a2 h 54 a2 Astrophysicssurveyingthestars

8/12/2019 a2 h 54 a2 Astrophysicssurveyingthestars

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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 = 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.



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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

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a2 h 54 a2 Astrophysicssurveyingthestars

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