5A-2 Astrophysics Surveying The Stars

1. 5A-2 AstrophysicsSurveying The Stars Astrophysics booklet pages 28 to 53 March 8th 2011

2. AQA A2 Specification

3. Astronomical distances 100 270 to 460X 3 36 000X 120 380 000X 11 150 000 000X 400 4 500 000 000X 30 40 x 1012X 9000 (40 000 000 000 000) (X 270 000 Sun) 80 x 1012X 2 260 x 1015X 3300 20 x 1018X 80 460 x 1021X 23 000

4. 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 took three days to reach the Moon. Distances in km: Moon: 380 000 km Sun: 150 000 000 km Proxima Centuari: 40 x 1012 km (a) speed = distance / time becomes: time = distance / speed = 380 000 km / 100 kmh-1 = 3 800 hours = 158 days (about 5 months) (b) Apollo speed: = 380 000 km / 3 days = 128 000 km day-1 time = distance / speed (i) = 150 000 000 / 128 000 to the Sun = 1 170 days (3.2 years) (ii) = 40 x 1012 km / 128 000 = 313 000 000 days to Proxima Cent. = 860 000 years Question

5. The light year One light year is the distance light travels through space in 1 year. Question: Calculate the distance of one light year in metres. distance = speed x time = 3.0 x 108 ms-1 x 1 year = 3.0 x 108 ms-1 x (365.25 x 24 x 60 x 60) s = 9.47 x 1015 m (9.47 x 1012 km) Also used: light second (e.g. the Moon is 1.3 light seconds away) light minute (e.g. the Sun is 8.3 light minutes away)

6. The Astronomical Unit (AU) This is the mean radius of the Earth’s orbit around the Sun. 1 AU = 150 000 000 km (150 x 109 m)

7. 1. Calculate the distance to Proxima Centuari in Astronomical Units, distance to PC = 40 x 1012 km AU = 40 x 1012 / 150 x 106 Distance to Proxima Centuari = 267 000 AU 2. How many AUs are there in one light year? 1 AU = 150 x 109m 1 lyr = 9.47 x 1015 m Ratio = 9.47 x 1015 / 150 x 109 = 63 000 AU lyr-1 Questions on AU

8. Earth - June 2θ nearby star View from the Earth: Earth - December distant stars JUNE DECEMBER Stellar parallax This is the shifting of nearby stars against the background of more distant ones due to the orbital movement of the Earth about the Sun. Measurement of the angle 2θ can yield the distance to the nearby star.

9. Earth - June R θ d Earth - December The parsec (pc) nearby star tan θ = R d becomes: d = R / tan θ angle θ is always VERY small and so tan θ = θ in radians and so: d = R / θ

10. 1 parsec is defined as the distance to a star which subtends an angle of 1 arc second to the line from the centre of the Earth to the centre of the Sun. 1 arc second = 1 degree / 3600 as 360° = 2π radians 1 arc second = 2 π / (360 x 3600) = 4.85 x 10-6 radian

11. Distance measurement in parsecs distance in parsecs = 1 / parallax angle in arc seconds 1.00 2.0 10 100 With ground based telescopes the parallax method of distance measurement is acceptably accurate for distances up to 100 pc.

12. Calculate the distance of 1 parsec measured in (a) AU (b) metres (c) light years. 1 AU = 150 x 109 m 1 light year = 9.47 x 1015 m (a) AU: d = R / θ R= 1AU d = 1 / (4.85 x 10-6 rad) 1 parsec = 207 000 AU (b) metres: d = R / θ R= 1AU = 150 x 109 m d = (150 x 109 m) / (4.85 x 10-6 rad) 1 parsec = 3.09 x 1016 m (c) light years: = 3.09 x 1016 / 9.47 x 1015 1 parsec = 3.26 light years Question 1

13. 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 (a) distance in parsecs = 1 / parallax angle in arc seconds = 1 / 0.25 distance to star = 4 parsecs (b) 1 parsec = 3.26 light years distance = 4 x 3.26 distance to star = 13 light years

14. Luminosity Luminosity is the power output of a star. luminosity = power = energy output time unit: watt The brightness of a star depends on a star’s luminosity.

