dynamics of the solar system - universe- · pdf filegeometry of the solar system ... the first...

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 ofAthens

(Vatikan)

Raffael1510

EuklidPythagoras

Sokrates

Platon+

Aristoteles Zarathustra

Ptolemäus

R

The World of Ptolemäus

Aristoteles

• Earth: center of the Universe

• Everything else rotates in circles around Earth

Platon and Aristoteles

The Epicycles of Ptolemäus

(1473-1543)

(Nikolaus Kopernikus)

Whichworld modelis correct?

The Heliocentric Concept

The Problemwith Mercury

(1473-1543)

(Nikolaus Kopernikus)

Why a circle?

Johannes Kepler

(1571-1630)

No circle !

Tycho Brahe (1546-1601)

Kepler‘s Laws:

1. Planets move in ellipses with the Sun in one of their focal points.

Perihelion distance:

Aphelion distance:

Eccentricity: Ellipticity:

• Orbits with eccentricity e < 1 are bound ellipses

p

a

ba

Elliptical orbits:

Jupiter: Eccentricity: 0.05

Perihelion: 4.95 AU

Aphelion : 5.46 AU

Eccentricity distributionof extrasolar planets

Pluto:

Uranus:

a = 30.11AU e = 0.0090

The 5 moons of Pluto

Pluto: discovered 1840 on a 248 yrs orbit

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 1.002Venus 0.723 0.615 0.007 1.001Earth 1.000 1.000 0.017 1.000Mars 1.524 1.881 0.093 1.000Jupiter 5.203 11.86 0.048 0.999Saturn 9.537 29.42 0.054 0.998Uranus 19.19 83.75 0.047 0.993Neptun 30.07 163.7 0.009 0.986Pluto 39.48 248.0 0.249 0.999

Kepler‘s laws cannot explain the distances of the planets .

But:

0.4 0.7 1 1.5 5.2 9.5 19.2 30. 40

Orbital radii in AU

The Titius-Bode Law:Planet Titius-Bode Observed Distance (AU) Distance

• A similar law works for many moons

The Moons of Uranus

c

Sir Isaac Newton (1642 – 1727)

m

• 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.

PlanetSun

angular momentum vector

The Orbital Plane:

• The planet moves in one plane. Its orbit is then completely specified by its radius r und its position angle

x

y

r

Sun

Planet

period angular velocity

Circle

This is true for all radial forces!!!

Axisymmetry: h_z conserved, no orbital plane

Triaxial: no h conservation no orbital plane

e = 0The second Keplerian Law

Circular orbit: (e = 0 and r = a):

3rd Keplerian Law:

e = 0.5The second Keplerian Law

e = 0.97The second Keplerian Law

+

+ +

e = 0

e = 0.5 e = 0.97

(Keplers third law) • No dependence on e or h• Closed ellipses• Only true if

f (r) ! GM (r)r2

= GM 0

r2

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

b

Ionisation:

Exchange:

No energy conservation for each particle!!!!!

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.

The Chaotic Moon

• 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

• Poincare showed that the 3-body problem in general has no analytical solution and can only be solved approximately.

• He also showed that the solutions can be rather complex and often chaotic.

However analytical solutions exist.

Lagrange-system

Henry Poincare (1854-1912)

Chenciner –Montgomery system

The large Moons of Saturn

(vulcanism)

(Cassini-Huygens)

chaoticrotation

The dance of the moons Janus and Epimetheus:

• Almost on the same orbit meet every 4 years

e = 0.97

+

+ +

e = 0

e = 0.5 e = 0.97

(Keplers third law)

Given T and a we can determine M+m

Searching for extrasolar planets

Dynamics of binaries

System from above (unlikely)

System from the side (more likely)

(G. Marcy und P. Butler, San Francisco)

The spectroscopic method

(G. Marcy und P. Butler, San Francisco)


System from above (unlikely)

System from the side (more likely)

-10

-20

0

20

10

0 302010

m/s

Years

The solar system

In 1995 Marcy & Butler had their spectrograph optimized enoughto detect velocity shifts of order a few m/s..


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

-60

m/s

Days 0 3

The hit by the Swiss

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)

!""#$%&'#(#)"*&#(+%"

,-./0&123&45'56&45(758#)"

9:"$;"'&")&+'<=

>?"&@5A"6&5B&-*+7CD"&/7C48

0.05 0.025 0 �0.025 �0.05 �0.075

0.05

0.025

0

0.025

0.05

0.075

0.15

0.125

0.15

0.175

0.1

R.A. �''�

Dec�''�

!"#$%&'($

)**+$

+,,+$

+,))$)**-$

./$ 01&'(2&$

034$.5$$6274&8$$

Keplerian Orbit

!"#$%&'(&)(*'+(,--,.(,--/.(0#&1(&)(*'+(,--/.(,--2.(,--3.(04''&55&6(&)(*'+(,--7*.8.(0&61&'(&)(*'+(,-9-(

M• = 4.3± 0.35 !106M!

The Bondi Radius

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

• Orbits with eccentricity 0 < e < 1 are bound ellipses

Major axis:

p

a

ba

Elliptical orbits:

The result of Newton‘s law of gravity is more general.

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)

Two free parameters: p and e

Unbound particles can get very close to gravitating object!

y = k ! x2

The gravitational potential:

(conservation of energy)

The Orbital Energy:

What does this mean, physically?

ddt

12!"r 2!

