5A-2 AstrophysicsSurveying The StarsAstrophysics booklet pages 28 to 53March 8th 2011

AQA A2 Specification

LessonsTopics1 to 9 1.3 Classification of StarsClassification by luminosityRelation between brightness and apparent magnitude.Apparent magnitude, mRelation between intensity and apparent magnitude. Measurement of m from photographic plates and distinction between photographic and visual magnitude not required.Absolute magnitude, MParsec and light year. Definition of M, relation to m: m M = 5 log (d / 10)Classification by temperature, black body radiationStefans law and Wiens displacement law.General shape of black body curves, experimental verification is not required.Use of Wiens displacement law to estimate black-body temperature of sources maxT = constant = 2.9 10-3 mK.Inverse square law, assumptions in its application.Use of Stefans law to estimate area needed for sources to have same power output as the sun. P = AT4Assumption that a star is a black body.Principles of the use of stellar spectral classesDescription 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 diagramGeneral 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 holesDefining 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 distances100270 to 460 X 336 000 X 120380 000 X 11150 000 000 X 4004 500 000 000 X 30 40 x 1012 X 9000(40 000 000 000 000) (X 270 000 Sun)80 x 1012 X 2260 x 1015 X 330020 x 1018 X 80460 x 1021 X 23 000

distance in kmmultiplierTop of the atmosphere100International Space Station270 to 460X 3Geostationary satellite36 000X 120The Moon380 000X 11The Sun150 000 000X 400Neptune (from the Sun)4 500 000 000X 30Proxima 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 1012X 2Centre of the Milky Way260 x 1015X 3300Andromeda Galaxy20 x 1018X 80Furthest object observed(GRB as of April 23rd 2009)400 x 1021X 20 000

QuestionCalculate 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 kmSun: 150 000 000 kmProxima Centuari: 40 x 1012 km(a) speed = distance / timebecomes: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-1time = distance / speed(i) = 150 000 000 / 128 000to the Sun = 1 170 days (3.2 years)(ii) = 40 x 1012 km / 128 000= 313 000 000 daysto Proxima Cent. = 860 000 years

The light yearOne 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 Earths orbit around the Sun.

1 AU = 150 000 000 km (150 x 109 m)

Questions on AU1. 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 AU2. How many AUs are there in one light year?

1 AU = 150 x 109 m1 lyr = 9.47 x 1015 mRatio = 9.47 x 1015 / 150 x 109 = 63 000 AU lyr-1

Stellar parallaxThis is the shifting of nearby stars against the background of more distant ones due to the orbital movement of the Earth about the Sun.Measurement of the angle 2 can yield the distance to the nearby star.

The parsec (pc)nearby startan = Rdbecomes:d = R / tan angle is always VERY small and so tan = in radiansand so: d = R /

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 = 2 radians1 arc second = 2 / (360 x 3600)= 4.85 x 10-6 radian

Distance measurement in parsecsdistance in parsecs = 1 / parallax angle in arc seconds1.002.010100With ground based telescopes the parallax method of distance measurement is acceptably accurate for distances up to 100 pc.

Parallax angle / arc secondsDistance / parsecs1.001.000.502.00.10100.01100

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 / R = 1AUd = 1 / (4.85 x 10-6 rad)1 parsec = 207 000 AU(b) metres:d = R / R = 1AU = 150 x 109 md = (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 yearsQuestion 1

Question 2Calculate 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 / 0.25distance to star = 4 parsecs

(b) 1 parsec = 3.26 light yearsdistance = 4 x 3.26distance to star = 13 light years

LuminosityLuminosity is the power output of a star.

luminosity = power = energy output time

unit: watt

The brightness of a star depends on a stars luminosity.

Intensity of radiation ( I )intensity = power of radiation area

unit: W m -2

Example:At the Earths surface the average intensity of sunlight is about 1400 W m -2

Suns Luminosity QuestionCalculate 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 surface area of a sphere, radius R is given by:A = 4 R2The surface area of a sphere of Earth radius:= 4 x (150 x 109 m)2= 4 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 1023Suns luminosity = 3.85 x 1026 W

The Suns radiation spreads out spherically.

Apparent magnitude, mThe 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.

