astrophysics practice questions

8/20/2019 Astrophysics practice questions

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Teacher Resource Bank

GCE Physics A

Other Guidance:

• Astrophysics

By P. Organ


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Teacher Resource Bank / GCE Physics A / Astrophysics / Version 1.0

2 Copyright © 2009 AQA and its licensors. All rights reserved. %&'

AQA Physics Specification A Astrophysics

Preface

This booklet aims to provide background material for teachers preparing students for the A2Optional Unit 5A Astrophysics; AQA Physics Specification A. It provides amplification ofspecification topics which teachers may not be familiar with and should be used in associationwith the specification. The booklet is not intended as a set of teaching notes.

Contents Page

Introduction 3

Chapter 1 Lenses and Optical Telescopes (Specification Reference A.1.1) 4

A Lenses 4

B Astronomical telescope consisting of two converging lenses 10

C Reflecting telescope 12

D Resolving power 15

E Charge coupled device 18

Chapter 2 Non-optical Telescopes (Specification Reference A.1.2) 19

A Single dish radio telescopes 19

B I-R, U-V and X-ray telescopes 20

Chapter 3 Classification of Stars (Specification Reference A.1.3) 22

A Classification by luminosity 22B Apparent magnitude, m 22

C Absolute magnitude, M 23

D Classification by temperature, black body radiation 24

E Principles of the use of stellar spectral classes 28

F The Hertzsprung-Russell diagram 30

G Supernovae, neutron stars and black holes 31

Chapter 4 Cosmology (Specification Reference A.1.4) 33

A Doppler effect 33B Hubble’s law 35

C Quasars 37

Appendices 38

A Astrophysics specification 38


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%&' Copyright © 2009 AQA and its licensors. All rights reserved. 3

Introduction

The Astrophysics option is intended to give students an opportunity to study an area of Physicsmany find fascinating. It includes a study of the main methods of acquiring information, how thatinformation is used to analyse and catalogue astronomical objects and modern theories on

cosmology. The topic provides a grounding in both the practical and theoretical study ofastrophysics.

In the first chapter, students see how some of the optics topics they have studied earlier in thespecification are applied to the design of telescopes. The relative merits of the two basicmethods – reflecting and refracting – are discussed as well as the influence the design of thetelescopes has on how well an image can be produced. Finally, the workings of a very commondevice used to collect optical images, the Charge Coupled Device (CCD), is studied.

Chapter 2 develops the ideas of chapter 1 by studying how information from some of the otherparts of the electromagnetic spectrum can be gathered. Comparisons are made between thedevices in terms of their design and application.

Having seen how the information is obtained, in chapter 3 students have the opportunity to seehow it is used to describe stars. The idea of apparent magnitude is introduced, with itsmathematical relationship to absolute magnitude and distance. This allows for a discussion ofthe light year and parsec. How the temperature of a star is measured is then described, beforecombining luminosity and temperature in the Hertzsprung-Russell (H-R) diagram. The differenttypes of stars – and other astronomical objects – is then put in a context of observableproperties rather than theoretical behaviour. How supernovae can be used to measuredistances and the implications these measurements are having on the ideas of dark energy,provide an opportunity for students to experience some of the stranger theories in moderncosmology. This chapter concludes with black holes and how their ‘size’ can be calculated.

The final chapter puts astrophysics into a wider context. The analysis of motion using theDoppler effect is described, and applied to various situations, including Hubble’s Law and theage of the Universe. This allows for a discussion of evidence for the Big Bang. Finally, thediscovery of Quasars, and some of their properties is investigated.

Astrophysics is an area undergoing constant change and development. This course attempts toreflect this by including very recent ideas such as dark energy. It is also an opportunity to putscientific theory into the context of discovery, as better techniques provide more informationabout exotic and fascinating phenomena.

As with the other options and the core A2 topics, students are expected to use their AS

knowledge and understanding in topics which make use of such knowledge and understanding.Each of the sections may be linked with AS and A2 core topics, thereby allowing core topics tobe reinforced. For example, the work on resolving power links closely to the work on diffractionin the AS topic, and the work on black holes can be related to the A2 topic on gravitational fields.


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Chapter 1 Lenses and Optical Telescopes

A Lenses

The Converging Lens

There is some evidence to suggest that the earliest lenses may have been made over 3000years ago, largely being used to focus sunlight in order to start fires. It was not until the 17th century that lenses were being used to make telescopes, however.

The diagram shows different shapes of lenses. This specification only considers in detail thebehaviour of light passing through convex lenses.

The process of refraction was covered in unit 2. When light passes from air into the lens, itslows down. Light which is not incident normally gets bent towards the normal to the boundary. As the light leaves the lens and enters the air, it will bend away from the normal as it speeds upagain.

lightenteringlens

lightleavinglens

normal

normal

biconvex piano-convex

positivemeniscus

negativemeniscus

piano-concave

biconcave


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A convex lens is designed to have the correct curvature needed to bring parallel light to a pointfocus. This diagram shows light entering a lens parallel to an axis of symmetry called theprincipal axis. The point on the axis where the rays cross is called the principal focus, F.The distance from the centre of the lens to the principal focus is called the focal length, f.

principalfocus

F

focal length, f

principalaxis

The refraction of the light passing through the lens is shown in this diagram. When drawing raydiagrams the situation is simplified by assuming that the lens is very thin so that the ray is onlyshown bending once, in the middle of the lens.

The other advantage of assuming that the lens is thin is that a ray of light passing through thecentre of a thin lens effectively doesn’t bend:

This means that its path is undeviated. This is useful in ray diagrams because the path ofthis ray can be drawn as a straight line.


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Ray diagrams

We can use this behaviour of the light passing through the lens to predict how an image wouldbe formed. This can be done in the form of a diagram, or by calculation.

Students may be required to draw the ray diagram for a converging lens where the object isbetween the focal point, F, and the lens, between F and 2F, at 2F and beyond 2F. They mayalso need to describe the properties of the image formed in each case.

Whichever diagram is needed, the ideas are the same: use a pencil and ruler. Start by drawinga principal axis, and drawing the lens as a line or very thin ellipse. Mark the principal foci, andplace the object. Next, two construction rays are needed. The first shows a ray of light travellingparallel to the principal axis and passing through the principal focus, F.

Fprincipalaxis

The second shows a ray of light passing through the centre of the lens and being undeviated.

Fprincipalaxis

All ray diagrams can be drawn using these two rays. There is a third alternative – (a ray of lightpassing through the principal focus, leaves the lens parallel to the principal axis) – which can betaught if preferred.


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Object beyond 2F

FF

object

image

The image is real, inverted and diminished

Object at 2F

FF

object

image

The image is real, inverted and the same size.


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Object between F and 2F

FF

object

image

The image is real, inverted and magnified.

Object between F and the lens

FF

objectimage

The image is virtual, upright and magnified. Notice how the dotted lines show where the imageappears to be – the feature of a virtual image. Students should be warned not to place theirobjects too near F, and to leave enough room for the image on the left. There is no need to

draw an ‘eyeball’.

