PHYS-633: Problem set #0 Solutions . 1. Angles, magnitudes, inverse square law: a. How far from the Earth would the Sun have to be moved so that its apparent angular diameter would be 1 arc second? (Express your answer in cm, solar radii, and AU.) First convert arcsec to radians: 1 arcsec= π=(180×60×60)= 4.8×10−6 radians. Then 2R 2R D = = = 4:1 × 105R = 2:9 × 1016 cm = 1900 AU : (1) α 4:8 × 10−6 Another way to look at it is to recall the actual angular diameter of the Sun is about ∼ 0:5o ≈ 1800 arcsec. Thus to get to 1 arcsec, the Sun would have to move away by a factor 1800, or to 1800 AU. b. How far away would a Frisbee of diameter 30 cm have to be to subtend the same angle? By above, 30 D = = 6:3 × 107 cm = 63 km : (2) 4:8 × 10−6 c. At the distance you calculated in (a), what would the Sun's flux here on Earth be (i.e. what would the solar constant be)? 2= 6 2 The actual solar constant is F = L =4πau 1:4 × 10 erg=s=cm . If we now move the Sun 1800 times further away, then by the inverse-square law, the flux (which is what we mean by the solar constant) would decrease by a factor 1=18002. Thus 1:4 × 106 F = = 0:43 erg=s=cm2 (3) ;new 18002 d. What would the Sun's apparent magnitude be? (Use m = -26.7 for the actual Sun, the one that's at 1 AU.) { 2 { The difference in apparent magnitude between two stars is just 2.5 times the log of the ratio of the flux. Remembering that a lower flux gives a larger magnitude (i.e. dimmer stars have bigger m), we have 2 m ;new = m + 2:5 log(F =F ; new) = −26:7 + 2:5 log(1800 ) = −10:4 : (4) Remembering that the brightest stars are around magnitude zero, we see that the sun would still be a very bright star, about 10,000 times brighter than the brightest actual star! (Since m=-10 is 10 magnitudes brighter than m=0, and each difference of 5 in magnitude represents a factor 100 in brightness, so 1002=10,000). 2. The star Dschubba (δ Sco) has a parallax p = 8 mas: Assuming it is a spherical blackbody with radius R = 7:5 R and surface temperature T = 28; 000 K, compute Dshuba's a. Luminosity, in erg/s, and L ; By Stefan-Boltzmann law: 2 4 38 L = 4πR σT = 1:2 × 10 erg=s = 30; 000 L (5) b. Absolute magnitude: By definition of absolute magnitude normalized to the sun's value Msun = +4:8, we have M = 4:8 − 2:5 log(30; 000) = −6:4 : (6) c. Apparent magnitude; A parallax of α = 8 mas 0.008 arcsec implies a distance of D = 1=0:008 = 125 pc. Thus the apparent magnitude is m = M + 5 log(D=10pc) = −6:4 + 5 log(12:5) = −0:9 : (7) d. Distance modulus; m − M = 5 log(12:5) = 5:5 (8) e. Radiant flux at the star's surface, in CGS, and relative to surface flux of the Sun; { 3 { L 30; 000 × 4 × 1033erg=s F (R) = = = 3:4 × 1013 erg=cm2=s : (9) 4πR2 4π(7:5 × 7 × 1010)2cm2 Relative to the flux at sun's surface, we have F (R) L=L 30000 = 2 = 2 = 533 : (10) F (R ) (R=R ) (7:5) f. Radiant flux at the Earth's surface, and the ratio of this to the solar irradiance; L 30; 000 4 × 1033erg=s F = = = 6:8 × 10−5 erg=cm2=s : (11) 4πD2 4π(1253 × 1018)2cm2 Relative to the solar flux at earth's orbit, we have F L=L 30000 −11 = 2 = 2 = 4:5 × 10 : (12) F (D=au) (125 206; 000) g. Peak wavelength λmax. By Wien's displacement law, 2:9 × 106nm K 2:9 × 106 λ = = nm ≈ 100nm : (13) max T 2:8 × 104 3. Parallax of Mars: In 1672, an international effort was made to measure the parallax angle of Mars at opposition, when it was on the opposite side of the Earth from the Sun, and thus closest to Earth. a. Consider two observers at the same longitude but one at latitude of 45 degrees North and the other at 45 degrees South. Work out the physical separation s between the observers given the radius of Earth is RE ≈ 6400 km. Viewed from the center of the Earth, the two observers at ±45o are separated by 90o, thus forming a right angle. The radius lines to each observer thus form the two lengths of an isosceles triangle with the observers separated by the base, with length p 3 s = 2Re = 8:8 × 10 km : (14) { 4 { b. If the parallax angle measured is 25 arcsec, what is the distance