Pogsons 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 StarsOther ObjectsSirius- 1.47Sun- 26.7Vega0Full Moon- 12.6Betelgeuse+ 0.58Venus- 4.6 (max)Deneb+ 1.25Jupiter- 2.9Polaris+ 2.01Andromeda Galaxy+ 3.4Dimmest star visible from Addlestoneabout + 4Neptune + 7.8Dimmest star visible from darkest skyabout + 6Faintest object observable by HST+ 31.5

QuestionCalculate how much brighter Sirius (m = -1.47) is compared with Polaris (m = 2.01)

Magnitude difference = 2.01 (-1.47)= 3.48If 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, MThe 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 1Calculate the absolute magnitude of the Sun if its apparent magnitude is 26.71 parsec = 207 000 AU

Distance from the Sun in parsecs: = 1 / 207 000d = 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)Suns 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 2Sirius 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 Suns apparent magnitude of -26.7.Sirius distance = 8.3 lyr1 parsec = 3.26 light years1 parsec = 207 000 AU

Distance to Sirius in parsecs: = 8.3 / 3.26d = 2.55 pc

Siriuss absolute magnitude by the method of question 1:Siriuss M = +1.50

m M = 5 log (d / 10)rearranged:log (d / 10) = [(m M) / 5]= [(-26.7 1.50) / 5] = - 28.2 / 5= - 5.64antiloging: = 2.29 x 10-6 = d / 10d = 2.29 x 10-5 pc= (2.29 x 10-5 x 207 000) AU= 4.74 AUThis 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

StarlightStars 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

Wiens displacement lawThe wavelength at peak power, max , is inversely proportional to the absolute temperature, T of the surface of a black body.max T = a constant

The constant is equal to 0.0029 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.

Star colour and temperatureBLUE stars are hotter than RED stars

Question 1Calculate the peak wavelength emitted by the Sun if its surface temperature is 6000 K.

Assuming that the Sun is a black body radiator, applying Weins displacement law:max T = 0.0029 mKmax x 6000K = 0.0029 mKmax = 0.0029 / 6000

Peak wavelength = 4.83 x 10-7 m (483 nm)

Question 2Red 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.

Weins displacement law:max T = 0.0029 mKbecomes:T = 0.0029 mK / max

For Betelgeuse: T = 0.0029 mK / 828 nm= 0.0029 mK / 8.28 x 10-7 m= 3 500 K

For Rigel:T = 0.0029 mK / 2.63 x 10-7 m= 11 000 K

Question 3A very large black body has a thermal temperature of 2.7K. Calculate its maximum power wavelength.

Weins displacement law:max T = 0.0029 mKmax x 2.7K = 0.0029 mKmax = 0.0029 / 2.7

Peak wavelength = 0.0011 m (1.1 mm)This is the wavelength of microwaves.The very large black body is the Universe.This cosmic microwave background radiation (CMB) was first detected by Penzias and Wilson in 1965 and is one of the main pieces of evidence that supports the Big Bang theory of the origin of the Universe.

Stefans lawThe 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 = A T4

Where is a constant known as Stefans constant. = 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 1Calculate 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 = 4 R2A = 4 x (1.39 x 106 km / 2) 2= 4 x (1.39 x 109 m / 2) 2= 4 x (6.95 x 108 ) 2A = 6.070 x 1018 m2

Assuming that the Sun is a black body radiator, applying Stefans law:P = 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 Suns Luminosity and it agrees closely with the Luminosity calculation performed earlier based on the 1360Wm-2 sunlight intensity data.

Question 2Calculate 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 Stefans law:P = A T4 4.09 x 1031 W = (5.67 x 10-8 W m-2 K-4) x A x (3500K)4A = (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 = 4 R2R = (A / 4) = (4.81 x 1024 / 4)= (3.825 x 1023)R = 6.18 x 1011 mRadius of Betelgeuse = 6.18 x 1011 m (618 000 000 km)

This approximately 4X the radius of the Earths orbit, about of the orbit of Jupiter. This is why Betelgeuse is called a SUPER-GIANT.

Stellar spectraThe 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 stars spectrum with the known absorption spectra for different elements and compounds.

Stellar spectral classesStars 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.

