 $ 1 ( ) W 1,2 = m! f d! x = m!! "#d! x = m # 1! # 2 We can travel to all points with (energy conservation)

Point mass potential The potential in spherically symmetric systems: r Escape velocity: Rotational velocity: Point mass:

Forces and the potential of extended, spherical systems Given a density distribution: Calculate from the gravitational potential and the gravitational force Newton s 1st law: The gravitational force within a spherical shell of homogeneous density is zero.

Newton s 2nd law The gravitational force of a spherical shell of homogeneous density and mass dm onto an object outside of the shell is equal to the gravitational force of a point mass dm in the center of the shell. Gravitational force at r r

The Milky Way

The Milky Way in Visible Light

A simple model of the Milky Way For r < R : Edge of the dark halo at r = R :

Escape velocity: Solar neighborhood Total mass:

The Magellanic Clouds

Tangential velocity: Radial velocity: Distance: Bound Orbit:

More sophisticated determination of Energy equation: Total visible mass: Note: The cosmic baryon fraction is 0.2. Where are most of the baryons?

Rotation curves of galaxies NGC 3198

Dark matter distribution in NGC 3198 Rotation curve Mass distribution M(r) Dark matter Visible matter

Orbits in extended, spherically symmetric mass distributions In spherically symmetric systems the gravitational acceleration is: a! (r! ) = f (r)!e! r In this case, point masses move in a fixed orbital plane. The angular momentum vector is constant: y We are looking for a solution in the equatorial plane of x!! r = f (r) + L2 r 3 L = r 2!

Fundamental orbital equation for spherically symmetric gravitational potentials!! r = f (r) + L2 r 3 L = r 2! In general this equation has to be integrated numerically. Exceptions: Keplerian potential Harmonic potential

r min,max (E, L) Loss of information about initial state Orbit area filling as change in angle!" over one orbital period is in general an irrational number Gas particles would dissipate random motion There are two and only two potentials with bound closed orbits: Physics of Rosette or loop orbits 1. Keplerian potential: 2. Harmonic potential:

Angular momentum and the effective potential Effective Potential:

Orbits with constant angular momentum L 0 Unbound hyperbolic orbits (comets) Bound orbits -1 0 Plummer profile 1 2 3 4 r The angular momentum acts like a centrifugal barrier

Orbits with constant angular momentum L Pericenter circle 0 E -1 0 Plummer profil 1 2 3 4 r Apocenter For given L the orbit with the lowest energy is a circle.! eff = E v R = 0

Why do cosmic particle systems appear relaxed?

M87 Essential Astrophysics SS 2014

Dynamics of a relaxed particle system

Essential Astrophysics SS 2014

CDM simulation

Dynamics of a relaxed particle system

Dynamics of a relaxed particle system v rad r

Dynamics of a relaxed particle system E r

Dynamics of a collapsing particle system

Dynamics of a collapsing particle system v rad r

Dynamics of a collapsing particle system E r

Phase Mixing Consider a disk of stars that move on circular orbits with angular velocity Assumption: All stars initially are at system is highly ordered initially end Stars

Phase Mixing of Stellar Systems For all stars are out of phase and equally distributed. Mixing timescale: