l>Astronomy 101 Problem Set #6 Solutions — Fall 2005

Astronomy 101 Problem Set #6 Solutions |

This Problem Set is due by 1:00 pm on Thursday, 13 OctoberProblem #1: Pluto”s moon Charon has aradius of 635 km, and a mass of 1.8 x 1021 kg. What isits average density?Based on this result, what kind of material do you think Charon ismade of? (Hint: You can find the densities of various materials onthe 30 September web page.)Solution: OK, we”ve got mass and radius,which is pretty much all we”ll need to a density calculation. Recallthat:density = mass/volumeThe volume of Charon is:volume = 4/3 pi r3 = 4/3 x 3.14 x (635 km)3 =1.07 x 109 km3Note that the units of the radius were cubed along with the value, soour volume has units of cubic kilometers.Now we can calculate the density:density = mass/volumedensity = 1.8 x 1021 kg / 1.07 x 109km3density = 1.67 x 1012 kg/km3Now this is a perfectly valid answer with perfectly validunits. Unfortunately, it”s not in the same units as the table I gaveyou on the 30 September web page,nor is it in the units given in the back of the book. If you”reinterested in comparing this density to the ones I gave you, you”llneed to convert: 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/m3Note in this conversion I had to multiply by 1 km/1000 m threetimes to change the km3 units to m3.Now this density lies between that of rock and water, so it might bereasonable to assume that Charon is made from a mixture ofthese. Really, what the average density tells you is that it”s prettyunlikely that Charon has a large iron core, or that it”s entirely rockand denser materials.Charon is in fact an amalgam of rock and ice (it”s really far away fromthe Sun, so the water is in the form of ice).Problem #2: A 60-watt light bulb emits 60joules/sec of energy ( 1 Joule/sec == 1 watt). Pretend for a momentthat all of this energy is emitted in the form of photons withwavelength of 600 nm (this isn”t true, of course — as a blackbody, a light bulb emits photons of a wide range in wavelengths). Calculatehow many photons per second are emitted by such a light bulb.Solution: This problem asks you tocalculate how many photons per second are emitted from a 60-wattlightbulb. Since 60 watts is 60 Joules per second, we know that weneed enough photons to carry 60 Joules of energy each second. So how much energy does one 600nm photon carry? We”ll need to use therelation between energy and wavelength Ephoton = h c / lambda = (6.63 x 10-34 J s) x (3.00 x 108 m/s)/600 x 10-9 m = 3.32 x 10-19 JOK, so how many of these photons will we need to make 60 Joules?Let”s assume we need some number N of them. Then, 60 Joules = N x 3.32 x 10-19 Jand N = 60 J / 3.32 x 10-19 J = 1.8 x 1020So, in order to emit 60 Joules per second, the lightbulb must emit 1.8 x 1020 photons per second.

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(that”s 180,000,000,000,000,000,000 photons per second!)Problem #3: Read the ExecutiveSummary of the Department of Energy”s Report, Emissionsof Greenhouse Gases in the United States 2003 and answer thefollowing questions:a) How many metric tons of anthropogenic carbon dioxide were emittedfrom US sources in 2003?b) What is the percentage increase in the US-generated carbondioxide emissions since 1990?c) Which sector of the US economy (e.g., residential,commercial, industrial, or transportation) is responsible for thelargest fraction of 2003 carbon dioxide emissions?d) US emissions of which greenhouse gases actuallydecreased between 2002 and 2003?e) Based on the information in Table 3 of Chapter1 of this report, by what amount wouldwe have to reduce yearly anthropogenic carbon dioxide emissions toresult in no increase in carbon dioxide in our atmosphere?f) By what fraction of current worldwide human-made emissionslevels would we have to reduce to achieve this goal?Solution: This problem should have beenfairly straightforward, provided that you spent the time reading thereport. Skimming, on the other hand, likely didn”t work.a) 5870.2 million metric tons of gas (from the first sentencein the “Carbon Dioxide” section). Note that 6935.7 million metric tonslisted in the first sentence of the Executive Summary was the “carbondioxide equivalent,” which is a quantity that also includes theeffects of emissions from other greenhouse gases such as methane,nitrous oxide, and others.b) a 17.6 percent increase since 1990 (from the secondparagraph of the “Carbon Dioxide” section).c) commercial (from Figure ES3). I was surprised to learn thatindustrial emissions contributed so little in relative terms.d) nitrous oxide, hydrofluorocarbons (HFCs), perfluorocarbons(PFCs), and Sulfur Hexafluoride(SF6) (from firstparagraph in the “overview”).e) Table 3 of Chapter 1 (not Table 3 of the ExecutiveSummary) shows that the total human made carbon dioxideemissions total 23100 million metric tons. It also shows that thetotal natural emissions produce 770000 million metric tons and thatnatural absorption accounts for 781400 million metric tons. Given theimbalance between natural emission and absorption, we can add 781400- 770000 = 11400 million metric tons into the atmosphere and it willbe absorbed by the ecosystem. Thus, if we were to reduce our emissionsto that level, the carbon dioxide concentration would remain constant.How much would we need to reduce? We”re currently emitting 23100million metric tons, and the ecosystem can absorb 11400 million metrictons, so it looks like we”ll have to reduce by 23100 – 11400 = 11700million metric tons of carbon dioxide.f) The fraction is just the amount we have to reduce by overthe total amount we”re currently emitting:11700/23100 = 0.506, or 51%.That means we would have to halve our carbon dioxide emissions toreach the goal of not increasing the CO2 in theatmosphere. Perhaps this gives you a sense of the magnitude of thisproblem. It won”t be solved with energy efficient cars and lowerwinter thermostat settings. To really halve our CO2production will require a massive change in how we live.(One last note: the data in Table 3 you used for parts e) andf) is actually based on emissions in the 1990s; today, nearly adecade later, the problem is even more severe because we”ve actuallyincreased the production of anthropogenic carbon dioxide. This meansthat the numbers you calculated for this part of the problemunderestimate the amount by which we would need to cut back now.)