Properties of an image

The three properties of the image required compare it with the object and relate to its size(magnified, diminished, same size), orientation (upright or inverted) and nature (real or virtual).If asked for properties in the examination, no credit will be given for position, colour etc.


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The lens formula

Although scale diagrams will not be asked for in the examination, it is useful to get pupils to usethese diagrams to predict the position, size and orientation of an image for a real practicalsituation, and then test it out.

The diagrams can also be used to verify the lens formula – the equation linking the focal length (f ),and the distances along the principal axis from the object to the lens (u) and the image to the lens (v ).

FF

object

image

v

u

f

Using similar triangles, it can be shown that

f v

h

f

h I o

−= and

v

h

u

h I o =

where ho and hI are the heights of the object and image respectively.

from which you get

vu f

111+=

which is the lens formula. For this equation to work, a couple of rules need to be followed

• converging lenses have positive focal lengths, diverging lenses have negative focal lengths. As the only lenses asked about in this specification are converging, this is easy to follow

• distances to real images are positive, and to virtual images are negative.

Students do not need to be able to reproduce a derivation of the lens formula, but it is probablyworth doing with them.

ho

hI


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B Astronomical telescope consisting of two converging lenses

The required ray diagram is given below.

FoFe

virtual

image atinfinity

light fromobject atinfinity

constructionline

objectivelenseyepiecelens

The telescope is in normal adjustment because the image is formed at infinity, ie the light raysleave the telescope parallel. This means that the focal points of the two lenses coincide, makingthe distance between them equal to the sum of the focal lengths.

The diagram includes the intermediate image (at the point where the two focal points coincide),but isn’t essential. Common problems with the diagram include

• the rays bending at the intermediate image (at Fo, Fe)

• the rays not leaving the eyepiece parallel to each other

• the rays not leaving the eyepiece parallel to the construction line

• the central ray bending at the objective lens.

Another difficulty some students have with this diagram is understanding that the three parallelrays come from the same point on the object, ie the rays may not be parallel in reality – but aretoo close to being parallel as to make any difference.

The angular magnification is based on the angles subtended by the object and image (clearly, ifthey are both at infinity, the magnification can not be based on their heights). It is important,

therefore, that students have an idea of radians to help them deal with this. The unit isimportant – even though it is a ratio, it still has the units of radians (rad).


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The angle, in radians, subtended by an object of height, h, a distance, d , away is given by:

d

h=θ

A useful thing to show is that the Sun and the Moon both subtend approximately the same angle

at the Earth – this is why solar eclipses are so spectacular. You can use the data to verify this.

Mean distance to Moon = 380,000 km

Mean diameter of Moon = 3500 km

Mean distance to Sun = 150,000,000 km

Mean diameter of Sun = 1,400,000 km

The light leaving the telescope causes the image to subtend a bigger angle than the object – it ismagnified. The definition of angular magnification is

angle subtended by image of eye Angular magnification, M , = angle subtended by object at unaided eye.

The use of M here can be misleading, as M is also used for the absolute magnitude later.

Looking at a simplified ray diagram,

FoFe

objectivelens

eyepiecelens

α αβ

βh

fo

fe

It should be clear from this diagram that the angular magnification is

β

α .

For small angles, fe

h=α and

fo

h= β , so

fe

fo=

β

α . Again this derivation is not needed but is

straight forward, and it does make use of the idea of the ‘small angle approximation’ which isalways useful to emphasise.

From this formula, and the ray diagram, it is clear that a high magnification telescope requires along objective focal length and short eyepiece focal length. Refracting telescopes are relativelyeasy to make in the lab using a couple of lenses, a metre ruler and plasticine and thisrelationship can be easily investigated.

The equations M = fe fo and distance from objective to eyepiece = fe fo + can be used together

to help design particular telescopes. It is fairly common to see past questions ask about this.


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C Reflecting telescope

The ideas behind the reflecting telescope start with the simple law;

Angle of incidence (i ) = angle of reflection (r )

This can be applied to curved surfaces too.

Leading to the idea that a concave mirror will bring

parallel light to a point focus. The focal point, F, istaken to be at half the distance to the centre ofcurvature. Unfortunately, circular (or spherical)mirrors are not so simple. Off-axial rays are broughtto a focus closer to the mirror.

This means that the focal point is different for

different rays of light. This results in a blurred imageand is called spherical aberration. This ray diagramis obtained fairly easily by following the reflectionlaw, having drawn a circular mirror on a piece ofpaper.


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The Cassegrain telescope

The problem of spherical aberration is overcome usinga parabolic mirror (the more mathematical studentsmay recognise a parabola as an ellipse with one focusat infinity).

There have been several designs of reflecting telescopes, but the specification requiresknowledge of a Cassegrain arrangement. Parallel light enters the telescope from a distantobject. The large parabolic primary reflector is used to collect the light and bring it to a focus. A secondary convex reflector is used to reflect the collecting light out through a hole in theprimary parabolic mirror. The light then arrives at an eyepiece. The whole arrangement isshown below. Notice that only two rays are required, and that they are drawn initially parallel to

the principal axis.

There are several common errors that are worth pointing out

• students often draw the two halves of the primary as two separate mirrors – ie their curvedoes not look like a continuous parabola

• the secondary reflector is often drawn plane, or even concave

• the eyepiece is sometimes left out

• The rays are sometimes drawn crossing before they hit the secondary mirror.

Although it is difficult to do using simple software, it is worth including the shading in all of thesediagrams to show which side is not the reflecting reflector.


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Relative merits of reflectors and refractors

The largest (and therefore best) telescopes are reflectors. There are several reasons for thedominance of reflectors in modern telescopes.

Reflectors can be made much larger than refractors because a mirror can be supported frombehind, whereas a lens must be supported at the edge. A large lens is likely to break under itsown weight. The size is important for two reasons

• a larger objective means a much greater collecting area (it is proportional to the diameter 2),this means fainter objects can be seen

• the larger the diameter of the aperture the greater the resolving power – ie the image willshow more detail, this is dealt with later.

Mirrors do not refract light and therefore do not suffer from chromatic aberration. The edge ofa lens can be seen as a slightly curved triangular prism. White light is dispersed – ie split into its

different colours – resulting in different focal points for different colours or wavelengths.The focus for blue (or violet) light is closer to the lens than that for red light. This can be seen inthese diagrams

The resulting images tend to have multi-coloured blurred edges.

Spherical lenses suffer from spherical aberration too. It has already been shown that, using aparabolic mirror, this can be illiminated in reflectors.

There are some problems with reflectors. The secondary mirror and the ‘spider’ holding it in

place both diffract the light as it passes, leading to a poorer quality image. Also, there is somerefraction eventually, in the eyepiece used to view the final image. The objective mirror is‘exposed’, but the objective lens in a refractor is protected as it is inside a closed container.Overall however, it is clear that the reflector has the edge.

