Introduction to Astronomy - 2005 Quiz with Solutions

Quiz 2 Name Solutions (please print) Last First 1. 2. 3. 4. 5. 6. Work any 7 of the 10 problems ­ indicate clearly which 7 you want graded. Spend 10 or 15 minutes reviewing the problems, and select 7. Immediately after this you have two continuous hours to work the problems. All problems are worth 14 points; everyone gets 2 points just for taking the exam. Closed book exam; you may use two pages of notes, and a calculator. Wherever possible, try to solve the problems using general analytic expressions. Plug in numbers only as a last step. 7. If you have any questions, e­mail the Instructor. 8. Turn exam in during lecture, Friday, April 15. Time Started: Time Stopped: Signature Problem 1 2 3 4 5 6 7 8 9 10 Total Grade Grader APPROXIMATE VALUES OF USEFUL CONSTANTS Constant c (speed of light) G (gravitation constant) k (Boltzmann’s constant) h (Planck’s constant) mproton eV (electron Volt) M� (solar mass) L� (solar luminosity) R� (solar radius) σ (Stefan­Boltzmann cons) Å(Angstrom) km (kilometer) pc (parsec) kpc (kiloparsec) Mpc (megaparsec) year day AU 1� (arc minute) 1�� (arc second) cgs units 3 × 1010 7 × 10−8 1.4 × 10−16 6.6 × 10−27 1.6 × 10−24 1.6 × 10−12 2 × 1033 4 × 1033 7 × 1010 6 × 10−5 10−8 105 3 × 1018 3 × 1021 3 × 1024 3 × 107 86400 1.5 × 1013 1/3400 1/200, 000 cm/sec dyne­cm2 /g2 erg/K erg­sec g erg g erg/sec cm erg/cm2 ­sec­K4 cm cm cm cm cm sec sec cm rad rad mks units 3 × 108 7 × 10−11 1.4 × 10−23 6.6 × 10−34 1.6 × 10−27 1.6 × 10−19 2 × 1030 4 × 1026 7 × 108 6 × 10−8 10−10 103 3 × 1016 3 × 1019 3 × 1022 3 × 107 86400 1.5 × 1011 1/3400 1/200, 000 m/sec N­m2 /kg2 J/K J­sec kg J kg J/sec m J/m2 ­sec­K4 m m m m m sec sec m rad rad Problem 1 A star in the Andromeda galaxy yields a bolometric flux at the Earth of F = 1.0 × 10−13 ergs cm−2 sec−1 (1.0 × 10−16 Watts m−2 ). It has a B­V color index of ­0.24. Take the distance to Andromeda to be 1 Mpc. Make use of the table below, where appropriate, to answer the following questions (interpolate between entries using any interpolation scheme that is reasonable). The reference flux for a bolometric magnitude of 0.0 is F0 = 2.5 × 10−5 ergs cm−2 sec−1 (2.5 × 10−8 Watts m−2 ). a. Find the bolometric magnitude, Mbol of the star. � � 1 × 10−13 Mbol = −2.5 log � 21 2.5 × 10−5 b. What is the approximate effective temperature, Teffective , of the star. Te � 17250 K, based on a linear interpolation from the table; or ∼ 16850 K, based on a logarithmic interpolation. Let’s adopt Te � 17, 000 K. c. Calculate the approximate radius of the star. F = L 4πσR2 T 4 = � 1 × 10−13 4πd2 4πd2 � F d2 R= � 6.4 R� σT 4 Table 1: Abbreviated Table of Main­Sequence Star Properties Spectral Type B­V (color) Mass (M� ) Teffective (◦ K) O5 B0 B5 A0 A5 F0 F5 ­0.45 ­0.31 ­0.17 0.00 +0.16 +0.30 +0.45 40 17 7 3.5 2.1 1.8 1.4 35,000 21,000 13,500 9,700 8,100 7,200 6,500 Problem 2 The binary system called “Cygnus X­1” consists of a black hole (Star 1; for this problem considered to be a point mass) orbiting a normal star (Star 2) with a period Porb = 5.6 days. Optical astronomers measure the orbital motion of the normal star via Doppler shifts and determine a “projected” velocity of v2 sin i = 75 km sec−1 . This can be combined with the orbital period to determine the mass function: f (M ) = 0.25 M� . The spectral type of the normal star is found to be B0. Other studies have led to a deter­ mination of the orbital inclination angle which turns out to be i = 30◦ . Use the information above to determine the mass of the black hole. A table of stellar properties is given in Problem 1; use this to help you determine the mass of Star 2. If you have done the problem correctly you will end up with the equivalent of a cubic equation in the mass of the black hole, MBH . You can either solve this with Mathematica or some other similar computer program, or simply plug in a handful of trial solutions and “zero in” on an approximate answer (10% is good enough). MB sin i3 f (M ) = � �2 � 0.25 M� Mc 1 + MB MB � 1+ Mc MB �2 � 2 M� From the table, the companion­star mass is Mc � 17 M� . So, MB � 1+ 17 M� MB �2 � 2 M� After several trial­and­error guesses for MB , we find that the left hand side of the equation equals 2 M� , only when MB � 11.8 M� . Problem 3 Following are HR diagrams for the globular