Exercise 4 (Stars and the universe) Suggested answers 1. (a) The star is nearly a blackbody, the spectrum of a star can be approximated as a blackbody radiation curve. On the curve, there is a peak which shifts to shorter wavelength when the temperature of the blackbody increases. From the position of the peak, astronomers can deduce the surface temperature of the star. In general, the spectra of hot stars peak at shorter wavelengths and therefore hot stars appear bluer. (More precisely, according to Wien’s displacement law, λmax and T of a black body are related by max T 2.90 10 3 m K . Thus, we can calculate the surface temperature of a star from the spectrum.) (b) When the spectrum of a star is examined closely, specific absorption lines can be found. The positions (i.e. wavelength) and strengths of the absorption lines in the spectrum are “atomic fingerprints” of the elements and are unique for each element, thus the chemical composition of the atmosphere of the star can be obtained. 2. (a) Luminosity L of a celestial body is the actual amount of electromagnetic radiation emitted by the body in each second. (b) (i) Since intensity I is the radiant power received per unit area at a position with a radial distance d from the source, we have I L L . total surface area 4πd 2 L , the intensity of Proxima Centauri measured on the Earth is 4 πd 2 6.53 10 23 I 3.23 10 11 W m 2 15 2 4π 4.24 9.46 10 (ii) Applying I 3. (a) Since a large portion of the spectrum of star X lies around the wavelength 700 nm, the star appears red. (b) From the graph, the spectrum of the Sun peaks at 5 × 10−7 m while the spectrum of star X peaks at 7 × 10−7 m. Applying Wien’s displacement law to the spectrum of the Sun and the star X, max T (5 10 7 ) (5800) 7 10 7 T T 4142.8 4140 K (c) The radiant flux or intensity I of star X is given by I T 4 (5.67 10 8 ) (4142.8) 4 1.67 107 W m 2 (d) The student may not be correct. Luminosity L of a star is given by L = 4R2 T4, that is, the luminosity depends on both the surface temperature of the star and the size of the star. Although the Sun has a higher surface temperature, the radius of star X is unknown, in other words, star X may be larger than the Sun. Therefore, there is not enough information to determine whether the Sun is more luminous than star X or not. 4. (a) Applying L = 4R2 T4 = 4 x (7 x 108 )2 x 5.67 x 10-8 x 60004 = 4.52 x 1026 W The luminosity of star A is 4.52 x 1026 W. (b) The rate of their luminosities is 2 2 4 7 108 6000 L A R A TA 2 4 4 9 LB RB TB 1.4 10 3000 4 Star A’s luminosity is 4 times that of star B. 5. (a) luminosity / L 10 000 red giants mainsequence stars 1 Sun 0.000 1 white dwarfs 30 000 6000 3000 surface temperature / K (b) (i) Spectral class shows the surface temperature of a star. The spectral classes are arranged as O, B, A, F, G, K, M if stars are arranged from the highest surface temperature to the lowest surface temperature (1A). W is of the class B while X and Y are of the same class K. Thus, W has a higher surface temperature while X and Y are of the same lower surface temperature (1A). (ii) (Note the arrow on the axis of the absolute magnitude: it increases downward). A smaller absolute magnitude of a star implies the star has a higher luminosity (1A). Thus, Y has the highest luminosity while W and X have the same lower luminosity. From (b)(i), Y and X have the same lower surface temperature. Thus, Y should be the largest according to the equation L = 4R2 T4 (L the largest, but T the smallest)(1A). Since W and X have the same luminosity and W has a higher surface temperature than X, thus W should be the smallest (1A). 6. (a) By IA = 2.512 m⊙ m A , I⊙ IA = 2.512 m⊙ m A I☉ = 2.512 26.740.77 I☉ = 9.90 1012 I☉ The brightness (or intensity) of Altair is 9.90 1012 I☉. (b) Altair must be farther away from us. It is because Altair has lower absolute magnitude than the Sun, so Altair must be more luminous, however it has a higher apparent magnitude, in other words, it is apparently dimmer than the Sun, so it must be farther away than the Sun. d (c) By M = m 5 log , 10 m M = 10 5 0.77 2.21 = 10 5 d 10 10 = 5.15 pc The distance between Altair and the Earth is 5.15 pc. (d) By I = T4, 1.285 x 108 = 5.67 x 10-8 T4 T = 6900 K The temperature of Altair is 6900K. (e) RA = LA 2 = 10.6 1.64 R☉ (6900 / 5800) 2 TA The radius of Altair is 1.64 R☉. 7. (a) Orbital speed of the Earth = 2π (1.5 10 11 ) 2 πr = = 29 900 m s1 T 365 24 60 60 Minimum speed of the Earth as seen by the space traveller = 100 000 29 900 = 70 100 m s1 Maximum speed of the Earth as seen by the space traveller = 100 000 + 29 900 = 129 900 m s1 (b) By λ v r , λ c for minimum wavelength, = λ vr 129 900 = 700 = 0.301 nm c 3 10 8 Minimum wavelength = 700 0.301= 699.699 nm For maximum wavelength, = λ vr 70 100 = 700 = 0.164 nm c 3 10 8 Maximum wavelength = 700 + 0.164 = 700.164 nm 8. (a) d = 1/p = 1/0.545 = 1.83 pc 0 v (b) By 0 0 c v 106.8 103 656.28 = - 0.234 nm 0 c 3 108 Negative sign means that the observed wavelength is smaller than the wavelength observed if the object is at rest. This corresponds to blue shift. Thus the star must be approaching. (c) Speed of the star = 902 106.82 140 km s -1 (d) To find the change in angular position, transverse direction should be considered. Observed change in position s = vt = 90 km s-1 x 365 x 24 x 3600 = 2.838 x 1012 m Required change in angular position = s/r = 2.838 x 1012 / (1.83 x 3.09 x 1016) = 5.02 x 10-5 rad 9. (a) Time / h 76 85 94 100 / 1015 m 114 77.9 15.4 30.6 Radial velocity / m s1 52.1 35.6 7.04 14.0 (b) (c) The graph repeats in 96 hours. Therefore the time T for the planet to make one complete revolution is 96 hours. (d) By T2 = 4π 2 r 3 , GM GMT 2 r = 4π 2 1 1 11 30 2 3 3 = (6.67 10 )( 4 10 )(96 3600 ) = 9.31 109 m 4π 2 Radius of the planet’s orbit r is 9.31 x 109 m. 10. (a) Star Sun Barnard’s Star Sirius A Vega Arcturs L 1 0.0045 23 51 160 T 1 0.56 1.6 1.6 0.78 R 1 0.21 1.9 2.8 21 Spectral Class G M A A K 11. (a) The recessional speed increases with distance. (1A) (b) Red shift / examined line emission spectra / Doppler effect (c) Galaxies moving away from each other (1A) suggests that they started out from a single point / explosion. (d) 1000 million light years MC 1-5 A A C B A 6-10 A D D B D Type main sequence star main sequence star main sequence star main sequence star giant (1A) (1A) (1A) 11-13 C A D Explanations to selected mc 2. d = 1/p = 1/0.001 = 1000 pc 4. Recall: L = 4R2 T4 9. Statement 2, the rotational velocities in the outer part of the galaxy are greater than expected. Refer to the rotation curve. 11. Recall: L = 4R2 T4. The luminosity is proportional to (temperature)4. 34 = 81.