Lab4 Magnitudes (Complete)

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Sam Houston State University *

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1403

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Astronomy

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Dec 6, 2023

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Magnitudes L AB 4 THE M AGNITUDE S CALE P URPOSE The purpose of this lab is to learn more about an alternate measurement astronomers use to designate the luminosity and apparent brightness of stars. After doing this lab, you should be able to explain the concept of apparent and absolute magnitudes, how they relate to the brightness of a star, and how this system of measurement works. You will also gain experience in relating apparent versus absolute magnitude to the distances of celestial objects. demonstrating the use of units of measurement in astronomy, including Astronomical Units, light years and parsecs. using essential vocabulary of the discipline being studied. B ACKGROUND I NFORMATION W HAT ARE M AGNITUDES ? When the Greek scientist Hipparcos determined first studied the brightnesses of stars, he did it "by eye." The first stars that "came out" at night were of first importance or first "rank", and today we call those stars "first magnitude stars". The dimmest stars that he could observe were called “sixth magnitude stars” with other stars of varying brightness being classified somewhere in between. Today, with measuring instruments, we know that what the Greeks called a difference of 5 magnitudes (from 1st magnitude to 6th magnitude) is close to a 100 times difference in apparent brightness. This difference now defines the magnitude scale. With this new definition, the magnitude scale was broadened to the very brightest stars the Greeks saw, and some magnitudes became negative. With light-meter instruments capable of discerning small differences in brightness, some magnitudes became fractional, much like earthquake magnitudes. And some objects that could be seen only with telescopes could now have magnitudes attached to them, and those magnitudes were numbers larger than 6, the naked eye (unaided) limit for most people.
Magnitudes Magnitudes and brightnesses are related to each other, but the magnitude scale is a DIFFERENCE (subtractive) scale and the apparent brightnesses (how bright the objects appear) make up a MULTIPLICATIVE scale. Thus if two stars have NO difference in their apparent magnitudes (i.e. they look exactly the same brightness), you multiply the brightness of one star by 1.0 to get the brightness of the second. On the other hand, if two stars have a difference of 1.0 in their apparent magnitudes, the brighter one (lower magnitude) will appear 2.512 times brighter than the dimmer one. Table 1 shows how the differences in magnitude correspond to changes in apparent brightness: T ABLE 1: T HE R ELATIONSHIP B ETWEEN M AGNITUDE C HANGES AND B RIGHTNESS C HANGES Apparent Magnitude Difference Brightness Multiplier 0.0 1.0 1.0 2.512 2.0 6.310 3.0 15.85 4.0 39.81 5.0 100.0 6.0 251.2 7.0 631.0 8.0 1585 9.0 3981 10.0 10,000 Each magnitude difference corresponds to 100^(0.2) or 100^(1/5) times in brightness. A magnitude difference of "m" corresponds to a change of 100^(0.2m) times in brightness. A little algebra will show that this is the same as 10^(0.4m). If we wanted to know what a 20 magnitude difference would be, we could calculate it as: 10^(0.4*20) = 10^8, or 100,000,000 times! Between the brightest object we see from Earth (the Sun, magnitude -26.8) and the faintest stars we see from Earth (with the Hubble Space telescope & the Keck Foundation telescope on Mauna Kea in Hawaii, magnitude = +30), there is a difference of 30-(-26.8)= 56.8 magnitudes or 10^(0.4*56.8)= 10^(22.72)= 5.25x10 22 times difference in brightness! B RIGHTNESS & M AGNITUDES : You may recall from the previous lab that for two objects of the same luminosity, if one is 10 times farther away then it will appear 100 times dimmer (due to the inverse square law of light). A factor of 100 in brightness is ALSO the same as making an object 5 magnitudes fainter! So there is a way of relating the apparent brightness to changes in magnitude and distance. 2
Magnitudes Distance Factor Brightnes s Factor Magnitude Difference 1.0 1.0 0.000 2.0 4.0 1.505 4.0 16.0 3.010 10.0 100.0 5.000 To get column 2, each distance factor is squared to show by what factor an object’s apparent brightness and magnitude are changed by changes in distance. For example, if an object of magnitude 3.5, for example, gets twice as far away, then it becomes 1/4th the brightness (i.e., 4 times dimmer), and its magnitude will become 1.5 magnitudes dimmer , and the apparent magnitude will change from 3.5 to 5.0. How did we get the last column? For 10^(0.4*x)=4 times dimmer, x turns out to be close to 1.5. You've probably heard of "a light-year" -- how far light travels in one year. Well, a parsec is how far away an object has to be so that it has a PARallax of one SECond of arc, or 1/3600 degree. In other words, as our Earth travels around the Sun, a distant object will seem to shift back and forth because the Earth is moving. To have a shift of one second of arc, an object has to be 3.26 light-years away. At a distance of 10 parsecs (pc), an object is 32.6 light-years away. If we were to see our Sun from 10 pc, our Sun would be one of the dimmest stars we could see with the naked eye, approximately 5th magnitude! This distance is a special one, and a star's apparent brightness as seen from 10 pc is called the ABSOLUTE MAGNITUDE of the star. If we imagine artificially moving stars to this distance, then the difference in brightness seen must be due to a difference in the energy output of the star. In other words, the absolute magnitude tells us something of the star’s actual wattage. As you can see, there is a relationship between apparent magnitude, absolute magnitude, and distance. The exact relationship is: Where m is the apparent magnitude, M V is the absolute magnitude (in visible light), and d is the distance to the star in parsecs. We are now in a position to find the absolute magnitudes of stars. Recall from our previous lab that we determined the distances to a number of stars based on their parallax. In that lab, we used HD 14802 as our example. For HD 14802, m = 5.19 and the distance we found was 21.93 pc. Thus: M V = 5.19 – 5log (21.93) + 5 M V = 5.19 – 5(1.341) + 5 3
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Magnitudes M V = 5.19 – 6.705 + 5 M V = 3.48 The absolute magnitude of HD 14802 is 3.48. Remember that the SMALLER the number, the greater the wattage. HD 14802 therefore has a smaller wattage than Sirius – it would be dimmer if you placed these two stars side by side. 4
Magnitudes P RE -L AB Q UESTIONS These questions must be answered before you show up to your lab class. You will not be able to start the lab exercise without the answers to these questions. A quiz based on these questions will be given at the beginning of the lab session. You should use your textbook (or any other reference source) to help answer these questions. 1. Star A has an apparent magnitude of 1.0. Star B has an apparent magnitude of 6.0. Which star appears brighter? By what factor? Star A. 2. At what distance from Earth is the absolute magnitude of a star equal to its apparent magnitude? 10 parsecs. 3. What is the apparent magnitude of our Sun? the absolute magnitude? The apparent magnitude of our Sun is -26.8. The absolute magnitude is 4.7. 4. If we could fly in a spaceship towards another star, as we traveled toward the star, how would its apparent and absolute magnitudes change? The star would seem brighter and the value of the apparent magnitude would change. The absolute magnitude would stay the same. 5. What is the absolute magnitude of Sirius, the brightest star in the night sky? -1.5 5