AST Homework I - Solutions
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1 AST 34 - Homework I - Solutions TA: Marina von Steinkirch, steinkirch@gmail.com State University of New York at Stony Brook September 3, 200 Halley s Orbit. ( point Halley s orbital period (P 76 years = s Halley s eccentricity (e a What is the semimajor axis (a? When calculating in Astronomic Units (AU and years, instead of meters and seconds (SI, this problem is very simple. The third Kepler s law states the relation between period (in years of the orbit and the average distance (ā, in AU, P 2 = ā 3, On the other hand, the relation to the semimajor axis is P 2 ka 3, with a proportionality constant k that depends on the units. When averaging the distance over the eccentric anomaly, one has ā = a, therefore we can apply directly the third law, P 2 = ka 3. (. Since the proportionality k is the same for all bodies orbiting around the sun, we have k = P 2 a 3 = P 2 Earth a 3 Earth = constant. Since P Earth = = P Earth, we have k = and equation. gives a = 7.83 AU = m.
2 b Estimate the mass of the Sun. In the same fashion as the problem 2.3 of the textbook, we have M = 4π2 G a 3 P 2. (.2 Now we need to put everything in SI since we want to find the mass in kilograms, M = 4π 2 ( ( kg. c Calculate the perihelion and the aphelion. From equation 2.3 of the textbook, we have In perihelion, θ = 0 o and r = a( e2 + e cos θ. (.3 r p = a( e = 7.83 ( = 0.58AU = m. In aphelion, θ = 80 o and r a = a( + e = 7.83 ( = 35.08AU = m. d Determine the orbital speed (v at perihelion, aphelion, the semiminor axis (b. The orbital speed at perihelion (v p is given by the equation 2.33 in the the textbook v 2 p = GM r p Plugging in the values, in SI units, ( + e = GM a ( + e e vp 2 = ( = (m/s 2, (.4 v p = m/s. The orbital speed at aphelion (v a is given by equation 2.34 in the the textbook, v 2 a = GM r a Plugging in the values, in SI units, ( e = GM a ( e. (.5 + e va 2 = ( = (m/s 2,
3 v a = m/s. The orbital speed at the semiminor axis (v b is given by using b 2 = a 2 ( e 2 (.6 on any of the previous equation. v b = m/s. e Compare the kinetic energy (K of the comet at perihelion and aphelion. K p K a = 2 µv2 p = v2 p 2 µv2 a va Earth as a Blackbody. ( point Internal heat Flow (F i W/m 2 Albedo (a 39% Earth s actual Temperature 280 K Luminosity of Sun (L at AU W Distance (d AU = m First of all, the intensity from the sun that reaches the outer atmosphere of the Earth is the solar constant or solar irradiance, and it can be easily calculate as in the example From the equation 3.2 of the the textbook, we have F = L 4πr 2, (2. F = 365 W m 2. However, we need to take account the albedo in this intensity of flux, I abs = ( af = = W m 2. Now, we calculate the total power absorbed and the the total power radiated, in and from the surface. The sunlight strikes the Earth on just one side, therefore the total power being absorbed is P abs = I abs πr 2 Earth, while the amount of power radiated away is P rad = I rad 4πR 2 Earth, 3
4 where I rad is the intensity being emitted. This is given by the Stefan-Boltzmann law for a blackbody, where T is the temperature of the blackbody and σ = W/m 2 K 4 a constant, Finally, the thermal equilibrium is given by resulting in I rad = σt 4. (2.2 P rad = P abs I abs πr 2 Earth = I rad 4πR 2 Earth, I rad = I abs 4 = = W m 2, 4 and putting back in 2.2, we find T 4 = = 246K This value is different of the actual value 280K due the fact that the temperature on Earth is influenced by the presence of an atmosphere, as we can see from effects such as the greenhouse.. One secondary explanation is that Earth is not a perfect blackbody. A last observation is that we did not use the value for the internal flux in problems of this kind. For future references, we can use the equation for the equilibrium temperature: ( a/4 T equilibrium = 280K. (2.3 d /2 3 Wien Law for a Blackbody. (0.5 point We derive Wien law from the equation of the blackbody spectrum (the Planck function given by the equations 3.22 and 3.24 of the textbook, in terms of wavelength of the radiation of the blackbody, B λ (T, λ = 22 λ 5 e λkt = 8π λ 5 e λkt, (3. or its description in terms of frequency, B ν (T, λ = 2hv3 c 2 e λkt = 8πhν3 c e Making use of any of them, the derivation starts by doing B λ = 0. λkt. (3.2 Defining a dimensionless variable x = hν λkt for the first equation or x = kt for the second, we obtain a equation of the form xe x = 5(e x or xe x = 3(e x. 4
