1	Astro 1050 Homework #1 Solutions given here on these pages.
2	Homework #1 1. What do you get when you multiply together 3 x 10³², 2 x 10⁵⁶, 0.5 x 10¹⁸, and 3 x 10¹⁰⁰ ? Many calculators won’t do this! Here is the power of scientific notation. Just multiply the leading numbers: 3 x 2 x 0.5 x 4 = 9. Add the exponents: 32 + 56 + 18 + 100 = 206. The product is therefore 9 x 10²⁰⁶.
3	Homework #1 2. If sunlight takes 8 minutes to travel from the Sun to the Earth, how long does moonlight take to travel from the Moon to the Earth? To find this, use information about the relative distance of the moon and the sun which you can find in your text. I’ve asked you to set this up in terms of light travel time. We know that the distance from the Earth to the sun is 1 AU, or about 150,000,000 km. The distance from the Earth to the moon is in appendix A table A-14, and is 384,000 km. 150 million divided by 384 thousand is 391. So, light will take 1/391 time to travel between the moon and Earth as it does from sun to Earth. So the travel time is 8min x 60 sec/min / 391 = 1.2 seconds.
4	Homework #1 3. What is the ratio of the mass of Earth to the mass of the moon? (Hint: look in appendix A of the text for these values.) Just take the ratio of masses. From table A-6, we have 6x10²⁴ kg/7x10²² kg. 10²⁴/10²² is 10² = 100. 6/7 is just under 1. So the answer is close to, but less than 100, without needing a calculator. Scientific notation makes it easy, and often in astronomy we’re not concerned with super precise answers. Remember that Earth is about 100 times more massive than the moon, not 81 times, or 60 times, or whatever. This approach lets you find the best multiple choice answer quickly.
5	Homework #1 4. How many suns would it take, laid edge to edge, to reach the nearest star? Basically, all you have to do here is divide the distance by the diameter of the sun. The distance is 4.28 light-years (table A-7). Convert to meters by multiplying by 9.46x10¹⁵m per light-year (table A-6). The diameter of the sun is 2 x radius = 14x10⁸m (table A-6). So, the number of suns is 4 x 10¹⁶ m/14 x 10⁸ m = 2.9 x 10⁷ m.
6	Homework #1 5. Is this class making you feel small, or large? 6. Everyone looked up the constellations Sagittarius and Scorpius in the text. This question encourages you to look at the figures, not just read the text.
7	Homework #1 7. One star is 2.512 times brighter than another. Therefore, in terms of magnitudes, its magnitude is what? An increase by a factor of 2.512 is a change of on magnitude. Magnitudes are “backwards” so an increase in brightness by 2.512 times means a decrease in magnitude by one.
8	Homework #1 8. If we have two stars in fictional constellations, and the two stars are named Alpha Buffalo and Beta Cowboy, which is brighter and by what factor? The apparent visual magnitude of Alpha Buffalo is 3.2 and the the apparent visual magnitude of Beta Cowboy is 2.7. Magnitudes are “backwards” so the star with the smaller magnitude (Beta Cowboy) must be brighter. By what factor? Use the equation I_A/I_B = 2.512 ^(m^B^-m^A⁾ = 2.512^-0.5 = 0.63 So I_B/I_A = 1/0.63 or about 1.6.
9	Homework #1 9. If you were kidnapped and transported to a distant location on Earth, but noticed that Polaris was 20 degrees above the horizon, what could you figure out about your whereabouts? c. The latitude of my location is 20 degrees north. Think about the celestial sphere and where Polaris would be in the sky if you were on the equator (0 degrees latitude) or on the North Pole (90 degrees latitude).
10	Homework #1 10. Which of the following sets of stars have ALL been the "north star" (or pretty nearly) at one time or another? c. Polaris, Vega, and Thuban. The precession figure in the textbook allows you to answer this one.