Astronomy 150 K. Nordsieck Spring 2000 Exam 1 Solutions True or False (Circle T or F) 1. ( T F ) In Madison the North Star, Polaris, is situated almost exactly at the zenith. False. Polaris is near the North Celestial Pole. The zenith is straight up; The NCP is up only at the North Pole! 2. ( T F ) Viewed from above the North Pole, the Earth rotates counterclockwise. True. Standing on the north pole, the earth rotates from right to left, causing the stars to appear to move from left to right. 3. ( T F ) During ionization an atom becomes a different element. False. In ionization, electrons are lost. The element depends on the number of protons in the nucleus. 4. ( T F ) A certain absorption line occurs in the sun at a wavelength of 500 nm. If the same line occurred in a star half as hot as the sun (and not moving with respect to the sun) it would be at 1000 nm. False. The wavelength of an absorption line for an atom depends only on the energies of the energy states of the absorbing atom, which does not depend on temperature. It is true that in a star half as hot as the Sun, the continuous radiation which gives it its color has its peak at twice the wavelength, but the positions of lines coming from a particular atom do not move (although some lines of certain atoms might disappear and others from different atoms might appear). 5. ( T F ) A photon of red light with a wavelength of 640 nm has twice as much energy as an ultraviolet photon at 320 nm. False. photon energy = h frequency = h c / wavelength. The 640 nm photon has 1/2 the energy.
Multiple Choice (circle A, B, C, or D) 6. 1 arc minute is A. 1/60 arc second B. 1/60 of a full circle C. 1/60 degree D. 60 degrees 7. The two ranges of light for which the Earth's atmosphere is reasonably transparent, are A. visible and far infrared radiation C. X-rays and visible radiation B. visible and radio radiation D. UV and radio waves 8. Two stars that have the same surface temperature have the same A. mass B. luminosity C. radius D. spectral type 9. The luminosity of a star depends upon radius and temperature. If a star had the same luminosity as the sun, but had a temperature of 10,800 K, how would this star's radius compare with the solar radius? (T sun = 5400 K) A. 2 times larger D. 4 times larger B. 4 times smaller E. the same C. 2 times smaller L/L sun = (R/R sun ) 2 (T/T sun ) 4 = 1 (R/R sun ) 2 = 1/ (T/T sun ) 4 = 1/2 4 = 1/16 R/R sun = 1/4 10. The size of a red giant is about A. 1 light year B. 1 solar radius C. the radius of the earth's orbit
Short Answers 11. Draw a starmap like the one you drew for the observing lab, putting in the compass directions, the zenith, and the observer's meridian. Now imagine that you are on the Earth's equator and sketch how the stars move in the sky. Mark in the celestial poles. 12. A big advantage of orbiting infrared telescopes is that they can be cooled. Explain why this helps, using Wien's Law for blackbody radiation. Infrared telescopes are observing at the wavelength where blackbodies shine at room temperature (300 K): (max) = 540 nm 5400 K / (300 K) = 10,000 nm The telescope structure is opaque and shines like a blackbody. To get rid of this, the only thing one can do is cool the whole telescope: Since the blackbody luminosity is proportional to T 4, just decreasing it by a factor of 2 (to 150 K) reduces the problem by 16! 13. An inhabitant of a planet around Sirius A observes the apparent angular motion of the Sun relative to the distant stars. Recall that Sirius A is in an orbit with a radius of 13 AU and a period of 80 years around the system center of mass, and the distance to the Sun is 2.6 pc. Sketch the apparent motion of the Sun., giving the scale in arcsec; support this with a figure. This is just the same as heliocentric parallax from the Earth, with the Earth's orbit around the Sun (1AU size, 1 year period) replaced by the orbit of Sirius A around the system center of mass (13 AU size, 80 year period (I am ignoring the presumed orbit of the hypothetical planet around Sirius A). The parallax triangle looks like this:
So the Sun will appear to move in a circle with a radius of 5 arcsec and a period of 80 years. If we would include the (unknown) orbit of the planet around Sirius A, there would also be a "loopthe loop" with a radius of (orbit size)/2.6 arcsec. 14. Vega (apparent magnitude 0) has a distance of 25 pc and a luminosity 50 times that of the Sun. How far away can we see a star with the same luminosity as the Sun with the naked eye (faintest apparent magnitude 5)? The question is, How far away would the Sun have be to be apparent magnitude 5? The apparent magnitude gives us the flux ratio with Vega. We have the luminosity ratio, so use the flux- luminosity- distance relationship to get the distance: Want F(Star) / F(Vega) = 10 -m(star) / 2.5 = 10-2 But F(Star) = L(Star) / (4 dist(star) 2 ) F(Vega) = L(Vega) / (4 dist(vega) 2 ) F(Star) / F(Vega) = L(Star) / L(Vega) / (dist(star) /dist(vega)) 2 dist(star) / dist(vega) 2 = L(Star) / L(Vega) / (F(Star) / F(Vega)) = 1 / 50 10-2 = 2 dist(star) = 1.4 dist(vega) = 35 pc The Sun would be only just visible at a distance of a mere 35 pc! Note: I gave you the wrong distance for Vega! It is 8 pc, not 25. Re-doing the problem, the Sun can only be seen by the naked eye if it is closer than 11 pc. For those who happen know that the absolute magnitude of the Sun is 4.8 (recall that the absolute magnitude, a weird way of expressing luminosity, is the apparent magnitude an object would have at a distance of 10 pc), this makes more sense.
15. Spectral classification of stars depends on the fact that the strength of the absorption lines of each ion can depend strongly on surface temperature. Using an energy level diagram for neutral Hydrogen, explain why the Hydrogen Balmer lines are weak for O stars (20,000 K), strong for A stars (10,000 K), and weak again for G stars (5000 K), even though Hydrogen is the most common element in the atmospheres of all these stars. The Hydrogen Balmer absorption lines are formed when an H atom in the second energy level absorbs a photon. The strength of the lines depends not just on the amount of Hydrogen, but specifically on the number of H atoms that are in the second level. For cool stars (G stars), all the H is in the lowest level, since collisions with neighboring atoms are not energetic enough to raise them to the second level, and the lines are weak. For A stars (about 10,000 K), collisions are strong enough that many H atoms are in the second state, so the lines are strong. For O hot stars (> 20,000 K), the collisions are so strong that most of the H atoms are ionized, and not capable of absorbing photons at all. 16. Draw an H-R diagram, carefully labeling the axes (include numerical values of luminosity and temperature!). Sketch in the position of the Sun, the Main sequence, the White dwarfs, red giants and the supergiants. Accurately draw in lines of constant radius for 1 R sun,, 0.01 R sun, and 100 R sun.