Exam TechniqueExam Technique

READ THE QUESTION!!READ THE QUESTION!! make sure you understand what you are

being asked to do make sure you do everything you are asked

to do make sure you do as much (or as little) as you

are asked to do [implicitly, by the number of marks]

Answer the question, the whole question, Answer the question, the whole question, and nothing but the questionand nothing but the question

Exam TechniqueExam Technique

Read the whole paper through before you Read the whole paper through before you startstart if you have a choice, choose carefully whether or not you have a choice, do the

easiest bits firstthis makes sure you pick up all the “easy” marksthis makes sure you pick up all the “easy” marks

PHY111 PHY111 do all of section A (20 questions, 40%) do 3 from 5 in section B (3 questions, 30%) do 1 from 3 in section C (1 question, 30%)

Last Year’s Exam, Section BLast Year’s Exam, Section B

Answer any 3 of 5 short questionsAnswer any 3 of 5 short questions

5 marks each5 marks each exam is out of 50

i.e. 120/50=2.4 minutes per marki.e. 120/50=2.4 minutes per mark

hence each question should take ~12 minutes to answer

do not let yourself get bogged down, butdo not let yourself get bogged down, but

do not write 2 sentences for 5 marks!do not write 2 sentences for 5 marks!

Question B1Question B1

Arcturus is a red giant star which is approximately Arcturus is a red giant star which is approximately 100 times as bright as the Sun in visible light.100 times as bright as the Sun in visible light. We call stars like Arcturus “giants” because they have

radii which are much larger than those of main sequence stars like the Sun. Explain how we know that this is so.

As measured relative to the Sun, Arcturus is moving at about 120 km/s. Do you think that Arcturus is part of the Milky Way’s stellar disc? If you do, explain why; if not, explain why not.

What fusion reaction is most probably powering Arcturus’ luminosity, and where in the star is fusion taking place?

B1 AnswerB1 Answer

Size of Arcturus:Size of Arcturus: red star → cooler than Sun cooler than Sun → less light per square metre but much brighter overall → much larger

Is it part of the disc?Is it part of the disc? Sun’s orbital velocity: 200 km/s (given) 120 km/s comparable to 200 km/s (not ~10x smaller,

as typical for disc stars so, probably not part of disc

Power source?Power source? it’s clearly a red giant → hydrogen to helium in shell

around helium core (helium-burning giant phase is much shorter, so unlikely)

Question B2Question B2

The diagram shows the The diagram shows the Hertzsprung-Russell diagram for Hertzsprung-Russell diagram for nearby stars whose parallaxes nearby stars whose parallaxes were accurately measured by the were accurately measured by the HIPPARCOS satellite.HIPPARCOS satellite. The colour index B – V of the star

measures, as the name suggests, the star’s apparent colour. What physical property of the star determines its colour?

What features of this diagram show that the solar neighbourhood contains stars of different ages, including stars which are younger than the Sun?

B2 AnswerB2 Answer

Colour index is determined by surface temperature

Presence of both upper main-sequence stars and a long red-giant branch

HIPPARCOS had a relatively small HIPPARCOS had a relatively small telescope. What differences would telescope. What differences would you expect to see in this diagram if you expect to see in this diagram if HIPPARCOS had been equipped HIPPARCOS had been equipped with a larger telescope? with a larger telescope? more lower-main-sequence

stars more white dwarfs

Question B3Question B3

Briefly describe the various processes by which Briefly describe the various processes by which elements heavier than helium are made in stars. elements heavier than helium are made in stars. Include in your description an explanation of the Include in your description an explanation of the type of star in which the process in question type of star in which the process in question might take place, e.g. main sequence stars, red might take place, e.g. main sequence stars, red giants, supernovae, etc. giants, supernovae, etc. Key terms to be included in your account:

p-process, r-process, s-process, α-process, neutron-rich, neutron-poor.

B3 Answer B3 Answer

Heavy elements are produced either directly by Heavy elements are produced either directly by fusion or indirectly by the addition of fusion or indirectly by the addition of neutronsneutrons to to fusion products. fusion products.

