Download Stars Part 1

The Birth & Death of
Stars Part 1
•Are Stars similar to our
Sun?
•How far away are they?
•What do they do?
Astronomy 12
The Birth & Death of
Stars Part 1
Learning Outcomes (Students will…):
• Describe and apply classification
systems and nomenclature used in the
classification of stars
• Use parallax to determine distance
• Calculate luminosity
• Describe nuclear fusion (what stars
“do”)
Stars – What are they?
Stars are great balls
of hot gas held
together by their
own gravity.
If stars were not so
hot, the gas
would collapse
into a small dense
body.
Stars – Why are they important?
Life on Earth depends
on the Sun. Nearly all
of our energy comes
from the Sun, even
coal and oil. This
could be true for
other extrasolar
planets!
Stars – What do they do?
Near the center of the
Sun nuclear fusion
generates energy and
this energy flows
outward towards the
surface where it is
radiated into space as
heat and light and other
forms of
electromagnetic
radiation.
Nuclear Fusion and the Sun




Stars create energy through nuclear fusion.
In this process, 2 nuclei (mostly hydrogen)
combine to produce a single, larger nucleus
which creates energy.
In large enough quantities, nuclear fusion can be
used as an energy source. To get large enough
quantities, atoms must be heated to 1.5 x 107K
(the heat of the Sun)!
1.5 x 107K = 15,000,000 K OR 14,999,726°C
Nuclear Fusion




The Sun radiates energy at the rate of 3.9x1026W
(watts) and has been doing so for several billion
years.
The fusion reaction in the Sun is a multistep
process in which (typically) hydrogen is burned
into helium.
Hydrogen is the “fuel”
Helium is the “product”
Nuclear Fusion

2 protons (1H) combine to form a deuteron
(2H), a positron (e+) and a neutrino (v).
 1H

+ 1H  2H + e+ + v
The positron quickly reacts with an electron
(e-) and both particles annihilate, their mass
energy appearing as two gamma-ray
photons (γ).

e+ + e-  γ + γ
Nuclear Fusion

The deuteron (2H) collides with another proton
(1H) and forms a 3He nucleus and a gamma ray
(γ).
 2H

+ 1H  3He + γ
Two 3He nuclei may eventually (within ten
thousand years) find each other.
 3He
+ 3He  4He + 1H + 1H
The Sun’s Nuclear Energy – When will it
run out???


The Sun has been creating nuclear energy for
about 5 billion years.
There is enough Hydrogen left to continue this
process for another 5 billion years.
What about nuclear fusion that creates
other atoms???



If the core temperature increases to about 1 x
108K then energy can be produced by burning
helium to make carbon.
As a star evolves and becomes hotter than 1 x
108K, other elements can be formed by other
fusion reactions.
This only works for elements up to the iron-56.
At this point, more energy would be consumed
than produced.
Nuclear Fusion Review Questions





1) What is the fuel for nuclear fusions (in most
cases)?
2) What is the product?
3) What is the largest atom that can be used in
nuclear fusion? Why?
4) What is the overall equation for nuclear
fusion?
5) How long will the Sun’s life be in total?





1) Hydrogen
2) Helium
3) Iron- 56 because at this point, more energy
would be consumed than produced.
4) ** See slide 10
5) 10 billion years
There are lots of stars in the
universe…and a lot of different stars!
Panorama view of the sky
Classifying Stars:
The Four Basic Characteristics of a Star
Luminosity
Size
Mass
Surface
Temperature
To discover these characteristics, distance
to a star must be determined…use parallax!
The Geometry
of Parallax
p
1 (AU)
d (in Parsecs) =
p (in arcseconds)
1 parsec (pc) = 3.26 ly
http://www.astronomynotes.com/starpr
op/s2.htm
Parallax from a Different Planet
If we lived on Mars, orbiting 1.5 times farther
away from the Sun, the parallax would be
1.
the same as from Earth
2.
1.5 times smaller than from Earth
3.
1.5 times bigger than from Earth
Journal Formula
1
d
arc sec
Symbols
d = distance in parsecs
p = angle in arcseconds
Units
pc = 1 parsec = 3.26 light years (ly)
1 light year = the distance that light travels in a
vacuum in one Julian year (365.25 days)
= 9,460,730,472,580.8 km
Journal: Example 1
Suppose a star has a parallax of 0.01 arc
seconds. How many parsecs away is it?
distance (in parsecs) = 1 / p
d=
1 (AU)
0.01
Answer: 100 parsecs
Journal: Example 2
Suppose a star has a distance of 12 parsecs.
How many arcseconds is its angle?
1
d 
p
Re arrange
1
p 
d
1
p 
12
p  0.08333
Answer: 0.083333 arcseconds
Category 1
 LUMINOSITY

