Astronomy 101 Problem Set #1 Solutions -- Fall 2005 10/2/19, 8)18 AM Bucknell Unversity Astronomy 101 Problem Set #1 Solutions Problem #1: Some of the oldest Moon rocks found by the Apollo missions are estimated to be about 4.45 billion years old. Calculate their age in seconds, and express the result in scientific notation (Use the relation 1 year = 365.25 days). Solution: The tricky part about most problems usually lies in figuring out how to apply what you already know to the problem at hand. In this case, we know lots of conversions between units of time (how many seconds in a minute, for example), but we don't know the requested conversion. So, let's start by writing down the conversions we know: 1 year = 365.25 days (remember leap year?) 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds 1 class = 52 minutes The last conversion is certainly correct, but it probably won't help us solve the problem at hand. Now let's try to solve the problem in parts. We already know that 1 day = 24 hours, so how many hours are in 365.25 days ( = 1 year)? 1 day = 24 hours 365.25 days = 365.25 days x (24 hours / 1 day) = 8766 hours For the next step, we do exactly the same thing, only this time, we make use of our knowledge of the conversion from hours to minutes. If there are 60 minutes in 1 hour, how many minutes are in 8766 hours? 1 hour = 60 minutes 8766 hours = 8766 hours x (60 minutes / 1 hour) https://www.eg.bucknell.edu/physics/astronomy/astr101/prob_sets/ps1_soln.html Page 1 of 13 Astronomy 101 Problem Set #1 Solutions -- Fall 2005 10/2/19, 8)18 AM = 525,960 minutes The last step is yet another repeat of the same process, using our knowledge of the conversion from minutes to seconds. If there are 60 seconds in a minute, how many seconds are in 525,960 minutes? 1 minute = 60 seconds 525,960 minutes = 525,960 minutes x (60 seconds / 1 minute) = 31,557,600 seconds Now, to finish it off, we multiply the number of seconds in a year by the number of years in the age of the oldest Moon rocks. 4,450,000,000 years x 31,557,600 seconds/year = 140,430,000,000,000,000 seconds Now we have to express this answer in scientific notation. In order to do that, we need to express our answer as a number between 1 and 10 times a power of ten. To do that, we use the technique of moving the decimal point by noting that: 140,430,000,000,000,000 = 14,043,000,000,000,000 x 10 = 1,404,300,000,000,000 x 10 x 10 = 140,430,000,000,000 x 10 x 10 x 10 = 14,043,000,000,000 x 10 x 10 x 10 x 10 = 1,404,300,000,000 x 10 x 10 x 10 x 10 x 10 = 140,430,000,000 x 10 x 10 x 10 x 10 x 10 x 10 = 14,043,000,000 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,404,300,000 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 140,430,000 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 14,043,000 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,404,300 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 140,430 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 14,043 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,404.3 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 140.43 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 14.043 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1.4043 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 We can consolidate this number by noting that https://www.eg.bucknell.edu/physics/astronomy/astr101/prob_sets/ps1_soln.html Page 2 of 13 Astronomy 101 Problem Set #1 Solutions -- Fall 2005 10/2/19, 8)18 AM 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1017 and get for an answer 1.4043 x 1017 seconds Note that we moved the decimal point by seventeen places in our number and multiplied by 1017. This is an easy way to convert big numbers to scientific notation without having to write down all those 10 x 10 x ... factors. However, we're still not finished here. The last step of this part is to express our answer with the appropriate number of significant digits. Since you were given the age of the oldest Moon rocks as 4.45 x 109 years, and not 4.4 x 109 years or 4.4537809 x 109 years, we should assume that this number is only accurate to three significant digits. That is, the age of the oldest Moon rocks is somewhere between 4.445 x 109 years and 4.455 x 109 years old. Well, our translation of the age from years to seconds certainly hasn't added any accuracy to the result, so the answer should still be expressed with only three significant digits. The easiest way to do this is to just to truncate our decimal number from 1.4043 to 1.40. So, our final answer is 1.40 x 1017 seconds Problem #2: For many of you, the metric system is fairly unfamiliar. Since ALL of science (and most commerce outside the US) makes use of the metric system, it would be a really good idea for you to have some comfort with the units. With that in mind, let's consider a few familiar objects and activities. For each question below, estimate the answer, and express that answer in metric units. How tall are you? How far away is the LC from where you are right now (tell me where you are, too)? How fast do you walk? How fast do you drive on the highway? Solution: For this problem, you'll likely come up with different answers, since you're not all the same height as me, and many of you probably drive pretty fast on the highways. Therefore, the grading on this problem reflects more about how you did the calculations, rather than the actual result. I'm 5 feet 11 inches tall, or a total of 71 inches. To express this height in metric units, we'll https://www.eg.bucknell.edu/physics/astronomy/astr101/prob_sets/ps1_soln.html Page 3 of 13 Astronomy 101 Problem Set #6 Solutions -- Fall 2005 10/2/19, 8)23 AM Bucknell Unversity Astronomy 101 Problem Set #6 Solutions This Problem Set is due by 1:00 pm on Thursday, 13 October Problem #1: Pluto's moon Charon has a radius of 635 km, and a mass of 1.8 x 1021 kg. What is its average density? Based on this result, what kind of material do you think Charon is made of? (Hint: You can find the densities of various materials on the 30 September web page.) Solution: OK, we've got mass and radius, which is pretty much all we'll need to a density calculation. Recall that: density = mass/volume The volume of Charon is: volume = 4/3 pi r3 = 4/3 x 3.14 x (635 km)3 = 1.07 x 109 km3 Note that the units of the radius were cubed along with the value, so our volume has units of cubic kilometers. Now we can calculate the density: density = mass/volume density = 1.8 x 1021 kg / 1.07 x 109 km3 density = 1.67 x 1012 kg/km3 Now this is a perfectly valid answer with perfectly valid units. Unfortunately, it's not in the same units as the table I gave you on the 30 September web page, nor is it in the units given in the back of the book. If you're interested in comparing this density to the ones I gave you, you'll need to convert: https://www.eg.bucknell.edu/physics/astronomy/astr101/prob_sets/ps6_soln.html Page 1 of 4 Astronomy 101 Problem Set #6 Solutions -- Fall 2005 10/2/19, 8)23 AM 1.67 x 1012 kg/km3 x (1 km/1000 m)3 = 1.67 x 1012 x 1/109 kg/m3 = 1.67 x 103 kg/m3, or 1670 kg/m3 Note in this conversion I had to multiply by 1 km/1000 m three times to change the km3 units to m3.