15. Intensity of radiation ( I ) intensity = power of radiation area unit: W m -2 Example: At the Earth’s surface the average intensity of sunlight is about 1400 W m -2

16. Calculate the luminosity of the Sun if the average intensity of sunlight at the Earth is 1360 W m -2. Distance from the Sun to the Earth = 150 x 106 km. The surface area of a sphere, radius R is given by: A = 4π R2 The surface area of a sphere of Earth radius: = 4π x (150 x 109 m)2 = 4π x 2.25 x 1022 = 2.83 x 1023 m2 Each of these m2 receives 1360 W. All of this power must come from the Sun. Therefore total power = luminosity = 1360 x 2.83 x 1023 Sun’s luminosity = 3.85 x 1026 W Sun’s Luminosity Question The Sun’s radiation spreads out spherically.

17. Apparent magnitude, m The apparent magnitude, m of a star in the night sky is a measure of its brightness which depends on the intensity of the light received from the star. Stars were in ancient times divided into six levels of apparent magnitude. The brightest were called FIRST MAGNITUDE stars, those just visible to the unaided eye in the darkest sky, SIXTH MAGNITUDE.

18. Pogson’s law (1856) In 1856, Norman Robert Pogson defined that the average 1st star magnitude was 100x brighter than the average 6th magnitude star. This means that for each change of magnitude star brightness changes by about 2.5x. (2.55 is about 100) This resulted in a few very bright stars (e.g. Sirius) in having NEGATIVE apparent magnitudes.

19. Examples of apparent magnitude

20. Calculate how much brighter Sirius (m = -1.47) is compared with Polaris (m = 2.01) Magnitude difference = 2.01 – (-1.47) = 3.48 If a change of magnitude of 1.0 results in a brightness change of 2.5 then Sirius is 2.5 3.48 times brighter than Polaris = 24.3 times brighter. Question

21. Absolute magnitude, M The absolute magnitude, M of a star is equal to its apparent magnitude if it were placed at a distance of 10 parsecs from the Earth. It can be shown that for a star distance d, in parsecs, from the Earth: m – M = 5 log (d / 10) NOTE: ‘log’ means BASE 10 logarithms

22. Calculate the absolute magnitude of the Sun if its apparent magnitude is – 26.7 1 parsec = 207 000 AU Distance from the Sun in parsecs: = 1 / 207 000 d = 0.000 00483 pc m – M = 5 log (d / 10) rearranged: M = m - 5 log (d / 10) = (-26.7) - 5 log (0.000 00483 /10) = (-26.7) - 5 log (0.000 000 483) = (-26.7) - 5 log (0.000 000 483) = (-26.7) - 5 x (- 6.316) = (-26.7) - (- 31.6) Sun’s absolute magnitude = 4.9 Question 1 Note: The Sun would appear to be a very faint star and it would probably not be visible from Addlestone!

23. Sirius has an apparent magnitude of – 1.47. Calculate the distance in AU it would need to be from the Earth to equal the brightness of the Sun’s apparent magnitude of -26.7. Sirius distance = 8.3 lyr 1 parsec = 3.26 light years 1 parsec = 207 000 AU Distance to Sirius in parsecs: = 8.3 / 3.26 d = 2.55 pc Sirius’s absolute magnitude by the method of question 1: Sirius’s M = +1.50 m – M = 5 log (d’ / 10) rearranged: log (d’ / 10) = [(m – M) / 5] = [(-26.7 – 1.50) / 5] = - 28.2 / 5 = - 5.64 antiloging: = 2.29 x 10-6 = d’ / 10 d’ = 2.29 x 10-5 pc = (2.29 x 10-5 x 207 000) AU = 4.74 AU Question 2 This is roughly the distance of Jupiter from the Earth

24. A wide field image of the sky including the Milky Way taken from Australia. Notice the different colours of the stars

25. 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 the pattern of a ‘black body radiator’ which is a perfect absorber (and emitter) of radiation.

