Milky Way Galaxy - people.physics.tamu.edupeople.physics.tamu.edu/papovich/courses/fall12/milkyway.pdf · Milky Way Galaxy Composite near-IR (2 micron) Image from the Two Micron All

Milky Way Galaxy

Composite near-IR (2 micron) Image from the Two Micron All Sky Survey (IPAC/Caltech/UMass)

Monday, August 20, 2012

Milky Way Galaxy

Our Galaxy is a collection of stars and interstellar matter - stars, gas, dust, neutron stars, black holes -

held together by gravity


Historical Models of the Milky Way Galaxy

In 1780 William Herschel produced the map below by counting stars in different directions. He concluded that the Sun is near the center of the Galaxy, and that the

dimensions along the plane were five times greater than the vertical thickness.

Herschel assumed (1) all stars have same luminosity (Absolute Magnitude), (2) Number density in space is roughly constant, and (3) there is nothing in space to obscure the Stars

(fainter stars are farther away)

Sun

In mid-1700s, Immanuel Kant (1724-1804) and Thomas Wright (1711-1786) proposed the Galaxy must be a disk of stars to

explain the circular distribution in the sky. They went further and suggested our Sun is one component in the Milky Way.

William Herschel (1738-1822)



Jacobus Kapteyn (1851-1922)

Jacobus Kapteyn (1851-1922) used star counting to confirm the Herschel model, but with much-improved methods. Now

called the Kapteyn Universe.

Galaxy consists of a flattened Spheroidal system with a decreasing stellar density with increasing distance from the center. His published self-titled “attempt” to describe the “Stellar system” (=Milky Way) appear in the year he died

(Kapteyn 1922, ApJ, 55, 302):



Numbers show where stellar density has declined by a factor of 2, 3, ... 10, from the central density.

Sun, y=650 pc, x=38 pc

Kapteyn Universe

Picture of the Galaxy:



Harlow Shapley (1885-1972)

From 1915-1919, Harlow Shapley estimated the distances to 93 globular clusters using RR Lyrae and W Virginis variable stars

(like Cepheids). Shapley found they are not uniformly distributed in the Galaxy, but are concentrated in the

constellation Sagittarius (where the center of Galaxy is). He determined these were 15,000 pc (15 kpc) away.

The most distant clusters he could measure were 70 kpc away. Shapley argued our Galaxy has a diameter of 100 kpc, close to 10x that of Kapteyn. Also as important, Shapley put

our Sun far from the center of the Galaxy. Kapteyn had the Sun near the center.



Who was right, Kapteyn or Shapley ?



Who was right, Kapteyn or Shapley ?

Neither ! They are both wrong, but for the same reason. They both ignored the effects of dust, which causes the extinction of light.

Kapteyn missed stars he could not see, could not see the most distant regions of the Milky Way.

Shapley’s variable stars were more luminous then he thought because their apparent magnitudes were extincted.

Similar to being on a boat and trying to see land through fog.


Star CountingTwo approaches:

Integrated star counts. Count the number of stars visible in an area of the sky, Ω, with magnitude < m.

Differential star counts: Count only those stars visible in Ω with magnitude between m and m+dm

Let the differential number of stars with absolute magnitudes between M and M+dM and attribute S within solid angle Ω at a distance r be given by

nM (M,S, Ω, r)dM

Units of nM are pc-3 mag-1.


Star Counting

nM (M,S, Ω, r)dM

Then the integrated number density of stars with attribute S at distance r in in a solid angle Ω is the integral:

n(S, Ω, r) =∞−∞ nM (M,S, Ω, r)dM

Units of n are pc-3.

The number density can be integrated over the volume of a cone extending from r=0 to r=d,

NM (M,S, Ω, d)dM = d

0 nM (M,S, Ω, r)Ωr2drdM

This is the Integrated star count


Star CountingStars with same absolute Magnitude have different apparent magnitudes because they have different distances.

Rewrite equations using apparent magnitude, m, and derive :

NM (M,S, Ω,m)dM

The change in the number of stars as the apparent magnitude is increased is given by

NM (M,S,Ω,m)dm dm

dM

Should this derivative be positive or negative ?

