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**Worked out for Geology/Physics 360**

Problems in Chapter 13 Worked out for Geology/Physics 360

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**Sirius Now (today) in March**

An Illustration of Parallax using HNSKY Sirius in Sept (6 months later)

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**Question: How would we find a closer star than Proxima Centauri?**

Problem 3. The Parallax of the red giant Betelgeuse is just barely measurable and has a value of about arc seconds. What is its distance? Suppose the measurement is in error by + or – arc seconds. What limits can you set on its distance. Answer: p(Betelgeuse) = arc seconds. Therefore its distance is d = 1/p = 200 pc. The error in this measurement is +/–0.003", so we have to add and subtract this figure from the parallax angle to give the lower and upper limits to the distance estimate. d(pc) = 1/0.008 = 125 pc, the larger angle corresponds to the closer distance d(pc) = 1/0.002 = 500 pc, the smaller angle corresponds to the farther distance

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**Problem 5. The star Rigel radiates most strongly at about 200 nm**

Problem 5. The star Rigel radiates most strongly at about 200 nm. What is its temperature? How does this compare to the Sun. 𝝀m for Rigel = 200 nm. Use Wien’s Law to find its temperature? T = 3 × 106 K nm / 𝝀m = 3 × 106 K nm / 200 nm = 15,000 K. This is much hotter than the surface of the Sun, which is about 6000 K. (2.5 times hotter). (Note that we have approximated the constant here as 3 x 106, it is closer to 2.9 x 106

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Problem 6. The bright southern star Alpha Centauri radiates most strongly at about 500 nm. What is its temperature? How does this compare to the Sun. . Alpha Centauri has its peak radiation at 500 nm. Applying Wien’s Law, T = 3 × 106 K nm / 500 nm = 6000 K. The temperature of this star is approximately the same as our Sun’s, so Alpha Centauri has a similar spectral type to the Sun (G2).

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Problem 7. Arcturus is about ½ as hot as the sun but is about 100 times more luminous. What is its radius compared to the Sun. . For Arcturus, T = To/2, where To is the temperature of the Sun. L = 100Lo. To find the star’s radius, use the Stefan-Boltzmann Law, which relates a star’s luminosity, L, to its temperature, T, and radius, R. L = 4𝜫R2 𝝈 T4 or LArc = 4𝜫R2Arc 𝝈 T4Arc and Lo = 4𝜫R2o 𝝈 T4o We want to find RArc in terms of Ro. 100 Lo = LArc = 4𝜫R2Arc 𝝈 T4Arc 100 (4𝜫R2o 𝝈T4o) = 4𝜫R2Arc 𝝈 T4Arc 100 (R2oT4o) = R2Arc T4Arc = R2Arc (To/2) 4 100 (R2oT4o) = R2Arc (T4o) / 24 100 (R2o) = R2Arc / 16 1600 (R2o) = R2Arc Therefore R Arc = (1600) 1/2 Ro = 40 Ro Arcturus is 40 times wider than the Sun.

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8 Stars … how I wonder what you are.. 8 Goals Stars are Suns. Are they: –Near? Far? –Brighter? Dimmer? –Hotter? Cooler? –Heavier? Lighter? –Larger? Smaller?

8 Stars … how I wonder what you are.. 8 Goals Stars are Suns. Are they: –Near? Far? –Brighter? Dimmer? –Hotter? Cooler? –Heavier? Lighter? –Larger? Smaller?

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