Exam (elaborations) TEST BANK FOR An Introduction to Modern Astrophysics By Carroll B.W. and Ostlie D.A. (Solution manual)

Exam (elaborations) TEST BANK FOR An Introduction to Modern Astrophysics By Carroll B.W. and Ostlie D.A. (Solution manual) CHAPTER 1 The Celestial Sphere 1.1 From Fig. 1.7, Earth makes S=P˚ orbits about the Sun during the time required for another planet to make S=P orbits. If that other planet is a superior planet then Earth must make one extra trip around the Sun to overtake it, hence S P˚ D S P C 1: Similarly, for an inferior planet, that planet must make the extra trip, or S P D S P˚ C 1: Rearrangement gives Eq. (1.1). 1.2 For an inferior planet at greatest elongation, the positions of Earth (E), the planet (P), and the Sun (S) form a right triangle (∠EPS D 90ı). Thus cos.∠PES/ D EP=ES. From Fig. S1.1, the time required for a superior planet to go from opposition (pointP1) to quadrature (P2) can be combined with its sidereal period (from Eq. 1.1) to find the angle ∠P1SP2. In the same time interval Earth will have moved through the angle ∠E1SE2. Since P1, E1, and S forma straight line, the angle ∠P2SE2 D ∠E1SE2  ∠P1SP2. Now, using the right triangle at quadrature, P2S=E2S D 1= cos.∠P2SE2/. P1 E1 S P2 E2 Figure S1.1: The relationship between synodic and sidereal periods for superior planets, as discussed in Problem 1.2. 1.3 (a) PVenus D 224:7 d, PMars D 687:0 d (b) Pluto. It travels the smallest fraction of its orbit before being “lapped” by Earth. 1.4 Vernal equinox: ˛ D 0h, ı D 0ı Summer solstice: ˛ D 6h, ı D 23:5ı Autumnal equinox: ˛ D 12h, ı D 0 ı Winter solstice: ˛ D 18h, ı D 23:5ı 1 2 Chapter 1 The Celestial Sphere 1.5 (a) .90ı  42ı/C 23:5ı D 71:5ı (b) .90ı  42ı/  23:5ı D 24:5ı 1.6 (a) 90ı  L < ı <90ı (b) L > 66:5ı (c) Strictly speaking, only at LD˙90ı. The Sun will move along the horizon at these latitudes. 1.7 (a) Both the year 2000 and the year 2004 were leap years, so each had 366 days. Therefore, the number of days between January 1, 2000 and January 1, 2006 is 2192 days. From January 1, 2006 to July 14, 2006 there are 194 days. Finally, from noon on July 14, 2006 to 16:15 UT is 4.25 hours, or 0.177 days. Thus, July 14, 2006 at 16:15 UT is JD .177. (b) MJD 53930.677. 1.8 (a) ˛ D 9m53:55s D 2:4731ı, ı D 2ı9016:200 D 2:1545ı. From Eq. (1.8),  D 2:435ı. (b) d D r  D 1:7  1015 m D 11,400 AU. 1.9 (a) From Eqs. (1.2) and (1.3), ˛ D 0:ı D 0:m and ı D 0:ı D 2:. This gives the 2010.0 precessed coordinates as ˛ D 14h30m29:4s, ı D 62ı43025:2600. (b) From Eqs. (1.6) and (1.7), ˛ D 5:46s and ı D 7:98400. (c) Precession makes the largest contribution. 1.10 In January the Sun is at a right ascension of approximately 19h. This implies that a right ascension of roughly 7h is crossing the meridian at midnight. With about 14 hours of darkness this would imply observations of objects between right ascensions of 0 h and 14 h would be crossing the meridian during the course of the night (sunset to sunrise). 1.11 Using the identities, cos.90ı  t/ D sin t and sin.90ı  t/ D cos t , together with the small-angle approximations cos   1 and sin  1, the expression immediately reduces to sin.ı C ı/ D sin ı C  cos ı cos : Using the identity sin.a C b/ D sin a cos b C cos a sin b, the expression now becomes sin ı cos ı C cos ı sinı D sin ı C  cos ı cos : Assuming that cos ı  1 and sinı  ı, Eq. (1.7) is obtained. CHAPTER 2 Celestial Mechanics 2.1 From Fig. 2.4, note that r 2 D .x  ae/2 Cy2 and r 02 D .x C ae/2 C y2: Substituting Eq. (2.1) into the second expression gives r D 2a  q .x C ae/2 Cy2 which is now substituted into the first expression. After some rearrangement, x2 a2 C y2 a2.1  e2/ D 1: Finally, from Eq. (2.2), x2 a2 C y2 b2 D 1: 2.2 The area integral in Cartesian coordinates is given by A D Z a a Z b p 1x2=a2 b p 1x2=a2 dy dx D 2b a Z a a p a2  x2 dx D ab: 2.3 (a) From Eq. (2.3) the radial velocity is given by vr D dr dt D a.1  e2/ .1 C e cos /2 e sin  d dt : (S2.1) Using Eqs. (2.31) and (2.32) d dt D 2 r 2 dA dt D L r 2 : The angular momentum can be written in terms of the orbital period by integrating Kepler’s second law. If we further substituteA D ab and b D a.1  e2/1=2 then L D 2a2.1  e2/1=2=P: SubstitutingL and r into the expression for d=dt gives d dt D 2.1 C e cos /2 P.1  e2/3=2 : This can now be used in Eq. (S2.1), which simplifies to vr D 2ae sin  P.1  e2/1=2 : Similarly, for the transverse velocity v D r d dt D 2a.1 C e cos / .1  e2/1=2P : 3 4 Chapter 2 Celestial Mechanics (b) Equation (2.36) follows directly from v2 D v2 r C v2  , Eq. (2.37) (Kepler’s third law), and Eq. (2.3). 2.4 The total energy of the orbiting bodies is given by E D 1 2 m1v2 1 C 1 2 m2v2 2  G m1m2 r where r D jr2  r1j. Now, v1 D Pr1 D  m2 m1 C m2 P r and v2 D Pr2 D m1 m1 C m2 P r : Finally, usingM D m1 C m2,  D m1m2= .m1 C m2/, and m1m2 D M, we obtain Eq. (2.25). 