One common misconception associated with the secondary mirror is that it blocks the light(which it does) resulting in the central portion of the image being missing (which is not the case).There will be a slight reduction in the amount of light, but the light from a distant source isparallel, so that even light from a source on the axis of the mirror will hit the secondary and becollected.

It is also worth noting that non-optical differences, such as cost, would not gain credit in the

examination.


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D Resolving power

The diffraction pattern produced by a circular aperture

When a wave passes through an aperture or past an obstacleit is diffracted. When monochromatic light passes through avery narrow slit this diffraction can result in interference fringes. Students should be familiar with the single slit diffractionpattern from unit 2 of the AS course.

Rotating that pattern about its centre produces the pattern for acircular aperture. The large central maximum is called the AiryDisc, and it is twice as wide as the further maxima in thepattern.

Students should be able to link the picture above with thegraph on the right. The vertical axis is intensity or brightness,and the horizontal axis is angle, ! , or sin ! . For small anglesthese two are approximately equal, if ! is in radians.

For a single slit, minima in the pattern are at angles given by

D

nλ θ =sin

where n is the ‘order’ of the minimum, λ is the wavelength ofthe light and D is the width of the slit.


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The Rayleigh Criterion

When light from an object enters a telescope it is diffracted, resulting in loss of detail in theimage. The aperture of the telescope is assumed to be the same diameter as the objective.

How much detail the telescope can show is called itsresolution.

It is easier to consider two objects which are very closetogether. Is the telescope capable of resolving them intotwo images, or will the diffraction patterns overlap so muchthat they will be seen as a single image?

This question is answered by considering the RayleighCriterion‘Two objects will be just resolved if the centre of thediffraction pattern of one image coincides with the first

minimum of the other’.

Using the equation from above gives a minimum angularseparation, ! , given by

D

λ θ ≈

where D is the diameter of the objective. The equation for acircular aperture is actually

D

λ θ 22.1=

which helps explain the approximately equals sign.

just resolved

not resolved


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The Problems with Resolving Power

When using a telescope, the better the resolution, the smaller the angle, ! , and therefore thegreater the detail which can be seen. In the examination, a question which asks the students tocalculate the resolving power in a particular situation requires the student to calculate ! .This angle should be referred to as the minimum angular resolution of the telescope.

Using the term ‘resolving power’ can cause students several problems.

• Some students expect the resolving power to be bigger for a smaller angle, which isreasonable. To achieve this, they invert their calculated angle. This is not penalisedprovided it is clear what the student is doing.

• The unit of ! is the radian, rad. Seeing the word ‘power’ encourages many students to give itthe unit watt, W.

• The wavelength and diameter have to have consistent units. Aperture diameters can begiven in metres, millimetres and even kilometres for some large baseline interferometrydevices. Wavelengths are often in nanometers, but can also be in micrometers, millimetresor metres. Problems with indeces are very common. Students should be encouraged topractise calculating the resolving power of a wide range of telescopes which use manydifferent parts of the electromagnetic spectrum.

The angle, ! , is a theoretical minimum. In reality other effects can make the calculated anglefairly irrelevant when considering the resolving power of a telescope, particularly at shorterwavelengths. These effects include the refraction of light as it passes through the atmosphere.It has also already been stated that the spider holding the secondary mirror in a Cassegrainarrangement causes diffraction problems.

In the examination, however, if the resolving power is asked for it is this angle which is needed.


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E Charged coupled device (CCD)

It is important to consider the role of the Charge-Coupled Device (CCD) in modern astronomybecause of the huge impact it has had. Students may also be familiar with their use in camerasand mobile phones.

The diagram above looks fairly complicated, and is beyond the requirements of the specification.Essentially the structure of the CCD is fairly straight forward. It can be thought of as a piece ofsilicon, which can be several centimetres along each side. This silicon chip is made up ofperhaps 16 million picture elements or pixels, formed into a two dimensional array. Electrodesconnected to the device help form potential wells which allow electrons to be first trapped, and

then moved in order for the information to be processed.

A telescope is used to form on image on the CCD. Photons striking the silicon liberate electronswhich are then trapped in the potential wells of the pixels. Exposure continues until sufficientelectrons have been trapped to allow the required image to be obtained.

The number of electrons trapped in each well is proportional to the number of photons hitting thepixel ie to the intensity of the light falling on the pixel. This means that the pattern of electrons inthe array is the same as the image pattern formed by the photons. The quantum efficiency,which is the percentage of incident photons which cause an electron to be liberated, can be 70%to 80%. This is very high compared to photographic film (about 4%).

When exposure is complete, the electrodes are used to ‘shuffle’ the electrons along the array so

that the contents of each well can be measured and the charge measurement used to create theimage.

CCDs have been used to detect several regions of the electromagnetic spectrum, not just visiblelight. An important part of their behaviour is their linear response so that very faint parts of arelatively bright image can be obtained. The resolution of the CCD itself is related to the size ofeach pixel – smaller pixels allow for more detail to observed as each pixel integrates the lightfalling on it.

In examinations on the previous specification, questions associated with the CCD were usuallyanswered well. Common problems tend to be linked to the process of electron liberation. Somestudents give detailed accounts of the photoelectric effect, which is understandable but not what

is required. Others can get confused with pair production, and write about the photons creating the electrons (and sometimes positrons). Clearly the language used to describe the process isimportant, and care must be taken to check content taken from unreliable sources, such ascertain websites.

Charge-Coupled Device

phase 1

phase 2

phase 3

electrodes

insulating layer

n-typesilicon

layer

trapped electrons potential wellp-type silicon layer


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Chapter 2 Non-optical telescopes

A Single dish radio telescopes

Ground-based observations are severely affected by the atmosphere for most of theelectromagnetic spectrum. Two of the least affected areas are visible, dealt with earlier, andradio waves.

The design of a single dish radio telescope is very similar to that of a reflecting optical telescope.The parabolic metal surface (reflecting dish) reflects the radio waves to an aerial without anyspherical aberration. There is no need for a secondary reflector as the aerial can be placed atthe focal point. Information is then transmitted for analysis.

A simple calculation would show that, despite their large size, the resolving power of radiotelescopes tends to be quite poor. In the early 1960s lunar occultation was used to identifysources of radio waves, and this led to the identification of the first quasar, 3C-273.

Man-made interference can cause a big problem for radio astronomy. Radio transmissions,mobile phones, radar systems, global navigation satellites, satellite TV and even microwaveovens all contribute to making the detection of extra-terrestrial objects difficult.

It is difficult to deal with interference from orbiting satellites, but the effect of microwave ovens isminimised by building radio telescopes away from centres of population. Linking telescopestogether also helps, as local interference can be removed from individual detectors, enhancingthe common signal from the distant sources.

One aspect of the design of radio telescopes is worth pointing out. Because the wavelengths ofthe radio waves are relatively long, the reflecting dish of the telescope can be made from meshrather than solid metal, decreasing the weight and cost. Provided the mesh size is smaller than

λ /20, radio waves are reflected rather than diffracted through the mesh.

