Transcript Chapter 3a powerpoint presentation

Chapter 3
Stars: Distances & Magnitudes
 Nick Devereux 2006
Revised 8/2012
Astrophysical Units & Constants
In addition to the usual list of physical constants – listed in
Appendix A (pg A-2), there is another list of astronomical
constants that we must be familiar with and these are listed in
on page A-1.
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The Sun as the “yardstick”
Since the distances and masses are so large in astronomy,
the basic units of measurement are expressed in terms of the
Sun.
The Astronomical Unit (AU) is the distance between the Earth and
the Sun,
1 AU = 1.496 x 1011 m
Which is perhaps more familiar to you as 93 million miles.
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The parsec
for larger distances,
there is the parsec (pc)
1 pc = 3.086 x 1016 m
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Angular units
Astronomers can measure the angular extent on the sky for
celestial objects, even if they don’t know how far away they
are, and therefore unable to attribute a linear size.
Radians
360o = 2 radians
So,
or,
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1o = 2  /360 radians
1o =  /180 radians
Angular Units (continued)
Arc seconds (‘’) and Arc Minutes (‘)
1‘ = 60 ‘’
1o = 60 ‘ = 3600 ‘’
From previous page…
1o =  /180 radians
So,  /180 radians = 3600 ‘’
 /(180 x 3600) radians = 1‘’
Or,
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1‘’= 4.8 x 10-6 radians
and 206265 ‘’ = 1 radian
Getting back to the
pc…..
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The pc
(radian) = arc /radius
1‘’/ 206265 ( ‘’/ radian) = 1 AU / 1pc
So, 1 pc = 1 AU x 206265
1 pc = 1.496 x 1011 x 206265 m
Or,
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1 pc = 3.086 x 1016 m
In words, a parsec is the distance
at which the separation between
the Earth and the Sun could be
resolved with a medium sized
telescope…
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What does resolved mean?
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The Resolving Power of a
Telescope
Depends on both the size of the telescope mirror, D,
and the wavelength,  ,of the light under observation.
(radian) = 1.22  /D
with  and D in the same units.
For the Hubble Space Telescope D = 2.4m
and  = 0.05‘’ @  = 5500 Å .
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Question: How far away would
you have to hold a dime (2cm in
diameter) for it to subtend an
angle of 1 , 0.l ?
‘’
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‘’
Mass
Usually, the masses of stars, galaxies, clusters of galaxies
are given in terms of the mass of the sun,
1 M  = 1.99 x 1030 kg
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Measuring Brightness
Brightness is measured in a variety of ways;
eg. Magnitude, Flux, and Luminosity
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Luminosity of the Sun
The luminosity of the Sun,
1 L  = 3.9 x 1026 W
You may recall that Watts = Joules/sec, so the
luminosity of the Sun is a measure of the rate of
flow of energy through the surface of a star.
Concept: Think of luminosity as the rate at
which a star emits packets (photons) of
energy…
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Flux
The further you move away from the star, the flux of photons,
(measured in units of W/m2) passing through a 1m x 1m area
goes down as the reciprocal of the distance squared;
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Flux (continued)
Quantitatively,
the flux f = L/4D2
Units: W/m2
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Magnitudes
The magnitude scale dates back to the Greek astronomer
Hipparchus (200 BC).
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Definition of Magnitude
The human eye perceives, as linear, what are actually logarithmic
differences in brightness.
m = -2.5log f + c
m is the apparent magnitude
f is the flux
c is a constant related to the flux of a zero magnitude star
Note the –2.5, the brighter the star (f increases), the more
negative the magnitude (m decreases).
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Differences in magnitudes are
equivalent to ratio’s of fluxes
This obviates the need to know the constant c, or, the zero point,
of the magnitude scale because;
m1 = -2.5log f1 + c
m2 = -2.5log f2 + c
m1 – m2 = 2.5 log f2 / f1
Note the c’s cancelled (c – c = 0)
Also, beware the subscripts are reversed on either side f the
equals sign.
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Question: A binary star system
has one star that is 8 times
brighter than the other. What is
the magnitude difference
between the two stars?
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Absolute Magnitude
We are unable to tell just by looking at the night sky if one star
is fainter than another because it is intrinsically fainter (ie. lower
luminosity) or just further away.
To realistically compare stars on an equal basis we introduce the
concept of Absolute magnitude (M) which is the magnitude stars
have if they are all placed at the same reference distance of 10 pc.
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--------------------------------------------------------------------- * (2)
d(pc)
----------------------------* (1)
10 pc
f1 = L/4(10)2 and f2 = L/4  d2
M – absolute magnitude = -2.5log f1 +c
m – apparent magnitude = -2.5log f2 +c
Then,
M-m = 2.5log f2 / f1
M-m = 2.5log L 4 (10)2 /4  d2 .L
M-m = 2.5log 100/d2
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Distance Modulus
M-m = 5 – 2.5log d2
M-m = 5 – 5log d
So, the absolute magnitude,
M = m + 5 – 5log d
(remember, the distance d must be in pc)
On rearranging,
m - M= 5log d – 5
Where the quantity m – M is known as the distance modulus
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Trigonometric Parallax
Of course, to calculate the absolute magnitude of a star, we must
know it’s distance. Distances to nearby stars can be found using
Trigonometric parallax.
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Parallax Angle 
Question:
Given the definitions for angular units provided earlier
show that the parallax angle,  measured in arc seconds is
equal to the reciprocal of the distance to the star in pc.
So that,
 = 1/d
The nearest star a Centauri at a distance of 1.3 pc has a parallax
angle  = 1/1.3 = 0.77 . All other stars have even smaller
parallaxes.
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