D. 1 degree (60 arcminutes / 1 degree) (60 arcseconds / 1 arcminute) = 3,600 arcseconds

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1 ASTRONOMY 101 SECTION 2 EXAM 1 1. How many arcseconds are in a degree? A B C. 60 D. 3,600 D. 1 degree (60 arcminutes / 1 degree) (60 arcseconds / 1 arcminute) = 3,600 arcseconds 2. The sun is approximately how many Earth s across? A. 1 B. 10 C. 100 D. 1,000 C. From class notes on size and scale. 3. You watch a bright star rotate 90 degrees around the south celestial pole. How long have you been watching this star? A. 3 hours B. 6 hours C. 9 hours D. 12 hours B. 90 degrees (24 hours / 360 degrees) = 6 hours 4. All stars are circumpolar: A. On the equator B. At the north pole C. At the south pole D. B & C E. A & B & C D. Because Earth rotates, the stars appear to move around the north celestial pole in the northern hemisphere and the south celestial pole in the southern hemisphere. When standing at the north pole, the north celestial pole is directly overhead and when standing at the south pole the south celestial pole is directly overhead. Hence, at the north and south poles all of the stars appear to move around the point directly overhead, meaning that they do not rise or set, meaning that they are circumpolar. When standing on the equator, the north celestial pole is on the northern horizon and the south celestial pole is on the southern horizon. Hence, on the equator all of the stars appear to move around these two points on opposite horizons, meaning that all of the stars rise in the eastern sky and all of the stars set in the western sky, meaning that no star is circumpolar.

2 5. After how many months are constellations that were setting at midnight now rising at midnight? A. 3 B. 6 C. 9 D. 12 B. Constellations that were setting but are now rising at any fixed time have shifted 180 degrees in the sky: 180 degrees (12 months / 360 degrees) = 6 months 6. It is hotter in summer and colder in winter because: A. Earth is closer to the sun in summer. B. Earth s orbit around the sun is elliptical. C. Your hemisphere of Earth is tipped toward the sun in summer. D. A & B E. A & B & C C. The seasons are the result of how directly the sun s light strikes Earth: It strikes most directly the hemisphere that is tipped toward the sun, causing summer there, and least directly the hemisphere that is tipped away from the sun, causing winter there. 7. When it is winter in the northern hemisphere, it is what in the southern hemisphere? A. Winter B. Spring C. Summer D. Fall C. If it is winter in the northern hemisphere, the northern hemisphere must be tipped away from the sun, meaning that the southern hemisphere must be tipped toward the sun, meaning that it must be summer in the southern hemisphere. 8. The day is now: A. 6.5 hours long B. 9 hours long C hours long D. 14 hours long E hours long C. You are supposed to be keeping track of this, at least approximately. 9. The moon revolves around Earth once every 27.3 days. The moon rotates once every: A days B days C days D. The moon does not rotate. A. The moon rotates and revolves at the same rate, due to tidal locking. Otherwise, the side of the moon that faces Earth would change with time.

3 10. Which of the following statements are true? A. When lunar eclipses occur, the moon is new. B. When solar eclipses occur, the moon is new. C. When the moon is new, lunar eclipses occur. D. When the moon is new, solar eclipses occur. B. When solar eclipses occur, the moon is directly between the sun and Earth and consequently its phase must be new. However, solar eclipses do not occur every time the moon is new because the Earth-Moon plane is tipped, by about 5 degrees, with respect to the Sun-Earth plane, meaning that the moon, although new, is not usually directly in front of the sun, but slightly above or below it. 11. After how many eclipse years does the sequence of eclipses repeat? A. 0.5 B. 1 C. 18 D. 19 D. 19 eclipse years is very close to an integer number of lunar months, meaning that the sequence of eclipses will repeat every 19 eclipses years. (Since the line of nodes regresses, an eclipse year is shorter than a regular year.) 12. Perfected the heliocentric model of the universe: A. Aristarchus B. Copernicus C. Galileo D. Kepler E. Ptolemy D. Kepler perfected the heliocentric model with his three empirical laws, which were later explained on physical grounds by Newton. 13. Not discovered by Galileo: A. Lunar features B. Sunspots C. Jovian moons D. Venusian phases E. Stellar parallaxes E. Stellar parallaxes, which prove that Earth is moving around the sun, were not measured until the mid-19 th century.