As well as improving the resolving power of the telescope, increasing the dish diameter alsoincreases the collecting power ie the rate at which energy can be collected by the telescope.This clearly depends on the intensity of the incident radiation and the area of the dish. The areaof the dish is proportional to the square of the diameter, so doubling the diameter increases thecollecting power by a factor of 4.

aerial

masthead preamplifier –high frequency circuits

coaxial cable in control room

parabolic metal surface

amplifier, tunerand detector –low frequency

chart recorderor computerdata logger


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Ground based telescopes are largely restricted to the visible and radio regions of theelectromagnetic spectrum.

Molecular absorption in the atmosphere

100

80

60

40

20

percentage ofenergy arriving at

sea level %

0O3

0 0 . 2

0 . 4

0 . 6

0 . 8

1 . 0

1 . 2

1 . 4

1 . 6

1 . 8

2 . 0

2 . 2

2 . 4

2 . 6

visibleUV IR

This graph shows the effect that water vapour, carbon dioxide, oxygen and ozone have on theIR, visible and U-V spectrum as the radiation passes through the atmosphere. This explainswhy telescopes detecting these parts of the em spectrum are placed on balloons, in orbit, at thetop of high mountains or in very dry areas, such as the Atacama Desert.

Perhaps surprisingly, X-rays and Gamma rays are also absorbed by the atmosphere. Some of

the earliest satellites were used to carry X-ray and Gamma ray detectors. Detection of thesehigh-energy photons allows analysis of some of the most violent and exotic phenomena in theUniverse, and may provide data about its composition, age and expansion rate.

B I-R, U-V and X-ray telescopes

wavelength/!m


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Information and links for some typical telescopes are given below.

emspectrum

telescope position examples of use

UKIRT

3.8 m diameterhttp://www.jach.hawaii.edu/UKIRT/

4200 m above sea

level on Mauna Kea,Hawaii

Infra redSpitzer

0.85 m diameterhttp://spitzer.caltech.edu/

earth-trailing solarorbit

finding exoplanets

observing after-effectsof gamma ray bursts,

observing nebulae

GALEX0.5 m diameter

www.galex.caltech.edu/

low Earth orbit(ht 700 km)

ultravioletCOS

(2.4 m diameter)http://hubblesite.org/

on board the HSTlow Earth orbit

(ht 570 km)

star formation rate indistant galaxies

analysing quasars

Chandra‘grazing’ mirrors

http://chandra.harvard.edu/

Earth orbit(ht 140,000 km)

X-rayXMM-Newton

‘grazing’ mirrorshttp://xmm.esac.esa.int/

Earth orbit(ht 30000 km)

discoveringsupermassive black

holes

observing stellarremnants

It is worth noting that the STIS (Space Telescope Imaging Spectograph) analyses wavelengthsof light from IR, visible and UV. The light is collected by the 2.4 m Cassegrain reflector on theHubble Space Telescope. In fact it is only the X-ray telescopes which require a significantlydifferent design – the ‘grazing’ mirrors that can be found on the Chandra and XMM-Newton

websites.

Although they are not on the specification, it is worth mentioning Gamma ray telescopes at thispoint. The occurrence of Gamma ray bursts is yet another aspect of astronomy which haseluded explanation. Detection of the particles produced by these Gamma rays when they hit theupper atmosphere allows for ground-based observations. It is the acronyms used for the variousexperiments which students may find interesting too – for example, ‘CANGAROO’ (Collaborationof Australia and Hippon for a Gamma Ray Observatory in the Outback) can only be based in Australia.

The table shows that many modern telescopes are put into space. There are three mainreasons for this

• the absorption of the electromagnetic waves by the atmosphere

• the light pollution and other intereference at ground level

• the effect the atmosphere has on the path of the light as it passes through.


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•

Chapter 3 Classification of stars

A Classification by luminosity

A brief look at the night sky would show students that some stars look brighter than others.In about 120 BC the Greek astronomer Hipparchus produced a catalogue of over 1000 stars andtheir relative brightness. He used a 6 point scale, with 1 being the brightest and 6 the dimmest.

The modern version of this is called the apparent magnitude scale. In simple qualitative termsthis is the brightness of the star as seen from Earth. This is to avoid problems with the nonvisible parts of the spectrum which should not be included in this simpler definition. So termslike ‘luminosity’ or ‘intensity’ should not be used. Luminosity is the total power radiated by a star,and the intensity is the power per unit area at the observer.

In this sense apparent magnitude is simply a scale of brightness (as judged by an observer)which decreases as brightness increases, and which has a value of about 6 for the faintest stars

which can be seen with the naked eye on a good night. Students can be introduced to apparentmagnitude in this way, and that it is initially treated as a qualitative scale. You could imaginestudents being asked to look at the stars in, say, The Plough, and to write them down in order ofbrightness and use some sort of scale to compare them. It should be stressed that what thestudents are using is the ‘visible luminosity’. As a star’s luminosity is its total power output at allwavelengths it is often referred to as the ‘bolumetric luminosity’. It would not be helpful tocatalogue a star’s bolumetric luminosity for observers to use with optical telescopes.

The fact that some stars appear brighter than others can be due to two reasons; it could becloser, or it could be emitting more power at visible wavelengths. The brightest star in ThePlough, for example, happens to be the furthest away.

B Apparent magnitude, m

With modern measuring techniques (eg photography, CCD cameras) a more quantitativeapproach to apparent magnitude can be made. The devices allow the intensity of the light froma star to be measured. The apparent magnitude scale now takes on a more precise meaning.

The branch of astronomy that deals with this is called photometry, and towards the end of the1700s, William Herschel devised one simple but inaccurate method to measure the brightness ofstars. One key point that arose from his work was the fact that a first magnitude star deliversabout 100 times as much light to Earth as that of a sixth magnitude star.

In 1856, following the development of a more precise method of photometry, Norman Pogsonproduced a quantitative scale of apparent magnitudes. Like Herschel he suggested thatobservers receive 100 times more light from a first magnitude star than from a sixth magnitudestar; so that, with this difference of five magnitudes, there is a ratio of 100 in the light intensityreceived.

Because of the way light is perceived by an observer, equal intervals in brightness are actuallyequal ratios of light intensity received – the scale is logarithmic. Pogson therefore proposed thatan increase of one on the apparent magnitude scale corresponded to an increase in intensityreceived by 2.51, ie the fifth root of 100. Thus a fourth magnitude star is 2.51 times brighter thana fifth magnitude star, but 6.31 times brighter than a sixth magnitude star.

With the faintest stars having an apparent magnitude of 6, this scale allows the brightest stars tohave negative apparent magnitudes, the brightest star having an apparent magnitude of –26.


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Pogson’s scale allows the intensity, I, measured to be converted to a numerical apparentmagnitude

m = – 2.51 log (I/2.56 x 10 –6)

Where 2.56 × 10 –6 lux is the intensity for a magnitude 0 star, and the log is to base 10.