OBeAFineGirl or GuyKissMe

25 000 to 50 00011 000 to 25 0007 500 to 11 0006 000 to 7 5005 000 to 6 0003 500 to 5 000 2 500 to 3 500He+ He HHe HH (strongest) ionised metalsionised metalsionised and neutral metalsneutral metals and TiOneutral metals

spectralclassIntrinsiccolourTemperature(K)Prominentabsorption linesSpectrumOblue25 000to 50 000He+ He HBblue11000to 25 000He HAblue-white7500to 11 000H (strongest),ionised metalswhite6000 to 7500ionised metalsGyellow-white5000to 6000ionised andneutral metalsKorange3500to 5000neutral metalsMred 2500to 3500neutral metalsand TiO

Balmer absorption linesThe 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 stars spectrum.656 nm486 nm434 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-1h = 6.63 x 10-34 Js(a) c = f becomes:f = c / = (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 diagramO B A F G K M

The Hertzsprung-Russell diagramMAIN SEQUENCEMost 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.SUPERGIANTSVery rare.Stars that are about 1000x larger than the Sun.WHITE DWARFSMuch 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.

QuestionAn 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, Stefans law applies.P = A T4 For the orange giant: Po = Ao To4For the main sequence star: Pm = Am Tm4But both stars have the same power because they have the same absolute magnitude.that is: Po = Pm and so: Ao To4 = Am Tm4Ao To4 = Am Tm4Ao / Am = Tm4 / To4but the area of a sphere: A = 4 R2hence:(4 R2)o / (4 R2)m = Tm4 / To4R2o / R2m = Tm4 / To4Ro / Rm = Tm2 / To2Ro / Rm = (Tm / To ) 2Ro / Rm = (15000 K / 5000 K ) 2Ro / Rm = (3 ) 2Ro / Rm = 9QED

The evolution of a Sun like star1. NEBULA AND PROTOSTARA star is formed as dust and gas 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 becomes high enough for nuclear fusion of hydrogen to helium to occur in its core. A star is formed.

2. MAIN SEQUENCEThe 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.

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 stars absolute magnitude and surface temperature gradually increase. In about two billion years time the Earth will become too hot to sustain life.gradual warming

3. RED GIANTOnce 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.

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 Earths orbit.

The RED GIANT phase lasts for about one fifth of the MAIN SEQUENCE stage.

4. PLANETARY NEBULA AND WHITE DWARFWhen 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 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 Supergiant StarsIf 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.

SupernovaeA 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.

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

TypeSpectrumLight outputOriginIano hydrogen lines; strong silicon linedecreases steadilywhite dwarf attracts matter and explodesIbno hydrogen lines; strong helium linedecreases steadilysupergiant collapses then explodesIcno hydrogen or helium linesdecreases steadilysupergiant collapses then explodesIIstrong hydrogen or helium linesdecreases unsteadilysupergiant collapses then explodes

Supernovae as standard candlesType 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)

QuestionIn 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= 5.4antiloging: = 251 000 = d / 10d = 2.51 x 106 pcDistance to galaxy = 2.5 Mpc

= (2.51 x 106 x 3.26) light yearsDistance = 8.2 million light years

Neutron starsA 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-3Neutron 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

QuestionEstimate 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 / volumegives: mass = density x volumemass = (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 holesIf 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

Schwarzchild radius, RsThe 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, Rswhere: 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.

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 1030 kgG = 6.67 x 10-11 Nm2 kg-2c = 3.00 x 108 x ms-1(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 mSchwarzchild radius = 8.89 km

(b) density = mass / volumevolume of a sphere: = 4/3 R3= 4/3 (8.89 x 103)3= 4/3 x (7.03 x 1011)volume = 2.94 x 1012 m3and 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 centresSupermassive 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.

QuestionCalculate 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.

Suns mass = 2.0 x 1030 kgG = 6.67 x 10-11 Nm2 kg-2c = 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 mSchwarzchild radius = 7.71 million km

(b) density = mass / volumevolume of a sphere: = 4/3 R3= 4/3 (7.71 x 109)3= 4/3 x (4.58 x 1029)volume = 1.92 x 1030 m3and 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 LinksBlack-body radiation curves - NTNUResolution from two circular apertures - NTNUThe Life History of a Star - Powerpoint presentation by KTSequential Puzzle on Star Evolution order- by KT - Microsoft WORD

Core Notes from Student Guide pages 28 to 53What 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 Weins and Stefans 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?

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.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?

Notes from the Student Guide pages 28 to 332.1 Star MagnitudesWhat 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.

Repeat the worked example on page 32 this time for a star of apparent magnitude 8.0Try the summary questions on pages 32 & 33

Notes from the Student Guide pages 34 to 392.2 Classifying StarsCopy Figure 1 on page 34 and explain the shapes of the curves.Define Weins and Stefans 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 472.3 The Hertzsprung-Russell DiagramCopy 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 532.4 Supernovae, Neutron Stars & Black HolesOutline 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

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