"#$%& =!!"r ' !"r = -G M +m

r3"r ' !"r = - d(

dr" ' dr"

dt= - d(

dt

At Perihelion:

With: and

Parabola:

Hyperbola:

Ellipse:

No dependence on h!!

The geometry isdeterminedjust by E.

(see also cosmology) !!

vp ! rp

Hyperbolic universe (k < 0): E > 0Parabolic universe (k = 0): E = 0Elliptical universe (k > 0): E < 0

Total energy:

Friedman equation

80

Halley‘s comet(74-78 period)

2024


( virial theorem )

3rd Keplerian Law:


3rd Keplerian Law:

The virial theorem applies to allbound stellar systems in equilibrium.

If orbital energy is extracted the kinetic energy increases and the satellitemoves onto an orbit with smaller radius: v a

For elliptical orbits the virial theorem is valid only when averaging over one orbit.

with

The Virial TheoremTotal energy:

Reason:

Note:

E = Ekin + Epot

2 !Ekin = "Epot

(negative specific heat)

!E =12!EpotEpot = !GM

r

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


F(r) ! r!(n!1)

Circular orbit: F(rc ) =vrot2

rc= L2

rc3

Small radial perturbation x, keeping L constant: r = rc + x

!!r = !F(r)+ L

2

r3Newton‘s law inspherical symmetry:

!!rc + !!x = !F(rc )! x

dFdr rc

+ L2

rc3 ! 3x

L2

rc4

!!x + x dFdr rc

+ 3F r( )r rc

!

"#

$

%& = 0

harmonicoscillator


F(r) = k / rn!1

Solution: x(t) = x0 exp i!t( ) ! 2 = dFdr rc

+ 3F(r)r rc

Stable solution: ! 2 > 0

Suppose:

dFdr

+ 3Fr

= ! krn!2

n !1( ) + 3 krn!2

= krn!2

3! (n !1)( ) > 0

!!x + x dFdr rc

+ 3F r( )r rc

!

"#

$

%& = 0

harmonicoscillator

n < 4

The dynamics of particle systems

Why stars do not collide

Voyager 1

(1977)

62,000 km/hSun: 800,000 km/h

4 light years

Proxima Centauri

74,000 yrsfor

Voyager I

?

5 cm

5000 km

5 cm

• Galaxies are collisionless system.

• During the collision of a galaxy, the stars do however not collide with each other.

• Galaxy-galaxy interactions however occur frequently

Text

Gravitation is a Long-Range Force

Stars are test particles within the potential of all the other stars

Newton: m

M

For N point masses we have:

The gravitational potential

Newton: m

M

For N point masses we have:

The gravitational acceleration is independent of the mass of the test particle.

The gravitational potentialtest 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

W = !E = m

!fd!r = "m #$

#r! dr!

%

r!

& = m $% " $ r!( )( ) = "m '$ r

!( )%

r!

&

The time independent gravitational potential is conservative

The required work W for a point mass m is independent of its detailed orbit.

(energy conservation)We can travel to all points with

W1,2 = m

!f d!x = m !

!"#d!x = m #1 !#2( )

1

2

$1

2

$

W=m

The potential in spherically symmetric systems:

r

Escape velocity:

Rotational velocity:Point mass:

Point mass potential

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.

Forces and the potential of extended, spherical systems

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.

r

Gravitational force at r

The Milky Way

The Milky Way in Visible Light

• For r < R :

• Edge of the dark halo at r = R :

A simple model of the Milky Way

Total mass:

Escape velocity:

Solar neighborhood

The Magellanic Clouds

Tangential velocity:

Radial velocity:

Bound Orbit:

Distance:

Energy equation:

More sophisticated determination of

Total visible mass:

Note: The cosmic baryon fraction is 0.2.

Where are most of the baryons?

NGC 3198

Rotation curves of galaxies

Rotation curve Mass distribution M(r)

Dark matter distribution in NGC 3198

Dark matter

Visible matter

In spherically symmetric systems the gravitational acceleration is:

In this case, point masses move in a fixed orbital plane.

The angular momentum vector is constant:

Orbits in extended, spherically symmetric mass distributions

x

yWe are looking for a solution in the equatorial plane of

a!(r!) = f (r) !e

!r

!!r = f (r)+ L2

r3

L = r2 !!

Fundamental orbital equation for spherically symmetric gravitational potentials

In general this equation has to be integrated numerically.

Exceptions: Keplerian potential

Harmonic potential

!!r = f (r)+ L2

r3

L = r2 !!

Physics of Rosette or loop orbits

rmin,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:

1. Keplerian potential:

2. Harmonic potential:

Effective Potential:

Angular momentum and the effective potential

Orbits with constant angular momentum L

Plummer

profile

1 2 3 4 r0

0

-1

The angular momentum actslike a centrifugal barrier

Unbound hyperbolic orbits(comets)

Boundorbits

Orbits with constant angular momentum L

Plummer

profil

1 2 3 4 r0

0

-1

ApocenterPericenter

E

circle

For given L the orbit with the lowest energy is a circle.

vR = 0!eff = E

Why do cosmic particle systems appear relaxed?

Essential Astrophysics SS 2014

M87

Dynamics of a relaxed particle system

Essential Astrophysics SS 2014

CDM simulation

r

vrad


E

r


Dynamics of a collapsing particle system

r

vrad


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

Stars

initially end

Phase Mixing of Stellar Systems

For all stars are out of phase and equally distributed.

Mixing timescale:

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