PHYS-333: Problem set #1 Solutions VERSION of February 10, 2018. 1. Energy flux and magnitude: a. Suppose two objects have energy fluxes of f and f + f, where f f. Derive an approximate expression for the magnitude difference m between these two objects. (Hint: note that ln(1 + x) x for x 1.) Note that your answer will be in terms of f as well as f. b. What magnitude difference do you get for a flux difference of 10%. From basic definition of magnitude difference, we have ( ) f + f m = 2.5 log = 2.5 log(1 + f/f) = 2.5 log e ln(1 + f/f) (1) f Thus using the expansion hint for ln(1 + x) x, we find m 2.5 0.43 f/f = 1.1 f/f. (2) So a flux difference of 10% gives m 0.11. 2. 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 AU.) First convert arcsec to radians: 1 arcsec= π/(180 60 60)= 4.8 10 6 radians. Then d = 2R α = 2R 4.8 10 6 = 4.1 105 R = 2.9 10 14 m = 1800AU. (3) Another way to look at it is to recall the actual angular diameter of the Sun is about 0.5 o = 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 (in km) away would a Frisbee of diameter 30 cm have to be to subtend the same angle? By above, d = 30 4.8 10 6 = 6.3 106 cm = 63 km. (4)
2 c. At the distance you calculated in (a), by what factor would the solar flux at earth be reduced? If we move the Sun 1800 times further away, then by the inverse-square law, the flux would decrease by a factor 1/1800 2. Thus F,new = 1.4 103 1800 2 = 4.3 10 4 watt/m 2 = 3 10 7 F (5) 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 m,new = m + 2.5 log(f /F, new) = 26.7 + 2.5 log(1800 2 ) = 10.4. (6) 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 100 2 =10,000). 3. Galaxies: distance, magnitude, and solid angle: a. What is the apparent magnitude of a galaxy that contains 10 11 stars identical to the Sun (i.e., assume its luminosity is equal to 10 11 L ) if it s at a distance of 10 million parsecs? Using eqn. (3-10) in DocOnotes-stars1.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 6 = +7.3. (7) b. 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, in both radians and degree? α = s d = 5 104 pc 10 7 pc = 5 10 3 rad = 0.3 (8)
3 c. What solid angle does it subtend (in steradians and in square degrees)? Ω = (π/4)(0.3) 2 degree 2 = 0.07 degree 2 (9) Ω = (π/4)(5 10 3 ) 2 radian 2 = 2.0 10 5 steradian (10) e. How does the galaxy s surface brightness (energy/time/area/solid angle) compare to the Sun s (express this as a ratio)? B gal B = F gal F Ω = L ( ) 2 ( ) 2 ( gal au α = 10 11 Ω gal L d gal α gal 1 2 10 12 ) 2 ( ) 2 0.5 = 7 10 14. 0.3 (11) 4. 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. Equating the solar flux intercepted by the Earth s cross-sectional area πr 2 e to the total blackbody emission over its surface area 4πR 2 e, we find σt 4 e 4πR 2 e = πr 2 e L 4πa 2 e (12) Thus T e = ( ) 1/4 L = 281 K = 8 C. (13) 16πa 2 e σ Alternatively, noting that L = σt 4 4πR 2, we can write this as T e = T 2 2R au 5800/2 107 280 K, (14) This has the advantage of not requiring use of the Stefan-Boltzmann constant σ, while relating the earth s temperature to that of the sun, reduced
4 by a factor equal to half the square root of the sun s angular diameter, 2R /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 has an albedo of a = 0.633, meaning the fraction of received light that is reflected 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 1 a. 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 0.367 1/4 = 0.78, reducing the above equilibrium temperature now to T e = 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 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! 5. 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 E 6400 km. Viewed from the center of the Earth, the two observers at ±45 o are separated by 90 o, thus forming a right angle. The radius lines to each observer thus form the two lengths of an isosceles triangle with the observers separated
5 by the base, with length s = 2R e = 8.8 10 3 km. (15) b. If the parallax angle measured is 22 arcsec, what is the distance to Mars? Give your answer in both km and AU. d = s α = 8.8 10 3 km 22arcsec/206000 (arcsec/radian) = 8.24 107 km = 0.55 AU. (16)