5A-2 Astrophysics
Surveying The Stars
Astrophysics booklet pages 28 to 53
March 8th 2011
AQA A2 Specification
Lessons
1 to 9
Topics
1.3 Classification of Stars
Classification by luminosity
Relation between brightness and apparent magnitude.
Apparent magnitude, m
Relation between intensity and apparent magnitude. Measurement of m from photographic plates and distinction between
photographic and visual magnitude not required.
Absolute magnitude, M
Parsec and light year. Definition of M, relation to m: m – M = 5 log (d / 10)
Classification by temperature, black body radiation
Stefan’s law and Wien’s displacement law.General shape of black body curves, experimental verification is not required.
Use of Wien’s displacement law to estimate black-body temperature of sources λmaxT = constant = 2.9 × 10-3 mK.
Inverse square law, assumptions in its application.
Use of Stefan’s law to estimate area needed for sources to have same power output as the sun. P = σAT4
Assumption that a star is a black body.
Principles of the use of stellar spectral classes
Description of the main classes. Temperature related to absorption spectra limited to Hydrogen Balmer absorption lines: need for
atoms in n = 2 state.
The Hertzsprung-Russell diagram
General shape: main sequence, dwarfs and giants.
Axis scales range from -15 to 10 (absolute magnitude) and 50 000 K to 2 500 K (temperature) or OBAFGKM (spectral class).
Stellar evolution: path of a star similar to our Sun on the Hertzsprung-Russell diagram from formation to white dwarf.
Supernovae, neutron stars and black holes
Defining properties: rapid increase in absolute magnitude of supernovae; composition and density of neutron stars; escape
velocity > c for black holes.
Use of supernovae as standard candles to determine distances. Controversy concerning accelerating Universe and dark energy.
Supermassive black holes at the centre of galaxies.
Calculation of the radius of the event horizon for a black hole Schwarzschild radius ( Rs ) Rs = 2GM / c2
Astronomical distances
Top of the atmosphere
International Space Station
Geostationary satellite
The Moon
The Sun
Neptune (from the Sun)
Proxima Centuari
(nearest star to the Sun)
Sirius (brightest star)
Centre of the Milky Way
Andromeda Galaxy
Furthest object observed
(GRB as of April 23rd 2009)
distance in km
multiplier
100
100
270
270toto460
460
X3
36 000
000
36
380
380 000
000
X 120
X 11
150000
000000
000
150
44500
500000
000000
000
12
40
40 xx 1012
X 400
X 30
XX 9000
9000
(40
(40 000
000 000
000 000
000 000)
000)
(X
(X 270
270 000
000 Sun)
Sun)
80
80xx10
101212
1515
260
260x x10
10
X 22
X
X 3300
3300
X
20
20xx10
101818
X 80
80
X
21
400x x10
10
21
460
X
000
X 20
23 000
Question
Calculate the time taken to:
(a) travel to the Moon at 100 kmh-1
(63 m.p.h.)
(b) (i) travel to the Sun and
(ii) Proxima Centuari using the Apollo
spacecraft that took three days to
reach the Moon.
Distances in km:
Moon: 380 000 km
Sun: 150 000 000 km
Proxima Centuari: 40 x 1012 km
(a) speed = distance / time
becomes:
time = distance / speed
= 380 000 km / 100 kmh-1
= 3 800 hours
= 158 days (about 5 months)
(b) Apollo speed:
= 380 000 km / 3 days
= 128 000 km day-1
time = distance / speed
(i) = 150 000 000 / 128 000
to the Sun = 1 170 days (3.2 years)
(ii) = 40 x 1012 km / 128 000
= 313 000 000 days
to Proxima Cent. = 860 000 years
The light year
One light year is the distance light travels
through space in 1 year.
Question: Calculate the distance of one light year in
metres.
distance = speed x time
= 3.0 x 108 ms-1 x 1 year
= 3.0 x 108 ms-1 x (365.25 x 24 x 60 x 60) s
= 9.47 x 1015 m (9.47 x 1012 km)
Also used:
light second (e.g. the Moon is 1.3 light seconds away)
light minute (e.g. the Sun is 8.3 light minutes away)
The Astronomical Unit (AU)
This is the mean radius of the
Earth’s orbit around the Sun.
1 AU = 150 000 000 km (150 x 109 m)
Questions on AU
1. Calculate the distance to
Proxima Centuari in
Astronomical Units, distance
to PC = 40 x 1012 km
AU = 40 x 1012 / 150 x 106
Distance to Proxima
Centuari = 267 000 AU
2. How many AUs are there
in one light year?
