wksht2_soln - Blackbody Telescopes and Earths Atmosphere...

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Blackbody, Telescopes and Earth’s Atmosphere Jessamyn Allen Section 108, 120 Wksht + Answers 1 BLACKBODY 1a. What mathematical formulae characterize a blackbody? There are 2 main ones so far we’ve talked about in class - I want names formulae. 1b. What are the units of each of the variables? Wien’s Law: λ peak T = constant (1) When the constant 2 . 9 × 10 7 ˚ AK , λ peak is in ˚ A and T (which usually refers the object’s surface temperature) is in Kelvin. If you have a wavelength given in nanometers, for example, you’d need to convert to its length in ˚ A if you use a value for the constant that’s given in ˚ A . Stefan-Boltzmann Law: ± = σT 4 (2) Here, ± is in energy emitted per square centimeter per second ( energy cm 2 × s and T , the surface temperature, is in Kelvin. 2. If you take the spectra of a star (essentially a blackbody), what physical quantities can you calculate? What about its motion? From the blackbody spectrum, you can measure the peak wavelength which tells you how hot (i.e. the temperature) the star is. From the Stefan-Boltzmann law you can tell how much energy the star is emitting per area per second (to get the total energy emitted per second you have to multiply ± by the star’s surface area and to know the surface area, you have to know how big it is, i.e. you need to know its radius). Also, looking at the spectrum 1
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you can measure the Doppler shift (if there is one) and tell whether the star is moving away or towards Earth and how fast. DiFerent types of motion can cause Doppler shift. If the star is in a binary system, it will orbit around a point is space or another object. During its orbit, it is sometimes moving towards us and sometimes moving away from us, which could cause a Doppler shift. 3. Rigel, a supergiant star ( M Rigel = 17 M sun ), has a surface temperature, T s = 11000 K . What part of the spectrum does it peak at? What color do you think it appears and why? Using Wien’s Law, we calculate λ peak = 2636 ˚ A which is in the UV part of the EM spectrum. Rigel will appear blue because it contains an excess of blue and UV light. True, Rigel emits at all the visible wavelengths (red, orange, green, blue, etc) so those colors do combine to form white light, but Rigel emits more heavily in the blue-purple portion of the visible spectrum so it appears more blue than white. 4. We’ll discuss two stars: Vega and Arcturus. Vega has a surface temperature
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This note was uploaded on 09/29/2010 for the course ASTRONOMY 05989 taught by Professor Filippenko during the Spring '10 term at Berkeley.

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wksht2_soln - Blackbody Telescopes and Earths Atmosphere...

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