5 These are the Lambert s log Functions and can only be solved numerically. Putting in some program such as Mathematica, we obtain x = and x = , respectively. Adjusting units on Kelvin, putting meters to nanometers ( nm = 0 9 m and the values of the constants c, h and k, we finally have λ max T k x = mk. 4 A model of the star Dschubba as a Blackbody. (2 points Surface Temperature of Dschubba(T 28,000 K Radius (r m Distance from Earth (d 23 pc pc a Find the Luminosity. From equation 2. we have L = F 4πr 2 = F 4π ( Approximating Dschubba as a blackbody, then the radiant flux at the surface can be calculated from the temperature from 2.2, F = σ T 4 = , = W m 2. plugging it in the first equation, we have the bolometric luminosity (luminosity integrated over all wavelengths of Dschubba, L = = 30, 374 L. b Find the Absolute Bolometric magnitude (M. From equation 3.8 of the textbook, we can find the absolute bolometric magnitude of the star from its flux and from comparing it to the absolute magnitude and the luminosity of the Sun, M = M 2.5 log 0 ( L L. (4. Luminosity bolometric of Sun (L W Absolute bolometric magnitude of Sun (M Resulting in M =
6 c Find the Aparent bolometric magnitude. From equation 3.6 of the textbook ( d m M = 5 log 0 (d 5 = 5 log 0, (4.2 0 pc we have m =.02. d Find the Distance modulus. From the same 3.6, the distance modulus is m M = e Find the Radiant flux at the star s surface. From 2. we make r = R, F = 4π ( = W m 2 Another way of finding the same result is If we approximate Dschubba as a blackbody, then the radiant flux at the surface can be calculated from the temperature from 2.2, F = σ T 4 = , = W m 2. Note that we had already found this result in item a. f Find the Radiant flux at Earth s surface, compare with the solar irradiance. Now we do r = d and let us assume that the flux received at the Earth s surface is not attenuated by its atmosphere. We then have F = L 4πd 2 = π( = W m 2. g Peak wavelength λ max. Using equation 3.5 of the textbook, the Wiens displacement law gives λ max = T = 03.2 nm. 5 The Planck Function B λ. (.5 points a Show that the Rayleigh-Jeans equation the Planck function for enough large λ. The Plank spectrum function, equation 3. or 3.22 of the textbook B λ (T, λ = 22 λ 5 e λkt, (5. can be writte as a first-order Taylor expansion of the exponential for λ kt or kt λ as e kt λ = kt λ, resulting in B λ (T = 22 λ 5. kt λ 6
7 From equation 3.20, the Rayleigh-Jeans equation is B λ 2ckT λ 4, which is exactly the previous result. We clearly see that it is a classical result since the h was canceled. b Plot the Planck Function and the Rayleigh-Jeans for the Sun. 6 Blackbody radiation and the Stefan-Boltzmann constant. (2 points a Integrate over all λ to obtain the total luminosity of a blackbody model star. Let us make A = kt in equation 3.20 and then integrate it on the whole space, L λ = 8π 2 R 2 2 λ 5 e A/λ dλ. Now we make the change of variable λ = A u L λ = 8π 2 R 2 2 From the straigthfoward integral with du dλ = a λ 2, obtaining du (A/u 3 e u. ( A u 3 du e u = π4 5, we finally have ( kt 4 L λ = 8π 2 R 2 2 π 4 5 = 2 π 5 k 4 5 h 3 c 2 4πR2 T 4 b Compare to the Stefan-Boltzmann equation and show σ. The Stefan- Boltzmann equation for the luminosity (from 2.2 and 3.7 in the textbook is L = σ4πr 2 T 4. (6. Comparing to item a, we have which is exactly the answer. σ obt = 2 π 5 k 4 5 h 3 c 2, c Compare the value of this expression to the value in Appendix A. Comparing to the appendix A of the textbook, 2 π σ 5 k 4 obt 5 h 3 c 2 σ W m 2 K 4 7
8 7 UBV Colors to Shaula star. (2 points Surface Temperature(T 22,000K Visible color (V.62, blue-white U-B B-V 0.23 Parallax angle a Estimate U B and B V. Compare to the measured values. From the equation 3.33 of the textbook we have ( Fλ d λ U B = 2.5 log 0 + C U B, (7. Fλ d λ and we can approximate these functions to the Planck functions, whose depends on λ and T. We write, for example, ( B365 λ U U B = 2.5 log 0 + C U B. (7.2 B 440 λ B Substituting the values for each bandwidth, from page 75, and making use U B=365 nm λ= 68 nm B B= 440 nm λ=98 nm V B=550 nm λ=89 nm of C U B = 0.87 and C B V = 0.65, we find and U B =.08, B V =.36. 8
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