1 AU = 150 x 109 m
1 lyr = 9.47 x 1015 m
Ratio =
9.47 x 1015 / 150 x 109
= 63 000 AU lyr-1
Stellar parallax
This is the shifting of nearby stars against the background
of more distant ones due to the orbital movement of the
Earth about the Sun.
Earth - June
2θ
nearby star
Earth - December
distant stars
View from
the Earth:
DECEMBER
JUNE
Measurement of the angle 2θ can yield the distance to
the nearby star.
The parsec (pc)
Earth - June
R
Earth - December
d
θ
nearby star
tan θ = R
d
becomes:
d = R / tan θ
angle θ is always VERY small and so
tan θ = θ in radians
and so: d = R / θ
1 parsec is defined as the distance to a
star which subtends an angle of 1 arc
second to the line from the centre of the
Earth to the centre of the Sun.
1 arc second = 1 degree / 3600
as 360° = 2π radians
1 arc second = 2 π / (360 x 3600)
= 4.85 x 10-6 radian
Distance measurement in parsecs
distance in parsecs
= 1 / parallax angle in arc seconds
Parallax angle
/ arc seconds
1.00
Distance
/ parsecs
0.10
1.00
1.00
2.0
2.0
10
10
0.01
100
100
0.50
With ground based telescopes the parallax method of
distance measurement is acceptably accurate for
distances up to 100 pc.
Question 1
Calculate the distance of
1 parsec measured in
(a) AU
(b) metres
(c) light years.
1 AU = 150 x 109 m
1 light year = 9.47 x 1015 m
(a) AU:
d=R/θ
R = 1AU
d = 1 / (4.85 x 10-6 rad)
1 parsec = 207 000 AU
(b) metres:
d=R/θ
R = 1AU = 150 x 109 m
d = (150 x 109 m)
/ (4.85 x 10-6 rad)
1 parsec = 3.09 x 1016 m
(c) light years:
= 3.09 x 1016 / 9.47 x 1015
1 parsec = 3.26 light years
Question 2
Calculate the distance to a star of parallax angle 0.25 arc
seconds in (a) parsecs and (b) light years.
1 parsec = 3.26 light years
(a) distance in parsecs
= 1 / parallax angle in arc seconds
= 1 / 0.25
distance to star = 4 parsecs
(b) 1 parsec = 3.26 light years
distance = 4 x 3.26
distance to star = 13 light years
Luminosity
Luminosity is the power output of a star.
luminosity = power = energy output
time
unit: watt
The brightness of a star depends on a star’s
luminosity.
Intensity of radiation ( I )
intensity = power of radiation
area
unit: W m -2
Example:
At the Earth’s surface the average intensity of
sunlight is about 1400 W m -2
Sun’s Luminosity Question
Calculate the luminosity of
the Sun if the average
intensity of sunlight at the
Earth is 1360 W m -2.
Distance from the Sun to
the Earth = 150 x 106 km.
The Sun’s radiation spreads out
spherically.
The surface area of a sphere,
radius R is given by:
A = 4π R2
The surface area of a sphere of
Earth radius:
= 4π x (150 x 109 m)2
= 4π x 2.25 x 1022
= 2.83 x 1023 m2
Each of these m2 receives 1360 W.
All of this power must come from
the Sun.
Therefore total power = luminosity
= 1360 x 2.83 x 1023
Sun’s luminosity = 3.85 x 1026 W
Apparent magnitude, m
The apparent magnitude, m of a star in the
night sky is a measure of its brightness
which depends on the intensity of the
light received from the star.
Stars were in ancient times divided into six levels of
apparent magnitude. The brightest were called
FIRST MAGNITUDE stars, those just visible to the
unaided eye in the darkest sky, SIXTH
MAGNITUDE.
Pogson’s law (1856)
In 1856, Norman Robert Pogson defined that the
average 1st star magnitude was 100x brighter than
the average 6th magnitude star.
This means that for each change of magnitude
star brightness changes by about 2.5x.
(2.55 is about 100)
This resulted in a few very bright stars (e.g. Sirius)
in having NEGATIVE apparent magnitudes.