Fusion products include the Fusion products include the α-process elementsα-process elements such as carbon-12, oxygen-16, neon-20 etc., such as carbon-12, oxygen-16, neon-20 etc., which can be produced by successive addition which can be produced by successive addition of α particles (helium nuclei) during the helium of α particles (helium nuclei) during the helium fusion stage of stellar evolution, as well as fusion stage of stellar evolution, as well as elements such as Si, S and Fe produced during elements such as Si, S and Fe produced during fusion of heavy elements in pre-supernova stars. fusion of heavy elements in pre-supernova stars.

B3 AnswerB3 Answer

NeutronsNeutrons are produced in helium-fusing stars and are produced in helium-fusing stars and can easily combine with nuclei because of the lack can easily combine with nuclei because of the lack of any electrostatic repulsion. If neutrons are rare of any electrostatic repulsion. If neutrons are rare and therefore are added to nuclei slowly, any and therefore are added to nuclei slowly, any unstable nucleus formed will decay before another unstable nucleus formed will decay before another neutron hits it. This neutron hits it. This s-processs-process produces nuclei produces nuclei close to the line of maximum stability. close to the line of maximum stability.

In supernovae, neutrons can be added to nuclei In supernovae, neutrons can be added to nuclei very rapidly (very rapidly (r-processr-process), producing highly ), producing highly neutron-neutron-richrich unstable nuclei which subsequently β-decay unstable nuclei which subsequently β-decay to to neutron-richneutron-rich stable nuclei. stable nuclei.

B3 AnswerB3 Answer

Neutron-poorNeutron-poor nuclei are formed by the nuclei are formed by the p-processp-process, , which is now believed to be, not proton addition, which is now believed to be, not proton addition, but knocking out of neutrons by high energy but knocking out of neutrons by high energy photons. photons.

Note: need all 5 points, in about this much detail, Note: need all 5 points, in about this much detail, for 5 marksfor 5 marks

Question B4Question B4

Draw a labelled diagram of the “Hubble tuning Draw a labelled diagram of the “Hubble tuning fork” system of galaxy classification. fork” system of galaxy classification.

Question B4Question B4

Explain briefly how galaxies are classified Explain briefly how galaxies are classified according to this scheme. according to this scheme.

ellipticals by shape: E0 – circularE6 - elongated

S(B)a→c by• decreasing size and brightness of bulge• increasingly loosely wound armsS(B)0: no spiral arms

E: ellipticalS: spiralSB: barred spiralIrr: irregular

Question B5Question B5

Explain what is meant by the term Explain what is meant by the term cosmic cosmic microwave backgroundmicrowave background.. Cosmic microwave background: blackbody (3K)

radiation observed to come equally from all directions in the universe (isotropic).

How do we believe the cosmic microwave How do we believe the cosmic microwave background is generated? background is generated? Believed to be generated when early universe

comprises a hot dense plasma in thermal equilibrium, and then “fossilised” when electrons and protons combine to form neutral hydrogen, rendering universe much more transparent to radiation.

Question B5Question B5

Why does this explanation support the “Hot Big Why does this explanation support the “Hot Big Bang” model of the early Universe?Bang” model of the early Universe? Supports “Hot Big Bang” theory of universe because this

theory naturally expects the early universe to be a hot dense plasma; other theories, especially the Steady State, have no such expectation.

What property of the CMB is best explained by the What property of the CMB is best explained by the idea of idea of inflationinflation? ? Extreme isotropy of early universe is difficult to explain

in Big Bang because radiation does not have time to traverse whole of presently observable universe before emission of CMB, hence hard to explain why regions on opposite sides of the sky are at the same temperature.

Question B5Question B5

Why is inflation needed to explain this property? Why is inflation needed to explain this property? Inflation explains this by postulating early period of

extremely rapid expansion, which means that whole currently visible universe originates from a single causally connected region of the pre-inflation universe; there’s no other way to ensure that the early universe reaches thermal equilibrium (exchanges photons)

(Note that during inflation universe expands faster than light – this is perfectly OK because it’s space that’s expanding, not the galaxies that are moving)

Last Year’s Exam, Section CLast Year’s Exam, Section C

Answer any 1 of 3 long questionsAnswer any 1 of 3 long questions15 marks each, ~36 minutes’ work15 marks each, ~36 minutes’ workQuestion C3 is on the seminars:Question C3 is on the seminars: Write short essays on any three of the

followingbinary starsbinary starsblack holesblack holesthe search for dark matterthe search for dark matterextrasolar planetsextrasolar planets

Note that you know this is coming, so more detail expected in answers!

