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Lecture 8

# Introduction to Astronomy-Lecture 8 Notes!.docx

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

Description
Unit # 5: Measuring the Properties of Stars:  Recall the distance units used in astronomy. 15  Light-year = 1ly = Distance that light travels in one year = 1ly=9.5 x 10 m.  Parsec = pc = 1 pc = 3.26 ly.  Astronomical Unit = AU = Average distance between the Earth and Sun. Distance via Parallax Method:  Sin 0 = Opposite = 1 = 0 (for small angles). Hypotenuse d  d = 1, if 0 is in seconds of are (”), then d is in parsecs (pc). 0”  d = 3.26, if we multiply by 3.26, then d is in light-years (ly). 0”  This method is good for stars at d < ~50 pc (163 ly).  *NOTE: “0” is “Parallax Angle”. Apparent Magnitude (m): nd  Hipparchus (Hippocras) 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 as bright as m = 2.  m = 2 is 2.5 times as bright as m = 3 etc.  Note that magnitude has No S.I. Units. Determination of Apparent Magnitude:  Star = Starlight = Photocell = Current = Meter = Computer  * NOTE: The Computer assigns the magnitude; the more negative the number the brighter the Sun. -26.5 m of Sun.  *NOTE: Most star names are Arabic, and one can tell how bright a star is by their name. A Magnitude Problem: *NOTE: QUESTION ON TEST # 2, DIFFERENT NUMBERS!  Which Star appears to be brighter? How MUCH brighter?  m = 8 is fainter than the barely-visible m = 6.  Formal has a smaller magnitude than W peg, so Formalhaut is brighter.  Triangle m = 8 – 1 = 7 Triangle m 7  Ratio of brightness = R = 2.5 = 2.5 =  ~ 610 times brighter. On test Absolute Magnitude (M):  Absolute magnitude (M) of a star is the apparent magnitude (m) that it would have if placed at a precise distance of 10 parsecs (32.6 ly) from us.  *NOTE: Test Question: Precise distance of 10 parsecs (32.6 ly) from us.  The equation which converts apparent magnitude to absolute magnitude is  M = m + 5 – 5log (d)  Where the distance, d, MUST be in parsecs. KNOW An Absolute Magnitude (M) Problem: * NOTE: Question on Test # 2, DIFFERENT NUMBERS! ONE OF 5 MATH-RELATED QUESTION ON TEST # 2! -6  What is the absolute magnitude of the Sun if its distance from us is 4.7 x 10 pc and its apparent magnitude is -26.5?  The Sun is the brightest celestial object.  M = m + 5 – 5log (d) -6  M = -26.5 + 5 – 5log ( 4.7 x 10 )  M = 5.14, with today’s, Smog and air pollution, the sun would be bare light-year (ly) visible if located 10 pc from us.  *NOTE: (SIGMA SIGN), “0” = 1/d = 1/1.35 = “0.74” R. Luminosity = L:  Luminosity, L, is the total amount of energy emitted by the star each second.  L = 4 (3.14) R (SIGMA SIGN) T where SIGMA SIGN = 5.67 x 10 W 8 2 4
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