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Astronomical distancesdistance in km multiplier
Top of the atmosphere 100
International Space Station 270 to 460 X 3
Geostationary satellite 36 000 X 120
The Moon 380 000 X 11
The Sun 150 000 000 X 400
Neptune (from the Sun) 4 500 000 000 X 30
Proxima Centuari
(nearest star to the Sun)
40 x 1012
(40 000 000 000 000)
X 9000
(X 270 000 Sun)
Sirius (brightest star) 80 x 1012 X 2
Centre of the Milky Way 260 x 1015 X 3300
Andromeda Galaxy 20 x 1018 X 80
Furthest object observed
(GRB as of April 23rd2009)
400 x 1021 X 20 000
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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 tookthree days to reach the Moon.
Distances in km:Moon: 380 000 km
Sun: 150 000 000 km
Proxima Centuari: 40 x 1012km
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The light yearOne light year is the distance light travels
through space in 1 year.
Question: Calculate the distance of one light year inmetres.
distance = speed x time= 3.0 x 108ms-1x 1 year
= 3.0 x 108ms-1x (365.25 x 24 x 60 x 60) s
= 9.47 x 1015m (9.47 x 1012km)
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)
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The Astronomical Unit (AU)
This is the mean radius of the
Earths orbit around the Sun.
1 AU = 150 000 000 km (150 x 109m)
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Questions on AU
1. Calculate the distance to Proxima Centuari inAstronomical Units, distance to PC = 40 x 1012km
2. How many AUs are there in one light year?
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Stellar parallaxThis is the shifting of nearby stars against the background
of more distant ones due to the orbital movement of theEarth about the Sun.
distant stars
nearby starEarth - December
Earth - June
View
from the
Earth in
June:
2
Measurement of the angle 2can yield the distance to
the nearby star.
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The parsec (pc)
nearby star
Earth - December
Earth - June
Rd
tan = Rd
becomes:
d = R / tan
angle is always VERY small and so
tan = in radians
and so: d = R /
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1 parsec is defined as the distance to astar which subtends an angle of 1 arc
second to the line from the centre of theEarth to the centre of the Sun.
1 arc second = 1 degree / 3600
as 360= 2radians
1 arc second = 2 / (360 x 3600)= 4.85 x 10-6radian
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Distance measurement in parsecsdistance in parsecs
= 1 / parallax angle in arc seconds
Parallax angle
/ arc seconds
Distance
/ parsecs
1.00 1.000.50 2.0
0.10 10
0.01 100
With ground based telescopes the parallax method ofdistance measurement is acceptably accurate fordistances up to 100 pc.
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Calculate the distance of 1 parsec measured in
(a) AU
(b) metres
(c) light years.
1 AU = 150 x 109m
1 light year = 9.47 x 1015m
Question 1
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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
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Luminosity
Luminosity is the power output of a star.
lum inosi ty = power = energy ou tput
t ime
unit:watt
The brightness of a star depends on a starsluminosity.
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Intensity of radiation ( I )
intensity = power of radiationarea
unit: W m-2
Example:
At the Earths surface the average intensity ofsunlight is about 1400 W m -2
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Suns Luminosity QuestionCalculate the luminosity of the Sun if the average
intensity of sunlight at the Earth is 1360 W m -2. Distancefrom the Sun to the Earth = 150 x 106km.
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Apparent magnitude, m
The apparent magnitude, mof a star in the
night sky is a measure of its brightnesswhich 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.
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Pogsons law (1856)
In 1856, Norman Robert Pogson defined that theaverage 1ststar magnitude was 100x brighter thanthe average 6thmagnitude star.
This means that for each change of magnitudestar brightness changes by about 2.5x.(2.55is about 100)
This resulted in a few very bright stars (e.g. Sirius)in having NEGATIVE apparent magnitudes.
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Examples of apparent magnitude
Nightime Stars Other ObjectsSirius - 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 AndromedaGalaxy
+ 3.4
Dimmest star visible
from Addlestoneabout + 4 Neptune + 7.8
Dimmest star visible
from darkest skyabout + 6 Faintest object
observable by HST+ 31.5
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Question
Calculate how much brighter Sirius (m = -1.47) iscompared with Polaris (m = 2.01)
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Absolute magnitude, M
The absolute magnitude, Mof a star is equal toits apparent magnitude if it were placed at adistance of 10 parsecs from the Earth.
