2005 Quiz 1 with Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Earth, Atmospheric, and Planetary Sciences Department Astronomy 8.282J­12.402J March 21, 2005 Quiz 1 ­ Solutions Name (please print) Last 1. 2. 3. 4. First Be sure to attempt all problems. The point values of the problems are indicated at the top of each page. Closed book exam; you may use one page of notes. Wherever possible, try to solve the problems using general analytic expressions. Plug in numbers only as a last step. Problem 1 2 3 4 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 cm/sec 7 × 10−8 dyne­cm2 /g2 1.4 × 10−16 erg/K 6.6 × 10−27 erg­sec 1.6 × 10−24 g 1.6 × 10−12 erg 2 × 1033 g 4 × 1033 erg/sec 7 × 1010 cm 6 × 10−5 erg/cm2 ­sec­K4 10−8 cm 105 cm 3 × 1018 cm 3 × 1021 cm 3 × 1024 cm 3 × 107 sec 86400 sec 1.5 × 1013 cm 1/3400 rad 1/200, 000 rad mks 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 units 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 (25 points) Orbital Dynamics A binary system consists of two stars in circular orbit about a common center of mass, with an orbital period, Porb = 10 days. Star 1 is observed in the visible band, and Doppler measurements show that its orbital speed is v1 = 20 km s−1 . Star 2 is an X­ray pulsar and its orbital radius about the center of mass is r2 = 3 × 1012 cm = 3 × 1010 m. (a) Find the orbital radius, r1 , of the optical star (Star 1) about the center of mass. v1 = r1 = 2πr1 Porb Porb v1 = 2.75 × 1011 cm 2π (b) What is the total orbital separation between the two stars, r = r1 + r2 ? r = r1 + r2 = 2.75 × 1011 + 3 × 1012 = 3.3 × 1012 cm (c) Compute the total mass of the system, M1 + M2 = Mtot . � �2 GMT 2π = r3 Porb � �2 3 2π r MT = = 2.8 × 1034 g = 14 M� Porb G (d) Compute the individual masses of Star 1 and of Star 2. M1 r2 3 × 1012 = = = 10.9 M2 r1 2.8 × 1011 M1 + M2 = MT = 14 M� ⇒ M2 = 1.2 M� ; M1 = 12.8 M� Problem 2 (25 points) Geometric and Physical Optics (a) Compare the angular resolution of the The Very Large Array (VLA), the Hubble Space Telescope, and the Spitzer Space Telescope. The effective diameters of the three instruments are 36 km, 2.5 m, and 85 cm, respectively, while typical wavelengths used for observation might be 6 cm, 0.6 µm, and 5 µm, respectively. Θmin � VLA : λ D 6 cm = 0.33�� 36 × 105 cm 0.6 × 10−6 m = 0.05�� 2.5 m 5 × 10−6 m Spitzer : = 1.2�� 0.85 m HST : (b) Each of the two Magellan telescopes has a diameter of 6.5 m. In one configuration the effective focal length is 72 m. Find the diameter of the image of a planet (in cm) at this focus if the angular diameter of the planet at the time of the observation is 45�� . Start with: s = αf , where s is the diameter of the image, f the focal length, and α the angular diameter of the planet. For the values given in the problem: s= 45 π 7200 = 1.6 cm 3600 180 (c) A prism is constructed from glass and has sides that form a right triangle with the other two angles equal to 45◦ . The sides are L, L, and H, where L is a leg and H is the hypotenuse. A parallel light beam enters side L normal to the surface, passes into the glass, and then strikes H internally. The index of refraction of the glass is n = 1.5. Compute the critical angle for the light to be internally reflected at H. Does it do so in the geometry described here? From Snell’s law we have: ng sin(θg ) = nair sin(θair ) 1 sin(90◦ ) ⇒ θcrit = 41.8◦ 1.5 But, from the geometry, θincident for the light striking side H is 45◦ . Thus, since θincident > θcrit , the light does undergo total internal