cluster, M3, and the open cluster ­ the Pleiades. a. Use these HR diagrams to find the ratio of the distances between M3 and the Pleiades. At B­V = 0.5, for example, the main­sequence magnitudes from the two diagrams are: VM3 � 20, and VPl � 9.5 But, we can write: � � L V = −2.5 log 4πd2 Fref We assume that stars on the main sequence with B­V = 0.5 all have the same L. Thus, � 2 � � � dM3 dM3 VM3 − VPl = 2.5 log = 5 log � 10.5 dPl d2Pl Thus, we find dM3 = 125 × dPl . b. If we take the absolute visual magnitude of a star with color B­V=0 to be MV = 0, find the actual distance to the Pleiades. V = MV + distance modulus, and I estimate V � 6.5 for B­V = 0. Thus, the distance modulus is ∼ 6.5, which implies a distance of ∼ 200 pc. c. Explain why the two HR diagrams look so different. The Pleiades is sufficiently young that most of the main­sequence is still in existence. By contrast, the track of stars running nearly vertically upward in the HR diagram for M3 indicates that all but the least massive stars (i.e., with M � 1M� ) have evolved away. d. How much intrinsically more luminous are stars at the top of the HR diagram for M3 than those at the bottom? Vmin � 22, Vmin − Vmax and Vmax � 13; ΔV � 9 � � Lmax Lmin = −2.5 log ⇒ � 4000 Lmax Lmin Problem 4 The image below is of the galaxy NGC 4565. Assume that this is a typical spiral galaxy seen nearly edge on. Credit: Russell Croman a. Identify as many generic features of this galaxy as you can. You may write on the white space provided and draw arrows to the appropriate places on the figure. bulge disk dust lanes no globular clusters visible halo could be identified, though not visible b. Indicate a likely size scale beside the drawing. How thick would you guess this galaxy is half way out from the center? ∼ 40 kpc in diameter ∼ 1 kpc in thickness c. Indicate approximately where the Sun would be located if this were the Milky Way. About half way out in the disk, near the midplane Problem 5 The disk of a galaxy can be modeled as a uniform slab of material of mass density, ρ, that is of (full) thickness, 2H, in the zˆ direction, and is effectively infinite in the xˆ and yˆ directions. Assume that the mass density for z > H is zero. a. Compute the effective gravity, �g , at an arbitrary distance, z, inside and above the disk. Sketch �g (z) for all z (i.e., for + and − values of z). Hint: make use of Gauss’ law for gravity � � = −4πGM . �g · dA � � = −4πGM �g · dA For a cylindrical “pillbox” of end area, A: −g(z)2A = −4πGρ2Az (inside the disk) −g(z)2A = −4πGρ2AH (outside the disk) or g(z) = 4πGρz (inside) g(z) = 4πGρH (outside) pointing toward the midplane b. Find the speed, vz that a star must have, starting at the middle of the disk, to get above height H, i.e., just outside of the mass distribution. Express your answer in terms of ρ, G, and H. � φ= z g(z)dz = 2πGρz 2 (inside) 0 In order for a star to reach z = H from the midplane, its kinetic energy must exceed the potential energy which is equal to the result found in part (a) evaluated at z = H, and multiplied by the mass of the star: 1 2 mv � 2πGρH 2 m 2 or, v 2 � 4πGρH 2 Problem 6 A galaxy is found to have a rotation curve, v(r), given by � � v(r) = � r r0 1+ r r0 �3/2 v0 , where r is the radial distance from the center of the galaxy, r0 is a constant with the dimension of length, and v0 is another constant with the dimension of speed. The rotation curve is defined as the orbital speed of test stars in circular orbit at radius r. a. Find an expression for ω(r), where ω is the angular velocity. What Oort A coefficient � � ] would an astronomer living in this Galaxy at r = r0 measure? [Recall: A = − 12 r0 dω dr 0 ω = v/r ⇒ A= ω(r) = 1 v0 r0 1 + r/r0 )3/2 3 v0 1 4 r0 (1 + r/r0 )5/2 which, when evaluated at r = r0 (from the definition of A), yields: A= 3 v0 29/2 r0 b. Find an expression for the mass, M (< r), contained in this galaxy inside of radius r. Assume a spherically symmetric mass distribution. From F = ma, we have: GM (< r) v2 = r2 r M (< r) = ⇒ M (< r) = r3 v02 r02 (1 + r/r0 )3 G v2r G Problem 7 Short answer questions: a. The density of stars in a particular globular star cluster is 106 pc−3 . Take the stars to have the same radius as the Sun, and to have an average speed of 10 km sec−1 . Find the mean free path for collisions among stars. Find the corresponding mean time between collisions. (Assume that the stars move in straight­line paths, i.e., are not deflected by gravitational interactions.) 