This defines the differential star count:

AM (M,S, Ω,m)dMdm ≡ NM (M,S,Ω,m)dm dMdm


Star Counting

Consider the case of an infinite Universe with uniform stellar density with no interstellar extinction (A=0), in which case

nM (M,S, Ω, r) = nM (M,S) = constant

Consider the case of an infinite Universe with uniform stellar density, in which case

NM (M,S, Ω, d) = nM (M,S)Ω d

0r2dr =

Ωd3

3nM (M,S)

Using m - M = 5 log(d/10 pc), we can write d in terms of m and M:

NM (M,S, Ω,m) =Ω3

nM (M,S)103(m−M+5)/5

=Ω3

nM (M,S) exp[ln 103(m−M+5)/5]

=Ω3

nM (M,S) exp[3(m−M + 5)/5] ln 10


Star Counting

Differentiating we get the differential count formula:

NM (M,S, Ω,m) =Ω3

nM (M,S) exp[3(m−M + 5)/5] ln 10

AM (M,S, Ω,m) =dNM (M,S, Ω,m)

dm

=ln 10

5Ω nM (M,S)103(m−M+5)/5

=3 ln 10

5NM (M,S, Ω,m)

If either AM or NM are known from the data, then the equations can be solved for nM(M,S) dM.

Problem is that when you integrate these equations to determine the total amount of light received on Earth, the answer diverges (infinite light). This is a result of the “constant” stellar density and implies an infinite amount from infinitely far away.

First pointed out by Heinrich Olbers (1758-1840), called Olber’s Paradox.


Galactic Coordinates

The Galactic Coordinate system is a spherical coordinate system defined by two angular coordinates ℓand b. b ∈ [-90o, +90o], and ℓ ∈ [0o, 360o].

Center is at the Earth. Galactic Plane is the plane of the Galactic Disk.

NGP = North Galactic Pole

Galactic center has coordinates ℓ=0o and b=0o.


Galactic CoordinatesThe Galactic Coordinates are related to the Equatorial Coordinates, defined by the rotational axis of the Earth.

Great circles through the North and South Earth Poles have constant Right Ascension, α.

Lines parallel to the Earth’s equator have constant Declination, δ.


Galactic CoordinatesThe Galactic Coordinates are related to the Equatorial Coordinates, defined by the rotational axis of the Earth.

Great circles through the North and South Earth Poles have constant Right Ascension, α.

Lines parallel to the Earth’s equator have constant Declination, δ.


Galactic CoordinatesBy IAU standards, the J2000.0 equatorial coordinates for the NGP (b=90o) are:

α=12h51m26.28s, δ=+27d7m41.7s

The Galactic center has equatorial coordinates (ℓ0=0o, b0=0o) :

α=17h45m37.20s, δ=-28d56m9.6s


The Morphology of the GalaxyThe solar Galactocentric distance, R0, is still debated. In 1985 the

International Astronomical Union (IAU) adopted R0 = 8.5 kpc. Recent studies find R0 = 8 kpc (Eisenhauer 2003).

The Galaxy is composed of a bulge, a thin and thick disk, and a halo.

Most stars are in disk components. Disk contains lots of gas and dust.

Halo has low density and it contains many globular clusters.


Distances in the Galaxy

Trigonometric Parallax - purely geometric effect, independent of physical assumptions.

Earth Earth6 months later

Sun

r

D= tan p ≈ p

r = 1AUr = 1AU

D

r

D =

1

p

pc


Distances in the GalaxyPhotometric Distance: Extinction and Reddening

If star’s luminosity class known from spectroscopy, then can derive distances using Absolute Magnitude (M) and apparent magnitude (m).

m−M = 5 log(D/10 pc)

Serious problem is extinction and reddening from dust. Can

overcome this for star clusters. Identify MS and “shift” it to fit unreddened main sequence.



Consider Radiative Transfer Equation for pure absorption and scattering.

dIν

ds= −κνIν rewrite as d ln Iν = dIν/Iν = −κνds

and integrate ln Iν(s)− ln Iν(0) = − s

0dsκν(s) ≡ −τν(s)

Where “τ” is the optical depth.

The specific flux, S, is therefore reduced by a factor e-τ

Sν = Sν(0)e−τν(s)

because m = -2.5 log S + const, Sν

Sν(0)= 10−0.4(m−m0) = e−τν(s) = 10− log(e)τν

or, define “extinction” Aν [units of magnitudes]

Aν = m−m0 = 2.5 log(e)τν = 1.086τν



Because the Extinction is dependent on frequency (alternatively wavelength), it is associated with a change in color, defined by the “color excess” as follows.