Question C1Question C1

The supergiant star Sanduleak −69 202, The supergiant star Sanduleak −69 202, about 160 thousand light years away in about 160 thousand light years away in the Large Magellanic Cloud, became the Large Magellanic Cloud, became famous in February 1987 when it was famous in February 1987 when it was seen to explode as a supernova – the first seen to explode as a supernova – the first visible to the naked eye since 1604.visible to the naked eye since 1604.

C1(a)C1(a)

Do you think that when it exploded Sanduleak Do you think that when it exploded Sanduleak −69 202 was (i) much older than the Sun, (ii) of −69 202 was (i) much older than the Sun, (ii) of a similar age to the Sun, or (iii) much younger a similar age to the Sun, or (iii) much younger than the Sun? Explain your reasoning clearly.than the Sun? Explain your reasoning clearly. Much younger Only stars much more massive than the Sun go

supernova, so Sk −69 202 was massive Massive stars have short lifetimes, because they

exhaust their fuel supply much faster The Sun is halfway through its main-sequence life, so

Sk −69 202 was much younger than the Sun when it exploded

C1(b)C1(b)

What would Sanduleak −69 202 have looked like What would Sanduleak −69 202 have looked like when it was on the main sequence? when it was on the main sequence? Very bright and very blue (top left of HRD)

Describe the evolution of Sanduleak −69 202 from Describe the evolution of Sanduleak −69 202 from its arrival on the main sequence to its eventual its arrival on the main sequence to its eventual demise. Your account should include an demise. Your account should include an explanation of the nuclear reactions taking place explanation of the nuclear reactions taking place in the star at each stage in its life (and where they in the star at each stage in its life (and where they are taking place), its likely location on the are taking place), its likely location on the Hertzsprung-Russell diagram, and the Hertzsprung-Russell diagram, and the approximate fraction of its lifetime spent in that approximate fraction of its lifetime spent in that stage. Include a brief account of the supernova stage. Include a brief account of the supernova explosion itself. explosion itself.

C1(b) answerC1(b) answer

EvolutionEvolution Main sequence, fusing hydrogen in core, at top left of

HR diagram. Star spends ~80% of its life here. When hydrogen exhausted in core, star shrinks until

hydrogen fusion starts outside helium core. Star will become a red (super)giant, at top right of HR diagram. This will last ~10% of the main sequence lifetime. Hydrogen fusion outside heats and enlarges the core until

helium fusion begins in the core. Star will get bluer again, moving left on the HRD. When helium exhausted in core, fusion moves outside core and star will return to the red. This whole period lasts <10% of the star’s lifetime.

C1(b) answerC1(b) answer

Evolution continuedEvolution continued Because the star is massive, it will go on to fuse

elements up to iron. This lasts a comparatively short time and the star may move back and forth on the HRD. An onion-like structure develops.

Eventually an iron core forms. Iron is stable against fusion, so collapse of iron core under gravity is not stopped by onset of fusion. Eventually a neutron star forms, and the collapsing stellar envelope bounces off the neutron star surface, creating a shockwave which powers the supernova explosion.

C1(c)C1(c)

How might we detect the post-explosion remnant How might we detect the post-explosion remnant of Sanduleak −69 202? Suggest a reason why of Sanduleak −69 202? Suggest a reason why we might we might notnot detect it. detect it. If remnant is a neutron star, it might be detected as a

pulsar: rapid, very regular pulses of radio emission. If remnant not detected, “lighthouse beam” from

pulsar might not be pointing our way, or core might have become a black hole rather than a neutron star.

C1(d)C1(d)

Briefly explain why the Sun is not destined to end Briefly explain why the Sun is not destined to end its days in the spectacular fashion adopted by its days in the spectacular fashion adopted by Sanduleak −69 202. What will the Sun eventually Sanduleak −69 202. What will the Sun eventually evolve into? evolve into? Sun is not massive enough to fuse elements heavier

than helium (core never gets that hot), therefore it will not form an iron core (it is also not a close binary, so it will not produce a Type Ia supernova).