2.5 Following a procedure similar to Problem 2.4, L D m1r1  v1 C m2r2  v2 D m1   m2 m1 Cm2  r    m2 m1 C m2  v Cm2  m1 m1 C m2  r   m1 m1 C m2  v D r  v D r  p 2.6 (a) The total orbital angular momentum of the Sun–Jupiter system is given by Eq. (2.30). Referring to the data in Appendicies A and C, Mˇ D 1:989  1030 kg, MJ D 1:899  1027 kg, M D MJ C Mˇ D 1:991  1030 kg, and  D MJMˇ= .MJ CMˇ/ D 1:897  1027 kg. Furthermore, e D 0:0489, a D 5:2044 AU D 7:786  1011 m. Substituting, Ltotal orbit D  q GMa  1  e2  D 1:927  1043 kg m2 s1: (b) The distance of the Sun from the center of mass is aˇ D a=Mˇ D 7:426  108 m. The Sun’s orbital speed is vˇ D 2aˇ=PJ D 12:46 m s1, where PJ D 3:743  108 s is the system’s orbital period. Thus, for an assumed circular orbit, LSun orbit D Mˇaˇvˇ D 1:840  1040 kg m2 s1: (c) The distance of Jupiter from the center of mass is aJ D a=MJ D 7:778  1011 m, and its orbital speed is vJ D 2aJ =PJ D 1:306  104 m s1. Again assuming a circular orbit, LJupiter orbit D MJ aJ vJ D 1:929  1043 kg m2 s1: This is in good agreement with Ltotal orbit  LSun orbit D 1:925  1043 kg m2 s1: (d) The moment of inertia of the Sun is approximately Iˇ  2 5 MˇR2 ˇ  3:85  1047 kg m2 and the moment of inertia of Jupiter is approximately IJ  2 5 MJR2 J  3:62  1042 kg m2: Solutions for An Introduction toModern Astrophysics 5 (Note: Since the Sun and Jupiter are centrally condensed, these values are overestimates; see Section 23.2.) Using ! D 2=P, LSun rotate D 1:078  1042 kg m2 s1 LJupiter rotate D 6:312  1038 kg m2 s1: (e) Jupiter’s orbital angular momentum. 2.7 (a) vesc D p 2GMJ =RJ D 60:6 km s1 (b) vesc D p 2GMˇ=1 AU D 42:1 km s1. 2.8 (a) From Kepler’s third law (Eq. 2.37) with a D R˚ C h D 6:99  106 m, P D 5820 s D 96:9 min. (b) The orbital period of a geosynchronous satellite is the same as Earth’s sidereal rotation period, or P D 8:614104 s. From Eq. (2.37), a D 4:22107m, implying an altitude of h D aR˚ D 3:58107m D 5:6 R˚. (c) A geosynchronous satellite must be “parked” over the equator and orbiting in the direction of Earth’s rotation. This is because the center of the satellite’s orbit is the center of mass of the Earth–satellite system (essentially Earth’s center). 2.9 The integral average of the potential energy is given by hUi D 1 P Z P 0 U.t/ dt D  1 P Z P 0 GM r .t/ dt: Using Eqs. (2.31) and (2.32) to solve for dt in terms of d, and making the appropriate changes in the limits of integration, hUi D  1 P Z 2 0 GM2r L d: Writing r in terms of  via Eq. (2.3) leads to hUi D  GM2a  1  e2  PL Z 2 0 d 1 C e cos  D  2GM2a  1  e21=2 PL : Using Eq. (2.30) to eliminate the total orbital angular momentum L, and Kepler’s third law (Eq. 2.37) to replace the orbital period P, we arrive at hUi D G M a : 2.10 Using the integral average from Problem 2.9 hri D 1 P Z P 0 r .t/ dt: Using substitutions similar to the solution of Problem 2.9 we eventually arrive at hri D a 2  1  e2 5=2 Z 2 0 d .1 C e cos /3 : (S2.2) It is evident that for e D 0, hri D a, as expected for perfectly circular motion. However, hr i deviates from a for other values of e. This function is most easily evaluted numerically. Employing a simple trapezoid method with 106 intervals, gives the results shown in Table S2.1. 6 Chapter 2 Celestial Mechanics Table S2.1: Results of the numerical evaluation of Eq. (S2.2) for Problem 2.10. e hr i =a 0.00000 1. 0.10000 1. 0.20000 1. 0.30000 1. 0.40000 1. 0.50000 1. 0.60000 1. 0.70000 1. 0.80000 1. 0.90000 1. 0.95000 1. 0.99000 1. 0.99900 1. 0.99990 1. 0.99999 1. 1.00000 0. 2.11 Since planetary orbits are very nearly circular (exceptMercury and Pluto), the assumption of perfectly circular motion was a good approximation. Furthermore, since a geocentric model maintains circular motion, it was very difficult to make any observational distinction between geocentric and heliocentric universes. (Parallax effects are far too small to be noticeable with the naked eye.) 2.12 (a) The graph of log10 P vs. log10 a for the Galilean moons is given in Fig. S2.1. (b) Using the data for Io and Callisto, we find a slope of 1.5. (c) Assuming that themass of Jupiter ismuch greater than themasses of any of the Galilean moons, Kepler’s third law can be written as logM C 2 logP D log  42 G C 3 loga; or logP D 3 2 log a C 1 2 log  42 G  1 2 logM: D mlog a C b where the y-intercept is b D 1 2 log  42 G  1 2 logM: Solving for logM we have logM D log  42 G  2b: Taking the slope as m D 3=2 and using the data for any of the Galilean moons we find b D 7:753 (in SI units). Solving givesM D 1:900  1027 kg, in good agreement with the value given in Appendix C and Problem 2.6. Solutions for An Introduction toModern Astrophysics 7 Figure S2.1: log10 P vs. log10 a for the Galilean moons. 