E(B − V ) = AB −AV = (B −B0)− (V − V0) = (B − V )− (B − V )0

The ratio AB/AV depends on the properties of the dust. Normally write the color excess in terms of this ratio:

E(B − V ) = AV

AB

AV− 1

= AV R−1

V

From sightlines to many stars in the

Milky Way, we have

RV = 3.1± 0.1


Distances in the Galaxy



Now, the relation between apparent, absolute magnitude, and distance includes extinction:

m−M = 5 log(D/10 pc) + A

Pulsating Stars

Well-defined relationship between period and luminosity for pulsating stars: δ Cepheid Stars (“classical Cepheids”), W Virginis stars (sometimes called “Pop

II” Cepheids), RR Lyrae stars (also “Pop II”).

This gives the Absolute Magnitude (luminosity).

Can then derive distance using apparent magnitude (and estimate for extinction),

5 log(D / 10 pc) = m - M


Pulsating Stars

Time (Period is 332 days)

Mag

nitu

de v

arie

s fr

om 3

.5 t

o 9.

Time Variability of o Ceti (in constellation Cetus) over past 150 years)


Log P (days)

MI

MV

MB

MB=(-2.76) Log P -0.472

MV=(-3.14) Log P -0.826

MV=(-3.41) Log P -1.325

from Tammann, Sandage, and Reindl (2003, A&A,

404, 423)


Pulsating Stars

Astronomers decrease the scatter in this relation using infrared data. Why ? Because at infrared wavelengths dust obscuration is less severe. In the infrared:

H = -3.234 log10 Pd + 16.079

Can reduce the scatter further by incorporating color index information:

H = -3.428 log10 Pd + 1.54〈 J - KS〉+ 15.637

Persson et al. 2004, AJ, 128, 2239


The Morphology of the GalaxyThe solar Galactocentric distance, R0, is still debated. In 1985 the

International Astronomical Union (IAU) adopted R0 = 8.5 kpc. Recent studies find R0 = 8 kpc (Eisenhauer 2003).

The Galaxy is composed of a bulge, a thin and thick disk, and a halo.

Most stars are in disk components. Disk contains lots of gas and dust.

Halo has low density and it contains many globular clusters.


The Morphology of the Galaxy

Structure of Thin and Thick Disks

We define the size of the components using the scale height. (We don’t have a way of defining the “edge” of the galaxy or its components ? )

Let n be the number density of stars in the disk, and z be the vertical distance above the Galactic midplane, then the number density, n(z), depends on the scale height, h:

the scale height is the point where the number density has dropped by a factor of e.

n(z) = n(0) exp− |z|

h

There are three distinguishing components to the Galactic Disk:

1. The Young Thin Disk, h~100 pc. (Sun is a member, with z~30 pc.)

2. The old Thin Disk, h~300 pc.

3. The Thick Disk, h~1.5 kpc.

(1b. Molecular gas, from which new stars are born, has h~65 pc.)



The scale heights provide information on the average velocity dispersion of the different components.

To 1st order, stars orbit around Galactic center on circular orbits, but they have some random motion. This leads to the velocity dispersion.

The number of stars (of a given population) at a fixed location with the i-th velocity component in the interval [vi, vi+dv] and d3v = dv1dv2dv3.

f(v)d3v :

The mean velocity of the population is then:

v = n−1

vf(v)d3v

The velocity dispersion of the i-th component is given by :

σ2i = v2

i − vi2 = n−1

(v2

i − vi2)f(v)d3v

The larger σ2, the broader the velocity distribution of a population.

Good analogy: in the atmosphere, the wind speed is the mean velocity of air particles. The velocity dispersion is the thermal motion from the temperature of the air.



The scale heights provide information on the average velocity dispersion of the different components.

The random motion of stars in the azimuthal direction (perpendicular to the disk) yields a larger velocity dispersion, and this yields larger “scale

heights” (thickness) to those components. This has the effect of pressure.

For example, in the Solar neighborhood, Stars with ages < 3 Gyr have σz ~ 16 km s-1(and h~200 pc). Stars with ages > 6 Gyr have σz ~ 25 km s-1 (h~350 pc).

Total Density of stars obtained from star counting and distance determinations yields

n(R, z) = n0

e−|z|/hthin + 0.02e−|z|/hthick

e−R/hR

With hthin = 325 pc (combining both old and young thin disks), hthick~1.5 kpc and hR=3.5 kpc.

Normalized such that n = 0.2 pc-3 in the Solar Neighborhood for stars with 4.5 < MV < 9.5 mag.



Age-Metallicity Relation

Thin and thick disks have different scale heights, stellar densities, and metal fractions and ages !

Recall that stars have different metal fractions, different Populations.