Questions will not be set on this equation, although its important features should be taught.In particular, the negative relationship between intensity and apparent magnitude and the 2.51ratio for a difference in apparent magnitude of 1, and, therefore, the log scale used for apparentmagnitudes. Tables such as the one below are easy to obtain using sources on the internet.

object apparent magnitude, m

the Sun -26

the full Moon -19

Venus -4

Sirius -1.5

Vega 0.0

Altair 0.8

Polaris 2.0

C Absolute magnitude, M

Units of distance

It is clear that apparent magnitude tells you very little about the properties of the star itself,

unless you know how far away it is. Astronomical distances are extremely big – and severaldifferent units of distance exist to make the numbers smaller and more manageable.

The light year is the distance travelled by light in a vacuum in one year. It can be easilyconverted into metres by multiplying the speed of light in a vacuum (3 × 108 ms –1) by the numberof seconds in a year (365 × 24 × 3600) to get 9.46 × 1015 m.

The AU (astronomical unit) is the mean distance from the Sun to the Earth and has a value of1.5 × 1011 m.

The unit which causes students the biggest problem is the parsec: the distance from which 1 AUsubtends an angle of 1 arc second (1/3600th of a degree), this is most easily defined from a

diagram such as this

S E1 AU

1 p a r s e c

θ

" = 1 arc sec

= 1/3600°


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The parsec is an important unit because of the way distances to nearby stars can be determined – trigonometrical parallax. This involves measuring how the apparent position of a star, againstthe much more distant background stars, changes as the Earth goes around the Sun.

1 pc is approximately 3.26 light year.

Definition of absolute magnitude, M

The absolute magnitude of a star is the apparent magnitude it would have at a distance of 10 pcfrom an observer. It is a measure of a star’s inherent brightness. This means that absolute andapparent magnitudes are measured on the same logarithmic scale.

As light intensity reduces in proportion to the inverse square of the distance, the relationshipbetween apparent and absolute magnitude can be related to distance, d , by the equation:

10log5 d

M m =−

‘d ’ is measured in parsec, and the log is to base 10. A quick glance shows that

• Stars which are closer than 10 pc (about 33 light years) have a brighter (more negative)apparent magnitude than absolute magnitude. m-M < 0

• Stars further than 10 pc have a dimmer (more positive) apparent magnitude than absolutemagnitude. m-M > 0

• If m = M , the star must be 10 pc away.

star apparent magnitude m absolute magnitude M m-M distance/light year

Sirius -1.46 1.43 -2.89 9

Pollux 1.1 1.1 0 33

Castor 1.58 0.59 0.99 52

Care must be taken comparing magnitudes – does a bigger magnitude mean brighter (greaterintensity) or dimmer (bigger number). It is safer to talk about brighter or dimmer magnitudes.

D Classification by temperature, black body radiation

All the information we get about stars comes from the electromagnetic radiation we receive.

For example, an analysis of the intensity of the light at different wavelengths allows astronomersto measure the ‘black body’ temperature of a star. Using this and an estimation of the poweroutput of a star allows its diameter to be determined.

Black body radiation and Wien’s displacement law

We are used to the terms ‘red hot’ and ‘white hot’ when applied to sources of heat. We are alsoaware that hot objects emit radiation at infrared wavelengths. This behaviour of hot objects andthe em radiation they produce was investigated at the end of the 19th century and contributed tothe development of quantum theory by Max Planck.

A black body is one that absorbs all the em radiation that falls on it (in practical terms, it is thehole in the wall of an oven, painted black on the inside). Analysis of the intensity of the emradiation emitted from a black body at different wavelengths produces the following result.


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The three lines are for three different temperatures

The graph shows that the hottest object (P) has a peak at the shortest wavelength. In fact, the

relationship between the wavelength of the peak, λ max , and the temperature, T , is called Wien’sdisplacement law:

mK const T 3

max 109.2 −×==λ

λ max is measured in metres and T is the absolute temperature and is therefore measured inkelvin, K.

Hotter stars will therefore produce more of their light at the blue/violet end of the spectrum andwill appear white or blue-white. Cooler stars look red as they produce more of their light atlonger wavelengths. When analysing the light from stars in this way it is assumed that the staracts as a black body, and that no light is absorbed or scattered by material between the star andobserver.

Stefan’s Law

Stefan’s law relates the total power output, P , of a star to its black body temperature, T , and

surface area, A.4 AT ! σ =

where # is called Stefan’s constant and has the value 5.67 × 10 –8 W m –2 K –4. The law issometimes called the Stefan-Boltzmann Law as it was derived by Stefan experimentally andindependently by Ludwig Boltzmann from theory.

P

Q

R

intensity

wavelength/nm

0

0

500 1000 1500 2000


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Qualitative comparison of two stars

The law tells us that, if two stars have the same black body temperature (are the same spectralclass), the star with the brighter absolute magnitude has the larger diameter. This fact is veryimportant when interpreting the Hertzsprung-Russell diagram, which is considered shortly.

It is quite common for Stefan’s Law to be used qualitatively when comparing the size of stars.For example, Antares and Proxima Centauri are both M class stars. Although they haveapproximately the same black body temperature, Antares is one of the brightest stars, with anabsolute magnitude of –5.3, and Proxima Centauri is one of the dimmest, with an absolutemagnitude of 15. It is not surprising to work out, therefore, that the diameter of Antares isapproximately 5500 times that of Proxima Centauri.

This YouTube clip is one of several which shows a comparison of starshttp://www.youtube.com/watch?v=z7lHgvGhHo4&feature=related.

Quantitative use to determine the diameter of a star

Consider two stars of black body temperatures T 1 and T 2. If the ratio of their power output(P 1 /P 2 ) is known, Stefan’s law can be used to calculate the ratio of their diameters, d 1 /d 2

2

1

21

22

2

1

P

P

T

T

d

d =

This equation is also useful when considering how the size of a star must change as it goesthrough its life cycle.


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The inverse-square law

There are several inverse square laws in Physics. Essentially any property which reduces to aquarter of its original value when you double the distance follows an inverse square law

2

0

d

I

I =

Derivation of )10

log(5 d

M m =−

As has already been stated, when the inverse square law equation is combined with Pogson’sequation for apparent magnitude

)1056.2

log(5.26−×

−= I

m

the distance equation can be obtained. The specification does not require this derivation, but itis worth demonstrating at this point

I 0 is the intensity of a light source at 1 pc, and I is the intensity at distance, d . Therefore, whenwe substitute in the inverse square law, we get

)1056.2

log(5.226

0

d

I m

−×−=

When d = 10 pc, m = M , so:

)101056.2

log(5.226

0

−×−=

I M

Subtracting

)101056.2

log(5.2)1056.2

log(5.226

0

26

0

−− ×+

×−=−

I

d

I M m

Tidying

)10

1log(5.2)

1log(5.2

22 +−=−

d M m

And therefore)10log(5)log(5 −=− d M m

So

)10

log(5 d

M m =−

Other uses of the inverse square law

Using this equation, the intensity of a light source can be determined at different distances.For example, if the Sun is used to provide the energy for solar cells on a space probe, theequation can be used to determine what happens as the probe gets further from the Sun.