2.13 (a) Since the velocity and position vectors are perpendicular at perihelion and aphelion, conservation of angular momentum leads to rpvp D rava. Thus vp va D ra rp D 1 Ce 1  e ; where the last relation is obtained from Eqs. (2.5) and (2.6). (b) Conservation of energy at perihelion and aphelion gives 1 2 v2 a  G M ra D 1 2 v2 p  G M rp : Making use of Eqs. (2.5) and (2.6), and using the result of part (a) to replace vp leads to 1 2 v2 a  GM a.1 C e/ D 1 2 v2 a  1 C e 1  e 2  GM 1.1  e/ : After some manipulation, we obtain Eq. (2.34); Eq. (2.33) follows immediately. (c) The orbital angular momentum can now be obtained from L D rpvp D a.1  e/ s GM a  1 C e 1  e : Equation (2.30) follows directly. 2.14 (a) From Kepler’s third law in the form P2 D a3 (P in years and a in AU), a D 17:9 AU. (b) Since mcomet  Mˇ, Kepler’s third law in the form of Eq. (2.37) gives Mˇ ' 42a3 GP2 D 1:98  1030 kg: (c) From Example 2.1.1, at perihelion rp D a.1e/ D 0:585 AU and at aphelion ra D a.1Ce/ D 35:2 AU. 8 Chapter 2 Celestial Mechanics (d) At perihelion, Eq. (2.33) gives vp D 55 km s1, and at aphelion, Eq. (2.34) gives va D 0:91 km s1. When the comet is on the semiminor axis r D a, and Eq. (2.36) gives v D r GMˇ a D 7:0 km s1: (e) Kp=Ka D  vp=va 2 D 3650. 2.15 Using 50,000 time steps, r D q x2 C y2 ' 1 AU when t ' 0:105 yr. (Note: Orbit can be downloaded from the companion web site at 2.16 The data are plotted in Fig. S2.2. (Note: Orbit can be downloaded from the companion web site at −2 −1 0 1 2 −2 −1 0 1 2 x (AU) y (AU) 0.9 0.4 0.0 Figure S2.2: Results for Problem 2.16. 2.17 (Note: Orbit can be downloaded fromthe companionweb site at (a) See Fig. S2.3. (b) See Fig. S2.3. (c) Figure S2.3 shows that the orbit of Mars is very close to a perfect circle, with the center of the circle slightly offset from the focal point of the ellipse. Kepler’s early attempts at developing a model of the solar system based on perfect circles were not far off. 2.18 A modified Fortran 95 version of Orbit that works for this problem is given below. (a) The orbits generated by the modified Orbit are shown in Fig. S2.4. (b) The calculation indicates S D 2:205 yr. (c) Eq. (1.1) yields a value of S D 2:135 yr. The results do not agree exactly because the derivation of Eq. (1.1) assumes constant speeds throughout the orbits. (d) No, because the relative speeds during the partially-completed orbits are different. (e) Since the orbits are not circular, Mars is at different distances from Earth during different oppositions. The closest opposition occurs when Earth is at aphelion and Mars is at perihelion, as in the start of this calculation. Solutions for An Introduction toModern Astrophysics 9 −2 −1 0 1 2 −2 −1 0 1 2 x (AU) y (AU) Figure S2.3: Results for Problem 2.17. The dots designate the elliptical orbit of Mars, and the principal focus of the ellipse is indicated by the circle at .x; y/ D .0; 0/. The dashed line is for a perfect circle of radius r D a D 1:5237 AU centered at x D ae D 0:1423 AU (marked by the ). Figure S2.4: The orbits of Earth and Mars including correct eccentricities. The positions of two successive oppositions are shown. The first opposition occurs when Earth is at aphelion and Mars is at perihelion (the positions of closest approach). 10 Chapter 2 Celestial Mechanics PROGRAM Orbit ! ! General Description: ! ==================== ! Orbit computes the orbit of a small mass about a much larger mass, ! or it can be considered as computing the motion of the reduced mass ! about the center of mass. ! ! "An Introduction to Modern Astrophysics", Appendix J ! Bradley W. Carroll and Dale A. Ostlie ! Second Edition, Addison Wesley, 2007 ! ! Weber State University ! Ogden, UT ! !------------------------------------------------------------------- ! **************This version has been modified for Problem 2.17****************** USE Constants, ONLY : i1, dp, G, AU, M_Sun, pi, two_pi, yr, & radians_to_degrees, eps_dp IMPLICIT NONE REAL(dp) :: t, dt, LoM_E, LoM_M, P_E, P_M REAL(dp) :: Mstar, theta_E, dtheta_E, theta_M, dtheta_M, r_E, r_M INTEGER :: n, k, kmax INTEGER(i1) :: ios !I/O error flag REAL(dp) :: delta !error range at end of period CHARACTER :: xpause REAL(dp), PARAMETER :: a_E = AU, a_M = 1.5236*AU, e_E = 0.0167, e_M = 0.0935 ! Open the output file OPEN (UNIT = 10, FILE = "O", STATUS = ’REPLACE’, ACTION = ’WRITE’, & IOSTAT = ios) IF (ios /= 0) THEN WRITE (*,’(" Unable to open O. --- Terminating calculation")’) STOP END IF ! Convert entered values to conventional SI units Mstar = M_Sun ! Calculate the orbital period of Earth in seconds using Kepler’s