Examples of apparent magnitude
Nightime Stars
Other Objects
Sirius
- 1.47
Sun
- 26.7
Vega
0
Full Moon
- 12.6
Betelgeuse + 0.58
Venus
- 4.6 (max)
Deneb
+ 1.25
Jupiter
- 2.9
Polaris
+ 2.01
Andromeda
Galaxy
+ 3.4
Dimmest star visible
from Addlestone
about + 4
Neptune
+ 7.8
Dimmest star visible
from darkest sky
about + 6
Faintest object
observable by HST
+ 31.5
Question
Calculate how much brighter Sirius (m = -1.47) is
compared with Polaris (m = 2.01)
Magnitude difference = 2.01 – (-1.47)
= 3.48
If a change of magnitude of 1.0 results in a brightness
change of 2.5
then Sirius is 2.5 3.48 times brighter than Polaris
= 24.3 times brighter.
Absolute magnitude, M
The absolute magnitude, M of a star is equal to
its apparent magnitude if it were placed at a
distance of 10 parsecs from the Earth.
It can be shown that for a star distance d, in
parsecs, from the Earth:
m – M = 5 log (d / 10)
NOTE: ‘log’ means BASE 10 logarithms
Question 1
Calculate the absolute magnitude
of the Sun if its apparent
magnitude is – 26.7
1 parsec = 207 000 AU
Distance from the Sun in parsecs:
= 1 / 207 000
d = 0.000 00483 pc
m – M = 5 log (d / 10)
rearranged:
M = m - 5 log (d / 10)
= (-26.7) - 5 log (0.000 00483 /10)
= (-26.7) - 5 log (0.000 000 483)
= (-26.7) - 5 log (0.000 000 483)
= (-26.7) - 5 x (- 6.316)
= (-26.7) - (- 31.6)
Sun’s absolute magnitude = 4.9
Note: The Sun would appear to be a very faint star and it
would probably not be visible from Addlestone!
Question 2
Sirius has an apparent magnitude
of – 1.47. Calculate the distance
in AU it would need to be from the
Earth to equal the brightness of
the Sun’s apparent magnitude
of -26.7.
Sirius distance = 8.3 lyr
1 parsec = 3.26 light years
1 parsec = 207 000 AU
Distance to Sirius in parsecs:
= 8.3 / 3.26
d = 2.55 pc
Sirius’s absolute magnitude by the
method of question 1:
Sirius’s M = +1.50
m – M = 5 log (d’ / 10)
rearranged:
log (d’ / 10) = [(m – M) / 5]
= [(-26.7 – 1.50) / 5]
= - 28.2 / 5
= - 5.64
antiloging: = 2.29 x 10-6 = d’ / 10
d’ = 2.29 x 10-5 pc
= (2.29 x 10-5 x 207 000) AU
= 4.74 AU
This is roughly the distance of Jupiter from the Earth
A wide field image of the sky including the Milky Way taken
from Australia. Notice the different colours of the stars
Starlight
• Stars differ in colour as well as brightness.
• Colour differences are only really apparent when stars
are viewed through a telescope as they can collect more
light than the unaided eye.
• A star emits thermal radiation that is continuous across
the electromagnetic spectrum.
• However, each star has a wavelength at which it emits at
maximum power. In the case of the Sun this corresponds
to the wavelength of yellow light.
• The power variation versus wavelength follows the
pattern of a ‘black body radiator’ which is a perfect
absorber (and emitter) of radiation.
Black body radiation curves
power radiated at
each wavelength
2000 K
1250 K
1000 K
0
1
visible
range
2
3
4
5
wavelength / μm
Wien’s displacement law
The wavelength at peak power, λmax , is
inversely proportional to the absolute
temperature, T of the surface of a black body.
λmax T = a constant
The constant is equal to 0.0029 metre kelvin
BEWARE! The above equation is usually quoted: λmax T = 0.0029 mK
‘mK’ does NOT mean ‘milli-kelvin’.
This equation can be used to determine the temperature of the ‘surface’
(known as the photosphere) of a star.
Star colour and temperature
power
yellow
red
orange
white
Star colour: blue
Surface T: 20
5500
3500
4300
7500
000
KK
UV
0
IR
500
1000
wavelength / nm
λmax : 386
530
150 nm
830
670
yellow
infra-red
red
blue
ultraviolet
Capella
Antares
Arcturus
Altair
Example: Spica
BLUE stars are hotter
than RED stars
Question 1
Calculate the peak wavelength emitted by the Sun
if its surface temperature is 6000 K.
Assuming that the Sun is a black body radiator,
applying Wein’s displacement law:
λmax T = 0.0029 mK
λmax x 6000K = 0.0029 mK
λmax = 0.0029 / 6000
Peak wavelength = 4.83 x 10-7 m
(483 nm)
Question 2
Red giant Betelgeuse, peak
wavelength 828nm, and blue
supergiant Rigel, peak
wavelength 263nm, are both
in the constellation of Orion.
