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Solutions Set #2 1. What is the B-band brightness ratio of Betelgeuse to Rigel? This requires a bit of on-line sleuthing, particularly since I forgot that Betelgeuse is a variable star. When I consult SIMBAD, I find the following numbers: B = 2.35 for Betelgeuse, and B = 0.09 for Rigel. Now it may seem weird that Rigel has a lower number (and hence, is brighter) than Betelgeuse, when Betelgeuse is the brightest star in Orion! This is because Rigel is a blue star and Betelgeuse is a red star. In the B band, Rigel is putting out a lot more light than Betelgeuse is. If we had picked R band, the answer would be quite different. To continue: In order to get the brightness ratio, we use the defintion of the magnitude scale: m1 − m2 = −2.5 log (F1 /F2 ), (1) where F1 /F2 is the brightness ratio were looking for. So we first solve this equation for the ratio, and then we can simply plug the numbers in and find the answer. The brightness ratio is given by m2 −m1 0.09−2.35 F1 = 10 2.5 = 10 2.5 = 0.125. F2 (2) So Rigel is about eight times brighter than Betelgeuse, in the B band. 2. If one star has m1 = 8.4 and the other star has m2 = 5.2, then we know right away that star two is brighter than star one. We know that five magnitudes difference is about 100 times brighter, and one magnitude difference is 2.5 times brighter, but this is 3.2 magnitudes difference, so it will be between 2.5 and 100, but closer to 2.5. Using the magnitude equation (Equation 1), we can put 3.2 on the left and solve for F1 /F2 to get that star two is about twenty times brighter than star one. The ratio of F1 /F2 = 0.0525, to be exact. So if F2 is 5% of F1 , then both of them together must be 105% of F1 . Which is to say, adding star two to star one makes the brightness increase by 5%. So what would the magnitude change corresponding to a 5% increase be? Well, we use Equation 1 again, but now the left side is unknown, and the right side is −2.5 log (1.0525) = −.05555. This means adding star one will make the magnitude go down by 0.0556, compared to star two. This means the magnitude of both will be 5.1, which is indeed less than 5.2 (because 1 both stars are indeed brighter than one star), but not much less, because a 5% increase is not a huge change. You could get the same answer by using star one as the comparison. Then both stars together would be 20 times brighter than star one. (Star two is 19 times brighter than star one, so the sum is 20 times brighter than star one by itself.) Again using Equation 1, −2.5 log (20) = −3.255, which is the change from star one by itself. Since star one is 8.4, we subtract 3.255 to get 5.1, which is the same thing we got the other way! 3. This problem is simply an exercise in getting used to scale and to thinking about non-visible light, and how all the different parts of the electromagnetic spectrum compare. To compute the answer, you get the frequency of the light (use c = λν if you dont know it), and then use E = hν to get the energy. c = 3.00 × 108 m/s, and h = 6.63 × 10−34 Js. So if I get frequency in hertz (Hz), I will get energy in joules (J). a) WQFS: 6.03 × 10−26 J b) Microwave oven: 1.60 × 10−24 J c) Cell phone: 7 × 10−24 J d) Human body: 2 × 10−20 J e) Red photon: 3.1 × 10−19 J f) Green photon: 3.6 × 10−19 J g) Blue photon: 4.7 × 10−19 J h) Dental x-ray: 10−13 J i) Gamma ray: 10−10 J A couple patterns I hope you notice: first of all, the visible light spans a tiny range of energy, compared to all the other sources of light. The difference between red and blue is less than a factor of two!!! Note that radio waves carry 10 million times less energy than visible light, which in turn carries a billion times less energy than gamma rays! Overall, the total span of these relatively common sources of radiation spans seventeen orders of magnitude!!!! Just to put it in perspective, the ratio of the sun-earth distance to the thickness of a sheet of aluminum foil is ten times smaller than that! Another interesting point: note that the energy of a cell phone photon is actually more than three times higher than that of a microwave oven photon. And yet a microwave oven will cook your food, while a cell phone causes no harm. (The people who think cell phones cause brain cancer seem to be wrong.) I can think of two reasons for this: first, ovens use a LOT more power than a cell phone, which means although each photon carries less energy, there are a heck of a lot more of them. Second, the energy of a 2 microwave oven photon has been carefully chosen because water “likes” to absorb that particular energy, so that energy gets absorbed more easily (by water) than a cell phone photon. 