It can be shown that for a star distance d, inparsecs, from the Earth:
m M = 5 log (d / 10)
NOTE: log means BASE 10logarithms
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Question 1
Calculate the absolute magnitude of the Sun if its apparentmagnitude is26.7
1 parsec = 207 000 AU
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Question 2Sirius has an apparent magnitude of1.47. Calculate the
distance in AU it would need to be from the Earth to equalthe brightness of the Suns apparent magnitude
of -26.7.
Sirius distance = 8.3 lyr
1 parsec = 3.26 light years
1 parsec = 207 000 AU
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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 thepattern of a black body radiator which is a perfect
absorber (and emitter) of radiation.
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Black body radiation curvespower radiated at
each wavelength
wavelength / m
0 1 2 3 4 5
visible
range
2000 K
1000 K1250 K
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Wiens displacement law
The wavelength at peak power, max , is
inversely proportional to the absolutetemperature, Tof the surface of a black body.
max T = a constant
The constant is equal to0.0029 metre kelvin
BEWARE! The above equation is usually quoted: max 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.
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Question 1
Calculate the peak wavelength emitted bythe Sun if its surface temperature is 6000 K.
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Question 2Red giant Betelgeuse, peak wavelength 828nm,
and blue supergiant Rigel, peak wavelength263nm, are both in the constellation of Orion.Calculate the surface temperatures of thesestars.
Betelgeuse
Rigel
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Question 3
A very large black body has a thermal
temperature of 2.7K.
Calculate its maximum power wavelength.
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Stefans law
The total energy per second (power), Pemitted
by a black body at absolute temperature, Tisproportional to its surface area, A and to T4.
P = A T4
Where is a constant known as Stefansconstant.
= 5.67 x 10-8W m-2K-4
This equation can be used to determine the surface areaand diameter of a star.
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Question 1
Calculate the power output of the Sun if its
diameter is 1.39 x 106
km and its surfacetemperature 5800 K.
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Question 2
Calculate the surface area and radius of
Betelgeuse if its luminosity is 4.09 x 1031
Wand its surface temperature 3500 K.
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Stellar spectraThe photosphere of a star gives off acontinuous spectrum.
However, when this light passes through theoutmost layer of a star, the corona, some of thewavelengths are absorbed by the hot gases inthis region.
This causes dark lines to be seen in theotherwise continuous spectrum given out bythe star.The wavelengths of these dark lines arecharacteristic to the elements and compoundsfound in the corona of the star.
The chemical composition of the star can bedetermined by comparing a stars spectrum
with the known absorption spectra for differentelements and compounds.
corona
photosphere
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Stellar spectral classes
Stars can be classified bytheir spectra
Starting from the hotteststars 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 groupsare found red and brown dwarfstars.
OBe
A
Fine
Girl or Guy
Kiss
Me
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spectral
class
Intrinsic
colour
Temperature
(K)
Prominent
absorption
lines
Spectrum
Oblue 25 000
to 50 000
He+ He H
B blue11000
to 25 000
He H
A blue-white
7500
to 11 000
H (strongest),
ionised metalswhite 6000 to
7500
ionised
metals
G yellow-white5000
to 6000
ionised and
neutral metals
K orange 3500to 5000 neutral metals
M red 2500
to 3500
neutral metals
and TiO
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Balmer absorption lines
The hydrogen absorption lines found in the visible
spectrum of the hottest stars (O, B and A only) are calledBalmer lines.
In such stars hydrogen atoms exist with electrons in the n
= 2state.
When these atoms are excited by the absorption ofphotons from the photosphere their electrons change from
n = 2to higher levels.
When they do this they absorb particular Balmer series
light wavelengths.These wavelengths show up as dark lines in the stars
spectrum.