Calculate the surface
temperatures of these stars.
Wein’s displacement law:
λmax T = 0.0029 mK
becomes:
T = 0.0029 mK / λmax
For Betelgeuse:
T = 0.0029 mK / 828 nm
= 0.0029 mK / 8.28 x 10-7 m
= 3 500 K
Betelgeuse
Rigel
For Rigel:
T = 0.0029 mK / 2.63 x 10-7 m
= 11 000 K
Question 3
A very large black body has a This is the wavelength of
thermal temperature of 2.7K. microwaves.
Calculate its maximum power
The ‘very large black body’
wavelength.
is the Universe.
This cosmic microwave
Wein’s displacement law:
background radiation (CMB)
λmax T = 0.0029 mK
was first detected by
λmax x 2.7K = 0.0029 mK
Penzias and Wilson in 1965
λmax = 0.0029 / 2.7
and is one of the main
pieces of evidence that
Peak wavelength
supports the ‘Big Bang’
= 0.0011 m
theory of the origin of the
Universe.
(1.1 mm)
Stefan’s law
The total energy per second (power), P emitted
by a black body at absolute temperature, T is
proportional to its surface area, A and to T4.
P = σ A T4
Where σ is a constant known as Stefan’s
constant.
σ = 5.67 x 10-8 W m-2 K-4
This equation can be used to determine the surface area
and diameter of a star.
Question 1
Calculate the power output of the Sun if its diameter is 1.39 x 106 km
and its surface temperature 5800 K.
Area of a sphere: A = 4π R2
A = 4π x (1.39 x 106 km / 2) 2
= 4π x (1.39 x 109 m / 2) 2
= 4π x (6.95 x 108 ) 2
A = 6.070 x 1018 m2
Assuming that the Sun is a black body radiator, applying Stefan’s law:
P = σ A T4
P = (5.67 x 10-8 W m-2 K-4) x (6.070 x 1018 m2) x (5800K)4
= (5.67 x 10-8 ) x (6.070 x 1018 ) x (1.132 x 1015)
Power output of the Sun = 3.89 x 1026 W
This is also called the Sun’s Luminosity
and it agrees closely with the Luminosity calculation performed earlier
based on the 1360Wm-2 sunlight intensity data.
Question 2
Calculate the surface area and radius of Betelgeuse if its luminosity is
4.09 x 1031 W and its surface temperature 3500 K.
Assuming that Betelgeuse is a black body, applying Stefan’s law:
P = σ A T4
4.09 x 1031 W = (5.67 x 10-8 W m-2 K-4) x A x (3500K)4
A = (4.09 x 1031) / [ (5.67 x 10-8) x (3500)4 ]
= (4.09 x 1031) / [ (5.67 x 10-8) x (1.501 x 1014) ]
= (4.09 x 1031) / (8.509 x 106)
Surface area of Betelgeuse = 4.81 x 1024 m2
Area of a sphere: A = 4π R2
R = √(A / 4π) = √(4.81 x 1024 / 4π)
= √(3.825 x 1023)
R = 6.18 x 1011 m
Radius of Betelgeuse = 6.18 x 1011 m
(618 000 000 km)
This approximately 4X the radius of the Earth’s orbit, about ¾ of the
orbit of Jupiter. This is why Betelgeuse is called a SUPER-GIANT.
Stellar spectra
The photosphere of a star gives off a
continuous spectrum.
However, when this light passes through the
outmost layer of a star, the corona, some of the
wavelengths are absorbed by the hot gases in
this region.
This causes dark lines to be seen in the
otherwise continuous spectrum given out by
the star.
The wavelengths of these dark lines are
characteristic to the elements and compounds
found in the corona of the star.
The chemical composition of the star can be
determined by comparing a star’s spectrum
with the known absorption spectra for different
elements and compounds.
photosphere
corona
Stellar spectral classes
Stars can be classified by
their spectra
Starting from the hottest
stars the groups are:
O, B, A, F, G, K, M
There are two further groups
(not required in the exam)
called L and T. In these groups
are found red and brown dwarf
stars.
