1/5/011 The Milky Way Galaxy Distribution of Globular Clusters around a Point in Sagittarius About 00 globular clusters are distributed in random directions around the center of our galaxy. 1
1/5/011 Structure of the Milky Way Galaxy Disk: stars, gas, dust, O and B stars, emission nebulae, open clusters, high metal abundance young objects dominate. Halo: globular clusters, red dwarfs, low metal abundance old objects dominate. Disk Bulge about 75,000 ly Disk includes spiral arms, delineated by young objects such as O and B stars. Halo Bulge: old, low metal abundance stars. Stellar Populations Population I Population II Extreme Intermediate t Intermediate t Extreme Location Spiral Arms Disk Nuclear Bulge Halo Metals (%) 1.6 0. Less than 0. Orbit Shape Average Age (yr) Circular 0 Million or Less Slightly Elliptical Moderately Elliptical Highly Elliptical 0.- Billion - Billion -1 Billion What does this table suggest about the origin of the Milky Way Galaxy?
1/5/011 Composition of the the Disk Component and the Spherical Component of the Galaxy Spherical Component Includes the nuclear bulge. Globular clusters. Old stars. Low metal abundance. Little gas and dust. Orbits have high eccentricities and random orientations Population II Disk Component Includes spiral arms. Open clusters. Young stars. High metal abundance. Lots of gas and dust. Orbits nearly circular and all revolve in the same direction with approximately the same orbital orientation Population I Measuring the Structure of the Galaxy Using the Doppler Effect Spiral Arm Tracers Sun Zone of Avoidance Why is there a zone of avoidance?
1/5/011 Rotation Curve of the Milky Way Galaxy Example 1 The Sun moves in a nearly circular orbit around the center of the Galaxy, which is about 7,700 ly away. How long does it take to complete 1 revolution? 4 15 r=7,700 ly =.77 9.46 m =.6 0 m ( ) ( ) circumference = π r = π.6 0 m = 1.65 1 m circumference time = speed ( )( ) speed = 0 m / s 1 15 1.65 15 7.50 = s = 7.4 s = =.7 years 7 0.16 Example Using the data from Example 1, calculate the mass of the part of our galaxy that is inside of Earth s orbit. a M = if a is in AU's andpisinyears;this in gives the mass in solar masses. P 0 0.6 9 a =.6 m = AU = 1.75 AU 11 1.5 9 ( 1.75 ) (.7 ) M = 11 M P =.7 years 4
1/5/011 Example How much mass is inside the orbit of a star 16 kpc from the center of the Galaxy? ( )( ) 16 0 9 a = 16kpc = 16.09 m = 4.94 m =.9 AU 5 v = 0km / s =.0 m / s 0 ( π )( 4.94 ) πa v. 16 P = = = 1.41 s = 4.65 years 5 9 (.9 ) ( 4.7 ) M = = 11 1.65 M Most of the Galaxy s mass is evidently not visible. More than 90% of it may be part of the dark halo, also known as the galactic corona,but it is not yet known what kind of mass it is. Other galaxies also have a large dark matter component. Spiral Structure of Our Galaxy The Sun is about 7,000 ly from the center. The thickness of the disk at the location of the Sun is only about 00 pc What causes the spiral arms? What causes the spurs? 5
1/5/011 Evidence for Spiral Structure Spiral arm tracers are large, luminous objects that are concentrated in the spiral arms. This includes O and B stars, HI clouds, and HII regions. O and B stars are concentrated in spiral arms because that is where most stars are formed, and they are bright and shortlived. After birth, they spend most of their lives near their birthplaces in the spiral arms. HII regions are large and bright. They are created by the ionizing UV radiation from O and B stars. What causes a galaxy to have spiral arms? X-ray Image of The Center of Our Galaxy - Sagittarius A * The nucleus of our galaxy contains a million solar mass black hole. This fact is deduced from the motion of objects near the center. The lobes of hot gas have temperatures of about 0 million kelvins. High temperatures are presumably due to supernova shock waves and stellar wind from hot stars. http://chandra.harvard.edu/photo/00/00long/index.html.4 arcminutes 6
1/5/011 Radio Image of Sagittarius A Central Region of Sagittarius A pc 7
1/5/011 Near Infrared Image of Stars Near SagA * (Evidence for the Presence of a Supermassive Black Hole) Speeds as high as 1400 km/s imply that these stars orbit a black hole with a mass of about million solar masses. Example 4 One of the stars near SagA * has an orbital speed of 1400 km/s and is about 0.0057 pc from SagA *. Calculate the mass of the central object. km v = 1400 a = 0.0057pc s a = ( 0.0057)( 06, 65AU) = 1.1 AU 16 ( 0.0057)(.06 m) = 14 = 1.77 m πa P = v P = 5.yr M ( 11 1.1 ) ( 5.) 11 ( ) π 1.77 km = 1.4 km / s = 6 =.6 M = 7.94 s = 7.94 s 7.16 s / yr
1/5/011 Problem 14.47 1 5 d = 160000ly = 1.51 m v = 00km / s =.00 m / s (.00 5 m / s) ( 1.51 1 m) GM v d 4 1 v = M = = =.04 kg = 1.0 M d G 11 Nm 6.67 kg Problem 14.4 v = 00 m / s d = 0 6400 m G = 6.67 11 Nm kg vd 6 6 M 7.69 kg. M = = = G 9