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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 the2ndand 6thenergy levels. Calculate (a) the frequencyand (b) the energy of the absorbed photon.
c= 3.0 x 108ms-1
h= 6.63 x 10-34Js
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The Hertzsprung-Russell diagram
40 000 20 000 10 000 5000 2500
temperature / K
O B A F G K M
- 15
- 10
- 5
0
+ 5
+ 10
+ 15
absolute
magnitude
The Sun
supergiants
giants
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The Hertzsprung-Russell diagram
MAIN SEQUENCE
Most stars found in this region.Star masses vary from cool low power red dwarf stars ofabout 0.1x solar mass at the bottom right to very hot bluestars of about 30x solar mass at the top left.
GIANTSStars 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.
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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.
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The evolution of a Sun like star1. NEBULA AND PROTOSTAR
A star is formed as dust and gasclouds (nebulae) in space collapseunder their own gravitationalattraction becoming denser anddenser to form a protostar (a star inthe making).
In the collapse gravitational potentialenergy is converted into thermalenergy as the atoms and moleculesgain kinetic energy.
The interior of the protostarbecomes hotter and hotter.
If the protostar has sufficient mass(> 0.08 x Sun)the temperaturebecomes high enough for nuclearfusion of hydrogen to helium tooccur in its core. A star is formed.
absolute
magnitude
HIGH
temperature
LOW
temperature
protostar
nebula
collapsingand
warming
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2. MAIN SEQUENCE
The newly formed star reaches internalequilibrium as the inward gravitationalattraction is balanced by outward radiationpressure. The star becomes stable with a
near constant luminosity.
The greater the mass of the star, thehigher will be its absolute magnitude andsurface temperature but the shorter is thetime the star remains MAIN SEQUENCE.
The Sun is about half-way through its 10billion year passage. The largest starsmay only last for tens of millions of years.
While on the MAIN SEQUENCE the starsabsolute magnitude and surfacetemperature gradually increase. In abouttwo billion years time the Earth will
become too hot to sustain life.
absolute
magnitude
HIGH
temperature
LOW
temperature
gradual
warming
The
Sun
NOW
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3. RED GIANT
Once most of the hydrogen in the core ofthe star has been converted to helium, thecore collapses on itself and the outerlayers of the star expand and cool as aresult. The star swells out, moves off the
MAIN SEQUENCE and becomes a REDGIANT.
The temperature of the helium coreincreases as it collapses. This causessurrounding hydrogen to undergo fusion,which heats the core further.
When the core reaches about 108 Khelium nuclei undergo fusion. This formseven heavier nuclei principally beryllium,carbon and oxygen. The luminosity of thestar increases as the star expands. TheSun is expected to achieve a radiusroughly equal to the Earths orbit.
The RED GIANT phase lasts for aboutone fifth of the MAIN SEQUENCE stage.
absolute
magnitude
HIGH
temperature
LOW
temperature
red giant
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4. PLANETARY NEBULA
AND WHITE DWARF
When nuclear fusion in the core of a giantstar ceases, the star cools and its corecontracts, causing the outer layers of thestar to be thrown off.
The outer layers are thrown off as shells ofhot gas and dust to form a PLANETARYNEBULA.
The remaining core of the star is white hotdue to the release of gravitational energy.
If it is less than about 1.4solar masses,the contraction of the core stops as theelectrons in the core can no longer beforced any closer.
The star is now stable and has become aWHITE DWARF. This gradually cools toinvisibility over a few billion years.
absolute
magnitude
HIGH
temperature
LOW
temperature
red giant
The Cats Eye
Planetary
Nebula
white
dwarf
planetary
nebula
absolute
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absolute
magnitude
HIGH
temperature
LOW
temperature
red giant
white
dwarf
protostar
nebula
planetary
nebula
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Red Supergiant Stars
If a star is greater than 4solar
masses, the core becomes hotenough to cause energy release,through further fusion, to formnuclei as heavy as iron insuccessive shells.
The star now has an onion likeinternal structure.
Supergiant star Betelgeuseimaged in ultraviolet light by the
Hubble Space Telescope and
subsequently enhanced by NASA.
The bright white spot is likely one
of its poles.
S
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SupernovaeA supernovae can occur when the ironcore of supergiant is greater than about1.4solar masses.