The inverse square law can also be used to estimate the power output of different objects.For example, Doppler shift information suggests that some quasars are billions of light yearsaway. Using the inverse square law, it can be shown that the power output of a quasar must beequivalent to that of a whole galaxy.

Assumptions in its application

Use of the inverse square law assumes that no light is absorbed or scattered between thesource and the observer and that the source can be treated as a point.


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E Principles of the use of stellar spectral classes

When the light created within a star passes through its ‘atmosphere’ absorption of particularwavelengths takes place. This produces gaps in the spectrum of the light from the star resulting

in an absorption spectrum. The wavelengths are related to frequency (c = f λ ) and therefore to

particular energies ( $E = hf ). Electrons in the atoms and molecules of the star’s atmosphere areabsorbing the light, and therefore ‘jumping’ to higher energy levels. The difference in theseenergy levels are discrete (have particular values) and therefore the frequencies of the absorbedlight are discrete.

Some early attempts to classify stars according to their spectra used alphabetic labels.When the processes were better understood it transpired that a more logical order, related totemperature, was adopted. The original alphabetical order was then lost.

Spectral ClassIntrinsicColour

Temperature/KProminent Absorption

Lines

O blue 25000 – 50000 He+, He, H

B blue 11000 – 25000 He, H

A blue-white 7500 – 11000 H (strongest) ionised metals

F white 6000 – 7500 ionised metals

G yellow-white 5000 – 6000 ionised & neutral metals

K orange 3500 – 5000 neutral metals

M red < 3500 neutral atoms, TiO

Probably the most famous mnemonic for remembering the order is Oh be a fine girl, kiss me!

The relationship between temperature and spectra is due to the effect of the energy on the stateof the atoms or molecules. At low temperatures, there may not be enough energy to exciteatoms, or break molecular bonds, resulting in the TiO and neutral atoms in the M class spectra. At higher temperatures atoms have too much energy to form molecules, and ionisation can takeplace (as seen in classes F and G). The abundance of hydrogen and helium in the atmosphereof the hottest stars means that their spectral lines start to dominate. The Hydrogen Balmerseries is a useful indicator here.

The spectrum above shows absorption lines at particular wavelengths in the visible spectrum.They are due to the absorption of light by excited hydrogen. The diagram below shows theelectron energy levels of a hydrogen atom.

656486434

410

wavelength/nm

intensity


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n=6

n=5

n=4

energyincreasing n=3

n=2

n=1

The electrons start in the n=2 state. To achieve this with a significant proportion of thehydrogen, the atmosphere of the star must be fairly hot. Any hotter than about 10000 K,however, and the hydrogen starts to be ionised. Absorbing the correct amount of energy excitesthe electrons even further, raising them to n=3, 4 etc. When the electrons de-excite, the light isemitted in all directions, and the electrons may do it in several steps or miss out n=2 and jump

down to n=1. This results in the gap, or reduction in intensity in the direction of the observer.The name given to the spectrum produced when transitions from n=2 are made is the HydrogenBalmer Absorption Spectrum. Chemists will be familiar with other series such as Lyman andPaschen.

In terms of the Balmer series, the following points can be noted:

Spectral Class Prominence of Balmer lines Explanation

O weak

B slightly stronger

star’s atmosphere too hothydrogen likely to be ionised

A strongesthigh abundance of hydrogen

in n=2 state

F weaktoo cool, hydrogen unlikely to

be excited

G

K

M

very weak/nonetoo little atomic hydrogen, far

too cool to be excited

When searching for the properties of stars, it will be clear that the spectral classes have beensubdivided, so that each spectral class has a 10 point scale. The Sun, for example is a class G2star.

Evidence from stellar absorption spectra supports temperature calculations based on black bodyradiation. As has been seen, temperature and absolute magnitude can give some indication ofthe size of a star. It is this relationship that is dealt with in the next section.


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F The Hertzsprung-Russell (H-R) diagram

There are many versions of the H-R diagram, but the one required on this specification is shownbelow. It is an x-y scattergraph, with stars as the points on the graph. The y -axis is absolutemagnitude, from +15 to –10, whilst the x -axis can be spectral class, as shown, or temperature

(from 50 000 K to 2500 K). For simplicity, the diagram is reduced to just three regions, althoughthere are others, such as the supergiants.

Stefan’s law can be used to explain why some stars are referred to as giants or dwarfs.For example, a dim (absolute magnitude 12) star in spectral class B must be much smaller thana much brighter star (absolute magnitude –10) in the same spectral class (ie at the sametemperature).

It is also useful to be able to follow a star,such as the Sun, as it evolves on the H-Rdiagram. The Sun is currently a G class star

with an absolute magnitude of 5. This puts iton the main sequence. As it uses up itshydrogen fuel and starts fusing helium its corewill get much hotter and it will expand. At thispoint it will leave the main sequence andbecome a giant star. Its diameter is likely toextend beyond the orbit of the Earth. As itexpands, the outer surface of the Sun willcool. Helium fusing continues but eventuallythe outer material of the Sun will be pushedaway to form a planetary nebula, leaving theextremely hot but small core as a white dwarf.Fusion has finished, and the white dwarf coolseventually to a brown dwarf.

absolute magnitude

spectral class

-10

-5

0

5

10

15

giants

dwarfs

main sequence

O B A F G K M

absolute magnitude

spectral class

-10

-5

0

5

10

15

giants

dwarfs

mainsequence

O B A F G K M


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G Supernovae, neutron stars and black holes

The evolution of a star is related to its mass. The Sun is relatively small, and its demise will beunspectacular. Larger stars are more likely to have a catastrophic ending.

Supernovae

Supernovae are objects which exhibit a rapid and enormous increase in absolute magnitude.Type 1a supernovae are useful because they reach the same peak value of absolute magnitude,making them standard candles, ie the absolute magnitude is known, the apparent magnitudecan be measured, so the distance can be determined.

The diagram shows the light curve of a typical type 1a supernova.

The peak value of absolute magnitude is –19.3, and occurs after about 20 days from the start ofthe increase in brightness. Conventionally time is measured from the peak.

Type 1a supernovae are due to an exploding white dwarf star, which is part of a binary system.The white dwarf increases in mass as it attracts material from its companion. Eventually thewhite dwarf reaches a size which allows fusion to start again, (the fusion of carbon whichrequires very high pressures and temperatures), which causes the star to explode.This occurs when the star reaches a critical mass, and produces a very consistent light curve.The other advantage of using Type 1a supernovae as standard candles is that they are verybright. This means they can be used to measure the distance to the furthest galaxies.Recent measurements of this kind have led to the idea that the expansion of the Universe maybe accelerating, and that the controversial ‘dark energy’ may be driving this expansion.

This is dealt with later.

