Download Stars Part One

Stars Part One:
Brightness and Distance
Concept -1 – Temperature
λmax (metres) = 2.90 x 10-3 m k
T (Kelvin)
λmax = Peak black body wavelength
T = The star’s surface temperature in Kelvins
λmax = 2.90 x 10-3 m K
T
λmax
From Jay Pasachoff’s Contemporary Astronomy
Put this in your notes: A star has a surface temperature of
5787 K, what is its λmax?
λmax = (2.90 x 10-3 m K)/T
= (2.90 x 10-3 m K)/5787 = 501 nm
501 nm
A star has a λmax of 940 nm, what is its surface temperature?
λmax = (2.90 x 10-3 m K)/T,
T = (2.90 x 10-3 m K)/ λmax
= (2.90 x 10-3 m K)/ (940E-9) = 3100 K
3100 K
Concept 0 – Total power output
Luminosity L = σAT4
L = The star’s power output in Watts
σ = Stefan Boltzmann constant = 5.67 x 10-8W/m2K4
A = The star’s surface area = 4πr2
T = The star’s surface temperature in Kelvins
Put this in your notes: Our Sun has a surface temp of
about 5787 K, and a radius of 6.96 x 108 m. What is its
Luminosity?
L = σAT4 = 5.67x10-8(4π(7x108 ) 2 )(5787)4 = 3.87x1026 Watts
3.87x1026 Watts
A star has a radius of 5.0 x 108 m, and Luminosity of 4.2 x 1026
Watts, What is its surface temperature?
Luminosity L = σAT4 ,
T =(L/(σ4(5x108)2)).25 =6968 K = 7.0x103 K
7.0x103 K
Concept 1 – Apparent Brightness
Apparent Brightness b = L
4πd2
b = The apparent brightness in W/m2
L = The star’s Luminosity (in Watts)
d = The distance to the star
L is spread out over a sphere..
Apparent Brightness b = L/4πd2
The inverse square relationship
From Jay Pasachoff’s Contemporary Astronomy
Put this in your notes: Our Sun puts out about 3.87 x 1026
Watts of power, and we are 1.50 x 1011 m from it. What is
the Apparent brightness of the Sun from the Earth?
b = L/4πd2 = 3.87E26/ 4π1.50E112
= 1370 W/m2
1370 W/m2
Another star has a luminosity of 3.2 x 1026 Watts. We measure an
apparent brightness of 1.4 x 10-9 W/m2. How far are we from it?
b = L/4πd2 , d = (L/4πb).5 = 1.3x1017 m
1.3x1017 m


8 W
2.52x10 m 2 
m  2.5log10

Apparent Magnitude:
b




b = The apparent brightness in W/m2
m = The star’s Apparent Magnitude
Note that the smaller b is, the bigger m is.
 in b = -5 in m
Logarithmic scale: x100
Concept 2 – Apparent Magnitude
Put this in your notes: What is the apparent magnitude of a star
with an apparent brightness of 7.2x10-10 Wm-2? What is that of a
star with an apparent brightness of 7.2x10-12 Wm-2?
3.9, 8.9
From Jay Pasachoff’s Contemporary Astronomy
So apparent magnitude is a counter-intuitive scale
-27 is burn your eyes out,
1 is a normal bright star, and
6 is barely visible to a naked, or unclothed eye.
(You can’t see anything if you clothe your eyes!!)
A magnitude 1 star delivers 100 times more W/m2
than a magnitude 6 star.
What is the Apparent Magnitude of a star that has an apparent
brightness of 1.4 x 10-9 W/m2 ?
m = 2.5log10 (2.52 x 10-8W/m2/b) = 2.5log10 (2.52 x 10-8W/m2/ 1.4 x 10-9 W/m2) = 3.1
3.1
The Hubble can see an object with an apparent magnitude of 28.
What is the apparent brightness of such a star or galaxy?
m = 2.5log10 (2.52 x 10-8W/m2/b) ,
b = 2.52 x 10-8W/m2 /10(m/2.5)
= 1.6 x10-19 W/m2
1.6 x10-19 W/m2
Concept 3 – Absolute Magnitude
Absolute Magnitude:
m - M = 5 log10(d/10)
M = The Absolute Magnitude
d = The distance to the star in parsecs
m = The star’s Apparent Magnitude
The absolute magnitude of a star is defined as what its apparent
magnitude would be if you were 10 parsecs from it.
Absolute Magnitude:
m - M = 5 log10(d/10)
M = The Absolute Magnitude
d = The distance to the star in parsecs
m = The star’s Apparent Magnitude
Example: 100 pc from an m = 6 star, M = ?
(10x closer = 100x the light = -5 for m)
M = 6 - 5 log10(100/10) = 1
Put this in your notes: The Sun has an apparent
magnitude of -26.8, we are 1.5x108 km or 4.9 x 10-6 pc
from the sun. What is the sun’s absolute magnitude?
M = m - 5 log10(d/10)
= -26.8 - 5 log10(4.86 x 10-6 /10) = 4.7
4.7
You are 320 pc from a star with an absolute magnitude of 6.3. What
is its apparent magnitude?
M = m - 5 log10(d/10),
m = M + 5 log10(d/10) = 6.3 + 5 log10(320/10) = 14
14
Concept 4 – H-R diagrams
In 1910, Enjar Hertzsprung
of Denmark, and Henry
Norris Russell at Princeton
plotted M vs T in
independent research.
Main Sequence
From Douglas Giancoli’s Physics
Bright
Dim
Hot
From Douglas Giancoli’s Physics
Cooler
Bright
Dim
Hot
From Jay Pasachoff’s Contemporary Astronomy
Cooler
O
B
A
F
G
K
M
Oh (Hot)
be
a
fine
girl,
kiss
me!
(Cooler)
The atmosphere
The Star
Light from star is filtered by
atmosphere
From Jay Pasachoff’s Contemporary Astronomy
Spectral types:
O - 30,000 - 60,000 K, ionized H, weak H lines, spectral lines are
spread out. O types are rare and gigantic.
B - 10,000 - 30,000K, H lines are stronger, lines are less spread out
(Rigel, Spica are type B stars)
A - 7,500 - 10,000K, strong H lines, Mg, Ca lines appear (H and K)
(Sirius, Deneb and Vega are A type stars)
F - 6,000 - 7,500K, weaker H lines than in type A, strong Ca lines
(Canopus (S.H.) and Polaris are type F)
G - 5,000 - 6,000K, yellow stars like the sun. Strongest H and K
lines of Ca appear in this star.
K - 3,500 - 5,000K, spectrum has many lines from neutral metals.
Reddish stars (Arcturus and Aldebaran are type K stars)
M - 3,500 or less, molecular spectra appear. Titanium oxide lines
appear. Red stars (Betelgeuse is a prominent type M)
Suffixes 0 (hottest) - 9 (coolest) so O0, O1…O9, then B0, B1…
SO…
Hot stars are:
Big, Bright, Brief and Blue
Cool stars are:
Diminuitive, Dim, and Durable and um… reD
More about brief and durable next time…
Spectroscopic “parallax”
Bright
Spectrum = M
(from diagram)
Measure m
Find d
Dim
From Douglas Giancoli’s Physics
Put in your notes:
How far is an B0
that has an m of 8?
M = m - 5 log10(d/10)
5 log10(d/10) = m - M
log10(d/10) = (m - M)/5
d/10 = 10((m - M)/5)
d = (10 pc)10((8 - -4.3)/5)
d = 2884 pc
2884 pc
How far is an K0
that has an m of 14?
d = (10 pc)10((14 5.8)/5)
d = 437 pc
437 pc