Population I: high metal fractions, Z~0.02.

Population II: low metal fractions, Z~0.001.

Population III: zero metal fraction, Z~0. (hypothesized).

Astronomers commonly use the ratio of Iron (Fe) to Hydrogen (H) relative to that in the Sun to quantify the metal fraction. We call this the metallicity:

Stars with [Fe/H] > 0 have a higher metal fraction than the Sun. Stars with [Fe/H] < 0 have a lower metal fraction.



Age-Metallicity Relation

Stars with [Fe/H] > 0 have a higher metal fraction than the Sun. Stars with [Fe/H] < 0 have a lower metal fraction.

extremely metal-poor stars (Population II) have [Fe/H] ~ -5.4. Highest values are [Fe/H] ~ 0.6.

Studying Globular Cluster “turn-off” masses, younger clusters have high [Fe/H] then older clusters, which have low [Fe/H]. This is the

age-metallicity relation.


The Morphology of the GalaxyAge-Metallicity Relation

Rana 1991, ARAA, 29, 129

Time since formation of disk (Age - td, where td = 12 Gyr)

Solar Value



Thin Disk: typical iron-hydrogen ratios are -0.5 < [Fe/H] < 0.3.

Thick Disk: typical iron-hydrogen ratios are -0.6 < [Fe/H] < -0.4 (some as low as -1.6?!)

Which contains older stars ? Which “formed” first ?



Thin Disk: typical iron-hydrogen ratios are -0.5 < [Fe/H] < 0.3.

Thick Disk: typical iron-hydrogen ratios are -0.6 < [Fe/H] < -0.4 (some as low as -1.6?!)

Which contains older stars ? Which “formed” first ?

Appears that star formation began in thin disk about 8 Gyr ago, and is continuing today. This is supported by the cooling times of white

dwarfs in the thin disk.

Thick disk predated most of that of the thin disk by 2-3 Gyr, probably during the period 10-11 Gyr ago.


from Digital Sky LLC

Spiral Structure

Galaxy M 51Monday, August 20, 2012

Evidence for Spiral Structure


Milky Way Galaxy

Dust and gas sits in very narrow midplane. Fuel for new stars.

Composite near-IR (2 micron) Image from the Two Micron All Sky Survey (IPAC/Caltech/UMass)


Credit: Axel Aitoff


Milky Way Galaxy

The Galactic Bulge

COBE Satellite image of Milky Way at 1.2-3.5 micron.


The Galactic Bulge

Galactic Bulge: Independent component from disk. Mass of the bulge is believed to be ~1010 M⊙. Scale Height is ~100 to 500 pc, depending on whether younger stars are used (smaller scale heights) than older stars (higher scale heights).

Surface brightness (units of L⊙ pc-2 ) follows the “r1/4 law” distribution, discovered by Gerard de Vaucouleurs (1918-1995), also called the de

Vaucouleurs profile.

Our Bulge has an effective radius, re = ~0.7 kpc.

The Bulge is very difficult to observe because it is so centrally concentrated and there is a lot of dust and gas in the Galactic center. Must look in “windows”

with lower extinction (one is the so-called “Baade’s window”).

Stars in the bulge have -2 < [Fe/H] < 0.5. Possibly multiple metallicity groupings in bulge. One group is <200 Myr old, one is as old as 7-10 Gyr.


The Galactic BulgeSurface Brightness, I, has units of Luminosity per pc2.

The effective radius is defined such that half the luminosity is emitted from within Re.

Re

0dR R I(R) =

12

∞

0dR R I(R)

Ie = I(Re) is the surface brightness at the effective radius.

I(R) = Ie exp−7.669

(R/Re)1/4 − 1

Alternatively, surface brightness give by the following, also called “de Vaucouleur’s profile” or the “r1/4-law”.

Our Bulge has an effective radius, re = ~0.7 kpc.


The Galactic Bulge

Bulge shows (at least) two populations. One with low [Fe/H] and high [α/Fe], and one with high [Fe/H] and low [α/Fe].

McWilliam 1997


The Galactic Bulge

Stars with low [Fe/H] have high [α/Fe]. (Oxygen is an alpha element). Early metal production occurred from core-collapse Supernovae, which produce more Oxygen (and

other α-elements) compared to Fe (which comes from Type Ia Supernovae).

Gilmore et al. 1989



Thin Diskcontains Spiral Arms

Thick Disk


The Galactic HaloGalactic (“Stellar”) Halo is composed of globular clusters (GCs) and field stars.