-16

-18

-12

-14

0 300200100

time/days

absolute magnitude


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Other supernovae also exist. When the core of giant stars stop generating energy by fusion,gravitational collapse can take place. The core may form a neutron star or black hole, and thegravitational energy released can cause the stars outer layers to heat up and be expelled.

Neutron stars

When a giant star has exploded its outer layer, the core may be too massive to become a whitedwarf and further collapse can take place. A core with a mass of about twice that of the Sun willform a neutron star. As its name suggests, it is believed to largely consists of neutrons, althoughsome theories suggest it may have a ‘crust’ made of atomic nuclei and electron fluid, withprotons and neutrons making up the interior. Due to the complex nature of its structure, otherproperties are considered to be its defining features.

A neutron star is believed to have the density of nuclear matter and be relatively small(about 12 km in diameter). Due to conservation of angular momentum, they tend to be spinningvery rapidly, although they do lose energy and slow down with time. They can be very powerfulradio sources due to their extremely strong magnetic fields, and this, combined with theirspinning, produce pulsars – which behave like radio ‘light houses’. An example can be found in

the Crab Nebula.

Black Holes

More massive cores can continue to collapse and form black holes. The defining feature of ablack hole is that its escape velocity is greater than the speed of light. The boundary at whichthe escape velocity is equal to the speed of light is called the Event Horizon. Anything within theEvent Horizon of a black hole cannot escape, not even light.

The radius of the Event Horizon is referred to as the Schwarzschild radius, R s. The equation isquite easily obtained from Newtonian gravitational theory. The escape velocity of an object ofmass M is found by considering the minimum KE needed to move from the surface of the objectto infinity, where the GPE is zero.

KE lost = GPE gained

R

GMmmv =221

rearranging and cancelling for the mass of the object escaping, m:

2

2

v

GM R =

For a black hole, with an escape velocity equal to the speed of light, c ,

2

2

c

GM R s =

It is important to note that, unlike a neutron star, the density of the black hole is not necessarilyvery large. It is believed that supermassive black holes, who’s mass may be several milliontimes the mass of the Sun, exist at the centre of most galaxies. The density of these black holesmay be similar to that of water. The radius of a black hole is proportional to its mass, whichmeans that the volume is proportional to mass3. As

volume

mass=density

the density of a black hole must be proportional to the inverse square of its mass, ie the moremassive the black hole, the less dense it is.

It is believed that a quasar is produced as a supermassive black holes at the centre of an activegalaxy consumes nearby stars.


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Chapter 4 Cosmology

A Doppler effect

Most people are familiar with the change in pitch that occurs as an ambulance moves past withits siren going. This is due to the Doppler effect. As the source of the sound waves approaches,the sound waves are effectively ‘bunched together’, reducing the wavelength and thereforeincreasing the frequency of the sound (higher pitch). As the source recedes the waves are‘stretched out’, increasing the wavelength and reducing the frequency (lower pitch).

The same effect occurs with light. However, as the size of the effect depends on the ratio of thespeed of the object to the speed of the wave it is less noticeable with light unless the source ismoving extremely quickly.

As with sound, if the source of light is moving away from the observer the wavelength isincreased. This is often referred to as ‘redshift’ as red is at the longer wavelength end of the

visible spectrum. The red shift is given the symbol z and can be calculated from the equation:

c

v

f

f z =

∆=

Where v is the speed of the source and $f is the change in frequency.

In terms of wavelength

c

v−=

∆

λ

λ

Notice the negative sign. This reminds us that the wavelength decreases if the source isapproaching.

Binary Star Systems

A typical application of the Doppler effect in cosmology is the study of eclipsing binary starsystems, ie two stars rotating about a common centre of mass. In our line of sight one starpasses in front of the other, causing the eclipse. A simplified light curve from an eclipsing binarysystem is shown below.

3.25

3.30

3.35

3.40

3.45

3.50

3.55

3.60

3.65

time / days

A p p a r e n t M a g n i t u d e


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Notice that the apparent magnitude scale increases going upwards – ie the dips in the graphcorrespond to the light dimming. In this example, the two stars have different brightness’s.When the apparent magnitude is 3.30, the system is at its brightest as both stars can be seen.The deeper dips are caused by the dimmer star passing in front of the brighter star – theapparent magnitude is 3.60. The shallower dips, to 3.45, occur when the dimmer star passes infront of the brighter.

If the light from one star is analysed in more detail, the shift in wavelength of one particularspectral line can be measured as the star follows its circular orbit.

656.20

656.22

656.24

656.26

656.28

656.30

656.32

656.34

656.36

0 2 4 6 8 10 12 14

time/days

w a v e l e n g t h / n m

The peak in the graph (at 656.35 nm) occurs when the star is receding from our point of view, atits maximum velocity, ie the two stars are next to each other. The eclipses occur when the staris moving at right angles to our line of sight and the wavelength is 656.28 nm.The orbital period is 8 days.

The Doppler equation can be applied to calculate the maximum recessional velocity, which isequal to the orbital speed of the star.

c

v−=

∆

λ

λ

rearranging and substituting

(656.35 – 656.28) × 3 × 108 656.28

= 3.2 × 104 m s-1

With the orbital speed and time period, the diameter of the orbit can be calculated using thecircular motion equation.

circumference of orbit = orbital speed × orbital period

= 3.2 × 104 × 8 × 24 × 3600

= 2.21 × 10

10

mdiameter = 2.21 × 1010/3.14 = 7.04 × 109 m


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Hubble’s law

Observation of distant galaxies showed significant red shifts. Further analysis suggested thatthe further away the galaxy is, the greater the red shift (see section A, Doppler Effect).

The Doppler equation was used to calculate the recessive velocity of galaxies. This led to thedevelopment of Hubble’s Law. The law deals with the relationship between recessive velocity, ",and distance, d , for galaxies

" = Hd

where H is Hubble’s constant, the currently accepted value of which is 65 km s-1 Mpc-1. Noticethat the unit of velocity is therefore km s-1, and the unit of distance is the Mpc (mega parsec).

The data from some typical galaxies is given in the table.

galaxy velocity/km s –1 distance/Mpc

NGC 1357 2094 28

NGC 1832 2000 31

NGC 5548 5270 67

NGC 7469 4470 65

A graph of velocity against distance using this data can be used to determine a value forHubble’s constant. Simply calculate the gradient of a line of best fit for these points whichpasses through the origin. The scatter in the points does give some indication of the

approximate nature of the relationship.

6000

5000

velocity/km s-1

4000

3000

2000

1000

010 20 30 40 50 60 700

distance/Mpc


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The data suggests that the Universe is expanding. It is like being an average runner in a race –those in front of you are faster than you and moving away from you. You are faster than thosebehind and you are moving away from them. From your point of view, everyone appears to bemoving away from you. Just like a race, Hubble’s Law suggests that there was once onecommon starting point. As the distance between the galaxies was zero at the start, the time

since the beginning of the expansion (the age of the Universe) can be found by dividing thedistance to a galaxy by its recessional velocity, ie the age of the universe = d /v = 1/H . This givesan age for the Universe of approximately 15 billion years.

