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Lecture 8

# Introduction to Astronomy-Lecture 8 Notes!.docx

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Ryerson University

Physics

PCS 181

Margaret Buckby

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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