to Mars? Give your answer in both km and AU. s 8:8 × 103 km D = = = 7:3 × 107 km = 0:48 au : (15) α 25arcsec=206000 (arcsec=radian) 4. Galaxies: distance, magnitude, and solid angle: a. What is the apparent magnitude of a galaxy that contains 1011 stars identical to the 11 Sun (i.e., assume its luminosity is equal to 10 L ) if it's at a distance of 5 million parsecs? Using eqn. (14) in DocOnotes1-stars.pdf, the apparent magnitude m of an object with luminosity L at a distance D is given by m = 4:8 − 2:5 log(L=L ) + 5 log(D=10 pc) = 4:8 − 2:5 × 11 + 5 × (5 + log(5)) = +5:7 ; (16) with the latter equality plugging in the above numbers to give m ≈ +3. b. What is the flux of this galaxy here on the Earth, in cgs units? L 1011 4 × 1033 erg=s F = = = 1:3 × 10−7 erg=cm2=s (17) 4πD2 4π (5 × 106 3 × 1018)2 cm2 c. If the galaxy is circular in shape, as seen from the Earth, and has a diameter of 50,000 pc, what is its apparent angular diameter? s 50 × 103 pc α = = = 10−2 rad = 2060 arcsec = 34 arcmin = 0:57 o (18) D 5 × 106 pc d. What solid angle does it subtend (in steradians and in square arc seconds)? Ω = (π=4)20602 arcsec2 = 3:2 × 106 arcsec2 (19) Ω = (π=4) α2 = 3 × 10−4 ster (20) e. How does the galaxy's surface brightness (energy/time/area/solid angle) compare to the Sun's (express this as a ratio)? { 5 { Assuming both the galaxy and the sun emit isotropically from their surface, then the ratio of surface brightness is just given by the ratio of their surface flux, 2 10 2 Igal Lgal=4πRgal 11 7 × 10 −14 = 2 = 10 4 18 = 2:2 × 10 : (21) I L =4πR 5 × 10 3 × 10 5. Equilibrium temperature of blackbody near a star: A spherical blackbody of radius a is at a distance d from a star with an effective tem- perature Teff and radius R. a. Assuming the blackbody has a rotation that over time exposes all parts of its surface to the star's light, compute its equilibrium temperature T . How does this depend on the blackbody's radius a? 4 2 Noting that the stellar luminosity is L = σTeff 4πR , we can write the star's 2 4 2 radiative flux at distance d as F (d) = L=4πd = σTeff (R=d) . The blackbody intercepts an energy πa2F (d), which in equilibrium it must re-radiate over its entire spherical surface area 4πa2. The equilibrium blackbody temperature T is given by equating this absorbed and emitted energy: R2 σT 4 πa2 = σT 44πa2 ; (22) eff d which can be readily solved to give r R T = T : (23) eff 2d Note that the blackbody radius a has cancelled, and so this equilibrium temperature is independent of the absorbing body's size. b. Derive an expression for the ratio, λmax;bb/λmax;∗, of the peak wavelength of emission for the blackbody to that of the star. r λ T 2d max;bb = eff = : (24) λmax;∗ T R c. Compute values for parts (a) and (b) in the case of a blackbody at 1 AU from the sun. Recalling that au/R ≈ 215, with T ≈ 5800 K, we find 5800 T (1 au) = p K = 280 K: (25) 430 { 6 { λmax;bb p = 430 = 20:7 ! λmax;bb ≈ 20:7 × 500 nm = 10:4 µm : (26) λmax; 6. Equilibrium Temperature of Earth: a. Modeling earth as a blackbody illuminated by the sun, compute its equilibrium tem- perature, and compare this to the temperature on a moderate spring day in Delaware. From part c. of previous problem, r R p T = T ≈ 5800 430 ≈ 280 K: (27) e 2 au On moderate spring day in Delaware, temperature is a bit higher that this, ca. 20 C (68 F). But given the approximations, this is pretty close to the characteristic temperature computed for a simple blackbody! b. According to http://en.wikipedia.org/wiki/Earth, Earth's \albedo", meaning the fraction of received light that is absorbed is only ca. 0.367, with the rest (63.3% ) being refected by, e.g. clouds, snow, etc., without contributing any heat to Earth. So now redo the calculation in (a) reducing the solar input energy by this amount. If only a fraction 0.367 of Sun's luminosity is actually absorbed by Earth, then the equilibrium temperature should be reduced by a factor 0:3671=4 = 0.78, reducing the above equilibrium temperature now to Te = 219 K = -54 C.