λ = 650 nm = c = m s 1 f =? c = fλ f = c λ = ( m s 1 ) ( m) = = Hz T = 1 f 4.

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Chapter 13 Stars Section 13.1 Astronomical measurements Worked example: Try yourself 13.1.1 CALCULATING THE FREQUENCY AND PERIOD OF LIGHT The speed of light in a vacuum is approximately 3.0 10 8 m s 1. The wavelength of red light is 650 nm. a What is the corresponding frequency for this wavelength? Determine what is known and unknown from the question and convert the quantities to SI units. Rearrange the wave equation for light to make the unknown the subject. Substitute the known values and calculate f. Round the answer to two significant figures to match the leastaccurate figure supplied. c 3.0 10 8 m s 1 λ 650 nm 650 10 9 f? c fλ (3.0 108 m s 1 ) (650 10 9 m) 461 10 12 4.6 10 14 Hz m b What is the period of the red light wave, i.e. the time for one complete wave to pass a given point? Refer to the relationship between period and frequency. Substitute the value for the frequency and solve for T, taking care to use standard form correctly. T 1 f T 1 f 1 4.6 10 14 2.2 10 15 s Worked example: Try yourself 13.1.2 CALCULATING THE WAVELENGTH The frequency of a particular radio wave is 3 10 8 Hz. What is the wavelength of this light? Write the wave equation in terms of λ. c fλ λ c f Substitute the value for frequency and the speed of light. Solve for λ. λ 3 108 3 10 8 1 m

Worked example: Try yourself 13.1.3 FINDING DISTANCE FROM PARALLAX ANGLE Proxima Centauri is about 0.77 from Earth. a Convert this parallax angle into parsec. Recall the rule for converting from parallax angle to parsec. Apply the rule to find the distance. d 1 p d 1 0.77 1.3 pc b Given that 1 pc is approximately 3 10 16 m, how far from Earth is Proxima Centauri, in metres? Use the relationship 1 pc 3 10 16 m 1.3 pc 1.3 3 10 16 m 3.9 10 16 m Worked example: Try yourself 13.1.4 FINDING DISTANCE FROM ABSOLUTE AND APPARENT MAGNITUDE A large distant galaxy has an apparent magnitude of 15 and an absolute magnitude of 20. How far away is it? Recall the rule connecting the apparent and absolute magnitude. Apply the rule to find the distance. M m + 5 5 log D 20 15 + 5 5 log D log D 8 D 10 8 pc 100 Mpc Section 13.1 Review KEY QUESTIONS SOLUTIONS 1 c fλ 3 10 8 350 10 9 8.6 10 14 Hz 2 c fλ 3 108 8 10 9 3.75 10 16 Hz 3 T 1 f 1 3.8 10 16 2.6 10 17 s 4 The shift toward the red end of the spectrum in light from distant galaxies. 5 There are 3600 arc seconds in a degree: 60 minutes in a degree, and 60 seconds in a minute: 1 60 60 3600.

6 d 1 1 2.27 pc p 0.44 2.27 3 10 16 6.81 10 16 m 7 d 1 p 1 1.56 0.641 pc 8 61-Cygni is much closer to Earth than 10 pc. If 61-Cygni was 10 pc away it would have a magnitude of +7.49, which is dimmer than it actually appears (+5.21) so it must be much closer than 10 pc. 9 If Betelgeuse was 10 pc away it would have a magnitude of 7.2, which is much brighter than it actually appears (+0.5) so it must be much further away than 10 pc. 10 C. Rigel s apparent magnitude (m) is a little brighter than Betelgeuse even though the distances are the same and so the absolute magnitude (M) will also be a little higher (more negative). This rules out options A and B. : For Rigel: m 0.12 For Betelgeuse m 0.5 and M 7.2, rearranging and putting this into the equation: M m + 5 5 log D 5 log D m + 5 M 0.5 + 5 ( 7.2) 12.7 Rigel is about the same distance away, so the value for 5 log D can be re-used in the equation for Rigel: M 0.12 + 5 12.7 7.58 which is closest to C, 8.1 11 M m + 5 5 log D 4.83 14.83 + 5 5 log D log D 3 D 10 3 1000 pc or 1 kpc Section 13.2 Classifying stars Worked example: Try yourself 13.2.1 DETERMINING THE BRIGHTNESS OF A STAR If it is found that two stars have the same radius and are the same distance from Earth. One looks blue and the other yellow. Which one will be brighter? The colour is an indication of the surface temperature of the star, hotter stars being bluer. Because it is hotter, the blue star will be radiating considerably more energy. The blue star must be hotter. The blue star will appear significantly brighter. Section 13.2 Review KEY QUESTIONS SOLUTIONS 1 The H R diagram plots the luminosity of a star (which is derived from the absolute magnitude) against the spectral type of stars (from which the temperature of the star is derived). 2 Along the main sequence, luminosity increases with the surface temperature. 3 The continuous spectrum provides information about the surface temperature of the star. The absorption spectrum gives information about the elements present.