Apparent or Visual Brightness (m)…Brighter
objects have smaller magnitudes than fainter
objects.
Luminosity

Luminosity (absolute brightness) depends on
distance and apparent brightness

Luminosity is the amount of electromagnetic
energy a body radiates per unit of time.
1
I 
2
r
Luminosity

Luminosity depends on:
 1) size of the star
(bigger = more
luminous!)
 2) distance to the star
(closer = more
luminous!)
 3) intervening matter
(dust and gas can
absorb light – reduces
luminosity and
increases redness)
Calculating Luminosity

If we know the distance to the star we can
measure luminosity:
L = 4πfd2
where…
the distance d to the star (m),
 the Flux f of the star (W/m2) where flux measures
light intensity
 the luminosity L of the star (Watts)
 1 Parsec = 3.08568025 × 1016 meters

Example 1: Finding Luminosity


What is the luminosity of our Sun which has a flux
of 1360 W/m2 and a distance of 4.84813×10-6
parsecs?
Step 1: Change distance to m
 3.08568025 x10
(4.84813 10 pc)x 
pc

11
 1.495977899 10 m
-6
16
m



Step 2: L = 4πfd2

L = 4π (1360)(1.495977899 x 1011)2
L = 3.82 x 1026 Watts

Luminosity
By definition (using more accurate measurements):
 Lsun = 3.9 x 1026 W
 However, we can measure astronomical luminosity
in Solar luminosity units, where
Lsun = 1 Solar luminosity unit
Solar luminosity units = L© (this should be a circle
with a dot!)

Example 2: Changing into Solar
Luminosity Units

If a star has a luminosity of 1.3 x 1010 Watts, what is
its luminosity in Solar luminosity units?
 1 Solar luminosity unit 
Lstar  (1.3 10 Watts)  

26
3.9 10 Watts


L  3.3 10 17 units
10
There is a Big Range of Stellar
Luminosities Out there!
Star
Sun
Proxima Centauri
Rigel (Orion)
Deneb (Cygnus)
Luminosity (in
units of solar
Luminosity)
1
0.0006
70,000
170,000
Example 3: Finding Distance from
Luminosity

You have a 100 W lightbulb in your laboratory.
Standing at a distance of d from the lightbulb,
you measure flux of the lightbulb to be 0.1
W/m2. How can you use this information to
determine the distance from you to the
lightbulb?



1) Rearrange
2) Sub in values
3) Solve for d
d
2
d
2
d
2
L

4f
100

4 *  * 0 .1
 79.58
d 
79.58
d  8.92
Example 4: Finding Flux

Find the flux of the star Arcturus if it is 11.25 pc away from
Earth and its Solar luminosity is 114.

Step 1: Change luminosity into Watts


114 Solar luminosity units x (

4.446 x 1028
3.9x1026 Watts
)
1 Solar luminosity unit
Step 2: Change distance into metres


11.25 pc x (3.0856802 x 1016m)
1 pc
3.47 x 1017 m
Example 4: Finding Flux

Find the flux of the star Arcturus if it is 11.25 pc
away from Earth and its Solar luminosity is 114.

