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# 2.6.12 Unit 4.docx

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School
Ryerson University
Department
Physics
Course
PCS 181
Professor
Margaret Buckby
Semester
Winter

Description
PCS 181 2/6/2012 UNIT 4 Interstellar Matter  This dust and gas that occupy space between the stars.  Stars form in the dust/gas clouds in space. DUST  Dust is ~1% of the total visible mass in space.  Obscures our view of distant objects  Is a mixture of  Elongated tiny grains dirty ice  Grains of graphite (carbon)  Silicate particles  Complex molecules Interstellar Reddening  Light from a star passes through the dust  The dust particle are ~ same length as blue λ’s  The dust scatters blue λ’s more than red λ’s  More red photons pass unhindered through the dust and reach the observer  Star appears redder than it actually is Gas E ~ 100 times more gas mass than dust mass in the Universe Gas ÷ Dust ~ 100  The visible Universe contains  ~ 75% hydrogen  24% helium  <1% Other Hi Regions  These are clouds of neutral (= NOT ionized) Hydrogen  Contain ~ 1 M to 100 M o  The cold Hydrogen gas is in its ground state (n = 1 orbit)  The Hydrogen atom has TWO possible ground state configurations  The electron may undergo a spontaneous “spin flip”  From the aligned (= parallel) state of the  Opposed (= anti-parallel) state  Excess energy is released as tiny burst of radiation  = a photon of definite wavelength λ = 21 cm = 0.21 m  In a radio region of EM spectrum  The corresponding frequency is  F = 1.42 x 10 Hz  = 1.42 Ghz  = 1.42 GigaHertz And… To search for messages from extra-terrestrials Star birth Scenario  Consider a giant interstellar cold molecular H 2loud.  > ~ tens of ly across 4 6  Contains (10 – 10 ) M o  Gravitational force ∝ mass  Random particle fluctuations in the cloud act as gravitational foci to attract more material  Contracting condensates in cloud dissipate their gravitational energy as heat (infrared radiation)  Material falling onto a condensate   internal heat and shock waves  H mol2cules dissociate  This contracting object is a PROTOSTAR  Core of protostar = the central 10% - 15%  Core reaches 10 million K  = 1 x 10 K6 7  = 1 x 10 K   hydrogen-based nuclear reactions begin in the core  A STAR IS BORN A Main Sequence Star has H-Based nuclear reactions in the core The entire birth process for an average star takes 10 to 10 years The process is faster for massive stars The Main Sequence stage 6  T core 10 x 10 K   Hydrogen fusion nuclear reactions in the core  A star spends its life in a delicate balance between inward gravitational pull  ~ 90% - 95% of stars are Main Sequence…  Including our Sun Hydrogen Fusion Reactions = Proton-Proton Chain 1 1 2 + H+ H  H + e + v (deuterium, positron, neutrino) 2 1 3 H+ H  He + γ (helium isotope, radiation burst) 3He+ He  He + H+ H ( helium, hydrogen) Overall, four Hydrogen nuclei fuse to form one Helium nucleus and… A small amount of mass is released as energy (4.3 x 10 -12J). 38 BUT... a typical star has 10 such reactions each second! So the energy produced each second is (4.3 x 10 -12J) (1 x 10 ) = ~ 4 x 10 W6 This is the luminosity (energy per second) observed for our Sun Luminosity = L = 4 π R σ T = energy/ s Life on the main sequence  Star fuses Hydrogen nuclei into Helium nuclei deep in its core via p-p chain 6  Core temperature must be at least 10 x 10 K = 10 million K  4 H  1 He 38  ~ 10 such reactions occur each second The Sun  Massive stars only live for a few million years  Normal stars live for billions of years  Our sun is a normal star and will fuse Hydrogen in its core for a total of ~10 billion years  Our sun is now 4.5 billion years old Nemesis ~ 55% of stars have a companion Mass extinctions occur on Earth ~ every 25-30 million years Perhaps the Sun has a never-observed companion, which comes near to the Sun periodically. It disturbs the orbits of near comets, asteroids, meteors. These may crash into Earth and precipitate mass extinctions Characteristics of Nemesis If it exists, Mass ~ 0.01 M o Period ~ 26- 30 million years The Red Giant Phase Core runs low on Hydrogen fuel and contracts dramatically Core contraction  Temperature