O
Be
A
Fine
Girl or Guy
Kiss
Me
spectral
class
Intrinsic
colour
Temperature
(K)
Prominent
absorption
lines
O
blue
25
25 000
000
to 50
50 000
000
to
He+ He H
He+ He H
B
blue
11000
11 000
to25
25000
000
to
He H
He H
A
bluewhite
7 7500
500 to
to11
11000
000
H (strongest),
(strongest)
ionised
ionised metals
metals
white
6 000to
6000
to7500
7 500
ionised
ionised
metals
metals
G
yellowwhite
5000
5 000
to 6000
6 000
to
ionised and
neutral metals
K
orange
3 500
3500
to 5000
5 000
to
neutral metals
neutral metals
M
red
2 500
≈≈ 2500
to 3500
3 500
to
neutral metals
and
and TiO
TiO
Spectrum
Balmer absorption lines
The hydrogen absorption lines
found in the visible spectrum of the
hottest stars (O, B and A only) are
called Balmer lines.
In such stars hydrogen atoms exist
with electrons in the n = 2 state.
When these atoms are excited by
the absorption of photons from the
photosphere their electrons change
from n = 2 to higher levels.
When they do this they absorb
particular Balmer series light
wavelengths.
These wavelengths show up as
dark lines in the star’s spectrum.
n=5
n=4
n=3
n=2
434 nm486 nm
656 nm
Question (Revision of Unit 1)
A fourth, violet Balmer line
has a wavelength of 410 nm
and is due to the transition
of an electron between the
2nd and 6th energy levels.
Calculate (a) the frequency
and (b) the energy of the
absorbed photon.
c = 3.0 x 108 ms-1
h = 6.63 x 10-34 Js
(a) c = f λ
becomes:
f=c/λ
= (3.0 x 108 ms-1) / (410 nm)
= (3.0 x 108 ms-1) / (410 x 10-9 m)
photon frequency = 7.32 x 1014 Hz
(b) E = h f
= (6.63 x 10-34 Js) x (7.32 x 1014 Hz)
photon energy = 4.85 x 10-19 J
The Hertzsprung-Russell diagram
- 15
absolute
magnitude
supergiants
- 10
-5
0
giants
+5
+ 10
The Sun
+ 15
40 000
O
20 000
B
10 000
A
F
5000
G
K
2500
temperature / K
M
The Hertzsprung-Russell diagram
MAIN SEQUENCE
Most stars found in this region.
Star masses vary from cool low power red dwarf stars of
about 0.1x solar mass at the bottom right to very hot blue
stars of about 30x solar mass at the top left.
GIANTS
Stars that are between 10 to 100x larger than the Sun.
SUPERGIANTS
Very rare.
Stars that are about 1000x larger than the Sun.
WHITE DWARFS
Much smaller than the Sun but hotter.
The above illustration represents star classes with the colours
very close to those actually perceived by the human eye.
The relative sizes are for main sequence stars.
Question
An orange giant and a main
sequence star have the same
absolute magnitude of 0.
Their surface temperatures are
5000K and 15 000K respectively.
Show that the radius of the orange
giant is 9 times larger than that of
the main sequence star.
Assuming that both stars act like
black bodies, Stefan’s law applies.
P = σ A T4
For the orange giant:
Po = σ Ao To4
For the main sequence star:
Pm = σ Am Tm4
But both stars have the same power
because they have the same absolute
magnitude.
that is: Po = Pm
and so: σ Ao To4 = σ Am Tm4
Ao To4 = Am Tm4
Ao / Am = Tm4 / To4
but the area of a sphere: A = 4π R2
hence:
(4π R2)o / (4π R2)m = Tm4 / To4
R2o / R2m = Tm4 / To4
Ro / Rm = Tm2 / To2
Ro / Rm = (Tm / To ) 2
Ro / Rm = (15000 K / 5000 K ) 2
Ro / Rm = (3 ) 2
Ro / Rm = 9
QED
The evolution of a Sun like star
1. NEBULA AND PROTOSTAR
absolute
A star is formed as dust and gas
magnitude
clouds (nebulae) in space collapse
under their own gravitational
attraction becoming denser and
denser to form a protostar (a star in
the making).
In the collapse gravitational potential
energy is converted into thermal
energy as the atoms and molecules
gain kinetic energy.
The interior of the protostar
becomes hotter and hotter.
If the protostar has sufficient mass
(> 0.08 x Sun) the temperature
HIGH
becomes high enough for nuclear
temperature
fusion of hydrogen to helium to
occur in its core. A star is formed.
protostar
collapsing
and
warming
nebula
LOW
temperature
2. MAIN SEQUENCE
The newly formed star reaches internal
equilibrium as the inward gravitational
attraction is balanced by outward radiation
pressure. The star becomes stable with a
near constant luminosity.
absolute
magnitude
The greater the mass of the star, the
higher will be its absolute magnitude and
surface temperature but the shorter is the
time the star remains MAIN SEQUENCE.