Step 3: Rearrange equation, sub in values, solve for f
L
f 
4d 2
28
(4.446 10 )
f 
17
4 (3.47 10 )
8
f  2.936 10 W / m
2
Luminosity



The luminosity of stars ranges from 0.01 Lo to 1
x 106 Lo.
The Sun is not nearly as bright as the most
luminous stars but is brighter than most stars.
The Sun is gradually becoming more luminous
(about 10% every 1 billion years). The Sun used
to be fainter in the past, which is possibly the
reason life on Earth has only existed for about 1
billion years on land.
Category 2

SIZE!
Stellar Size


Stars are very spherical so we characterize a
star’s size by its radius.
We always compare to our Sun!
R
Stellar Radii vary in size
from ~1500 RSun for a
large Red Giant to
0.008 RSun for a White
Dwarf.
Size Equation

L = 4πR2s T4,





Where L is the luminosity in Watts
R is the radius in meters

8
5
.
67
x
10

s is the Stefan-Boltzmann constant

W 
2
4 
m K 
T is the star's surface temperature in Kelvin
NOTE: R is often given in Solar radii (radius of Sun)
Example 1








Find the Luminosity of Zeta Puppis which has a
temperature of about 38,000 K and a radius of 18 Ro.
Then change this into Solar luminosity units.
Step 1: Change units
R = 18 x (6.955 x 108 m) = 1.2519 x 1010 m
Step 2: Plug into equation
L = 4πR2s T4
L = 4π(1.2519 x 1010)2(5.67 x 10-8) (25000)4
L = 2.328 x 1032 W
L = 597,038 Lo
Example 2





Find the radius of a new star,
SBUTLER, that has a
luminosity of 5Lo and a
temperature of 25,000K.
Step 1: Change units
5Lo x (3.9 x 1026 Watts/1Lo)
= 1.95 x 10 27
Step 2: Rearrange equation
L = 4πR2s T4
R
L
4sT 4
1.95x10 27
R
4 (5.67 x10 8 )( 25000) 4
R  83702962m
Category 3

MASS!
Mass

Stars fall into a narrow mass range because
below 0.08 Solar masses, nuclear reactions cannot be
sustained
AND
 greater than 100 Solar masses stars are unstable.


High mass stars burn their fuel at higher rates
and live shorter lives than low-mass stars.
How do you weigh a star?
By observing the star and anything that
orbits it (maybe even another star)
 Use Kepler’s Laws of Planetary Motion
and Newton’s Law of Gravitation to
determine mass

Formula – A “metric” version of Kepler’s
Law!
4 a
M
2
(GP )  N
2 3

Where:
M = mass of star in kg
 N = mass of planet in kg
 a =distance from star to
planet in m (average)
 G = gravitational
constant
(6.673 × 10-11m3 kg-1 s-2)
 P = period of the orbit in
seconds

What do I have to know?

How to change from a mass in kg to a mass in
Solar mass units

Change 15 Mo into kg.
Change 1.52 x 1035 kg into Mo.