Third Law (Eq. 2.37) ! To be used to determine time steps P_E = SQRT(4*pi**2*a_E**3/(G*Mstar)) ! Enter the number of time steps and the time interval to be printed n = n = n + 1 !increment to include t=0 (initial) point kmax = 100 ! Print header information for output file WRITE (10,’("t, theta_E, r_E, x_E, y_E, theta_M, r_M, x_M, y_M")’) ! Initialize print counter, angle, elapsed time, and time step. k = 1 !printer counter theta_E = 0 !angle from direction to perihelion (radians) theta_M = 0 t = 0 !elapsed time (s) dt = P_E/(n-1) !time step (s) delta = eps_dp !allowable error at end of period ! Start main time step loop DO ! Calculate the distance from the principal focus using Eq. (2.3); Kepler’s First Law. r_E = a_E*(1 - e_E**2)/(1 - e_E*COS(theta_E)) !Earth starts at aphelion r_M = a_M*(1 - e_M**2)/(1 + e_M*COS(theta_M)) !Mars starts at perihelion ! If time to print, convert to cartesian coordinates. Be sure to print last point also. IF (k == 1 .OR. (theta_E - theta_M)/two_pi > 1 + delta) & WRITE (10, ’(9F10.4)’) t/yr, theta_E*radians_to_degrees, r_E/AU, & Solutions for An Introduction toModern Astrophysics 11 r_E*COS(theta_E)/AU, r_E*SIN(theta_E)/AU, & theta_M*radians_to_degrees, r_M/AU, & r_M*COS(theta_M)/AU, r_M*SIN(theta_M)/AU ! Exit the loop if Earth laps Mars. IF ((theta_E - theta_M)/two_pi > 1 + delta) EXIT ! Prepare for the next time step: Update the elapsed time. t = t + dt ! Calculate the angular momentum per unit mass, L/m (Eq. 2.30). LoM_E = SQRT(G*Mstar*a_E*(1 - e_E**2)) LoM_M = SQRT(G*Mstar*a_M*(1 - e_M**2)) ! Compute the next value for theta using the fixed time step by combining ! Eq. (2.31) with Eq. (2.32), which is Kepler’s Second Law. dtheta_E = LoM_E/r_E**2*dt theta_E = theta_E + dtheta_E dtheta_M = LOM_M/r_M**2*dt theta_M = theta_M + dtheta_M ! Reset the print counter if necessary k = k + 1 IF (k > kmax) k = 1 END DO WRITE (*,’(/,"The calculation is finished and the data are in O")’) WRITE (*,’(//,"Enter any character and press <enter> to exit: ")’, ADVANCE = ’NO’) READ (*,*) xpause END PROGRAM Orbit CHAPTER 3 The Continuous Spectrum of Light 3.1 (a) The parallax angle is p D 33:600=2 D 8:14  105 rad. For a baseline B D R˚ (Fig. 3.1), d D B tanp D 7:83  1010 m D 0:523 AU: (b) Suppose that the clocks of the two observers are out of sync by an amount t , causing the baseline for their distance measurement to be 2B C vrelt instead of 2B. Here, vrel is the relative velocity of Mars and the Earth at the time of opposition, vrel D 29:79 km s1  24:13 km s1 D 5:66 km s1: The calculated value (which is wrong) of the distance between Earth and Mars is dcalc D R˚ tan p ; while the actual value is dact D R˚ C 0:5vrelt tan p : If the distance to Mars is to be measured to within 10 percent, then dact  dcalc dact D 0:1: Inserting the expressions for dact and dcalc and solving for t results in t D 250 s. 3.2 From Example 3.2.1, the solar constant is 1365 Wm2. Using the inverse square law, Eq. (3.2), with L D 100 Wgives r D 0:0764 m D 7:64 cm. 3.3 (a) From Eq. (3.1), d D 1 p00 pc D 2:64 pc D 8:61 ly D 5:44  105 AU D 8:14  1016 m: (b) The distance modulus for Sirius is, from Eq. (3.6), m M D 5 log10.d=10 pc/ D 2:89: 3.4 From Example 3.6.1, the apparent bolometric magnitude of Sirius is m D 1:53. Then from the previous problem, M D m C 2:89 D 1:53 C 2:89 D 1:36 is the absolute bolometric magnitude of Sirius. Using Eq. (3.7) withMSun D 4:74 leads to L=Lˇ D 100.MSunM/=5 D 22:5: 12 Solutions for An Introduction toModern Astrophysics 13 3.5 (a) Converting the subtended angle to radians gives 0:00100 D 4:85  109 radians. Thus, the distance required is d D 0:019 m=4:85  109 D 3:92  106 m D 3920 km. (b) i. Assume that grows at a rate of approximately 5 cm=7 days D 8  108 m s1. (I suspect it is an order of magnitude faster in the author’s lawn!) ii.  D 0: corresponds to roughly 2  1011 radians. If SIM PlanetQuest is just able to resolve a change in the length of a blade of grass in one second, the spacecraft would need to be at a distance of d D 8  10 8 m s1 2  1011 s1 D 4 km: 3.6 Equations (3.6) and (3.8) combine to give m  5 log10  d 10 pc D MSun  2:5 log10  L Lˇ : Use the inverse square law, Eq. (3.2), to substitute L D 4d2F and Lˇ D 4.10 pc/2F10;ˇ. The distance terms then cancel, resulting in Eq. (3.9). 