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Download Astronomy 114 Problem Set # 7 Due: 30 Apr 2007 SOLUTIONS 1
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Astronomy 114 Due: Problem Set # 7 30 Apr 2007 SOLUTIONS 1 What is the smallest angle (in arc seconds) that could theoretically be resolved by the 10-m Keck telescope at a wavelength of 500 nm? The angular resolution of a telescope with diameter D, at some wavelength λ satisfies: λ radians θ = 1.22 D If λ = 500 nm, and for Keck, D = 10 m, we have θ = 1.22 5 × 10−1 m 500nm = 1.22× = 6.1×10−8 10m 10m radians = 0.0126 arcseconds Unfortunately, due to atmospheric seeing, the angular resolution is limited to a fraction of an arc second at best. Especially for ground-based telescopes, the main goal is collecting photons! 2 How big would a radio telescope observing at 20 cm wavelength have to be in order to resolve the same angle as the Keck telescope in the last problem? Since 20 cm is in radio wavelength, we expect a very large diameter of the telescope in order to resolve small angle, such as 0.0126 arc second. Use the equation as previous question, we have θ = 1.22 λ D radians, which can be used to calculate the diameter of the telescope by solving for D given θ. That is: 0.2m λ = 1.22 × = 4 × 106 meters −8 θ 6.1 × 10 radians = 4 × 103 km. D = 1.22 This is a large fraction of the Earth’s surface and certainly much larger than a common optical telescope, say, Keck. So radio astronomers use an interferometer. This consists of widely separated individual antennas whose signals are later combined to resolve such small angles. 3 The Sun orbits the center of the Galaxy at the speed of 220 km/s, 8500 pc from the center. What is the mass enclosed by the Sun’s orbit? Since the mass distribution of the Galaxy is roughly spherical symmetric, the Sun’s motion is determined by the mass enclosed by its orbit. Therefore The centrifugal force should be roughly balanced by the gravitational pull. Let M(r0 ) denote the mass enclosed, we have Fcentri = 2 M(r0 )M⊙ M(r0 )Vcirc =G = Fgrav . r0 r02 Thus the enclosed mass M(r0 ) = = = = = 2 r0 Vcirc G (8500pc) × (220km/s)2 G (2.6 × 1022 cm) × (2.2 × 107 cm/s)2 6.67 × 10−8 cm3 g −1s−2 1.89 × 1044 g 9.5 × 1010 M⊙ ≈ 1011 M⊙ (1) 4 The rotation curve for the Sa galaxy NGC 4378 is shown in Figure 26-28 in your book. Using the data from that graph, calculate the orbital period of stars 20 kpc from the galaxy center. What is the mass of the galaxy out to 20 kpc from its center? We can read from the plot that for NGC 4378, the circular velocity at 20 kpc is roughly 280 km/s. Therefore the period of stars at 20 kpc from the center takes 2 × π × 20kpc 2πR = Vcirc 280km/s 2 × π × 20 × 1000 × 3.09 × 1013 km = 280km/s 16 = 1.4 × 10 seconds 8 = 4.4 × 10 yrs P = to complete one entire orbit. ] When stars orbit about the center of a galaxy, their centrifugal force should be roughly balanced by the gravitational force by the galaxy. We can approximate the galaxy’s mass distribution as spherically symmetric, so the gravitational pull at some radius R is induced by the mass within R, we denote it M(R). For the centrifugal force, we have Fcent = 2 mstar Vcirc , R for gravitational force, we have Fgrav = GM(R)mstar . R2 Equal these two forces, we have M(R = 20kpc) = 2 Vcirc ×R = 2.6 × 1041 G kg = 3.65 × 1011 M⊙ [Alternatively, we could use the M(r) relationship from Newton’s Laws that we derived earlier.] This mass is much larger than determined from the luminous matter, so there should be a lot of invisible “dark” mass in the NGC 4378. 5 What types of galaxies are most likely to have new star forming? List observational evidence to support your answer. Irregular galaxies are possible sites for new star formation. There are often knots of star forming region, indicated by OB association and HII regions, seen throughout the irregular galaxies. Spiral galaxies are also sites of new star formation. The observational evidence for this includes HII regions in spiral arms, bluer color of stellar light on the leading edges of spiral arms, and the overall bluer color of spiral galaxies. By contrast, elliptical galaxies show little or no star formation and low relative gas fraction. Elliptical galaxies would be the worst place to look for new star formation. 6 Freedman & Kaufmann Chap. 26, Problem 28, pg. 605 The light collecting power of a telescope is proportional to its area; and area is proportional to the sqaure of its diameter. So by using larger telescope we can dramatically increase the ability to see very remote (dimmer) galaxies. Note that Hubble law is a function of distance, so the farther we observe, the better accuracy we can gain in the Hubble law. 7 Freedman & Kaufmann Chap. 26, Problem 29, pg. 605 The image shows very evident bright regions that may be sites of OB associations and HII regions (note the Hydrogen α color), which indicate the recent star formation. The inner part of the galaxy is rather blue, indicating a young population of stars. 8 Freedman & Kaufmann Chap. 26, Problem 30, pg. 605 The distance of galaxies are often measured using standard candles. However, these methods must be calibrated by accurately determining their properties from very nearby standard candles. One example is the Cepheid Variable that we often use to determine the distance of galaxies. By accurately measuring the parallax of nearby Cepheid Variables we can then determine the correlation between their luminosity and luminosity variability period. Using this as a standard candle, we can measure the distance of galaxies by observing the variability of the Cepheid variables in them. A new, improved calibration of the Cepheid period–luminosity relation shifted one of the “rungs” in the distance ladder and caused the revision of all the larger distance estimates that depended on Cepheids.