Problem 5 (20 points): Give the appropriate name for the following variable stars. Use a reason- able abbreviation for the name of the constellation. a. (10 pt): The 13th variable star discovered in the constellation Duntante. The first nine names are R through Z, and 10, 11, 12, and 13 are RR, RS, RT, and RU. Therefore, you would call this star “RU Dun” b. (10 pt): The 302nd variable star discovered in the constellation Iampetus? In this case, it is easiest to count backwards: Final 10 stars (Numbers 325 - 334) are QQ through QZ Previous 11 stars (Numbers 314 - 324) are PP through PZ Previous 12 stars (Numbers 302 - 313) are OO through OZ so the 302nd star is OO. Therefore, you would call this star “OO Iam” Problem 4 (15 points) a. (8 pt): Use Zeilik Figure 13-10 and the position of the RR Lyrae gap to estimate the distance to the globular cluster M3. Figure 13-10 shows the RR Lyrae gap to be at apparent magnitude m ≈ 15.5. These stars all have absolute magnitude M ≈ 0.5, regardless of their period. Therefore mM–== 15.0 5logd – 5 and so logd=or 4 d==104 pc 10 kpc which is the same answer you get with main sequence fitting. (See Exam 1.) b. (7 pt): What is the ratio of the periods for population I and population II Cepheids with 1000 times the luminosity of the sun? 2 4 From Zeilik Figure 18-3, midway between the 10 and 10 tick marks, you read log PI = 0.7 log PII = 1.3 so − = −0.4 log PI log PII = log (PI/PII) or = 10−0.4 = 0.4 PI/PII Problem 3 (25 points): Consider the X-ray binary system Cygnus X-1. a. (5 pt): What is the Schwarzschild radius of the black hole companion in Cygnus X-1. 2GM MBH R ==--------------------BH-3km×---------------- =48km BH 2 c MSUN where we used Eq.17-7b in Zeilik, and the 16 solar mass value for Cyg X-1. b. (15 pt): Matter falls from a large distance onto the black hole emitting 5% of its gravitational potential energy as X-rays. What rate of mass accretion (i.e. kg/sec) is necessary to emit with the X-ray luminosity observed for Cygnus X-1? Let a small amount of matter ∆m fall into the black hole in time ∆t. The amount of ∆[]()∆ ⁄ × energy radiated into X-rays in this time is just LtGM= BH m RBH 0.05 so we can write ∆ R m L × BH ()⁄2 -------∆ - ==---------- ----------------- 40L c t 0.05 GMBH 30 (Notice how much drops out in this equation!) With L=2×10 W (Zeilik Tab. 18-4 or lecture notes), this gives ∆m14 –8 --------9==×10 kg/sec 1.4×10 Solar Mass/year ∆t c. (5 pt): Is this consistent with the expected mass loss rate for the visible companion star observed in Cygnus X-1? (You might want to use the converstion that one solar mass per year 22 equals 6.3×10 kg/sec.) Explain. The companion to Cygnus X-1 is an O9 supergiant. The mass loss in such massive −7 stars is on the order of 10 solar masses, or more, per year. (See Zeilik page 309, or class notes.) This can easily meet the requirements as the mass source falling onto the black hole to produce X-rays. Problem 2 (15 points): Consider two white dwarf stars. Star #1 is twice as massive as star #2. The stars have the same surface temperatures. Ignoring relativistic effects, what is the ratio of their luminosities? The two important relations you need are LR∝2T4 and 1 R∝ -------------- M13⁄ So you can write 24 23⁄ ⁄ L 1 R 1 T1 M2 123 ------ ====------ ------ -------- --- 0.63 L 2 R 2 T2 M1 2 since T1=T2. Problem 1 (25 points): Consider a star identical to our sun. a. (20 pt): Assume the star converts 10% of its hydrogen to helium while on the main sequence, and then arrives on the “helium main sequence” as a red subgiant, burning helium in its core. If the star then burns all of this helium to carbon, estimate its lifetime in this stage. The atomic weights of 4He and 12C are 4.0026 and 12.000 times the mass of hydrogen, respectively. Refer to figures in Zeilik for other information you may need. 29 The mass available for helium burning is 10% of the solar mass or 2×10 kg. 29 −27 55 This translates to a number N=2×10 /(4×1.67×10 )=3.0×10 helium nuclei. In helium burning, three 4He are converted to one 12C releasing −27 2 −12 (3×4.0026−12.000)×1.67×10 ×c = 1.17×10 J of energy. The total available energy for helium burning is therefore 55 −12 43 (3.0×10 / 3) × 1.17×10 = 1.17×10 J Zeilik Figure 16-5 shows that the star’s luminosity at this point in its life is 100 28 times the solar luminosity, or 3.9×10 J/sec. Therefore, the star’s lifetime in this stage is just 43 28 14 1.17×10 / 3.9×10 = 3.0×10 sec = 10 million years b. (5 pt): List the final three stages of this star’s life, which follow the core helium burning stage. The final three stages of this star’s life, after exhausting helium in the core, are • Asymptotic red giant / variable star • Planetary nebula • White dwarf Exam #2 79205 Astronomy Fall 1996 NAME: Solution Key You have two hours to complete this exam. There are a total of five problems and you are to solve all of them. Note that not all the problems are worth the same number of points. You may use your textbook (Zeilik), workbook (Hoff), and class notes and handouts, or other books. You may not share these resources with another student during the test. Indicate any figures or tables you use in your calculations. Show all Work! GOOD LUCK! Problem Score Worth 1. 25 2. 15 3. 25 4. 15 5. 20 Total Score: 100.