midterm2-review

midterm2-review - Midterm #2 Review Optical depth: 7' : nal...

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Unformatted text preview: Midterm #2 Review Optical depth: 7' : nal optically thin: 7' << 1 7' = fraction of radiation absorbed only in this limit!). 10 11 I2 In In general, split the domain into thin slabs, such that d7 << 1 Change in radiation as we pass through one such slab is d] = —Id¢ Integrating, we find I : [De—T AH d7 Optical depth can be thought of as the number of mean free paths a photon must move until it reaches the surface of the star. (Carrolandoaie) Mean free path: L : 1/(na) AsT 203 (spring 2011) Midterm #2 Review 200310128 05:24 UT This can explain limb darkening: Consider equal optical depths along two lines of sight (one to the edge and one to the center) When we look toward the edge, we go through the atmosphere at an oblique angle, so we hit 7 2 lat a higher radius Since we see deeper layers when looking toward the center, we are seeing hotter material —> appears brighter. AsT 203 (spring 2011) Midterm #2 Review The most important reaction sequence in the Sun is the proton- proton chain. Slightly more massive stars use the CNO cycle. “6‘ w ‘H l l \ l’ r >V<\ A" fie? e’< o“ l v r3?v no [7' (“N 'eN ‘« \ r V 110 V $6 ’VK \W 0’}va ’Y Y 13 o Proton 'Y Gamma Ray y g 0 Neutron V Neutrino V a J O Positron ASTm (Spring 2011) Understand why high T is needed for fusion, why energy comes out... Midterm #2 Review For the p-p chain,4H —> 4He , the mass difference in this reaction is E : Am02 : 0.03 mp02 If we take the Sun to be 70% hydrogen and assume that all the hydrogen nuclei fuse, we have an energy release of 33 $ - 0.03(3 x 1010 cm s—l)2 : 9.45 x 1051 erg M E = 0.7— - 0.03 mp8 2 0.7 4771p The timescale that the Sun can stay on the main sequence is E 9.45 x 1051 erg 18 10 In reality, only about 10% or so of this will fuse. AST 203 (Spring 2011) Midterm #2 Review HSE: at every radius inside the star, P is such that it exactly balances the weight of the material above it. pressure ——> gravity 4— dP E * P9 GM 2 P N Understand how to use this to estimate the pressure in the center of a star or at the base of an accreted layer (see your (lomsenneue.a..) h o m Cunyrml em» human Emcelim Immorng a» my- Wsflev‘ AST 203 (Spring 2011) Gravitational Potential Energy Gravitational potential energy between two point masses: U 2 _ Gmlmg 74 For a spherical mass distribution, we add up (integrate) the gravitational potential energy between all particles: GM2 Uz—a R where a is ~ 1. For a layer of mass AM on the surface of a star of mass M)h gravitational potential energy of the layer is: GM*AM U:— R AST 203 (Spring 2011) Midterm #2 Review Summary of low Mass star evolution main sequence: core H burning giant branch: H shell burningl helium flash: explosive He core ignition (still H shell burning) horizontal branch: core He burning; H shell burning asymptotic giant branch: H and He shell burning AST 203 (Spring 2011) 105 ’0‘) E planetary nebula 1 V 10‘ ’4-'-——'—“ / / 10’ / / asymptotic giant A red :1? 102 _ / giant >‘ horizontal ’5 10 \ bramh subgiant ‘5 \ E 1 \ main-sequence 104 dwarf 10‘: white dwarf 10‘3 T. 4 I —J L _l 40,000 20,000 10,000 5,000 2,500 effective temperature (K) Figure 8.10. The complete evolution of a low-mass star from the main sequence to a white dwarf. The track from the asymptotic giant branch to a white dwarf (via a planetary nebula) is un- certain and is shown as a dashed curve. Midterm #2 Review Understand on H-R diagram: Lines of constant radius i_ i 2 1 4 Lo — Re To Main-sequence lifetimes LIEL : Masses How massive stars evolve, where 23:01? ~ 08 1715 ll white dwarfs are, 1071 .2 A l 10 white dwarfs ' 0‘21 U” 1073 , ‘ \ .1531 x 104 . 1‘ i . 1 40000 20000 10000 5000 2500 T00 AST 203 (Spring 2011) Midterm #2 Review High mass stellar evolution: When the carbon is spent, neon burning begins, and then oxygen and silicon burning. Eventually, we are left with an inert iron core Many layers of different types of shell burning. The envelope of the star expands in response to the luminosity, and the star becomes a red supergiant. H burning He burning C burning Ne burning O burning Si burning Fe core AST 203 (Spring 2011) Midterm #2 Review For an ordinary gas, the pressure is given by the ideal gas law: Average mass p P : nkT : per particle At large densities, electron degeneracy becomes important. Electrons are packed together so tightly that quantum mechanics ascribes them a large momentum: AxAp > h A degenerate gas exerts a pressure even at zero temperature. Electron degeneracy pressure is r 5/3 0/3 4/3 4/3 p 2 K é L p 2 K1 E L 14 1 g cn1’3 14 1 g cn1’3 non-relativistic relativistic Maximum white dwarf mass is set by the maximum weight that relativistic electron degeneracy can support. Chandrasekhar limit is 1.4 M AST 203 rep. ing 2011) o ' Midterm #2 Review Know how to convert between mass density and number density. Charge neutrality: n6 2 an Here, 112 is the number density of the nuclei Total mass density = mass density of the nuclei + the mass density of the electrons: p = Arnan + mane ~ Ampnz SO AST 203 (Spring 2011 :1 Midterm #2 Review You should be prepared to explain in a few sentences: Binding Energy of Nuclei / Fe-peak Doppler broadening Type II supernova Sunspots Type la supernova Structure ofthe Sun Classical nova Optical depth/mean free path X-ray burst Limb darkening Pulsar Stellar evolution Cepheid variable Hydrostatic equilibrium Population l/ll/lll stars Convection/radiation (random walk) Supernova remnant Regimes ofthe H-R diagram Planetary nebula Stellar lifetimes, masses, radii on H- R diagram White dwarf/Chandrasekhar limit Neutron Star Close binary evolution/Roche lobes Degeneracy pressure This list is not complete... AST 203 (Spring 2011) ...
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