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Lecture 5

# PCS 181 Lecture Notes - Lecture 5: Large Magellanic Cloud, Proper Motion, Stellar Kinematics

Department
Physics
Course Code
PCS 181
Professor
Raffi Karshafian
Lecture
5

This preview shows page 1. to view the full 5 pages of the document. Unit 5: Measuring the Properties of Stars
Distance Units
-lightyear=ly: distance that light travels in 1 year, 1 ly=9.5x1015m
-parsec=pc: 1pc=3.26 ly
-astronomical unit=AU: avg distance between Earth and Sun
Distance Via Parallax Method
-we introduce a parallax angle, θ
-sin θ= (for small angles
-d= if θ is in seconds of arc (“), then d is in parsecs (pc)
-d= if we multiply by 3.26, then d is in lightyears (ly)
-this method is good for stars at d < ≈50 pc (163 ly)
Apparent Magnitude (m)
-Hipparchus (Hipparcos) of Greece (2nd century BCE) divided VISIBLE stars into 6
groups
-m=1 (brightest star in Greek sky)
2
3
4
5
-m=6 (barely visible)
-magnitude of Sun is -26
-m=1 is 2.5x brighter than m=2
-m=2 is 2.5x brighter than m=3, etc
-note that magnitude has NO S.I units
Apparent Magnitude (m) Problem
-W Pegasi is an 8th magnitude (m=8) star
-Formalhaut is 1st magnitude (m=1)
-which star appears to be brighter?
-how MUCH brighter?
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Only page 1 are available for preview. Some parts have been intentionally blurred. -∆m=8-1=7
-ratio of brightness=R=2.5∆m=R2.57≈610x brighter
Absolute Magnitude (M)
-absolute magnitude (M)=apparent magnitude (m) of star AT a distance of 10 parsecs
(=32.6 ly) from us
-to convert m to M:
M=m+5-5log(d)
-distance, d, MUST be in parsecs
Absolute Magnitude (M) Problem
-what is absolute magnitude of the Sun?
-do=4.7x10-6 pc
-apparent magnitude is -26.5 (=brightest celestial object)
-M=m+5-5log(d)
-M=
-M=5.14
-today’s smog and air pollution the Sun would be barely visible at 10 pc from us
Luminosity=L
-luminosity, L, is total amount of energy emitted by star each second
-L=
-
-
-R=
-surface temperature of the Sun is 5800K.
-Ro in m?
-if L=4πR2σT4
-R=
-R=
-R=7x108 m
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