Shapley thought GCs were spherically distributed. Now appear to be two populations.

Older (~11-13 Gyr), metal-poor GCs have [Fe/H] < -0.8, spherical distribution, with n~r-3.5.

Younger clusters have [Fe/H] > -0.8, in Galactic plane

Zinn 1985, ApJ, 293, 424


The Components of the Galaxy

Neutral Gas Thin Disk Thick Disk Bulge Halo

Mass (1010 M⊙) 0.5 6 0.2-0.4 1 0.3

LB (1010 L⊙) 0 1.8 0.02 0.3 0.1

M/LB - 3 ~10 3 ~1-3

Radius (kpc) 25 25 25 4 >100

Scale Height (kpc)

<0.1 0.35 1 0.1-0.5 3

[Fe/H] >+0.1 -0.5 to +0.3 -2.2 to -0.5 -2 to 0.5 < -5.4 to -0.5

Age [Gyr] <~ 10 8 10 <0.2 to 10 11 to 13


Kinematics of the Galaxy

Π ≡ dR

dt, Θ ≡ R

dθ

dt, Z ≡ dz

dt

Define “dynamic” Local Standard of Rest (LSR), to be the rest-frame of the Sun, moving in a perfectly circular orbit about the Galactic Center.

Alternatively, can use “kinematic” LSR, defined as rest-frame the “average” motions of stars in the solar neighborhood.

ΠLSR = 0, ΘLSRΘ0, ZLSR = 0



Relative to the LSR, a star’s peculiar velocity is

V = (VR, Vθ, Vz) ≡ (u, v, w)u = Π−ΠLSR = Π

v = Θ−Θ0

w = Z

Assuming the Galaxy is axisymmetric, then averages in the radial (u) and azimuthal (w) velocities should be zero, <u>~0 and <w>~0.



Not true for v component, because stars could be on different orbits.

Consider orbits in the figure above. The average v velocity is linked to the distribution of the

radial velocity, u. Therefore,

v = −Cu2where C is positive and sign follows from the fact that the stronger the radial

velocity dispersion, the more the rotational velocity will deviate.


Kinematics of the GalaxyThe phenomenon of asymmetric drift implies that stars in the solar neighborhood will

“drift” away from the Sun as they all orbit. This gives us a way to measure the peculiar velocity of the Sun (relative to the LSR).

v⊙ =−∆u,−Cu2 − v,−∆w

Measurements yield:

v⊙ = (−10, 5, 7) km s−1

which means that relative to the LSR, the Sun is moving (1) toward the Galactic Center, (2) more rapidly in the direction of the Galactic center, and (3) north out of the Galactic plane.


Kinematics of the GalaxyDifferential Rotation and Oort’s Constants

In 1927, Jan Oort derived a series of relations used to this day to determine the rotation curve

of the Galactic Disk.

r = R

sin θcos θ

r =

D sin R0 −D cos

sin θ = (D/R) sin

cos θ = (R0/R)− (D/R) cos

V = r = V (R)− cos θsin θ

Assuming that V⊙ ≈VLSR then we have

∆V = V −V⊙ =

V (R0/R)− V (D/R) cos − V0

−V (D/R) sin

∆V =

R0(Ω− Ω0)− ΩD cos −DΩ sin

with this becomesΩ(R) =

V (R)R



The radial and tangential velocities then follow by the projection of ΔV

vr = ∆V ·

sin − cos

= (Ω− Ω0)R0 sin

∆V =

R0(Ω− Ω0)− ΩD cos −DΩ sin

vt = ∆V ·

cos sin

= (Ω− Ω0)R0 cos − ΩD

Oort considered rotation near Sun, for which D<<R0

Ω− Ω0 dΩdR

R0

(R−R0)



vr = ∆V ·

sin − cos

= (Ω− Ω0)R0 sin

vt = ∆V ·

cos sin

= (Ω− Ω0)R0 cos − ΩD

Ω− Ω0 dΩdR

R0

(R−R0)

Recalling that Ω=Θ/R, then we get

vr

dΘdR

R0

− Θ0

R0

(R−R0) sin

vt

dΘdR

R0

− Θ0

R0

(R−R0) cos − Ω0D



Where the Oort Constants are

Or:

vr AD sin 2

vt AD cos 2 + BD

A ≡ −12

dΘdR

R0

− Θ0

R0

,

B ≡ −12

dΘdR

R0

+Θ0

R0

And A=14.8 km s-1 kpc-1 and B=-12.4 km s-1 kpc-1.Lastly, this allows us to write relations for the Solar Neighborhood as

Ω0 =V0

R0= A−B,

dV

dR

R0

= −(A + B)




Kinematics of the GalaxyHydrogen 21 cm “Tangent Point” Method for Kinematics

Sun

1

2

3

4

5

6

7

8


Kinematics of the GalaxyHydrogen 21 cm “Tangent Point” Method for Kinematics


Rotation Curves of the Milky Way

Rotation curve for our Galaxy. Strange thing is.... rotation curve is flat beyond the Solar circle, R0 = 8.5 kpc.