26. power radiated at each wavelength 2000 K 1250 K 1000 K 0 1 2 3 4 5 visible range wavelength / μm Black body radiation curves

27. Wien’s displacement law The wavelength at peak power,λmax , is inversely proportional to the absolute temperature,Tof the surface of a black body. λmax T = a constant The constant is equal to0.0029 metre kelvin BEWARE! The above equation is usually quoted: λmax 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.

28. power red Star colour: 3500 K Surface T: UV IR λmax: 830 nm infra-red orange white yellow blue Antares Example: 0 500 1000 5500 K 7500 K 4300 K wavelength / nm 20 000 K 670 nm red 530 nm yellow 386 nm blue 150 nm ultraviolet Capella Altair Arcturus Spica Star colour and temperature BLUE stars are hotter than RED stars

29. Calculate the peak wavelength emitted by the Sun if its surface temperature is 6000 K. Assuming that the Sun is a black body radiator, applying Wein’s displacement law: λmax T = 0.0029 mK λmax x 6000K = 0.0029 mK λmax = 0.0029 / 6000 Peak wavelength = 4.83 x 10-7 m (483 nm) Question 1

30. Red giant Betelgeuse, peak wavelength 828nm, and blue supergiant Rigel, peak wavelength 263nm, are both in the constellation of Orion. Calculate the surface temperatures of these stars. Betelgeuse Rigel Question 2 Wein’s displacement law: λmax T = 0.0029 mK becomes: T = 0.0029 mK / λmax For Betelgeuse: T = 0.0029 mK / 828 nm = 0.0029 mK / 8.28 x 10-7 m = 3 500 K For Rigel: T = 0.0029 mK / 2.63 x 10-7 m = 11 000 K

31. A very large black body has a thermal temperature of 2.7K. Calculate its maximum power wavelength. Wein’s displacement law: λmax T = 0.0029 mK λmax x 2.7K = 0.0029 mK λmax = 0.0029 / 2.7 Peak wavelength = 0.0011 m (1.1 mm) This is the wavelength of microwaves. The ‘very large black body’ is the Universe. This cosmic microwave background radiation (CMB) was first detected by Penzias and Wilson in 1965 and is one of the main pieces of evidence that supports the ‘Big Bang’ theory of the origin of the Universe. Question 3

32. Stefan’s law The total energy per second (power), P emitted by a black body at absolute temperature, Tis proportional to its surface area, A and to T4. P = σAT4 Where σ is a constant known as Stefan’s constant. σ = 5.67 x 10-8 W m-2 K-4 This equation can be used to determine the surface area and diameter of a star.

33. Calculate the power output of the Sun if its diameter is 1.39 x 106 km and its surface temperature 5800 K. Area of a sphere:A = 4π R2 A = 4π x (1.39 x 106 km / 2) 2 = 4π x (1.39 x 109 m / 2) 2 = 4π x (6.95 x 108 ) 2 A = 6.070 x 1018 m2 Assuming that the Sun is a black body radiator, applying Stefan’s law: P = σAT4 P = (5.67 x 10-8 W m-2 K-4) x (6.070 x 1018 m2) x (5800K)4 = (5.67 x 10-8 ) x (6.070 x 1018 ) x (1.132 x 1015) Power output of the Sun = 3.89 x 1026 W This is also called the Sun’s Luminosity and it agrees closely with the Luminosity calculation performed earlier based on the 1360Wm-2 sunlight intensity data. Question 1

34. Calculate the surface area and radius of Betelgeuse if its luminosity is 4.09 x 1031 W and its surface temperature 3500 K. Assuming that Betelgeuse is a black body, applying Stefan’s law: P = σAT4 4.09 x 1031 W = (5.67 x 10-8 W m-2 K-4) x A x (3500K)4 A = (4.09 x 1031) / [ (5.67 x 10-8) x (3500)4 ] = (4.09 x 1031) / [ (5.67 x 10-8) x (1.501 x 1014) ] = (4.09 x 1031) / (8.509 x 106) Surface area of Betelgeuse = 4.81 x 1024 m2 Area of a sphere:A = 4π R2 R = √(A / 4π) = √(4.81 x 1024 / 4π) = √(3.825 x 1023) R = 6.18 x 1011 m Radius of Betelgeuse = 6.18 x 1011 m (618 000 000 km) This approximately 4X the radius of the Earth’s orbit, about ¾ of the orbit of Jupiter. This is why Betelgeuse is called a SUPER-GIANT. Question 2