3.7 (a) Use Eq. (3.13) for the force due to radiation pressure for an absorbing surface, and set  D 0. hSi is just the solar constant, hSi D 1365Wm2. From Newton’s second law with a D 9:8 m s1, the sail area is A D r 2 D cFrad hSi D cma hSi : Solving for the sail’s radius gives r D 91 km. (The spacecraft is already orbiting the Sun, so the Sun’s gravity is not included in this calculation.) (b) The radiation pressure force for a reflecting surface, Eq. (3.14), is twice as great as for part (a), so the sail’s radius is smaller by a factor of p 2: r D 64 km. 3.8 (a) From the Stefan–Boltzmann equation, Eq. (3.16), L D AT 4 D 696 W. (b) Wien’s displacement law, Eq. (3.15), shows that max D 9477 nm, in the infrared region of the electromagnetic spectrum. (c) Again from Eq. (3.16), L D 585 W. (d) The net energy per second lost is 696W  585 W D 111 W. 3.9 (a) The star’s luminosity is (Eq. 3.17) L D 4R2T 4 e D 1:17  1031 W D 30,500 Lˇ: (b) From Eq. (3.7), M D MSun  2:5 log10.L=Lˇ/ D 6:47; withMSun D 4:74 for the Sun’s absolute bolometric magnitude. (c) The star’s apparent bolometric magnitude comes from Eq. (3.6): m D M C 5 log10.d=10 pc/ D 1:02: (d) m M D 5:45. (e) Equation (3.18) gives the surface flux, F D T 4 e D 3:48  1010 Wm2: 14 Chapter 3 The Continuous Spectrum of Light (f) At the Earth, the inverse square law (Eq. 3.2) with r D 123 pc D 3:8  1018 m gives F D L 4r 2 D 6:5  10 8 Wm2: This is 4:7  1011 of the solar constant. (g) Wien’s law in the form of Eq. (3.19) shows that max D .500 nm/.5800 K/ 28,000 K D 104 nm: Figure S3.1: Results for Problem 3.10. The Planck function (solid line) and the Rayleigh–Jeans law (dashed line) are shown. 3.10 (a) When   hc=kT, we can approximate the denominator of the Planck function, Eq. (3.22), as ehc=kT  1 ' 1 C hc kT  1 D hc kT : Inserting this into the Planck function results in the Rayleigh–Jeans law, B.T / ' 2ckT=4. (b) As seen from Fig. S3.1, the Rayleigh–Jeans value is twice as large as the Planck function when  ' 2000 nm. 3.11 At sufficiently short wavelengths, the exponential term in the denominator of Planck’s function dominates, meaning that ehc=kT  1. Thus B   2hc2=5  e hc=kT ; which is just Eq. (3.21). 3.12 An extremum in the Planck function (Eq. 3.22), B D 2hc2=5 ehc=kT  1 ; Solutions for An Introduction toModern Astrophysics 15 occurs when dB=d D 0. dB d D 10hc2=6 ehc=kT  1  2hc2=5  ehc=kT  1 2   hc 2kT ehc=kT D 0; which simplifies to 5  ehc=kT  1  D  hc kT ehc=kT : Defining x  hc=kT, this becomes  1/ D xex which cannot be solved analytically. A numerical solution leads to x D hc=maxkT D 4:97. Inserting values for h, c, and k results in maxT D 0:0029 m K, which is Wien’s law (Eq. 3.15). 3.13 (a) An extremum in the Planck function (Eq. 3.24), B D 2h3=c2 eh=kT  1 ; occurs when dB=d D 0. dB d D 6h2=c2 eh=kT  1  2h3=c2  eh=kT  1 2  h kT eh=kT D 0; which simplifies to 3  eh=kT  1  D  h kT eh=kT : Defining x  h=kT , this becomes  1/ D xex which cannot be solved analytically. A numerical solution leads to x D hmax=kT D 2:82. Inserting values for h and k results in max=T D 5:88  1010 s1 K1. (b) Using Te D 5777 K for the Sun, max D 3:40  1014 Hz. (c) c=max D 883 nm, which is in the infrared region of the electromagnetic spectrum. 3.14 (a) Integrate Eq. (3.27) over all wavelengths, L D Z 1 0 L d D Z 1 0 82R2hc2=5 ehc=kT  1 d: Defining u D hc=kT, the luminosity becomes L D 82R2k4T 4 c2h3 Z 1 0 u3 du eu  1 : The value of the integral is 4=15, so L D 86R2k4T 4 15c2h3 : (b) By comparing the result of part (a) with the Stefan–Boltzmann equation (L D 4R2T 4, Eq. 3.17), we find that the Stefan–Boltzmann constant is given by  D 25k4 15c2h3 : 16 Chapter 3 The Continuous Spectrum of Light (c) Inserting values of k, c, and h results in  D 5:  10 8 Wm2 K4; in perfect agreement with the value listed in Appendix A. −0.5 0 0.5 1.0 1.5 2.0 1.5 1.0 0.5 0 −0.5 −1.0 B−V U−B Blackbody B0 A0 F0 G0 K0 M0 Sun Sirius Figure S3.2: Results for Problem 3.15. This shows the locations of the Sun and Sirius on a color–color diagram. 3.15 (a) From Appendix G, the Sun’s absolute visual magnitude is MV D 4:82. From Example 3.2.2, the Sun’s distance modulus is m M D 31:57. Thus V D MV  31:57D 26:74. (b) From Appendix G, B  V D 0:650 and U  B D 0:195 for the Sun. So B D V C0:650 D 26:1 and MB D B C 31:57 D 5:47, while U D B C 0:195 D 25:9 and MU D U C 31:57 D 5:67. (c) See Fig. S3.2. 3.16 To find the filter through which Vega would appear brightest, we can take ratios of the radiant fluxes received through the filters. For the U and B filters, Eq. (3.31) shows that we are interested in R1 0 FSU d R1 0 FSB d : The central wavelength of the U filter is U D 365 nm, and U D 68 nm is the bandwidth of the U filter. Similarly, for the B filter B D 440 nm with B D 98 nm, and for the V filter V D 550 nm with V D 89 nm. Assuming that S D 1 inside the filter bandwidth and S D 0 otherwise, and employing Eq. (3.29) for the monochromatic flux, the integrals can be approximated to obtain, R 1 0 FSU d R 1 0 FSB d '  B U 5 ehc=BkT  1 ehc=UkT  1  U B D 0:862 for T D 9600 K. In the same manner, R1 0 FSU d R1 0 FSV d ' 1:42 and R1 0 FSB d R1 0 FSV d ' 1:64: So Vega would appear brightest when viewed through the B filter and faintest through the V filter. Solutions for An Introduction toModern Astrophysics 17 3.17 From Eq. (3.32) and Eq. (3.29) for the monochromatic flux, mbol D 2:5 log10 Z 1 0 F d C Cbol mbol D 2:5 log10 R1 0 L d 4r 2 ! C Cbol mbol D 2:5 log10  Lˇ 4r 2 C Cbol: At r D 1 AU, Lˇ=4r 2 is just the solar constant. We have 26:83 D 2:5 log10.1365/C Cbol so Cbol D 18:99. 