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SOLUTIONS TO EXERCISE 11
(a) The collecting area of the radio telescope is A = pi*r^2 = 3.14 * (35 m /2)^2 = 962 sq. meter. So the luminosity on this telescope is the flux from Voyager 2 to Earth, times the collecting area of the dish: L = F * A = 1.59 * 10^(-24) W/m^2 * 962 m^2 = 1.53 * 10^(-21) Watts. The amount of energy received in one hour (remember that a watt equals a joule per second) is thus E = L * t = 1.53 * 10^(-21) W * 3600 sec = 5.51 * 10^(-18) joule. (b) Since flux is L/(4*pi*r^2), we can figure out the distance you would need to be away from the light bulb by using the flux on Earth from Voyager 2 and the luminosity of the light bulb. 1.59 * 10^(-24) W/m^2 = 1.3 W / (4 * pi * r^2) ; thus, r^2 = 1.3 W / (1.59 * 10^(-24) W/m^2 * 4 * 3.14) r = sqrt(6.51 * 10^22 m^2) = 2.55 * 10^11 m = 2.55 * 10^8 km. This is equivalent to (2.55 * 10^8)/1.61 = 158 million miles, slightly longer than the distance between Earth and the Sun. (c) How much luminosity from the light bulb arrives in your eye at that distance? A good estimate of the size of your pupil might be about half a centimeter (5 mm) in diameter at maximum dilation. In that case, L = F * A = 1.59 * 10^(-24) W/m^2 * 3.14 * (0.0025 m)^2 = 3.12 * 10^(-29) Watts. That's less than one visible light photon per century! |