35. photosphere corona Stellar spectra The photosphere of a star gives off a continuous spectrum. However, when this light passes through the outmost layer of a star, the corona, some of the wavelengths are absorbed by the hot gases in this region. This causes dark lines to be seen in the otherwise continuous spectrum given out by the star. The wavelengths of these dark lines are characteristic to the elements and compounds found in the corona of the star. The chemical composition of the star can be determined by comparing a star’s spectrum with the known absorption spectra for different elements and compounds.

36. Stellar spectral classes O Stars can be classified by their spectra Starting from the hottest stars the groups are: O, B, A, F, G, K, M There are two further groups (not required in the exam) called L and T. In these groups are found red and brown dwarf stars. Be A Fine Girl or Guy Kiss Me

37. 25 000 to 50 000 He+ He H 11 000 to 25 000 He H 7 500 to 11 000 H (strongest) ionised metals 6 000 to 7 500 ionised metals F 5 000 to 6 000 ionised and neutral metals 3 500 to 5 000 neutral metals ≈ 2 500 to 3 500 neutral metals and TiO

38. The hydrogen absorption lines found in the visible spectrum of the hottest stars (O, B and A only) are called Balmer lines. In such stars hydrogen atoms exist with electrons in the n = 2 state. When these atoms are excited by the absorption of photons from the photosphere their electrons change from n = 2 to higher levels. When they do this they absorb particular Balmer series light wavelengths. These wavelengths show up as dark lines in the star’s spectrum. n = 5 n = 4 n = 3 n = 2 Balmer absorption lines 434 nm 486 nm 656 nm

39. A fourth, violet Balmer line has a wavelength of 410 nm and is due to the transition of an electron between the 2nd and 6th energy levels. Calculate (a) the frequency and (b) the energy of the absorbed photon. c = 3.0 x 108 ms-1 h = 6.63 x 10-34 Js (a) c = f λ becomes: f = c / λ = (3.0 x 108 ms-1) / (410 nm) = (3.0 x 108 ms-1) / (410 x 10-9 m) photon frequency = 7.32 x 1014 Hz (b) E = h f = (6.63 x 10-34 Js) x (7.32 x 1014 Hz) photon energy = 4.85 x 10-19 J Question (Revision of Unit 1)

40. absolute magnitude - 15 supergiants - 10 - 5 MAIN SEQUENCE 0 giants + 5 white dwarfs + 10 The Sun + 15 40 000 20 000 10 000 5000 2500 temperature / K The Hertzsprung-Russell diagram OBA F GKM

41. The Hertzsprung-Russell diagram MAIN SEQUENCE Most stars found in this region. Star masses vary from cool low power red dwarf stars of about 0.1x solar mass at the bottom right to very hot blue stars of about 30x solar mass at the top left. GIANTS Stars that are between 10 to 100x larger than the Sun. SUPERGIANTS Very rare. Stars that are about 1000x larger than the Sun. WHITE DWARFS Much smaller than the Sun but hotter.

42. The above illustration represents star classes with the colours very close to those actually perceived by the human eye. The relative sizes are for main sequence stars.

43. 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. Assuming that both stars act like black bodies, Stefan’s law applies. P = σAT4 For the orange giant: Po= σAo To4 For the main sequence star: Pm= σAm Tm4 But both stars have the same power because they have the same absolute magnitude. that is: Po= Pm and so: σAo To4= σAm Tm4 Ao To4= Am Tm4 Ao / Am= Tm4 / To4 but the area of a sphere: A = 4π R2 hence: (4π R2)o / (4π R2)m= Tm4 / To4 R2o / R2m= Tm4 / To4 Ro / Rm= Tm2 / To2 Ro / Rm= (Tm / To ) 2 Ro / Rm= (15000 K/ 5000 K) 2 Ro / Rm= (3) 2 Ro / Rm= 9 QED Question