3.18 According to Eq. (3.33) and Example 3.6.2, U  B D 2:5 log10  B365 U B440 B C CUB and similarly for B  V . Here, B365 is the Planck function evaluated at the central wavelength of the U filter (U D 365 nm) and U D 68 nm is the bandwidth of the U filter. Similar expressions describe the B filter (centered at B D 440 nm with B D 98 nm) and the V filter (centered at V D 550 nm with V D 89 nm). From Example 3.6.2, the constants are CUB D 0:87 and CBV D 0:65. From Eq. (3.22) for the Planck function, we find for T D 5777 K U  B D 2:5 log10 " B U 5 ehc=BkT  1 ehc=UkT  1  U B # CCUB D 0:22; which is quite different than the actual value of U  B D C0:195 for the Sun. In the same manner, the estimated value of B  V is C0:57, which is in fair agreement with the actual value of B  V D C0:65 for the Sun. 3.19 (a) Following the procedure for Problem 3.18 with T D 22,000 K gives estimates of U  B D 1:08 (actual value is U  B D 0:90) and B  V D 0:23 (the value measured for Shaula). (b) A parallax angle of 0: implies a distance to Shaula of d D 1=p D 216 pc. From Eq. (3.6) and V D 1:62,MV D 5:05. CHAPTER 4 The Theory of Special Relativity 4.1 Inserting Eqs. (4.10)–(4.13) into Eq. (4.15) and simplifying results in .a2 11  a2 41c2/x2  2.a2 11u C a41a44c2/xt C y2 C z2 D .a2 11u2 C a2 44c2/t2: For this to agree with Eq. (4.14), x2 C y2 C z2 D .ct/2, requires that a2 11  a2 41c2 D 1 a2 11u C a41a44c2 D 0 a2 11u2 C a2 44c2 D c2 Solving these yields the values of a11, a44, and a41 given in the text. 4.2 Assume that events A and B occur on the x-axis. We can assume with no loss of generality that xB > xA and that event A caused event B [so light has time to travel from A to B: 0 < xB  xA < c.tB  tA/]. From Eq. (4.19), the time interval between events A and B as measured in frame S0 is tB 0  tA 0 D .tB  tA/  u=c2.xB  xA/ q 1  u2=c2 : As measured in frame S0, event B will occur after event A (tB 0  tA 0 > 0) if the numerator on the right-hand side is positive. numerator D .tB  tA/  u=c2.xB  xA/ D .tB  tA/ 1  u c   xB  xA c.tB  tA/  > 0 because u=c < 1 and 0 < xB  xA < c.tB  tA/. Thus tB 0  tA 0 > 0, and cause precedes effect; event A occurs before event B, as measured in frame S0. 4.3 As measured in frame S, in time t=2 the mirror moves forward a distance ut=2 while the light ray travels a path of length ct=2. From the Pythagorean theorem,  ct 2 2 D  ut 2 2 Cd2 while as measured in frame S 0, d2 D .ct 0 =2/2. Eliminating d gives  ct 2 2 D  ut 2 2 C  ct 0 2 2 : Solving for t gives Eq. (4.26), t D t 0 q 1  u2=c2 : Identifyingt D tmoving and t 0 D trest achieves the final result. 18 Solutions for An Introduction toModern Astrophysics 19 4.4 From Eq. (4.29), Lmoving=Lrest D q 1  u2=c2 D 1=2, so u=c D 1= p 2 D 0:866. 4.5 For this problem, p 1  u2=c2 D 0:6. (a) tP D .60 m/=.0:8c/ D 0:250 s. (b) The rider sees a train at rest of length Lrest D Lmoving p 1  u2=c2 D 60 m 0:6 D 100 m: (c) The rider sees a moving platform of length Lmoving D Lrest q 1  u2=c2 D .60 m/.0:6/ D 36 m: (d) tT D .100 m/=.0:8c/ D 0:417 s. (e) According to T , the train is 100 m long and the moving platform is 36 m long. The time T measures for the platform to travel the extra 64 m is .64 m/=.0:8c/ D 0:267 s. 4.6 For this problem, p 1  u2=c2 D 0:6. When distances are given in light years, it is handy to express c D 1 ly yr1. (a) The two events are the starship leaving Earth and arriving at ˛ Centauri. According to an observer on Earth, these events occur at different locations, so the time measured by a clock on Earth is .t/moving D 4 ly 0:8c D 5 yr: (b) The starship pilot is at rest relative to the two events; they both occur just outside the door of the starship. According to the pilot, the trip takes .t/rest D .t/moving q 1  u2=c2 D 3 yr: (c) Using Lrest D 4 ly, the distance measured by the starship pilot may be found from Lmoving D Lrest q 1  u2=c2 D 2:4 ly: (d) According to Eq. (4.31) with trest D 6 months and  D 0, the time interval between receiving the signals aboard the starship is tobs D trest.1 C u=c/ p 1  u2=c2 D 18 months: (e) The same as part (d). (f) From the relativistic Doppler shift, Eq. (4.35), obs D rest s 1 C vr =c 1  vr =c D .15 cm/ r 1C 0:8 1  0:8 D 45 cm: 20 Chapter 4 The Theory of Special Relativity 4.7 According to an observer on Earth, the signal sent from the starship just as it arrives