rises in core 6 At T = 100 x 10 K = 1 x 10 8 = 100 million K Helium Fusion Reactions = Triple α Process 4 4 8 He + He  Be + Y (Beryllium, radiation burst) 4 8 12 He + Be  C + Y (Carbon, radiation burst) Shell Fusion 6 H fusion: T core= 10 million Kelvins = 10 x 10 K He fusion T core 100 million Kelvins = 100 x 10 K 6 Core energy + Shell energy  increased radiation pressure outward. Star expands and surface reddens = a Red Giant A Red Giant star is NOT fusing Hydrogen in its core. A very massive star can undergo Several Red Giant phases. Successive core contractions can produce He, C, O Ne, Mg, Si Evolutionary Stages of a Normal Star  Our Sun is a normal star and is NOT massive  Our Sun and stars of similar mass can ONLY fuse H He and He  C  When our Sun has a core consisting of Carbon (and a little bit of Oxygen), it will DIE because our Sun has insufficient mass to perform C-based nuclear reactions needed to generate outward pressure and balance the inward pull of gravity End of Star formation Stars FORM in vast interstellar Hydrogen and Helium clouds As the Universe ages, Hydrogen is being transformed into heavier elements inside the stars The Hydrogen component of the Universe is always declining. The abundance of heavier elements are increasing UNIT 5 Distance via Parallax Method  We introduce a parallax angle Ɵ  Sin Ɵ = opposite/ hypotenuse = 1/d = Ɵ (for small angles)  d = 1/ Ɵ” if Ɵ is in seconds of arc (“), then d is in parsecs (pc)  d = 3.26/ Ɵ” if we multiply by 3.26, then d is in light years (ly)  This method is good for stars at d < ~ 50 pc (163 ly) Apparent Magnitude (m)  Hipparchus (Hipparcos) of Greece (2 century B.C.) divided the VISIBLE stars into 6 groups  M = 1 (brightest star in Greek Sky)  M = 2  M = 3  M = 4  M = 5  M = 6 (barely visible) M = 1 is 2.5 times brighter than m = 2 M = 2 is 2.5 times brighter than m = 3 etc. th W Pegasi is an 8 magnitude ( m = 8) star st Fomalhaut is 1 magnitude (m =1) Which star appears to be brighter? M = 6 is barely visible M = 8 is even fainter  Fomalhaut (m = 1) is brighter than W peg Absolute Magintude Absolute magnitude (M) = apparent magnitude (m) of star AT.. AT a distance of 10 parsecs (=32.6 ly) from us To convert m to M: M = m + 5- 5log (d) The distance, d MUST be in parsecs Problem: What is the absolute magnitude of the Sun? -6 D o 4.7 x 10 pc Apparent magnitude is -26.5 ( = brightest celestial object) M = m + 5 – 5log(d) M = -26.5 + 5 – 5log(4.7 x10 )-6 M = 5.14 Today’s smog and air pollution  the sun would be barely-visible at 10 pc from us. Absolute Magnitude Problem What is the absoulute magnitude of Alpha centauri? Parallax? DαCen = 4.4 ly Apparent magnitude = m = -0.01 We must convert the distance to parsecs 4.4 ly x 1 pc. 3.26 ly = 1.35 pc M = m + 5 -5log(d) M = -0.01 + 5 -5log(1.35) M = 4.34 αCen is dimmer if pushed out to 10 pc (=32.6 ly) Its parallax angle 0 (in arcsec = seconds of arc) is 0 = 1/d =1/1.35 = 0.74” Luminosity = L Luminosity, L is the total amount of energy emitted by the star each second 2 4 L = 4 π R σ T Where σ = 5.67 x 10 W/ m K 2 2 26 L o 3.9 x 10 W Radius Determination Problem The surface temperature of the Sun is 5800K R on m? 2 4 If L = 4 π R σ T , then R = [ L/ 4 π σ T ] 1/2 26 -8 4 1/2 R = [ 3.9 x 10 / 4 π (5.57 x 10 ) (5800) ] R = ~ 7 x 10 m = R o The HR Diagram A plot of stellar luminosities or magnitudes versus surface temperature or spectral type. The Harvard Classification System (= spectral type) is O B A F G F K M Spectroscopic “Parallax” This method enables us to determine the distance to farther stars. It is more inaccurate than the parallax angle method. Find the distance (in pc and in ly) to a main sequence star with apparent magnitude of -1.1 and a surface temperature of 10000 K HR diagram  absolute magnitude, M = ~2 Recall: M = m +5 -5log(d)  Log(d) = m –M +5 / 5 (m-M+5)/5 d= 10 d = 10 (-1.1 -2 +5)/5 Spectroscopic Parallax Problem 2 Spica is a main-sequence star, Spectral class B0 apparent magnitude 0.91 HR diagram  M ~ -3.5 D = 10 (m –M+5)/5 D = 10 (0.91 +3.4+ 5) /5 1.882 D = 10 D = ~76 pc Period luminosity curve for Cepheid variable stars A Cepheid variable s
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