1 1 = 6 −3 2 10 pc πR nσ 1 �� � 2 × 1027 cm −50 −3 3 × 10 cm × 1.5 × 1022 cm2 �� τcoll � 2 × 1027 cm � 2 × 1021 sec � 6 × 1013 years 6 10 cm/sec b. A white dwarf star composed entirely of carbon (6 C 12 ) reaches a mass of 1.4 M� and all the carbon burns rapidly to magnesium (12 M g 24 ). Compute the energy released in this reaction (you should consult the table of Atomic Mass Excess in Problem 10). Compare the nuclear energy released with the gravitational binding energy, U , of the white dwarf. For U you can use U � GM 2 /R, and choose some reasonable value for R. Is there sufficient nuclear energy to disrupt the white dwarf, i.e., to blow it apart? Atomic mass excess of 6 C 12 ≡ 0 Atomic mass excess of 12 M g 24 = −13.93 MeV Each reaction therefore liberates 13.93 MeV The number of Mg nuclei that can be made is: (1.4 M� /24 amu) � 7 × 1055 The total nuclear energy released = 1057 MeV = 1.5 × 1051 ergs The gravitational binding energy, U: U� 6.7 × 10−8 × 7.8 × 1066 GM 2 � � 1 × 1051 ergs 8 R 5 × 10 Thus, the two energies are rather comparable, with probably sufficient nuclear energy to blow the star apart. Problem 8 The equation of state for cold (non­relativistic) matter may be approximated as: P = aρ5/3 − bρ4/3 where P is the pressure, ρ the density, and a and b are fixed constants. Use a dimensional analysis of the equation of hydrostatic equilibrium to estimate the “radius­mass” relation for planets and low­mass white dwarfs whose material follows this equation of state. Specifically, find R(M ) in terms of G and the constants a and b. You should set all constants of order unity (e.g., 4, π, 3, etc.) to 1.0. [Hint: solve for R(M ) rather than M (R)]. You can check your answer by showing that for higher masses, R ∝ M −1/3 , while for the lower­masses R ∝ M +1/3 . dP = −gρ dr aρ5/3 − bρ4/3 ∼ R � GM R2 aM 5/3 bM 4/3 − ∼ R6 R5 � �� M R3 GM 2 R5 � � aM 5/3 − bM 4/3 GM ∼ R 2 aM 5/3 aM 1/3 R∼ � GM 2 + bM 4/3 GM 2/3 + b For small masses, R ∝ M 1/3 as for rocky planets, while for larger masses, R ∝ M −1/3 as for white dwarfs where the degenerate electrons are not yet relativistic. Problem 9 Once a star like the Sun starts to ascend the giant branch its luminosity, to a good approx­ imation, is given by: 105 L� 6 L= Mcore , M�6 where the symbol � stands for the solar value, and Mcore is the mass of the He core of the star. Further, assume that as more hydrogen is burned to helium – and becomes added to the core – the conversion efficiency between rest mass and energy is: ΔE = 0.007 ΔMcore c2 . a. Use these two expressions to write down a differential equation, in time, for Mcore . L≡ 0.007ΔMcore c2 ΔE 105 L� 6 = = Mcore Δt Δt M�6 b. Solve the differential equation for the core mass, Mcore (t), as a function of time. To make the problem easier, do not evaluate either L� or M� until the next step. dMcore 105 L� dt = 6 Mcore M�6 0.007c2 Integration yields: 1 5 Mcore,i − 1 5 Mcore,f = 5 × 105 L� t 0.007M�6 c2 , where the subcripts i and f stand for “initial” and “final” values, respectively. c. Find the time for the star to ascend the giant branch when its core mass increases from Mcore = 0.2 M� to Mcore = 0.5 M� . � � 1 1 0.007M� c2 t= − 5 × 105 L� 0.25 0.55 Finally, plug in values for M� and L� to find: t � 6 × 108 yr Problem 10 This problem relates to the principal nuclear burning chain that powers the Sun, the p − p chain. a. Write down the 3 nuclear reactions in the p − p chain. 1H 1 + 1 H 1 ⇒ 1 H 2 + e+ + ν 1H 2H 3 2 + 1H 1 ⇒ 2H 3 + γ + 2H 3 ⇒ 2H 4 + 2 1H 1 b. Use the table on the following page to compute the energy released from either reaction involving 2 He3 in part (a) – 3 significant figures are sufficient. (The table gives the atomic mass excesses, expressed in MeV.) 13.13 + 7.29 ⇒ 14.93 + �, for 2nd equation � � 5.49 MeV or 2 × 14.93 ⇒ 2.42 + 2 × 7.29 + �, for 3rd equation � � 12.9 MeV c. Compute how much energy is released, in total, from the conversion of 4 hydrogen nuclei into 1 helium nucleus (you may ignore the electrons). Hint: you may bypass intermediate reactions. 4 × 7.29 ⇒ 2.42 + � � � 26.7 MeV Table 4­1 from “Principles of Stellar Evolution and Nucleosynthesis” by Donald Clayton, published by McGraw­Hill.