A white dwarf (surrounded initially by a planetary nebula).

C2(a)C2(a)

Well over 100 planets have now been Well over 100 planets have now been discovered orbiting other stars.discovered orbiting other stars.Explain how the typical properties of these Explain how the typical properties of these extrasolar planets extrasolar planets differdiffer from the properties of from the properties of the planets in the solar system.the planets in the solar system. most discovered planets are gas-giant-sized, but in

orbits typical of our terrestrial planets (< 3AU) some planets are in orbits which are very small

indeed (<<1 AU), where our solar system has no planets at all

many have very eccentric (elliptical) orbits, whereas all planets in our solar system are in nearly circular orbits

C2(a)C2(a)

In what respects are the discovered planets In what respects are the discovered planets similarsimilar to those of the solar system? to those of the solar system? almost all systems have only one giant planet, and

very few indeed have more than 2 (cf. Jupiter and much smaller Saturn in solar system)

planets are discovered around stars with heavy element content similar to or higher than the Sun

spectral class is also similar to the Sun’s

C2(b)C2(b)

Most of these planets have been discovered by Most of these planets have been discovered by the Doppler shift method. Explain how this the Doppler shift method. Explain how this technique works and discuss which planets it technique works and discuss which planets it might most easily detect. might most easily detect. Doppler shift method measures velocity of star (in line

of sight) around system centre of mass. Expect periodic motion corresponding to elliptical orbit. Size of shift gives (lower limit to) mass of planet.

It is easiest to detect massive planets in close orbits edge-on to line of sight, because these produce the largest shifts.

C2(b)C2(b)

How does this relate to the typical properties of How does this relate to the typical properties of extrasolar planets you described in part (a)?extrasolar planets you described in part (a)? Properties in part (a) are definitely biased: Earth-sized

planets not detectable with current technology. Jupiter and Saturn are within range of masses detected. Our system has one (barely) detectable gas giant, so one or two planets per system is reasonable (Saturn not detectable with current technology).

C2(b) continuedC2(b) continued

Close orbits are also favoured by the technique, and also Close orbits are also favoured by the technique, and also by the fact that measurements have only been going on by the fact that measurements have only been going on for ~10 years (so even Jupiter would have completed not for ~10 years (so even Jupiter would have completed not quite one orbit). Therefore finding gas giants in “asteroid-quite one orbit). Therefore finding gas giants in “asteroid-belt” sized orbits but not farther out is likely due to bias. belt” sized orbits but not farther out is likely due to bias. However, finding gas giants in orbit with periods of a few However, finding gas giants in orbit with periods of a few days, though efficiency is biased, does demonstrate that days, though efficiency is biased, does demonstrate that such objects (unexpectedly) exist.such objects (unexpectedly) exist.High eccentricities: not obviously a biased result, though High eccentricities: not obviously a biased result, though for larger orbits it may be for larger orbits it may be −− large eccentricity gives higher large eccentricity gives higher peak velocity, hence is easier to detect.peak velocity, hence is easier to detect.High heavy element content of stars is not biased by High heavy element content of stars is not biased by technique (people have looked around low metallicity technique (people have looked around low metallicity stars), and is expected given theory that planets form stars), and is expected given theory that planets form from coagulated dust. Spectral class from coagulated dust. Spectral class isis biased: M class biased: M class stars are hard objects for high resolution spectroscopy.stars are hard objects for high resolution spectroscopy.

C2(c)C2(c)

What are the properties we think a planet needs to have if What are the properties we think a planet needs to have if life is to evolve on it? Briefly describe how future life is to evolve on it? Briefly describe how future astronomers might find evidence for life, astronomers might find evidence for life, notnot necessarily necessarily intelligent, in other planetary systems. intelligent, in other planetary systems. Orbiting in reasonably circular orbit, and not tidally

locked to star (no great extremes of temperature); around stable star (not close binary and not flare star) with lifetime in excess of 2 Gyr; with liquid water (surface water implies location within habitable zone, but subsurface water may not, cf. Europa).

Use space-based interferometer working in infra-red to get necessary resolution: look for ozone IR spectral features (terrestrial oxygen is biogenic). Assumes that photosynthesis is universal, and that not enough oxygen is produced abiogenically to make ozone layer