4 By determining what type of star it is from its spectrum, it can then be placed on the diagram according to its spectral type (OBAFGKM). This enables the luminosity (or absolute magnitude) to be found. The distance can be determined from a comparison of the luminosity with the apparent magnitude. 5 Less-broad spectral patterns indicate a large star. A peak in the red section of the visible spectrum coincides with a cooler star. A red giant would best fit these observations. 6 Betelgeuse; Vega; Sirius A; the Sun; Sirius B; Proxima Centauri 7 The size of a star cannot be seen even using the best telescopes and so must be determined indirectly. Once the temperature is known, the amount of power given off per square metre can be calculated. This is compared with the total luminosity of the star to find the total surface area and hence the radius. 8 C. Luminosity is approximately proportional to the cube of the mass. 2 3 8. 9 D. The H R diagram plots the luminosity of a star (which is derived from the absolute magnitude) against the spectral type of stars (from which the temperature of the star is derived). 10 A. Luminosity (vertical axis) does increase with temperature (horizontal axis) but not in a straight line, so option A is the best answer. Section 13.3 The life and death of stars Worked example: Try yourself 13.3.1 MASS ENERGY The visible portion of the energy the Sun is producing each second is approximately equal to 5.0 10 25 J s 1. At what rate is the Sun losing mass due to this energy loss? (Use c 3.0 10 8 m s 1.) The energy comes from the fusion of hydrogen into helium with a corresponding loss in the potential energy of the nuclei. This loss of energy will correspond to a mass loss given by Einstein s equation E m. Rearrange E m in terms of mass and solve. E m Worked example: Try yourself 13.3.2 SCHWARZSCHILD RADIUS E m 5.0 10 25 J c 3.0 10 8 m s 1 m? m E 5.0 1025 (3.0 10 8 ) 2 5.6 10 8 kg s 1 So the Sun is losing mass due to visible radiation at a rate of 5.6 10 8 kg s 1. A star 30 times the mass of the Sun collapses into a black hole. What would its Schwarzschild radius be in kilometres? (Use G 6.7 10 11 N m 2 kg 2, M 2.0 10 30 kg and c 3.0 10 8 m s 1.) Use the equation for calculating the Schwarzschild radius. Substitute the known values into the equation for the Schwarzschild radius and solve for r s. r s 2GM r s 2GM 2 6.7 10 11 2 10 30 30 (3.0 10 8 ) 2 89333 m 89 km

Section 13.3 Review KEY QUESTIONS SOLUTIONS 1 Nuclear fusion reactions in the Sun involve fusing hydrogen nuclei to produce helium nuclei. In contrast, hydrogen burning in oxygen involves only the electrons in the outer shell of the atoms. The energy involved in nuclear reactions is about 100 million times greater than chemical reactions. 2 C. The photosphere is where the rising convection gases radiate their energy and cool, emitting visible light which is detected as sunlight. 3 E m 6 10 9 (3.0 10 8 ) 2 5.4 10 26 J 4 protostar main sequence star red giant white dwarf black dwarf 5 planetary nebulae 6 The Sun is most likely to initially expand into a red giant before collapsing and becoming a white dwarf. 7 the Schwarzschild radius 8 r 1 2GM 2 6.7 10 11 25 2.0 10 30 (3 10 8 ) 2 74400 m 74 km 9 r 1 2GM 2 6.7 10 11 1.9 10 27 (3 10 8 ) 2 2.8 m 10 a red sequence b blue cloud c green valley Chapter 13 Review 1 c fλ 3.0 108 0.001869 1.6 10 11 Hz 2 T 1 f 1 1.6 10 11 6.2 10 12 s 3 B. Earth s change in position causes a change in apparent position of closer stars due to parallax. Stellar parallax is the term used as applied to using parallax in determining the distance to stars. 4 Canopus is many times further away from Earth than Sirius, which means that when viewed from Earth is will not appear as bright. 5 D. As Deneb is further away than Rigel, the apparent magnitude of Deneb will be less. Lower magnitudes have higher positive numbers. 6 a continuous b emission c continuous d absorption e emission 7 spectral class, surface temperature and chemical composition 8 Their spectra show just the same lines as our Sun and these lines correspond to the 98 known elements in our periodic table.

9 Barnard s Star, Tau Ceti, Vega, Aldebaran, Rigel 10 a white dwarfs b main sequence c supergiant stars d giant stars 11 Like all stars Rigel would have started from a dust and gas cloud collapsing to form a protostar. As Rigel starts to convert silicon to iron as the main fusion process, less energy will be produced than needed for the fusion process. It will begin to collapse. As a giant star, the core of Rigel can then be expected to heat to billions of degrees in a fraction of second. An explosive supernova results. The final stage will be either a neutron star or a black hole. 12 Stars are born into the middle (approximately) of the main sequence, after rapidly igniting once the protostar collapses. 13 The Sun is close to the centre of the H R diagram; that is, in terms of the overall range, it is of average temperature and average brightness. However, most stars are actually cooler and duller than the Sun. 14 r s 2GM 2 6.7 10 11 20 1.989 10 30 (3.0 10 5 ) 2 59 228 m 5.9 10 4 m 15 A black hole is also known as a singularity. 16 The observer will see light near the black hole slow down, theoretically to the point where an object falling past the event horizon of a black hole would appear frozen at the black hole s edge. 17 A, B and C. Star production will be low or will have ceased, a supermassive back hole will lie at its centre and it is the product of one or more merges of smaller galaxies. While astronomers would be inferring each of the conclusions, they are based on many observations of similar galaxies. 18 E m 4 10 9 (3 10 8 ) 2 3.6 10 26 J 19 r 1 2GM 2 6.7 10 11 6.0 10 24 (3.0 10 8 ) 2 8.9 10 3 m 8.9 mm 20 r 1 2GM M r1c2 2G 100000 (3.0 108 ) 2 2 6.7 10 11 6.7 10 31 kg This is approximately 34 M