PHYS-638-07f: 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⊙ 5 16 D = = = 4.1 10 R⊙ = 2.9 10 cm . (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)? 6 2 In class we noted that the solar constant is F⊙ = 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 6 1.4 10 2 F⊙ = × = 0.43 erg/s/cm (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.) 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⊙ = m⊙ + 2.5 log(F⊙/F⊙,new)= 26.7 + 2.5 log(1800 )= 10.4 . (4) ,new − − – 2 – 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. 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 50 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/10pc) = 4.8 2.5 11+5 6 log(5) = +10.8 , (5) − − × × with the latter equality plugging in the above numbers to give m +10.8. ≈ b. What is the flux of this galaxy here on the Earth, in cgs units? 11 33 L 10 4 10 erg/s 9 2 F = = × = 1.4 10− erg/cm /s (6) 4πD2 4π (5 107 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? 3 s 50 10 pc 3 α = = × = 10− rad = 206 arcsec = 3.4 arcmin (7) D 50 106 pc × d. What solid angle does it subtend (in steradians and in square arc seconds)? Ω=(π/4)2062 arcsec2 = 3.2 104 arcsec2 (8) × 2 α 2 10−3 Ω = 4π = 4π = 3 10−7 ster (9) 2π rad 2π × e. How does the galaxy’s surface brightness (energy/time/area/solid angle) compare to the Sun’s (express this as a ratio)? – 3 – 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 . (10) I⊙ L⊙/4πR⊙ 5 10 3 10 × × × 3. Equilibrium Temperature of Earth: a. Assuming Earth is a blackbody, use the known luminosity and distance of the Sun to estimate Earth’s average equilibrium surface temperature if the solar energy it intercepts is radiated to space according the Stefan-Boltzman law. Compare this to the temperature on a moderate spring day in Delaware. 2 Equating the solar flux intercepted by the Earth’s cross-sectional area πRe 2 to the total blackbody emission over its surface area 4πRe, we find 4 2 2 L⊙ σTe 4πRe = πRe 2 (11) 4πae Thus 1/4 L⊙ T = = 281 K = 8 C (12) e 16πa2 σ e On moderate spring day in Delaware, temperature is a bit higher, ca. 20 C (68 F). But given the approxmiations, this is pretty close to the character- istic temperature computed for 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. c. Which result seems more “reasonable”? Briefly discuss what other physics might be important to include to understand the actual surface temperature of Earth. This apparently more realistic model thus seems to give a temperature that is much lower than the typical temperature of the actual Earth. The key piece of physics missing is the “greenhouse effect”, which blocks the re-radiation of solar energy, forcing the surface of the Earth to be warmer – 4 – than it would otherwise be, much like a blanket at night keeps our skin at a higher temperature than it would otherwise be. Bottom line: the greenhouse effect and the albedo effect roughly cancel, making the simple blackbody temperature in part (a) come out about right! 4. 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 R 6400 km. E ≈ 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 s = √2R = 8.8 103km . (13) e × b. If the parallax angle measured is 25 arcsec, what is the distance to Mars? Give your answer in both km and AU. 3 s 8.8 10 km 7 D = = × = 7.3 10 km = 0.48 au . (14) α 25arcsec/206000 (arcsec/radian) × 5. Brightness of the full moon: a. Given the Sun’s radius R⊙ and distance to Earth ae, compute the Sun’s solid angle on the sky, both in steradians, and as a fraction of the full sky’s 4π steradians. 1 2 2 −5 Ω⊙/2π = dµ = 1 µ⊙ = 1 1 (R⊙/ae) (1/2)(R⊙/ae) 1.1 10 (15) µ⊙ − − − ≈ ≈ × Z p −5 Ω⊙ 6.8 10 ster (16) ≈ × −6 Ω⊙/4π 5.5 10 (17) ≈ × – 5 – b. By coincidence, the moon covers about this same fraction. Use this fact together with the moon’s albedo of 0.07 (meaning the moon reflects back 7% of the sunlight that hits its surface) to estimate how much dimmer the full moon appears in optical light compared to sunlight on Earth. 2 Since the solar flux declines as (R⊙/ae) , then a perfectly reflecting moon would have its surface brightness reduced by this fraction. With an albedo of 0.07, this surface brightness would be reduced by that extra factor. Fi- nally, since the solid angles are the same, the ratio of surface brightness is also the ratio of apparent brightness. 2 f R⊙ moon = 0.07 = 1.6 10−6 (18) f⊙ a × e c. Given that the Sun’s apparent magnitude is m⊙ = 26.7, use your answer for part − (c) to estimate the apparent magnitude of the full moon in optical light. m = 26.7 2.5 log(F /F⊙)= 12.2 (19) moon − − moon − d. Assume that the sunlight that is not reflected is re-radiated as if the moon were a blackbody. Compute the average equilibrium temperature of the moon. Surface flux of the moon’s thermal emission is given by: 2 2 4 2 4 4 0.93πRmoonL⊙ 0.93 4πR⊙σT⊙ 0.93 R⊙σT⊙ Fem = σTmoon = 2 2 = 2 = 2 (20) 4πae 4πRmoon 16πae 4ae 0.25 R⊙ T = 0.93 T⊙ 0.048 T⊙ 273 K (21) moon 2a r e ≈ ≈ e. Estimate the peak wavelength at which the moon emits this blackbody radiation, com- paring it to the peak wavelength of the Sun’s spectrum. What part of the electromagnetic spectrum does this correspond to? T⊙Tmoon 500 nm λ = λ ⊙ 10, 400 nm (22) max,moon max, 0.048 ≈ ≈ This is in the infrared.