1) Provide approximate answers for the following questions about the appearance of the Moon.

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1 Astronomy 2110 Fall 2018 Exam 2 October 26, 2018 Part 1: Short qualitative/quantitative questions. Don't over-think these. Answers should be simple and straightforward. If you are spending more than two or three minutes on any one question you are probably over-doing it. Answer 5 of the 6 questions in this category. You receive no bonus or partial credit by doing more than five. If you start work on more than five then clearly indicate which five you want graded (otherwise we will grade the first five you attempted). Each question is worth 10 points. 1) Provide approximate answers for the following questions about the appearance of the Moon. Phase of the Moon that transits the Meridian at sunset: First Quarter Rising time of the Last (3 rd ) Quarter Moon: Midnight Percentage of the lunar surface visible from Earth: about 60% Declination of the First Quarter Moon in September: around -20 (occupies January Sun position) Number of hours between successive Meridian crossings of the Moon: about 25 (the moon takes around 30 days to orbit Earth. One day of orbit represents about 1/30 th of 24 hours, so about an extra hour) 2) Explain, using only the motivating words and concepts, where the concept of Jean s Mass comes from and what it has to do with star/solar system formation? The Jean s mass is the mass contained inside a Jean s radius/volume. The Jean s radius is the radius at which the gravitational free-fall time to the center is equivalent to the sound-speed propagation time to the center and thus defines the natural diameter for cloud instability agains gravitational collapse. For typical interstellar parameters the Jean s mass is about 0.2 solar masses, consistent with collapse of Jean s volumes to make stars (and associated planetary systems). Pretty cool. 3) What are the advantages of building larger and larger telescopes? Light collecting area increases as diameter squared and translates to sensitivity to fainter objects. Angular resolution improves as 1/D bigger telescopes see more detail. 4) How would the number of eclipses per year and their duration be different (or the same!) if the Moon s orbit were inclined 20 degrees to the ecliptic. Answer qualitatively. No calculation is required. Also, why have more people seen total lunar eclipses than have seen total solar eclipses? If the Moon s orbit was not inclined at all eclipses would be frequent, happening every new and full moon. A little bit of tip (the current situation) limits things to a few eclipse opportunities every 6 months when the Moon crosses the plane of the Earth s orbit. Even more inclination would make for even fewer opportunities. The duration would be the same because duration is largely determined by the moon s orbital speed relative to the Earth s diameter / Earth s umbral shadow diameter.

2 5) How many photons do you need to collect to reach a signal-to-noise ratio (SNR) of 100 if, during the observation, you also collect 5000 photons of background radiation? SNR = signal/noise = N / sqrt(n+b), so 100*100 = 10,000 = N 2 / N+5000 N 2 10,000*N 50,000,000 = 0 solving the quadratic gives 14,000 photons. 6) What is common about the derivation of the Bohr equation and the fundamental properties of planetary orbits? What is fundamentally different? Both rely on equating attraction (gravitational for planets, electrostatic for Bohr) to centripetal acceleration to find the velocity for an object (planet or electron) in orbit at a distance, r. A planet can orbit at any distance, but the Bohr atom has an additional constraint that the angular moment of the orbit must be an integer multiple, n, of h/2p. Part II: Do 2 of 3 of the following 20 point questions. Once again, you only get credit for two. If you try all three and don't indicate which are to be graded we will grade the first two. 7. Above is a spectrum of the star Vega. You can see the continuum emission from the star and the different absorption lines. The central wavelength of each absorption line is given in the black box right next to it.

3 a) Identify and mark Balmer-alpha and Balmer-gamma lines in this plot. Show your calculation for these wavelengths. λ =91.2nm ( 1 (lower) 2 1 2) 1 (upper) For Balmer transitions lower=2. Alpha is a transition from a level one higher, gamma from a level 3 higher. Using lower=2 and upper=3 one gets a wavelength of 656 nm. 5 and 2 gives 434 nm. Both of these absorptions appear and are labeled in the plot above. b) Estimate the peak wavelength of the continuum emission. What is the temperature of the star implied by this emission peak? What color does this correspond to in the visible spectrum? The peak is around 410 nm = 0.41 microns. Wien s law says T = 2900 um / lambda(peak) so T = 2900/0.41 K or 7000K. c) Vega is at a distance of 8 parsecs from Earth. What is the flux from Vega received at Earth given that the radius of Vega is three times the radius of the Sun, and given the temperature you derived above (assume a temperature for the Sun of 6000K). Flux = Luminosity / 4pD 2 D = distance = 8 pc = 8 * AU = 1.6x10 6 AU * 1.5x10 11 m/au = 2x10 17 m. Luminosity = 4pR 2 st 4 T=7000K from above R = 3 * 7x10 8 m so Flux = R 2 st 4 / D 2 = (2x10 9 ) 2 * 5.7x10-8 * (7000) 4 / (2x10 17 ) 2 = 4x10 18 * 6x10-8 * 7 4 * / 4x10 34 = 24x10 22 * 2500 / 4x10 34 noting that 7 squared is roughly 50 so 7 4 is about 2500 = 6x10-12 * 2.5x10 3 = 15 x 10-9 = 1.5x10-8 W/m 8. You have been asked to design a multi-telescope space interferometer that will work at a wavelength of 400 nm with the aim of collecting resolved images of Earth-sized planets around nearby stars. The target stars are all those within 5 parsecs of the Sun. What spacing between telescopes would you require so that you can see several (say 10) Airy disks across a planet the size of the Earth at 5 parsecs distance (i.e. the Airy disk corresponds to 1/10th of an Earth diameter at 5 pc distance)? The Earth is roughly 12,000 km in diameter, so we need to build a telescope with angular resolution capable of seeing 1/10th of this, or 1,200 km, at a distance of 5 parsecs. So in simple terms l/d = 1200 km / (kilometers in 5 parsecs) with l = 4x10-7 meters

4 D (in meters) = 4x10-7 meters * { (kilometers in 5 parsecs) / 1200 km } 5 parsecs is 10 6 AU (5 x 200,000) and there are 1.5x10 8 kilometers in an AU so (kilometers in 5 parsecs) = 1.5x10 14 km. D = 4x10-7 meters * (1.5x10 14 / 1.2x10 3 ) = 4x10 4 meters = 40 kilometers 9. A new extrasolar planet candidate is discovered orbiting one arcsecond away from a one solar mass star that is 25 parsecs from Earth. The starlight shows zero Dopper shift but spectroscopic observations of the planet (well beyond current technology by the way) reveals that the nm hydrogen line appears at a wavelength of nm. Evaluate the validity of the claim that this object is a planet in orbit around the neighboring star. A planet 1 away from a star that is 25 parsecs away is 25 AU from the star. Give the star s has mass equal to the Sun, planetary orbital speed is 1 30km/ s. So 6 km/s ( R( AU )) The Doppler shift is 1 nm so a little less than one part in /600th the speed of light is 500 km/s. If the object has a radial velocity of 500 km/s relative to the star it is certainly not a planet orbiting 25AU from the star.

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