The Sun is about half-way through its 10
billion year passage. The largest stars
may only last for tens of millions of years.
While on the MAIN SEQUENCE the star’s
absolute magnitude and surface
temperature gradually increase. In about
two billion years time the Earth will
become too hot to sustain life.
The
Sun
NOW
gradual
warming
HIGH
temperature
LOW
temperature
3. RED GIANT
Once most of the hydrogen in the core of
the star has been converted to helium, the
core collapses on itself and the outer
layers of the star expand and cool as a
result. The star swells out, moves off the
MAIN SEQUENCE and becomes a RED
GIANT.
absolute
magnitude
The temperature of the helium core
increases as it collapses. This causes
surrounding hydrogen to undergo fusion,
which heats the core further.
When the core reaches about 108 K
helium nuclei undergo fusion. This forms
even heavier nuclei principally beryllium,
carbon and oxygen. The luminosity of the
star increases as the star expands. The
Sun is expected to achieve a radius
roughly equal to the Earth’s orbit.
The RED GIANT phase lasts for about
one fifth of the MAIN SEQUENCE stage.
red giant
HIGH
temperature
LOW
temperature
4. PLANETARY NEBULA
AND WHITE DWARF
When nuclear fusion in the core of a giant
star ceases, the star cools and its core
contracts, causing the outer layers of the
star to be thrown off.
The Cat’s Eye
Planetary
Nebula
absolute
magnitude
The outer layers are thrown off as shells of
hot gas and dust to form a PLANETARY
NEBULA.
The remaining core of the star is white hot
due to the release of gravitational energy.
If it is less than about 1.4 solar masses,
the contraction of the core stops as the
electrons in the core can no longer be
forced any closer.
The star is now stable and has become a
WHITE DWARF. This gradually cools to
invisibility over a few billion years.
red giant
planetary
nebula
white
dwarf
HIGH
temperature
LOW
temperature
absolute
magnitude
red giant
planetary
nebula
protostar
white
dwarf
HIGH
temperature
nebula
LOW
temperature
Red Supergiant Stars
If a star is greater than 4 solar
masses, the core becomes hot
enough to cause energy release,
through further fusion, to form
nuclei as heavy as iron in
successive shells.
The star now has an ‘onion’ like
internal structure.
Supergiant star Betelgeuse
imaged in ultraviolet light by the
Hubble Space Telescope and
subsequently enhanced by NASA.
The bright white spot is likely one
of its poles.
Supernovae
A supernovae can occur when the iron
core of supergiant is greater than about
1.4 solar masses.
In this case the gravitational forces are too
great for the repulsive forces of electrons.
Electrons are forced to react with protons
to form neutrons.
The Crab Nebula
The remnant of a
supernova
observed in 1054
p + e - → n + ve
The sudden collapse of the core occurs
within a few seconds and its density
increases to that of atomic nuclei, about
1017 kgm-3
The core suddenly becomes rigid and
collapsing matter surrounding the core hits
it and rebounds as a shock wave
propelling the surrounding matter outwards
into space in a cataclysmic explosion.
The exploding star releases so much
energy that it can outshine the host galaxy.
A supernova is typically a thousand million
times more luminous than the Sun. Within
24 hours its absolute magnitude will reach
between -15 and -20.
Elements heavier than iron are formed by
nuclear fusion in a supernova explosion.
Their existence in the Earth tells us that
the Solar System formed from the
remnants of a supernova.
Types of supernova
Type
Spectrum
Light output
Origin
Ia
no hydrogen
lines; strong
silicon line
decreases
steadily
white dwarf
attracts matter
and explodes
Ib
no hydrogen
lines; strong
helium line
decreases
steadily
supergiant
collapses then
explodes
Ic
no hydrogen or
helium lines
decreases
steadily
supergiant
collapses then
explodes
II
strong hydrogen decreases
or helium lines
unsteadily
supergiant
collapses then
explodes
Supernovae as standard candles
Type 1a supernovae have a known peak luminosity
allowing them to be used as ‘standard candles’.
At their peak all of these supernovae have an absolute
magnitude, M of -19.3 ± 0.03.
