 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:

• 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.

Now this density lies between that of rock and water, so it might be reasonable to assume that Charon is made from a mixture of these. Really, what the average density tells you is that it's pretty unlikely that Charon has a large iron core, or that it's entirely rock and denser materials.

Charon is in fact an amalgam of rock and ice (it's really far away from the Sun, so the water is in the form of ice).

Problem #2: A 60-watt light bulb emits 60 joules/sec of energy ( 1 Joule/sec == 1 watt). Pretend for a moment that all of this energy is emitted in the form of photons with wavelength of 600 nm (this isn't true, of course -- as a blackbody, a light bulb emits photons of a wide range in wavelengths). Calculate how many photons per second are emitted by such a light bulb.

Solution: This problem asks you to calculate how many photons per second are emitted from a 60-watt lightbulb. Since 60 watts is 60 Joules per second, we know that we need enough photons to carry 60 Joules of energy each second.

So how much energy does one 600nm photon carry? We'll need to use the relation 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 J
OK, 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 J
and
• N = 60 J / 3.32 x 10-19 J
• = 1.8 x 1020
So, in order to emit 60 Joules per second, the lightbulb must emit 1.8 x 1020 photons per second. (that's 180,000,000,000,000,000,000 photons per second!)

Problem #3: Read the Executive Summary of the Department of Energy's Report, Emissions of Greenhouse Gases in the United States 2003 and answer the following questions:

a) How many metric tons of anthropogenic carbon dioxide were emitted from US sources in 2003?

b) What is the percentage increase in the US-generated carbon dioxide emissions since 1990?

c) Which sector of the US economy (e.g., residential, commercial, industrial, or transportation) is responsible for the largest fraction of 2003 carbon dioxide emissions?

d) US emissions of which greenhouse gases actually decreased between 2002 and 2003?

e) Based on the information in Table 3 of Chapter 1 of this report, by what amount would we have to reduce yearly anthropogenic carbon dioxide emissions to result in no increase in carbon dioxide in our atmosphere?

f) By what fraction of current worldwide human-made emissions levels would we have to reduce to achieve this goal?

Solution: This problem should have been fairly straightforward, provided that you spent the time reading the report. Skimming, on the other hand, likely didn't work.

a) 5870.2 million metric tons of gas (from the first sentence in the "Carbon Dioxide" section). Note that 6935.7 million metric tons listed in the first sentence of the Executive Summary was the "carbon dioxide equivalent," which is a quantity that also includes the effects of emissions from other greenhouse gases such as methane, nitrous oxide, and others.

b) a 17.6 percent increase since 1990 (from the second paragraph of the "Carbon Dioxide" section).

c) commercial (from Figure ES3). I was surprised to learn that industrial emissions contributed so little in relative terms.

d) nitrous oxide, hydrofluorocarbons (HFCs), perfluorocarbons (PFCs), and Sulfur Hexafluoride(SF6) (from first paragraph in the "overview").

e) Table 3 of Chapter 1 (not Table 3 of the Executive Summary) shows that the total human made carbon dioxide emissions total 23100 million metric tons. It also shows that the total natural emissions produce 770000 million metric tons and that natural absorption accounts for 781400 million metric tons. Given the imbalance between natural emission and absorption, we can add 781400 - 770000 = 11400 million metric tons into the atmosphere and it will be absorbed by the ecosystem. Thus, if we were to reduce our emissions to that level, the carbon dioxide concentration would remain constant.

How much would we need to reduce? We're currently emitting 23100 million metric tons, and the ecosystem can absorb 11400 million metric tons, so it looks like we'll have to reduce by 23100 - 11400 = 11700 million metric tons of carbon dioxide.

f) The fraction is just the amount we have to reduce by over the total amount we're currently emitting:

11700/23100 = 0.506, or 51%.

That means we would have to halve our carbon dioxide emissions to reach the goal of not increasing the CO2 in the atmosphere. Perhaps this gives you a sense of the magnitude of this problem. It won't be solved with energy efficient cars and lower winter thermostat settings. To really halve our CO2 production will require a massive change in how we live. (One last note: the data in Table 3 you used for parts e) and f) is actually based on emissions in the 1990s; today, nearly a decade later, the problem is even more severe because we've actually increased the production of anthropogenic carbon dioxide. This means that the numbers you calculated for this part of the problem underestimate the amount by which we would need to cut back now.) 