I Need help solving the table ? Part 2. True Rotation Rates Because…
Question Answered step-by-step I Need help solving the table ? Part 2. True Rotation Rates Because… I Need help solving the table ? Part 2. True Rotation RatesBecause each galaxy is at a different inclination relative to our point of view, the width of the emission line does not accurately describe how fast the galaxy is actually rotating, as shown in Figure 2. So, we need to adjust our measured widths depending on the orientation of the galaxy.Image transcription textMore observed rotation 6″ RNo observed Aux, rotation Allobserved rotation &a… Show more… Show moreFigure 2: Examples of observing a spiral galaxy at different inclinations To correct for the galaxies’ inclination:Image transcription textW. 50 W 50 (corrected = (Equation 1)sin(i… Show more where W50 is your measured line width from question 1 and i is the inclination of each galaxy, which is listed for you in the 2nd column of Table 1. WE RECOMMEND THIS WEBSITE FOR YOUR CALCULATIONS: https://www.wolframalpha.com/An example using Equation 1, you would type in: 200/sin(80), WHERE 200 IS W50 AND 80 IS iIF YOU USE A SCIENTIFIC CALCULATOR/YOUR PHONE, CHECK TO MAKE SURE YOU ARE IN DEGREES, NOT RADIANS.IF YOU GET A NEGATIVE NUMBER, YOU ARE IN RADIANS, STOP IT, BIG NO-NO! Calculate W50(corrected) for each galaxy and record your answers in column 3 of table 1. Round off your answer to 2 decimal places. Part 3. From Rotation to BrightnessThe most important application of the HI emission line width is its use as an extragalactic distance measurement method on one of the furthest-reaching rungs of the cosmic distance ladder. The Tully-Fisher relation (named after Brent Tully and Richard Fisher) is a correlation between a galaxy’s rotation rate and its intrinsic luminosity (how bright it actually is), much like the correlation between a Cepheid variable star’s period and its absolute magnitude (Leavitt’s Law, or The Period-Luminosity relation). So, the Tully-Fisher relation allows you to calculate each galaxy’s true luminosity using your W50(corrected) calculations. The relation is: Image transcription textL = W=(corrected 4 Equation2… Show more where L is the galaxy’s luminosity in units of solar luminosities. Calculate luminosities for each galaxy IN SCIENTIFIC NOTATION, ROUNDED OFF TO 2 DECIMAL PLACES, and record them in column 4 of Table 1. Now that you have the galaxies’ luminosities, you can calculate their absolute magnitudes as follows:Image transcription textM = -2.510910 (L) + 4.82 (Equation3)… Show more where M is the galaxy’s absolute magnitude and L is the luminosity value from Question 3. AGAIN, WE RECOMMEND THIS WEBSITE: https://www.wolframalpha.com/Example using Equation 3 : -2.5log10(1e10) + 4.82, WHERE 1e10 (1 x 1010) IS LIF YOU USE A SCIENTIFIC CALCULATOR/YOUR PHONE, MAKE SURE TO USE LOG BASE 10! Calculate the absolute magnitudes of your galaxies and record your results in column 5 of table 1. Round off your answer to 2 decimal places.Part 4. Distance!You now have ½ the pieces you need to calculate a distance. You have absolute magnitudes, and in order to calculate a distance modulus, you also need apparent magnitudes. The writer of this lab was nice enough to give those to you too! The galaxies’ m values are listed in column 6 of table 1. You can now calculate distances using the distance modulus equation: Image transcription textD = 10(m-M+5)/5 Equation4… Show more Note that D will be in parsecs, so divide your distance by 1,000,000 in order to convert from parsecs to megaparsecs (Mpc). Use the distance modulus equation to calculate distances for all 5 galaxies and record your results in the last column of Table 1. Note that D will be in parsecs, so divide your distance by 1,000,000 in order to convert from parsecs to megaparsecs (Mpc). Use the distance modulus equation to calculate distances for all 5 galaxies and record your results in the last column of Table 1. Table 1GalaxyW50(km/s)i(°)W50(corrected)(km/s)L(L?)M(mag)m(mag)D(Mpc)Adam 53.3 12.61 Michael 64.1 12.82 Val 45.8 14.00 Christian 71.5 15.28 Ben 88.2 14.95 Science Astronomy SCI 2240 Share QuestionEmailCopy link Comments (0)
