Introduction to Galaxies - Stony Brook Astronomy€¦ · Introduction to Galaxies ... •Note that...

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Introduction to Galaxies Basic Unit of the Universe = Galaxy Galaxy - gravitationally-bound aggregate of a) 10 6 -10 11 stars b) Gas (few 10s of % by mass) & dust (M gas /M dust ~ 200-500) c) Dark matter Note that stars form from gas, so the gas content of galaxies must have been higher in the past

Transcript of Introduction to Galaxies - Stony Brook Astronomy€¦ · Introduction to Galaxies ... •Note that...

Introduction to Galaxies

• Basic Unit of the Universe = Galaxy

• Galaxy - gravitationally-bound aggregate of

a) 106-1011 stars

b) Gas (few 10s of % by mass)& dust (Mgas/Mdust ~ 200-500)

c) Dark matter

• Note that stars form from gas, so the gas content ofgalaxies must have been higher in the past

Galaxies

• Composition = stars & gas are ~ 70% H, ~ 25% He, ~few % metals by mass

• Masses ~ 106 - 1012 Msun (1 Msun = 1.99x1030 kg)

• Sizes ~ 10 - 100 kpc

• Rotation Period of Galaxies ~ 10 - 100 Myr (I.e., theones that do rotate!)

• Average separation ~ 1 Mpc

Age of the Universe – Hubble Diagram (1926)

• v (km s-1) the Doppler motion Δλ / λ = v / c• R (Mpc), where f ~ R-2.• So, v ~ R, v = H0R

Age of the Universe –Hubble Diagram

(1926)

• H0 = 50-100 km/s/Mpc (we’re adopt H0 = 75 km/s/Mpc)• tH = R/v = 1/H0 = 1-2x1010 yr ago (~ Age of Universe)• Age of Earth = 4.5x109 yr• Note: We’ve ignored acceleration/deceleration for this

calculation

Typical spiral galaxy - Milky Way

• Number of stars ~ 1011

• Mass ~ 1012 Msun

• How many times has the Sun orbited the galactic center? Distance of Sun from the center of the galaxy ~ 8.5 kpc Time for one orbit t = 2πR / v = 2π 8.5 kpc / 250 km s-1 = 2x108 yr Thus, the Sun has made 4.5x109 / 2.0x108 ~ 20 turns

around the galactic center

How often do stars collide?

• The Number density of stars in the disk is,

• The mean cross section, σ, of stars is calculated byassuming every star is like the sun,

Volumedisk = (thickness)π (radius)2 = HπR2

= (3×1019m)π (3×1020m)2 = 8.5 ×1060m3

n =Nstars

V=

1011stars8.5 ×1060m3 =1.17 ×10−50m−3

σ = π (2Rsun )2 = 6.08 ×1018m2

Mean freepath λ =1nσ

=1.4 ×1028km

Stellar collisions (continued)• Given that vrandom = 40 km s-1,• Stars collide every,

• I.e., not very often• Note that considering the gravitational cross section only

lowers this time by a factor of 100.• The benefit of this is that we can treat stars as pointmasses, & galaxies as a continuous mass distributionresponding to the gravitational field in the galaxy.

tcollision ~λ

vrandom=1×1019 years

Stars• In actuality, every star is not like the sun. Stars range in

mass from 0.1 Msun to 100 Msun.• Objects with M < 0.1 Msun cannot ignite• Objects with M > 100 Msun would be blown apart by

radiation pressure

Interstellar Gas & Dust

• Normal ISM has gas number density n = 105 - 106 m-3 &temperature T ~ 100 K

• Dust has a size a ~ 0.1 µm (10-7 m) & a refractorycomposition (metals - e.g. SiO2)

Disk Gas Depletion Rate of the Milky Way• SFR ≈ 3 Msolar yr -1• M(H2) ≈ 2.5x109 Msolar• Depletion time = 2.5x109 / 3 = 8x108 yr

• The Sun takes 2x108 yr to orbit the galaxy, thus depletion timewill occur in 4 revolutions

• The universe is 15 Gyr, thus depletion time takes 5% the age ofthe universe

• Most spiral galaxies have several x108-9 Msolar of molecular gas

• Why does the MW have 2.5x109 Msolar of molecular gas? Whynot 0?

• Given the rapid depletion time & the age of the universe, whereare the dead disks (I.e., spiral galaxies with no gas in theirdisks)?

Gas Depletion

• The determinations of SFRs are incorrect• The determinations of the amount of molecular gas in the MW

and other Spiral Galaxies is incorrect• Molecular Gas is replenished by infalling atomic gas.

Possible Solutions