44. 1. NEBULA AND PROTOSTAR A star is formed as dust and gas clouds (nebulae) in space collapse under their own gravitational attraction becoming denser and denser to form a protostar (a star in the making). In the collapse gravitational potential energy is converted into thermal energy as the atoms and molecules gain kinetic energy. The interior of the protostar becomes hotter and hotter. If the protostar has sufficient mass (> 0.08 x Sun) the temperature becomes high enough for nuclear fusion of hydrogen to helium to occur in its core. A star is formed. protostar collapsing and warming absolute magnitude nebula MAIN SEQUENCE HIGH temperature LOW temperature The evolution of a Sun like star

45. 2. MAIN SEQUENCE The newly formed star reaches internal equilibrium as the inward gravitational attraction is balanced by outward radiation pressure. The star becomes stable with a near constant luminosity. The greater the mass of the star, the higher will be its absolute magnitude and surface temperature but the shorter is the time the star remains MAIN SEQUENCE. The Sun is about half-way through its 10 billion year passage. The largest stars may only last for tens of millions of years. While on the MAIN SEQUENCE the star’s absolute magnitude and surface temperature gradually increase. In about two billion years time the Earth will become too hot to sustain life. absolute magnitude The Sun NOW MAIN SEQUENCE HIGH temperature LOW temperature gradual warming

46. 3. RED GIANT Once most of the hydrogen in the core of the star has been converted to helium, the core collapses on itself and the outer layers of the star expand and cool as a result. The star swells out, moves off the MAIN SEQUENCE and becomes a RED GIANT. The temperature of the helium core increases as it collapses. This causes surrounding hydrogen to undergo fusion, which heats the core further. When the core reaches about 108 K helium nuclei undergo fusion. This forms even heavier nuclei principally beryllium, carbon and oxygen. The luminosity of the star increases as the star expands. The Sun is expected to achieve a radius roughly equal to the Earth’s orbit. The RED GIANT phase lasts for about one fifth of the MAIN SEQUENCE stage. absolute magnitude red giant MAIN SEQUENCE HIGH temperature LOW temperature

47. 4. PLANETARY NEBULA AND WHITE DWARF When nuclear fusion in the core of a giant star ceases, the star cools and its core contracts, causing the outer layers of the star to be thrown off. The outer layers are thrown off as shells of hot gas and dust to form a PLANETARY NEBULA. The remaining core of the star is white hot due to the release of gravitational energy. If it is less than about 1.4 solar masses, the contraction of the core stops as the electrons in the core can no longer be forced any closer. The star is now stable and has become a WHITE DWARF. This gradually cools to invisibility over a few billion years. The Cat’s Eye Planetary Nebula absolute magnitude red giant MAIN SEQUENCE white dwarf HIGH temperature LOW temperature planetary nebula

48. absolute magnitude protostar MAIN SEQUENCE nebula red giant white dwarf HIGH temperature LOW temperature planetary nebula

49. Supergiant star Betelgeuse imaged in ultraviolet light by the Hubble Space Telescope and subsequently enhanced by NASA. The bright white spot is likely one of its poles. Red Supergiant Stars If a star is greater than 4 solar masses, the core becomes hot enough to cause energy release, through further fusion, to form nuclei as heavy as iron in successive shells. The star now has an ‘onion’ like internal structure.

50. The Crab Nebula The remnant of a supernova observed in 1054 Supernovae A supernovae can occur when the iron core of supergiant is greater than about 1.4 solar masses. In this case the gravitational forces are too great for the repulsive forces of electrons. Electrons are forced to react with protons to form neutrons. p + e-→ n + ve The sudden collapse of the core occurs within a few seconds and its density increases to that of atomic nuclei, about 1017 kgm-3 The core suddenly becomes rigid and collapsing matter surrounding the core hits it and rebounds as a shock wave propelling the surrounding matter outwards into space in a cataclysmic explosion. The exploding star releases so much energy that it can outshine the host galaxy. A supernova is typically a thousand million times more luminous than the Sun. Within 24 hours its absolute magnitude will reach between -15 and -20. Elements heavier than iron are formed by nuclear fusion in a supernova explosion. Their existence in the Earth tells us that the Solar System formed from the remnants of a supernova.