at ˛ Centauri reaches Earth 5 yrC4 yr D 9 yr after the ship left Earth. The signals sent by the outbound starship arrive at 18 month intervals, when tE(months) D 0; 18; 36; 54; 72; 90; 108: During the remaining year of the ten-year roundtrip journey (as measured on Earth), the signals reach Earth more frequently. According to Eq. (4.31) with trest D 6 months and  D 180ı, the time interval between receiving the signals on Earth is tobs D trest.1  u=c/ p 1  u2=c2 D 2 months: The rest of the signals arrive on Earth at tE(months) D 110; 112; 114; 116; 118; 120: The Earth observer has received 12 signals (not including tE D 0), and so concludes that 6 years have passed onboard the starship during the round-trip journey to ˛ Centauri, although ten years have passed on Earth. According to the starship pilot, while outbound the starship receives a signal when tS(months) D 0; 18; 36: At that point, the starship immediately reverses direction and travels back toward Earth. Again according to Eq. (4.31) with trest D 6 months and  D 180ı, during the return trip the time interval between receiving the signals aboard the starship is 2 months. So the starship receives a signal from Earth when tS(months) D 38; 40; 42; 44; 46; 48; 50; 52; 54; 56; 58; 60; 62; 64; 66; 68; 70; 72: At that point, the starship arrives at Earth. The starship has received 20 signals (not including tS D 0), so the starship pilot deduces that 10 years have passed on Earth during the round-trip journey to ˛ Centauri, although the pilot has aged only six years. 4.8 The redshift parameter for the quasar Q2203C29 is (Eq. 4.34) z D obs  rest rest D 656:8 nm  121:6 nm 121:6 nm D 4:40: From Eq. (4.38), the speed of recession of the quasar is vr =c D 0:9337. 4.9 The redshift parameter for the quasar 3C 446 is z D 1:404. From Eq. (4.37), if the quasar’s luminosity is observed to vary over a time of tobs D 10 days, then the time for the luminosity variation as measured in the quasar’s rest frame is trest D tobs z C 1 D 4:16 days: 4.10 Taking a differential of Eqs. (4.16) and (4.19) gives dx 0 D dx  udt p 1  u2=c2 and dt 0 D dt  u=c2 dx p 1  u2=c2 : Solutions for An Introduction toModern Astrophysics 21 Dividing the first of these equations by the second, we find dx0 dt0 D dx  udt dt  u=c2 dx D dx=dt  u 1  .u=c2/.dx=dt/ : Substituting v0 x D dx0=dt0 and vx D dx=dt then produces Eq. (4.40). In a similarmanner, the differential version of Eq. (4.17) is dy0 D dy. Dividing this by the dt0 equation yields dy0 dt0 D dy p 1  u2=c2 dt  u=c2 dx D dy=dt p 1  u2=c2 1  .u=c2/.dx=dt/ : Substituting v0 y D dy0=dt0, vx D dx=dt and vy D dy=dt then produces Eq. (4.41). Equation (4.42) is obtained in the same way. 4.11 (a) As measured in frame S0 , the spacetime interval is .s 0 /2 D .ct 0 /2  .x 0 /2  .y 0 /2  .z 0 /2: Using Eqs. (4.16)–(4.19), we have .s 0 /2 D " c.t  ux=c2/ p 1  u2=c2 #2  x  ut p 1  u2=c2 !2  .y/2  .z/2: After some algebra, this becomes .s 0 /2 D .c2  u2/.t/2 1  u2=c2 C .u2=c2  1/.x/2 1  u2=c2  .y/2  .z/2: or .s 0 /2 D .ct/2  .x/2  .y/2  .z/2 D .s/2: (b) The time interval between two events that occur at the same location is the proper time interval,  . Setting x D 0, y D 0, and z D 0 and substitutingt D  in the expression for the interval, we find s D c . The first event could have caused the second event because light had more than enough time to travel the zero distance between the two events. (c) If .s/2 D 0, then ct D q .x/2 C .y/2 C.z/2: This says that light had exactly enough time to travel the distance between the two events. The first event could not have caused the second event unless light itself carried the causal information. (d) The proper length (which we will call L) of an object is obtained by measuring the difference in the positions of its ends at the same time. Settingt D 0 and substituting.L/2 D .x/2C.y/2C.z/2 in the expression for the interval, we find L D q  .s/2. 4.12 (a) This is a straightforward substitution. (b) From the velocity transformations, Eqs. (4.40)–(4.42) vx 0 D c sin  cos  u 1  u=c2.c sin  cos / vy 0 D c sin  sin p 1  u2=c2 1  u=c2.c sin  cos / vz 0 D c cos  p 1  u2=c2 1  u=c2.c sin  cos / 22 Chapter 4 The Theory of Special Relativity Substituting this into v0 D q vx 02 C vy 02 C vz 02 results in, after much manipulation, v 0 D c 1  u=  cos /   1  2u c sin  cos C u2 c2 sin2  cos2 1=2 ; which is just v0 D c. 4.13 Identify the frame S with Earth and frame S0 with starship A, so vA D u D 0:8c. The velocity of starship B as measured from Earth is vB D 0:6c. Then the velocity of starship B as measured from starship A may be found using the velocity transformation (Eq. 4.40) vB 0 D vB  u 1  uvB=c2 D 0:946c: The velocity of starship A as measured from starship