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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
Kepler‘s Laws:
1. Planets move in ellipses with the Sun in one of their focal points.
Perihelion distance:
Aphelion distance:
Eccentricity: Ellipticity:
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
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:
The Titius-Bode Law:Planet Titius-Bode Observed Distance (AU) Distance
• A similar law works for many moons
The Moons of Uranus
c
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 = 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
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)
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)
The spectroscopic method
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..
The spectroscopic method
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
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
!"#$%&'(&)(*'+(,--,.(,--/.(0#&1(&)(*'+(,--/.(,--2.(,--3.(04''&55&6(&)(*'+(,--7*.8.(0&61&'(&)(*'+(,-9-(
M• = 4.3± 0.35 !106M!
• Orbits with eccentricity 0 < e < 1 are bound ellipses
Major axis:
p
a
ba
Elliptical orbits:
The result of Newton‘s law of gravity is more general.
Circle Ellipse Parabola Hyperbola
Eccentricity e : 0 < 1 1 > 1
Perihelion : a p/(1+e) = a(1-e) p/2 p/(1+e) Aphelion : a p/(1-e) = a(1+e)
Two free parameters: p and e
Unbound particles can get very close to gravitating object!
y = k ! x2
The gravitational potential:
(conservation of energy)
The Orbital Energy:
What does this mean, physically?
ddt
12!"r 2!
"#$%& =!!"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
Circular orbit: (e = 0 and r = a):
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)
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 ?
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
What if we would live in higher dimensional space ?
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
• 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 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
Energy equation:
More sophisticated determination of
Total visible mass:
Note: The cosmic baryon fraction is 0.2.
Where are most of the baryons?
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:
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
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
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