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Astronomical distances

Astronomical distances. 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

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Astronomical distances

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  1. Astronomical distances

  2. 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 Question

  3. 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)

  4. 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)

  5. Calculate the distance to Proxima Centuari in Astronomical Units, distance to PC = 40 x 1012 km How many AUs are there in one light year? Questions on AU

  6. Earth - June 2θ nearby star Earth - December View from the Earth in June: distant stars 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.

  7. 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

  8. 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

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

  10. 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 Question 1

  11. 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

  12. 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.

  13. 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

  14. 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. Sun’s Luminosity Question

  15. 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.

  16. 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.

  17. Examples of apparent magnitude

  18. Calculate how much brighter Sirius (m = -1.47) is compared with Polaris (m = 2.01) Question

  19. 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

  20. Calculate the absolute magnitude of the Sun if its apparent magnitude is – 26.7 1 parsec = 207 000 AU Question 1

  21. 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 Question 2

  22. 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.

  23. 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

  24. Wien’s displacement law The wavelength at peak power, λmax , is inversely proportional to the absolute temperature, T of the surface of a black body. λmax T = a constant The constant is equal to 0.0029 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.

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

  26. 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

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

  28. Stefan’s law The total energy per second (power), P emitted by a black body at absolute temperature, T is proportional to its surface area, A and to T4. P = 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.

  29. Calculate the power output of the Sun if its diameter is 1.39 x 106 km and its surface temperature 5800 K. Question 1

  30. Calculate the surface area and radius of Betelgeuse if its luminosity is 4.09 x 1031 W and its surface temperature 3500 K. Question 2

  31. 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.

  32. 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

  33. F

  34. 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. Balmer absorption lines

  35. 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 Question (Revision of Unit 1)

  36. 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 O B A F G K M

  37. 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.

  38. 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. Question

  39. 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

  40. 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

  41. 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

  42. 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

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

  44. 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.

  45. 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.

  46. Types of supernova

  47. Supernovae as standard candles Type 1a supernovae have a known peak luminosity allowing them to be used as ‘standard candles’. At their peak all of these supernovae have an absolute magnitude, M of -19.3 ± 0.03. By noting their apparent peak magnitude, m such supernovae can be used to determine this distances to galaxies using the equation: m – M = 5 log (d / 10)

  48. In a distant galaxy a Type 1a supernova is observed to have an apparent magnitude of + 8.0. Calculate the distance to this galaxy in (a) parsecs and (b) light years if the supernova has an absolute magnitude of – 19. 1 parsec = 3.26 light years Question

  49. A neutron star is formed from the remnant core of a supernova. Gravitational forces cause electrons to react with protons to form neutrons. p + e-→ n + ve The star now has a density of atomic nuclei, about 1017 kgm-3 Neutron stars were first discovered in 1967 as a result of the radio beams that they emit as they rapidly rotate. They are also called ‘pulsars’ with frequencies of up to 30 Hz Neutron stars

  50. Question Estimate the mass of a tea-spoonful of neutron star. Take the density of a neutron star to be 1.0 x 1017 kgm-3

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