4 14. Every year, a star is observed to move 0.1 arcseconds with respect to background stars for six months, followed by a return to its original position over the course of the next six months. How far away is this star? A. 1,100 AU B. 69,000 AU C. 2,100,000 AU D. 4,100,000 AU D. distance = (baseline / 2 (360 degrees / angular shift. baseline = 2 AU. angular shift = 0.1 arcseconds (1 arcminute / 60 arcseconds) (1 degree / 60 arcminutes) = degrees. distance = (2 AU / 2 degrees degrees = AU 15. A new planet is observed to orbit the sun once every 1,000 years. What is the semi-major axis of its orbit? A. 100 AU B. 680 AU C. 4,600 AU D. 32,000 AU A. a = (P / 1 year) 2/3 AU = (1,000 years / 1 year) 2/3 AU = 100 AU 16. We put a satellite into a circular orbit around Earth at half the distance to the moon. What is the period of its orbit? A lunar months B lunar months C. 1.6 lunar months D. 2.8 lunar months A. P = (a / 1 Earth-Moon distances) 3/2 lunar months = (0.5 Earth-Moon distances / 1 Earth-moon distance) 3/2 lunar months = 0.35 lunar months 17. The frequency of 1-m radio waves is: A Hz B Hz C Hz D Hz B. = c / = ( m/s) / (1 m) = s -1 = Hz 18. The peak wavelength of a 10,000 K star is: A nm B. 290 nm C. 2.9 m D. 290 m peak = 2,900 nm / (T / 1,000 K) = 2,900 nm / (10,000 K / 1,000 K) = 290 nm

5 19. With the human eye, a 10,000 K star is: A. Too red to see B. Red C. Blue D. Too blue to see C. Although the peak wavelength is too blue to see with the human eye, the spectrum is broad and extends well past the visible band. Consequently, the star is visible to the naked eye and since more blue light is emitted than red, the star is blue in color. 20. Oven A is twice as hot as oven B, yet otherwise identical. How many times more energy does oven A emit than oven B? A. 2 B. 4 C. 8 D. 16 D. Since energy flux is proportional to T 4, energy flux A / energy flux B = (T A / T B) 4 = 2 4 = Light from a star passes though a cold hydrogen cloud on its way to Earth. We receive a: A. Lyman series absorption spectrum B. Balmer series absorption spectrum C. Lyman series emission spectrum D. Balmer series emission spectrum A. By Kirchhoff s Third Law, we receive an absorption spectrum. Since the cloud is cold, the hydrogen atoms begin in the ground state, resulting in a Lyman series absorption spectrum. 22. Aliens observe at this cloud from a completely different direction. They receive a: A. Lyman series absorption spectrum B. Balmer series absorption spectrum C. Lyman series emission spectrum D. Balmer series emission spectrum C. By Kirchhoff s Second Law, the absorbed light is then re-emitted, but in all directions, resulting in a Lyman series emission spectrum in every other direction.

6 23. A hydrogen atom is in an excited state! Its electron decays from the fourth excited state to the second excited state. What type of photon is released? A. Lyman alpha B. Balmer beta C. Balmer delta D. Paschen beta E. Paschen delta D. All transitions down to the second excited state result in Paschen series photons. The first transition in the series (third excited state to second excited state) is called Paschen alpha. The second transition in the series (fourth excited state to second excited state) is called Paschen beta. 24. Lyman-alpha light has a wavelength of 91 nm. A galaxy, moving away from us at a speed of 300 km/s, emits Lyman-alpha light. We detect this Lyman-alpha light at wavelengths that are: A. Shorter than 91 nm B. Equal to 91 nm C. Longer than 91 nm D. None of the above C. Since the galaxy is moving away from us, its light is redshifted, meaning that we detect the emitted light at longer wavelengths. 25. What is the change in wavelength of this light? A. 0 nm B nm C nm D. 91 nm C. = emitted (v / c) = 91 nm (300 km/s / km/s) = nm