By noting their apparent peak magnitude, m such
supernovae can be used to determine this distances to
galaxies using the equation:
m – M = 5 log (d / 10)
Question
In a distant galaxy a Type 1a
supernova is observed to
have an apparent magnitude
of + 8.0. Calculate the
distance to this galaxy in (a)
parsecs and (b) light years if
the supernova has an
absolute magnitude of – 19.
1 parsec = 3.26 light years
m – M = 5 log (d / 10)
rearranged:
log (d / 10) = [(m – M) / 5]
= [(8.0 – (-19) / 5]
= 27 / 5
= 5.4
antiloging: = 251 000 = d / 10
d = 2.51 x 106 pc
Distance to galaxy = 2.5 Mpc
= (2.51 x 106 x 3.26) light years
Distance = 8.2 million light years
Neutron stars
A neutron star is formed from the
remnant core of a supernova.
Gravitational forces cause
electrons to react with protons to
form neutrons.
p + e- → n + v e
The star now has a density of
atomic nuclei, about 1017 kgm-3
Neutron stars were first discovered in
1967 as a result of the radio beams that
they emit as they rapidly rotate.
They are also called ‘pulsars’ with
frequencies of up to 30 Hz
Question
Estimate the mass of a tea-spoonful of neutron star.
Take the density of a neutron star to be 1.0 x 1017 kgm-3
Estimated volume of the contents of a tea-spoon = 2 cm3
density = mass / volume
gives: mass = density x volume
mass = (1.0 x 1017 kgm-3 ) x (2 cm3)
= (1.0 x 1017 kgm-3 ) x (0.000 002 m3)
mass of a tea-spoonful of neutron star = 2 x 1011 kg
= 2 billion tonnes!
Black holes
If the core remnant of a
supernova is greater than about
3 solar masses the neutrons are
unable to withstand the
immense gravitational forces
pushing them together.
The core collapses on itself and
becomes so dense that not even
light can escape from it.
It is now a black hole.
What the core now consists of is
unknown. It is sometimes
referred to as a singularity.
Evidence for the existence of black
holes was first found in 1971 from
an X-ray source called Cygnus X-1
which was in the same location as
a supergiant star.
Star evolution summary
protostar
main
sequence
star
CORE MASS
red
supergiant
> 1.4 ʘ
supernova
red giant
nebula
MASS
< 0.23 ʘ
CORE MASS
planetary
nebula
brown dwarf
(failed star)
< 1.4 ʘ
MASS
0.23 to 4 ʘ
MASS
> 0.05 ʘ
MASS
< 0.05 ʘ
MASS
>4ʘ
ʘ = Sun
white
dwarf
<3ʘ
>3ʘ
neutron
star
black
hole
Schwarzchild radius, Rs
The size of a black hole is defined as
the distance from its centre at which
the escape speed is equal to that of
light.
This is known as the Schwarzchild
radius, Rs
where: Rs = 2GM / c2
The surface of the sphere defined by
the Schwarzchild radius is called the
‘event horizon’, because nothing
that occurs inside this boundary (any
event) can be observed on the
outside.
event horizon
Rs
singularity
Question
Calculate the for a black
hole of mass 3 x Sun.
(a) its Schwarzchild
radius,
(b) its mean density
inside its event horizon.
(a) Rs = 2GM / c2
= [2 x (6.67 x 10-11 Nm2 kg-2) x (3 x 2.0 x 1030 kg)]
/ (3.00 x 108 x ms-1)2
= (8.00 x 1020) / (9.00 x 1016 )
= 8.89 x 103 m
Schwarzchild radius = 8.89 km
Sun’s mass
= 2.0 x 1030 kg
G = 6.67 x 10-11 Nm2 kg-2
c = 3.00 x 108 x ms-1
(b) density = mass / volume
volume of a sphere: = 4/3 πR3
= 4/3 π (8.89 x 103)3
= 4/3 π x (7.03 x 1011)
volume = 2.94 x 1012 m3
and so:
density = (6.0 x 1030 kg) / (2.94 x 1012 m3)
Mean density inside the event horizon
= 2.03 x 1018 kgm-3
Galactic centres
Supermassive black holes are thought to
exist at the centres of most galaxies
including our own.
The mass of such black holes can be
estimated by measuring the orbital
speeds of stars near to the galactic
centre.
In the case of the Milky Way this is
estimated to be about 2.6 million solar
masses.
Sagittarius A – The location
of the supermassive black
hole at the centre of our
galaxy
Question
Calculate the for a black
hole at the centre of the
Milky Way of mass 2.6
million x Sun.