The equation assumes that the expansion rate is constant. Recent evidence suggests,however, that the rate of expansion may be increasing. This relates to the controversial ‘darkenergy’.

Dark energy

At the end of the last century, astronomers measured the distance to several very distantType 1a Supernovae, and measured the red shift of their host galaxies. They expected to find

that the rate of expansion was slowing down and that the supernovae would appear to brighterthan suggested by the extent of their red shift and Hubble’s Law. What they found was that thesupernovae were dimmer than expected, suggesting that the rate of expansion of the Universeis increasing. Further measurements since then have supported this idea, and in order toexplain this increasing expansion some astronomers have used the controversial idea of ‘darkenergy’.

At the moment there is no known mechanism for this expansion, or the Dark energy that drives it.Some astronomers believe it is linked to the cosmological constant introduced and then discardedby Einstein during his work on gravity. Some believe it is due to a ‘negative vacuum pressure’ ora quantum field effect. The important issue is that there is evidence for its existence, and that, asno one knows what it is, it is controversial.

The Big Bang

Dark energy is just the latest cosmological controversy in a long line. Ten years ago it wasquasars, and fifty years ago it was the Big Bang.

The theory suggests that, over the past 15 billion years or so, the Universe has expanded froman extremely hot and dense point, and is still expanding.

Evidence clearly comes from the Hubble relationship and observations of the red shift of distantgalaxies. Further supporting evidence came from the discovery of the cosmological microwave

background radiation. This is a ‘glow’ from all parts of the Universe, the spectrum of whichfollows a black body radiation curve with a peak in the microwave region. This peakcorresponds to a temperature of 2.7 K. It can be interpreted as the left over ‘heat’ from the bigbang, the photons having been stretched to longer wavelengths and lower energies (morecorrectly perhaps, it is the radiation released when the Universe cooled sufficiently for matterand radiation to ‘decouple’, with the combination of protons and electrons to form neutral atoms).The final piece of evidence comes from the formation of nuclear matter. The Big Bang theorysuggests that a very brief period of fusion occurred when the Universe was very young, resultingin the production of helium from fusing hydrogen. The Universe then expanded and cooled toorapidly for the creation of larger nuclei, resulting in a relative abundance of Hydrogen and Helium(in the ratio of 3:1) spread uniformly throughout the Universe, and a lack of larger elements.This is consistent with observation.


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Quasars

Red shift measurements suggest that quasars are the most distant and most powerfulobservable objects. Using the inverse square law, calculations suggest that they can produce

the same power output of several galaxies. Variations in their output, however, suggests thatthey may be only the size of a solar system. They appear in the centre of young (ie distant)galaxies, often obscuring the host galaxy because they are so bright.

Whilst they were once controversial, they are now believed to come from supermassive blackholes at the centre of active galaxies.

Quasars were first discovered as very powerful radio sources, and, due to the poor resolvingpower of radio telescopes, it was very difficult to associate a quasar with its optical counterpart.It was only with very large baseline interferometry that quasars could be pinpointed and we nowknow that quasars emit em radiation from the whole spectrum.


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Appendix A Astrophysics Specification

In this option, fundamental physical principles are applied to the study and interpretation of theUniverse. Students will gain deeper insight into the behaviour of objects at great distances fromEarth and discover the ways in which information from these objects can be gathered. Theunderlying physical principles of the optical and other devices used are covered and someindication given of the new information gained by the use of radio astronomy. Details ofparticular sources and their mechanisms are not required.

A.1.1 Lenses and Optical Telescopes

• Lenses Principal focus, focal length of converging lens.Formation of images by a converging lens.Ray diagrams.

f vu

111=+

• Astronomical telescope consisting of two converging lenses Ray diagram to show the image formation in normal adjustment. Angular magnification in normal adjustment.

M =eyeunaidedatobject bysubtendedangle

eyeatimage bysubtendedangle

Focal lengths of the lenses.

e

o

f

f M =

• Reflecting telescopes Focal point of concave mirror.Cassegrain arrangement using a parabolic concave primary mirror and convexsecondary mirror , ray diagram to show path of rays through the telescope as far as theeyepiece.Relative merits of reflectors and refractors including a qualitative treatment of sphericaland chromatic aberration.

• Resolving power Appreciation of diffraction pattern produced by circular aperture.Resolving power of telescope, Rayleigh criterion,

D

λ θ ≈

• Charge coupled deviceUse of CCD to capture images.Structure and operation of the charge coupled device: A CCD is silicon chip divided into picture elements (pixels).Incident photons cause electrons to be released.The number of electrons liberated is proportional to the intensity of the light.These electrons are trapped in ‘potential wells’ in the CCD. An electron pattern is built up which is identical to the image formed on the CCD.When exposure is complete, the charge is processed to give an image.Quantum efficiency of pixel > 70%.


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A.1.2 Non-optical Telescopes

• Single dish radio telescopes, I-R, U-V and X-ray telescopes Similarities and differences compared to optical telescopes including structure,positioning and use, including comparisons of resolving and collecting powers.

A.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 andvisual magnitude not required.

• Absolute magnitude, M Parsec and light year.Definition of M , relation to m

m – M = 5 log10

d

• Classification by temperature, black body radiationStefan’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

λ maxT = 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 asthe sun.

4 AT # P = Assumption that a star is a black body.

Principles of the use of stellar spectral classesDescription of the main classes:

SpectralClass

IntrinsicColour

Temperature (K) ProminentAbsorption Lines

O blue 25000 – 50000 He+, He, H

B blue 11000 – 25000 He, H

A blue-white 7500 – 11000 H (strongest)ionised metals

F White 6000 – 7500 ionised metals

G yellow-white 5000 – 6000 ionised & neutralmetals

K orange 3500 – 5000 neutral metals

•

M red < 3500 neutral atoms, TiO

Temperature related to absorption spectra limited to Hydrogen Balmer absorption lines:need for atoms in 2=n state.


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• The Hertzsprung-Russell diagram General shape: main sequence, dwarfs and giants. Axis scales range from –15 to 10 (absolute magnitude) and50 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; compositionand density of neutron stars; escape velocity > c for black holes.Use of supernovae as standard candles to determine distances. Controversy concerningaccelerating 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 ( s R )

2s

2

c

GM R ≈

A.1.4 Cosmology

• Doppler effect

z =c

v

f

f =

∆ and

c

v

$

$−=

∆

For v « c applied to optical and radio frequencies.

Calculations on binary stars viewed in the plane of orbit, galaxies and quasars.

• Hubble’s law Red shift

Hd v =

Simple interpretation as expansion of universe; estimation of age of universe, assuming

H is constant.Qualitative treatment of Big Bang theory including evidence from cosmologicalmicrowave background radiation, and relative abundance of H and He.

• Quasars Quasars as the most distant measurable objects.Discovery as bright radio sources.Quasars show large optical red shifts; estimation of distance.

astrophysics practice questions

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