Clemens 1985, ApJ, 295, 422


r

R0Let Mass of Galaxy have a constant surface density, Σ, for r < R. Velocity is

then just from Newton’s Laws:

R

with yields

Solving for v, gives: for r < R

For r > R, we have:

Solving for v, gives: for r > R



Rotation Curves of Spiral Galaxies

Observations !v ~ constant (r0)



Rotation curve for our Galaxy. Strange thing is.... rotation curve is flat beyond the Solar circle, R0 = 8.5 kpc.

Clemens 1985, ApJ, 295, 422


Rotation Curves of Spiral Galaxies

Rubin, Thonnard, & Ford, 1978, ApJ, 225, L107

Vera Rubin (b1928)

Responsible for most of the work

on the “galaxy rotation rate”

problem.


You can work out what the matter density profile should be to match the observed rotation curves of galaxies.

Assume it is spherical:

rdr

Consider a spherical shell of radius r and thickness dr. The mass in the shell is dMr

Take Newton’s laws for the force acting on a particle (a star) in this shell.

rearranging and differentiating

Let the mass in the shell be

Then this leads to:


Solving for the density gives

A slight variation keeps the density from diverging at r → 0 :

This is the Dark Matter distribution in galaxies. True for the Milky Way and others.

Julio Navarro, Carlos Frenk, and Simon White in 1996 ran a series of cold-dark

matter simulations, and they came up with a “Universal profile” used today:


This is the Dark Matter distribution in galaxies. True for the Milky Way and others.

Julio Navarro, Carlos Frenk, and Simon White in 1996 ran a series of cold-dark matter compute simulations, and they came up with a “Universal profile” used today:

Simon WhiteCarlos FrenkJulio Navarro

This seems valid over an very large range of a and ρ0. For the smallest galaxies to the largest galaxy clusters.


Simon WhiteCarlos FrenkJulio Navarro


The Galactic Center

Challenging to observe because of all the dust/gas !

But, in 15 million years, the Sun will be 85 pc above the Galactic midplane, we would presumably have a much better view then !


The Galactic Center

Astronomers use high angular resolution images in the near-IR (~2 micron) to help see through the dust. This is helpful because

there are large number of K and M giant stars (T ~ 4000 K) in the central part of the galaxy, and

these are brightest in at 2-micron.

Note that the nearest star to the Sun is ~1 pc away. The density of

stars is much higher in the Galactic Center !

From Schödel et al. 2002Monday, August 20, 2012

The Galactic Center

Astronomers use high angular resolution images in the near-IR (~2 micron) to help see through the dust. This is helpful because there are large

number of K and M giant stars (T ~ 4000 K) in the central part of the galaxy, and these are brightest in at 2-micron.

Astronomer group led by Rainer Schödel and Reinhard Genzel followed the orbits of K-giants near the Galactic center.

One star, S2, has a period of 15.2 yr with eccentricity e=0.87 and perigalacticon distance of 1.8 x 1013 m = 120 AU (a few times bigger than

Pluto’s orbit).

You can work out from Kepler’s laws that the mass interior to S2’s orbit is ~3.5 x 106 solar masses.


The Galactic Center


The Galactic Center


The Galactic CenterProf. Andrea Ghez’s UCLA group.


http://www.astro.ucla.edu/~ghezgroup/gc/pictures/orbitsMovie.shtml

http://www.astro.ucla.edu/~ghezgroup/gc/pictures/orbitsMovie.shtml

The Galactic Center


The Galactic Center

Nature, Vol. 419, p. 694 (2002)


The Galactic Center

Nature, Vol. 419, p. 694 (2002)


The Galactic Center

Degeneracy between distance to center of Galaxy and Mass

of supermassive blackhole


Documents

Milky Way Galaxy - people.physics.tamu.edupeople.physics.tamu.edu/papovich/courses/fall12/milkyway.pdf · Milky Way Galaxy Composite near-IR (2 micron) Image from the Two Micron All