In this case the gravitational forces are toogreat for the repulsive forces of electrons.Electrons are forced to react with protonsto form neutrons.
p + e- n + ve
The sudden collapse of the core occurswithin a few seconds and its densityincreases to that of atomic nuclei, about1017 kgm-3
The core suddenly becomes rigid andcollapsing matter surrounding the core hitsit and rebounds as a shock wavepropelling the surrounding matter outwardsinto space in a cataclysmic explosion.
The exploding star releases so muchenergy that it can outshine the host galaxy.
A supernova is typically a thousand milliontimes more luminous than the Sun. Within24 hours its absolute magnitude will reachbetween -15 and -20.
Elements heavier than iron are formed bynuclear fusion in a supernova explosion.Their existence in the Earth tells us thatthe Solar System formed from theremnants of a supernova.
The Crab Nebula
The remnant of asupernova
observed in 1054
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Types of supernova
Type Spectrum Light output Origin
Iano hydrogen
lines; strong
silicon line
decreases
steadily
white dwarf
attracts matter
and explodes
Ibno hydrogen
lines; stronghelium line
decreases
steadily
supergiant
collapses thenexplodes
Icno hydrogen or
helium lines
decreases
steadily
supergiant
collapses then
explodes
IIstrong hydrogen
or helium lines
decreases
unsteadily
supergiant
collapses then
explodes
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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 absolutemagnitude, Mof -19.3 0.03.
By noting their apparent peak magnitude, msuchsupernovae can be used to determine this distances togalaxies using the equation:
m M = 5 log (d / 10)
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Question
In a distant galaxy a Type 1a supernova is observed to
have an apparent magnitude of + 8.0. Calculate thedistance to this galaxy in (a) parsecs and (b) light years if
the supernova has an absolute magnitude of19.
1 parsec = 3.26 light years
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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 + ve
The star now has a density of
atomic nuclei, about 1017 kgm-3
Neutron stars were first discovered in1967 as a result of the radio beams thatthey emit as they rapidly rotate.
They are also called pulsars withfrequencies of up to 30 Hz
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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
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Black holesIf the core remnant of asupernova is greater than about3solar masses the neutrons areunable to withstand theimmense gravitational forcespushing them together.
The core collapses on itself and
becomes so dense that not evenlight can escape from it.
It is now a black hole.
What the core now consists of isunknown. It is sometimesreferred to as a singularity.
Evidence for the existence of blackholes was first found in 1971 froman X-ray source called Cygnus X-1which was in the same location asa supergiant star.
St l ti
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Star evolution summary
nebula
brown dwarf
(failed star)
red giant
white
dwarf
protostar
main
sequence
star
red
supergiant
MASS
> 0.05
MASS
< 0.05
MASS> 4
MASS
0.23 to 4
supernova
CORE MASS
< 1.4 > 1.4
CORE MASS
< 3 > 3
neutron
star
black
hole
= Sun
MASS
< 0.23
planetary
nebula
S h hild di
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Schwarzchild radius, RsThe size of a black hole is defined as
the distance from its centre at whichthe escape speed is equal to that oflight.
This is known as the Schwarzchild
radius, Rswhere: Rs= 2GM / c2
The surface of the sphere defined bythe Schwarzchild radius is called the
event horizon, because nothingthat occurs inside this boundary (anyevent) can be observed on theoutside.
Rs
singularity
event horizon
Q ti
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QuestionCalculate the for a black hole of mass 3 x Sun.
(a) its Schwarzchild radius,
(b) its mean density inside its event horizon.
Suns mass = 2.0 x 1030kg
G = 6.67 x 10-11Nm2 kg-2
c = 3.00 x 108x ms-1
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Galactic centres
Supermassive black holes are thought to
exist at the centres of most galaxiesincluding our own.
The mass of such black holes can be
estimated by measuring the orbital
speeds of stars near to the galacticcentre.
In the case of the Milky Way this is
estimated to be about 2.6 million solar
masses.
Sagittarius AThe location
of the supermassive blackhole at the centre of our
galaxy
Q ti
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QuestionCalculate the for a black hole at the centre of the Milky Wayof mass 2.6 million x Sun.
(a) its Schwarzchild radius,(b) its mean density inside its event horizon.
Suns mass = 2.0 x 1030kg
G = 6.67 x 10-11
Nm2
kg-2
c = 3.00 x 108x ms-1