= 4,462K T eff (B) =

Transcription

1 Homework 1 Solutions Problem 1: Star A emits most of its light in the orange, Star B in the gre en and Star C in the blue color range. What wavelengths are these most likely to be, and what effective temperature would each have? Answer: For Star A, orange light is near 650 nm; for Star B, green light is near 550 nm; and for Star C, blue light is near 450 nm. In general, T eff = /λ for λ in nm. T eff (A) = = 4,462K T eff (B) = = 5,273K T eff (C) = = 6,444K Problem 2: If Star 1 is a Sun-like with T eff = 5800K, Star 2 is a B-type star with T eff = 28000K and R = 7R, and Star 3 is a K-type star with T eff = 4900K and R = 0.8R, determine how much energy (in Watts) each star radiates in Johnson U, V, and I filters. For the U filter use a central wavelength of 350 nm with a width of 100 nm; for the V filter use a central wavelength of 520 nm with a width of 100 nm; and for the I filter, use a central wavelength of 800 nm with a width of 175 nm. (Use the formula on slide 20 of lecture 2.) (18 points) Answer: Use the following equation: L λ ( λ) = 8π2 R 2 hc 2 1 λ 5 e ( λ λκt) hc 1 Be sure you have converted all units to MKS! Here is an example: the first calculation for Star 1 for the U filter. L 350nm ( λ) S1 = 8π2 ( ) 2 ( )( ) 2 1 ( ) 5 [ e ( )( ) ( )( )5800 L 350nm ( λ) S1 = ( )( )( ) = W ] 1 ( ) 1

2 Star U V I Problem 3: Calculate the colors of the stars in Problem 2 in a simplified way. Take the root to the base of the luminosities above and then determine the U-V and V-I indices for all three stars. Which one is bluest? Which one is the most red? (14 points) Answer: Okay, my directions screwed up this question. Because of the backwards nature of magnitudes (brighter stars having smaller magnitudes), my suggested way won t work. It has to be done using the formula given in the lecture: m U m V = log ( LV L U using the following log conversion to get to base log A (X) = log 10(X) log 10 (A) Here are the numerical values of the pseudo-magnitudes which I directed you to do Star U V I which give the following colors Star U-V V-I Immediately, you notice that the bluest star has positive values while the reddest star has negative values, which is exactly backwards to how it should be. 2 )

3 Using the proper equation given above, you would arrive at the correct values Star U-V V-I Which truly shows the expected results, the hottest star, Star 2 is the bluest, and the coolest star, Star 3, is the reddest. So please come away with the idea that in real measurements, U V is negative for blue stars and positive for red ones. Problem 4: Star H has m V = 6.3 and Star J has m V = 11.3, which star is brighter and what is the brightness ratio? Answer: Star J is brighter (smaller magnitude) by 100 times (the magnitude difference is 5, which is a factor of 100). L ap (H) L ap (J) = 100(m J m H )/5 = 100 Problem 5: Star K is at 824 pc from us and has an apparent magnitude of What is its absolute magnitude? Answer: Use the distance modulus, solved for M. m M = 5log ( ) d M = log 10pc ( ) 824pc = pc Problem 6: What would be the angle of parallax for Star K? If the Gaia spacecraft can measure parallaxes with 1 micro-arc second precision, can it measure Star K s distance? Answer: Use the parallax formula, d = 1/p for d in parsecs and p in arcseconds. For Star K: p = 1/d = 1/824pc = Can Gaia measure this? Yes, Star K s parallax is greater than 1 microarcsecond. BTW: The distance corresponding to 1 micro-arc second (which is the precision of the Gaia spacecraft) is: 3

