ASTR 100: Homework 1 SolutionsMcGaugh, Fall 20081. Problem 1-15. “Our solar system is bigger than some galaxies.”This statement makes no sense. A galaxy is composed of hundreds of billions of stars, whileour solar system is ‘just’ the stuff orbiting one star, our sun. Our solar system is a tiny partof the Milky Way Galaxy, being much smaller than it.2. Problem 1-41. Distance by light:The problem asks you to compute the distance that light can travel in different intervals oftime, measured in kilometers and miles.d = vt: distance = speed x time.In thi s case, the speed is the spe e d of light: v = c. According to Appendix A of the boo k,c = 3 × 105km/s. The exact number is c = 299, 792 km/s; it is OK to use either value.a) 1 light-second = (3 × 105km/s)(1 s).The seconds cancel out (see Appendix C), so 1 light-second = 300, 0 00 km.1 km = 0.62 miles, so (300,000 km)(0.62 miles/km) = 186,000 miles.That’s a long way to go in one second!b) Similarly, 1 light-minute = (3 × 105km/s)(1 minute).Be consistent in the use of units. 1 minute = 60 seconds, so1 light-minute = (3 × 105km/s)(60 s). Now the units again cancel nicely, and1 light-minute = 1.8 × 107km = 1.1 × 107miles.c) 1 light-hour = (3 × 105km/s)(1 hour).1 light-hour = (3 × 105km/s)(60 minutes/hour)(60 seconds/minute)1 light-hour = 1.08 × 109km = 6.7 × 108miles.d) 1 light-day. There are 24 hours in a day, so1 day = (2 4 hours) (60 minutes/hour)(6 0 seconds/minute) = 86,400 seconds.1 light-day = (3 × 105km/s)(86, 400 s), so1 light-day = 2.6 × 1010km = 1.6 × 1010miles.For a little perspective, the Voyager 1 spacecraft was launched 31 years ago, in 1977 . It hasbeen traveling away from the earth e ver since, and i s now the most remote man-made objectat a distance of 1.6 × 1010km. Light the sun emits today will pass Voyager tomorrow.3. Problem 1-42. Moonlight and Sunlight:This simply inverts the problem above: Since d = vt, t = d/v.a) Light travel time from the moon to earth:t =dmoonc. dmoon= 3.8 × 105km (Textbook Table E.3).t =3.8×105km3×105km/st = 1.3 seconds.b) Light travel time from the sun to earth:t =dsunc. dsun= 1.5 × 108km (Table E.2).t =3×105km/s1.5×108kmt = 500 s = 8 minutes, 20 seconds.14. Problem 2-24. “If the Earth’s orbit were a perfect circle, we would not have seasons.”This statement is clearly f a lse. A perfectly circular orbit would mean that the earth is alwaysthe same distance from the sun. However, the primary cause of our seasons is the ti lt of theearth’s spi n axis with respect to its orbital plane. Eliminating the variation in distance fromthe sun would not a ffect thi s pri mary attribute.5. Problem 2-28. The north celestial pole (i.e., the North Star) is at an altitude of 35◦aboveyour northern horizon. This tells you that (a) you are at latitude 35◦N.6. Problem 2-32. If the sun rises precise ly due east, (b) it must be the day of the equinox (eitherspring or fall).7. Problem 3-13. “The Yankees are t he best baseball team of all time.”On its face, this is the assertion of a partisan fan, not a scientific claim. However, can weevaluate this claim scientifically? Is there some me asure by which the Yankees (or some otherteam) could be evaluated? Baseball is a statistics rich sport; there are many well -recordedpossibilities. Which team has the be st overall won-lost record? Which team has won themost pennants? Which has won the most world s e ri e s? These are all questions that can beanswered clearly and quantitatively. The trick is agreeing which statistic qualifies a teamas the ‘best of all time.’ I t often happens that even when we agree on the facts, we maynot agree on what they m e an. Does a team qua lify as the best-ever if it mee ts one of thecriteria above? Only if it meets all three? Or is there some other measure that would be moreappropriate? The problem returns us to the phrasing of the assertion: what do we mean by‘best?’ If we can agree on that, and can define a quantitative way to measure it, then we canturn this assertion into something accessible to science. Agreeing on what we mean by ‘ best’is not simple, which is why scientis ts prefer more specific, limited language. “The Yankee shave been the most succ essful franchise in winning pennants” would be a more measured (andmeasurable!) statement.8. Extra credit 2-44. The Sun’s diameter.θ = L/D. In this case, we want to find the siz e of t he sun (L) from its observed angul a rdiameter (θ = 0.5◦) and distance (D = dsun= 1.5 × 108km). HenceL = θD. This formula hol ds only when θ is in radians, so first convert units: 1 radia n =180/π degrees.θ = 0.5◦= 0.5 degrees(π/180)( radians/degree)θ = 0.0087 radians. NowL = (8.7 × 10−3)(1.5 × 108km)L = 1.31 × 106km, very close to the actual value of 1.39 × 106km.Indeed, we only miss it by a little because we have approximated the angular size of the sunas half a degree. This is close, but not exact. We can invert the procedure to find the exactvalue:θ = L / D = (1 .39 × 106km)/(1.5 × 108km) = 0.0093 radians =
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