reflection. sin(θcrit ) = Problem 3 (25 points) Doppler Effect and Bohr Atom (a) A binary system with circular orbit is viewed edge on (i.e., in the plane of the orbit), and the spectrum of one of the stars is recorded on a regular basis (the other star is undetectable). The orbital speed of the visible star is 300 km s−1 , and the orbital period is Porb . The center of mass of the binary system is moving toward the Solar System at 300 km s−1 . Compute an expression for the Doppler shift of the Hα line (at 6563 Å) as a function of time, and sketch Δλ as a function of orbital cycle. You may ignore the orbital speed of the Earth around the Sun. The time­dependent Doppler shift of the star is given by: Δλ v � − sin(θ) , λ c where θ = 0 corresponds to the source moving toward us. Thus, � � Δλ 300 300 2πt �− − sin + φorb , λ 3 × 105 3 × 105 Porb where the first term represents the constant center of mass motion for the binary, while the second term is the time­varying Doppler shift of the star being observed. The parameter φorb is the orbital phase. Finally, in terms of a wavelength shift, we have: � � �� 2πt −3 −3 ˚ . Δλ � −6563A 10 + 10 sin + φorb Porb � � �� 2πt Δλ � −6.5˚ A 1 + sin + φorb . Porb (b) If the Bohr energy levels scale as Z 2 , where Z is the atomic number of the atom (i.e., the charge on the nucleus), estimate the energy or wavelength of a photon that results from a transition from n = 3 to n = 2 in Fe, which has Z = 26. Assume that the Fe atom is completely stripped of all its electrons except for one. � � 1 1 hν = 13.6 Z 2 2 − 2 eV nf ni � � 1 1 2 − eV hν = 13.6 × 26 22 32 hν = 1280 eV = 1.28 keV ⇒ 9.6Å Problem 4 (25 points) Luminosity and Magnitudes (a) A white dwarf star has an effective temperature, Te = 50, 000 degrees Kelvin, but its radius, RWD , is comparable to that of the Earth. Take RWD = 104 km (107 m or 109 cm). Compute the luminosity (power output) of the white dwarf. Treat the white dwarf as a blackbody radiator. L = 4πR2 σTe4 = 4π(109 )2 (5.7 × 10−5 )(50, 000)4 ergs s−1 L � 4.5 × 1033 ergs s−1 � 1 L� (b) An extrasolar planet has been observed which passes in front of (i.e., transits) its parent star. If the planet is dark (i.e., contributes essentially no light of its own) and has a surface area that is 2% of that of its parent star, find the decrease in magnitude of the system during transits. The flux goes from a maximum of F0 , when the planet is not blocking any light, to 0.98F0 when the planet is in front of the stellar disk. So, the uneclipsed magnitude is: m0 = −2.5 log(F0 /Fref ) . When the planet blocks 2% of the stellar disk, the magnitude increases to: m = −2.5 log(F/Fref ) = −2.5 log(0.98F0 /Fref ) . Thus, the change in magnitude is: Δm = m − m0 = −2.5 log(0.98) � 0.022 magnitudes (c) If a star cluster is made up of 104 stars, each of whose absolute magnitude is −5, compute the combined apparent magnitude of the cluster if it is located at a distance of 1 Mpc. The absolute magnitude of one of the stars is given by: M = −2.5 log(L/Lref ) = −5 , where L is the stellar luminosity, and Lref is the luminosity of a zero magnitude star. This equation implies that L = 100 Lref . Armed with this fact, we can now compute the combined magnitude of the collection of 104 stars: MTOT = −2.5 log[(104 × 100 Lref )/Lref ] = −2.5 log(106 ) = −15 Finally, the distance modulus corresponding to 1 Mpc is 5 log(106 /10) = 25. Therefore, the apparent magnitude of the star cluster at this distance is: m = M + distance modulus ⇒ m = −15 + 25 = +10 .