4 d = 1/p = 1/ = pc = 1Mpc. Problem 7: Supernova 1987a, in the LMC has a distance modulus of How far away is Sn1987A in parsecs and in what year did the supernova actually occur for the light to reach us in 1987? Answer: Use the distance modulus for part 1 and then either use t = d/c or convert pc to ly for part 2. d = 10 (m M+5)/5 = 10 ( )/5 = 52,000pc = 169,000ly So SN 1987A occured 169,000 years ago, which would have been ,000+1 = B.C. (very roughly). Problem 8: The diameter of Betelgeuse is 700 times that of the Sun. At its distance of 197 pc, what is the angle across its diameter? (How big is it, in arcseconds?) Answer: From Lecture 2 Slide 21: R = α d (In fact we did this exact 2 problem in class!) and so α = 2R where R and d must be in the same units d (so they cancel out). α = = rads = Problem 9: The closest star to us, Proxima Centauri, has a planet, Proxima Centauri b. The orbit of Proxima Centauri b is days. How many orbits ago are we seeing Proxima Centauri b when we look at it from Earth? Answer: It takes light 1 year to travel 1 light year. So convert the distance to light years and that s how long ago we are seeing the star. Divide that by its orbital period and you have the number of orbits ago we are seeing it. Google says Proxima Centauri is light year away. Convert the orbital period to years: days/ = years / = orbits ago. Problem 10: The distance to a visual binary is 13pc. Assuming the binary is seen edge on, determine the masses of both stars using the following information: The maximum angular separation is 0.024, the ratio of the 4

5 angular separations is SA/SB=3.2 and the orbital period is days. What are the masses of stars A and B in this binary? Be very careful about units! (i.e. use radians, which are unitless.) Answer: From Lecture 2 Slide 32: M 1 + m 2 = 4π2 G ( ) 3 d α 3 cosi P 2 and from the ratio of the angular separations, we have the ratio of the masses (the more massive star moves less; this is from center-of-mass). Since SA/SB=3.2 M B M A = 3.2 M B = 3.2M A As Matt noted in class, there is a relationship between and pc that can be used to simplify this. From this last equation: (m 1 +m 2 ) = (m 1 +m 2 ) = ( ) 3 dpc αarcsec 3 cosi Pyr 2 ( ) (106/365.25) 2 = 0.36M Or for the top formula, we have to convert everything to MKS: = rads, d = m, P = seconds. M 1 + m 2 = 4π ( ) [( )( )] rads 3600 /deg 180 o /rad (106.0d 86400sec/day) 2 = ( (( )( ) = kg = 0.36M Since we know M B = 3.2M A, we can write M A + 3.2M A = 0.36 M A = 0.086M and M B = 0.274M. These would be pretty low mass stars! Problem 11: Analysis of an eclipsing, double-lined spectroscopic binary having a period of P=8.6yr shows that the maximum Doppler shift of the hydrogen Balmer H α (656.28nm) line is 0.072nm for Star A and nm for Star B. The orbits are nearly circular and the eclipse tells us that the orbital inclination is near 90o (edge-on). Determine the velocities (in km/s), orbital radii (semimajor axes; convert to AU), and masses (in solar units) of each star. 5

6 Answer: Start with velocities: We are given Doppler shifts. λ λ = v c v = c λ λ v A = 0.072nm nm km/s = 32.9km/s. v B = nm nm km/s = 3.1km/s. The semimajor axis are related to the circumference: d = 2πr where d is the distance the star travelled during one orbit: vt a A = v AP 2π = 32,900m/s 8.6yr s/yr 2π = m = 9.47AU a B = v AP 2π = 3,100m/s 8.6yr s/yr = m = 0.98AU 2π The semimajor axis of the system is then = 10.46AU Now for the masses: Remember that the ratio of the masses is the inverse ratio of their velocities (or Doppler shifts): M B = v A = λ A = 10.6 m A v B λ B or M B = 10.6 m A Now for the total mass: First the short (P 2 /a 3 )way: (M B +m A ) = a3 AU P 2 yr = = 15.47M then the long way (Lecture 2 Slide 40): For the individual masses: (M B +m A ) = P 2πG (v A +v B ) 3 = (8.6yr s/yr) 2π (32, ,100) 3 = = kg = 15.1M 10.6m A + m A = m A = 1.3M and M B = 14.1M 6