(a) its Schwarzchild
radius,
(b) its mean density
inside its event horizon.
Sun’s mass
= 2.0 x 1030 kg
G = 6.67 x 10-11 Nm2 kg-2
c = 3.00 x 108 x ms-1
(a) Rs = 2GM / c2
= [2 x (6.67 x 10-11 Nm2 kg-2) x (2.6 x 106 x 2.0 x 1030 kg)]
/ (3.00 x 108 x ms-1)2
= (6.90 x 1026) / (9.00 x 1016 )
= 7.71 x 109 m
Schwarzchild radius = 7.71 million km
(b) density = mass / volume
volume of a sphere: = 4/3 πR3
= 4/3 π (7.71 x 109)3
= 4/3 π x (4.58 x 1029)
volume = 1.92 x 1030 m3
and so:
density = (5.2 x 1036 kg) / (1.92 x 1030 m3)
Mean density inside the event horizon
= 2.71 x 106 kgm-3
NOTE: This is 1012 x lower than the low mass
(3 x Sun ) black hole.
Internet Links
• Black-body radiation curves - NTNU
• Resolution from two circular apertures
- NTNU
• The Life History of a Star Powerpoint presentation by KT
• Sequential Puzzle on Star Evolution
order- by KT - Microsoft WORD
Core Notes from Student Guide pages 28 to 53
1.
2.
3.
4.
5.
6.
7.
What is a light year?
Draw a diagram and explain the
meaning of the parsec.
What is meant by (a) apparent &
(b) absolute magnitude? Give
the equation that relates these
two quantities.
Copy Figure 1 on page 34 and
explain the shapes of the
curves.
Define Wein’s and Stefan’s and
give equations relating to these
laws.
Copy the table on page 37 and
explain the process of formation
of star spectra.
How are Balmer lines formed in
star spectra?
8. Copy the Hertzsprung-Russell
Diagram on page 42 and explain the
patterns shown.
9. Describe the evolution of a star similar
to the Sun with reference to the
Hertzsprung-Russell Diagram.
10. Outline the evolution of a high mass
star. Explain how the mass of a star,
or its core, determines what happens.
11. Explain how supernovae can be used
as ‘standard candles’ to determine
distance.
12. What is (a) a neutron star; (b) a pulsar
& (c) a black hole?
13. Explain what is meant by the
Schwarzschild radius.
14. What is the evidence for supermassive
black holes at the centres of galaxies?
Notes from the Student Guide pages 28 to 33
2.1 Star Magnitudes
1. What is a light year?
2. Draw a diagram and explain the meaning of
the parsec.
3. What is meant by (a) apparent & (b) absolute
magnitude? Give the equation that relates
these two quantities.
4. Repeat the worked example on page 32 this
time for a star of apparent magnitude 8.0
5. Try the summary questions on pages 32 & 33
Notes from the Student Guide pages 34 to 39
2.2 Classifying Stars
1.
2.
3.
4.
5.
6.
7.
Copy Figure 1 on page 34 and explain the shapes of the curves.
Define Wein’s and Stefan’s and give equations relating to these
laws.
Copy the table on page 37 and explain the process of formation of
star spectra.
How are Balmer lines formed in star spectra?
Repeat the worked example on page 35 this time for a star whose
peak intensity wavelength is at 150 nm.
Repeat the worked example on page 36 this time for a star whose
power output is 9.0 x 1029 W and a surface temperature of 5000 K.
It will have a different surface area.
Try the summary questions on page 39
Notes from the Student Guide pages 40 to 47
2.3 The Hertzsprung-Russell Diagram
1.
2.
3.
4.
Copy the Hertzsprung-Russell Diagram on page 42
and explain the patterns shown.
Describe the evolution of a star similar to the Sun with
reference to the Hertzsprung-Russell Diagram.
Repeat the worked example on page 41 this time for a
star that has a surface temperature of 5000 K.
Try the summary questions on page 46 & 47
Notes from the Student Guide pages 48 to 53
2.4 Supernovae, Neutron Stars & Black Holes
1.
2.
3.
4.
5.
6.
Outline the evolution of a high mass star. Explain how
the mass of a star, or its core, determines what
happens.
Explain how supernovae can be used as ‘standard
candles’ to determine distance.
What is (a) a neutron star; (b) a pulsar & (c) a black
hole?
Explain what is meant by the Schwarzschild radius.
What is the evidence for supermassive black holes at
the centres of galaxies?
Try the summary questions on page 52 & 53