4. You are monitoring a fading star with your telescope. You measure that the magnitude is increasing at a constant rate of 0.2 magnitudes every 24 hours. How would you describe the change in brightness of the star as a function of time? Usually, the concept of magnitudes is introduced in comparing two different stars. However, it works just as well for comparing the same star at different times. If ∆m = 0.2 in one day, then we can use the definition of magnitudes to get the brightness ratio between today and yesterday: ! mtoday − myesterday Ftoday . = 0.2 = −2.5 log Fyesterday (3) Solve for the brightness ratio: 0.2 Ftoday = 10− 2.5 = 0.832, Fyesterday (4) which is another way of saying that every time 24 hours pass, the star has dimmed to 83% of its brightness the previous day. Alternately, you could say that the brightness dims by 17% each day. 5. You have two 100-watt light bulbs, but one of them is eight times as far from you as the other. What is the difference between their magnitudes? If one light bulb is 8 times as far away as the other, it must look 82 = 64 times dimmer. Which means the ratio of brightnesses is 64. So I can plug that into the magnitude equation: m1 − m2 = −2.5 log (64) = 4.5. (5) A brightness ratio of 64 corresponds to a magnitude difference of 4.5, with the more distant (dimmer) light bulb having the larger magnitude number. Note that a change in brightness of 100 times corresponds to a magnitude change of 5, and you found in problem 2 that a change in brightness of 20 times corresponds to a magnitude change of 3.2, so it would make sense that a magnitude change of 4.5 would be from a brightness change between 20 and 100 times, which 64 is. 6. The sun puts out a total power of about 1026 joules of energy every second in the form of light. 3 a) If you assume it is all emitted as visible light, roughly how many photons are emitted by the sun every second? In problem 3, you found that red, blue and green light all have energies of about 10−19 J, so I am going to just assume that the energy of a single photon is 10−19 J. If the sun is putting out 1026 J every second, I divide those two numbers to get 1045 photons every second. b) Roughly how many photons per square meter are hitting the Earth from the sun? The Earth is 1.5 × 1011 m from the Sun. A sphere of radius r has a surface area of 4πr2 . That means the 1045 photons have to be spread out among 3 × 1023 m2 . That means for every square meter at the Earth, there are about 4 × 1021 photons passing through that square meter every second. c) How many photons enter your eye every second if you look directly at the Sun? How many photons go in your eye each second if you look at the Sun through an 8” telescope? The area of your pupil is roughly 2 × 10−5 m2 . If we multiply this by the answer from the previous problem, we get 8 × 1016 photons per second, or 1017 if we round off. So, if you look at the sun for a second, almost ten thousand billion photons hit your retina. That’s a lot. Of course, if you were to look at the sun, your iris would contract, shrinking the size of your pupil to block more photons, but not by a factor of 10! It still hurts!! The area of the telescope aperture is roughly 103 times bigger than the area of your eye. So a thousand times as many photons enter the telescope than enter your eye every second. That’s about 1019 photons entering the telescope every second. That’s why you never want to look through the telescope at the sun; its like your pupil just got a thousand times bigger. (Although the secondary blocks some of that.) d) The star Sirius is roughly 25 magnitudes dimmer than the Sun. When you look through the 8” telescope at Sirius, roughly how many photons enter your eye each second? By the magnitude formula in Equation 1, if the difference in magnitudes is 25, that means the brightness ratio is 10−10 . If 1020 photons enter the telescope every second from the sun, then 1010 photons will enter the sun every second from Sirius. Remarkable, when you think about it. Sirius is so unimaginably far away, and the telescope is only 8” across, and yet ten billion photons from Sirius would go through its opening every second. 4