Category 4

SURFACE TEMPERATURE
The Sun


Although nuclear fusion needs heat of 1.5 x
107K (core), the surface temperature of the Sun
is about 5,700 K.
The temperature of stars are between 3,000K to
30,000 K.
How do you measure the surface
temperature of a star?
•
Every element (when heated) will emit lines
that lie along the visible light spectrum a.k.a.
R-O-Y-G-B-I-V
•
Stars are mostly composed of hydrogen (and
some helium)
Spectral Types of Stars
Spectral types are defined by the:
•
•
existence of spectral lines belonging to various
elements, ions, & molecules in a star’s spectrum
the relative strengths of these lines
However, spectral type is not determined by a
star’s composition.
•
all stars are made primarily of Hydrogen &
Helium
Reason for Spectral Types
Spectral type is determined by a star’s surface
temperature.
Astronomers use Kelvin (K)
as temperature unit…
272 K = 0 oC
298 K = 30 oC
1000 K = 727 ºC
Spectral Type Classification System
O B A F G K M
Oh Be A Fine Girl/Guy, Kiss Me!
50,000 K
3,000 K
Temperature
Stellar Classification
Analogy
Make a plot that shows the general relationship
between height and weight for humans.
- now add to your plot the population of
basketball players who are very tall and very thin.
- now add the population of obese children
•The plot would show a cluster of people that would have similar
“middle-of-the-road” height/weight ratios
•It would also show a smaller cluster of “very tall and very thin”
AND a smaller cluster of “very short and very fat”
Height
Very Tall,
Very Thin
?
?
Very Short,
Very Fat
Weight
•The plot would show a cluster of people that would have similar
“middle-of-the-road” height/weight ratios
•It would also show a smaller cluster of “very tall and very thin”
AND a smaller cluster of “very short and very fat”
How can we classify stars
1) Collect information on
a large sample of stars.
2) Measure their
luminosities
3) Measure their surface
temperatures
(need their spectra)
The Hertzsprung-Russell Diagram
The Hertzsprung-Russell Diagram
BRIGHT
HOT
COOL
FAINT
The Hertzsprung-Russell Diagram
The Main Sequence
 ~90% of all stars
are in the main
sequence (MS)
 ~90% of all MS
stars are cooler
spectral types than
the Sun (i.e., at the
lower MS)
 All MS stars fuse
H into He in their
cores.
The Hertzsprung-Russell Diagram
Mass-Luminosity
Relation:
 L  M3.5
 For example, if
the mass of a star
is doubled, its
luminosity
increases by a
factor 23.5 ~ 11.
The relation is for
main sequence
stars only!
Mass of MS Star
L  M3.5
The Hertzsprung-Russell Diagram
Red Giants
- Red Giant stars
are very large, cool
and quite bright.
e.g., Betelgeuse is
100,000 times more
luminous than the Sun
but is only 3,500K on
the surface. It’s radius
is 1,000 times that of
the Sun.
The Hertzsprung-Russell Diagram
Supergiants
•Very bright
•Very hot
Size of Star:
The Hertzsprung-Russell Diagram
White Dwarfs
- White Dwarfs
are hot, but since
they are so small,
they are not very
luminous.
Main Sequence
Lifetime


All M-S stars have temperatures sufficient to
fuse H into He in their cores.
Luminosity depends directly on mass:
more mass = more pressure from upper layers
 fusion rates must be high to maintain equilibrium



Lifetime  (Amt of Fuel)/(Rate of Burning) 
M / L  M / M3.5  1 / M2.5
Higher mass stars have shorter lives!
The Hertzsprung-Russell Diagram
More mass,
more fuel,
very fast burning.
Shorter
Lifetime
of Star
Less mass,
less fuel,
slow, steady burning.
Think
SUV vs Honda Civic
Longer
Mass
(MSun)
Luminosity
(LSun)
0.10
3×10-3
0.50
Surface
Temperature
(K)
Radius
(RSun)
Main sequence
lifespan (yrs)
2,900
0.16
2×1012
0.03
3,800
0.6
2×1011
0.75
0.3
5,000
0.8
3×1010
1.0
1
6,000
1.0
1×1010
1.5
5
7,000
1.4
2×109
3
60
11,000
2.5
2×108
5
600
17,000
3.8
7×107
10
10,000
22,000
5.6
2×107
15
17,000
28,000
6.8
1×107
25
80,000
35,000
8.7
7×106
60
790,000
44,500
15
3.4×106
•As mass of a M-S star increases, luminosity, surface temperature and radius (size) increase
•As mass of a M-S star increases, life span on M-S decreases
1.
2.
3.
4.
5.
6.
7.
8.
Review Questions:
Where are most stars?
What is the common
characteristics of MS
stars?
What determines the
location of a star in
the MS?
Where do you find the
largest stars?
The smallest?
The most massive
one?
The coolest stars?
The H-R Diagram
How do we know the age
of a star?
1. MS, 2. H He, 3. M, 4. upperright, 5. lowerleft, 6. upperleft, 7. lowerright, 8. normally we don’t