B is then C0:946c. 4.14 According to Newton’s second law, taking a time derivative of the relativisticmomentum (Eq. 4.44) gives the force acting on a particle of mass m. dp dt D d dt mv p 1  v2=c2 ! : Writing v2 D v  v and using F D dp=dt, we find that F D mdv=dt p 1  v2=c2 C mv=c2  1  v2=c2 3=2  v  dv dt : To simplify this, it is useful to take the vector dot product of this equation with the velocity vector. F  v D mv  dv=dt p 1  v2=c2 C mv2=c2  1  v2=c2 3=2  v  dv dt ; which reduces to v  dv dt D  1  v2=c23=2 m F  v: Using this to replace v  dv=dt in the expression for F and solving for a D dv=dt leads to the desired result, a D F m  v mc2 .F  v/ : 4.15 (a) The particle starts at rest, so the velocity, force, and acceleration are in the same direction. In this case, the result of Problem 4.14 simplifies to a D dv dt D F m  1  v2 c2 D F m  1  v2 c2 3=2 : Now integrate Z v 0  1  v2 c2 3=2 dv D Z t 0 F m dt Solutions for An Introduction toModern Astrophysics 23 to find Ft m D v p 1  v2=c2 : Solving for v=c, we get v c D .F=m/t p .F=m/2t2 Cc2 : (b) Solving the above equation for the time t yields t D 1 F=m v p 1  v2=c2 : If F=m D g D 980 cm s2, then the times to achieve the given speeds are t.0:9c/ D 6:32  107 s (2:0 yr) t.0:99c/ D 2:15  108 s (6:8 yr) t.0:999c/ D 6:84  108 s (21:7 yr) t.0:9999c/ D 2:16  109 s (68:6 yr) t.1:0c/ D1s (never!) 4.16 From Eq. (4.45), setting the relativistic kinetic energy equal to the rest energy mc2 gives mc2. 1/ D mc2, so  D 2. Solving for velocity shows that v=c D 1= p 2 D 0:866. 4.17 From Eq. (4.45), the relativistic kinetic energy is K D mc2 1 p 1  v2=c2  1 ! : At low speeds, v=c  1, and we can use the first-order Taylor series .1  x2/1=2 ' 1 C x2=2. Thus, for low speeds, K ' mc2  1 C 1 2 v2 c2  1  D 1 2 mv2: 4.18 Begin with Eq. (4.46) for the total relativistic energy, E D mc2. Squaring both sides and subtracting m2c4 from each side results in E2  m2c4 D m2c4   2  1  D mc2. C 1/OEmc2.  1/: From Eq. (4.48), the left-hand side is equal to p2c2, while the term on the right-hand side in square brackets is the relativistic kinetic energy (Eq. 4.45). So p2c2 D mc2. C 1/K; Solving for K gives K D p2 .1 C  /m : 4.19 According to Eqs. (4.46) and (4.44), the total relativistic energy and relativistic momentum are E D mc2 and p D mv. Multiplying the momentum by c, then squaring both equations and subtracting produces E2  p2c2 D  2m2c4   2m2v2c2 D  2m2c4  1  v2 c2 : Because  2 D 1=.1  v2=c2/, this reduces to E2 D p2c2 C m2c4, which is the desired result. CHAPTER 5 The Interaction of Light and Matter 5.1 (a) The rest wavelength for the H˛ spectral line is rest D 656:281 nm. From Eq. (5.1), the radial velocity of Barnard’s star is vr D c .obs  rest/ rest D 113 km s1: (b) From Eq. (3.1), d D .1=p00/ pc D 1:82 pc. The transverse velocity of Barnard’s star is (Eq. 1.5) v D d  D 89:4 km s1. (c) v D q v2 r C v2  D 144 km s1. 5.2 (a) The distance between the grating’s lines is d D 3:33  106 m. Using d sin  D n with n D 2, the angles are 20:695ı for  D 588:997 nm, and 20:717ı for  D 589:594 nm. The angle between the second-order spectra of these two lines is therefore  D 0:0219 A° . (b) From Eq. (5.2), the number of lines that must be illuminated is N D  n D 494; where the average wavelength  D 589:296 nm was used. 5.3 Using the values provided in Appendix A, it is found that hc D 1239:8 eV nm ' 1240 eV nm. 5.4 From Eq. (5.4), D hc   Kmax D 1240 eV nm 100 nm  5 eV D 7:4 eV: 5.5 Referring to Figure 5.3, conservation of momentum gives pi D pf cos  C pe cos 0 D pf sin   pe sin ; where pi and pf are the initial and final photon momenta, respectively, and pe is the momentum of the recoiling electron. Rewriting these equations so that only terms involving the angle are on the right-hand side, and then squaring both equations and adding produces p2 i  2pipf cos  C p2 f D p2 e : Now multiply each side by c2 and add m2e c4 to both sides. Using Ei D pi c and Ef D pf c (Eq. 5.5) for the initial and final photon energies, we find m2 ec4 C E2 i  2EiEf cos  C E2 f D p2 e c2 C m2 ec4 D E2 e ; where the last term on the right comes from Eq. (4.48) for the total relativistic energy of the electron. However, the expression for the conservation of the total relativistic energy, Ei C mec2 D Ef C Ee 24 Solutions for An Introduction toModern Astrophysics 25 can be used to replace Ee in the preceding equation, m2 ec4 C E2 i  2EiEf cos  C E2 f D .Ei  Ef Cmec2/2: This simplifies to 1 Ef  1 Ei D 1 mec2 .1  cos /: Using Eq. (5.5) for the photon energy, E D hc=, results in  D f  i D h mec .1  cos /: 5.6 The change of wavelength is given by the Compton scattering formula, Eq. (5.6), but with the electron mass replaced by the proton mass,  D f  i D h mpc .1