Wk12-GalaxyRotationActivity

Your Name: Lab TA: Zoom Group Partners: Astronomy 1101 Galaxy Rotation and Dark Matter Lab Activity Worksheet Galaxy Rotation Figure 1, below, shows a visible-light image of the galaxy NGC3198, a galaxy that is roughly half the size and half the total luminosity of the Milky Way: Figure 1 : Visible-light image of galaxy NGC 3198 – a galaxy is located in the constellation Ursa Major. This image was taken as part of Advanced Observing Program (AOP) program at Kitt Peak Visitor Center during 2014. Credit: KPNO/ NOIRLab/ NSF/ AURA/ John Vickery and Jim Matthes/ Adam Block 1. How would you classify NGC 3198 using the Hubble’s Galaxy Classification scheme from the book? I would say that the N GC 3198 is a spiral/spaghetti 2. Based just on this image , do you think the galaxy is rotating clockwise (as seen from our vantage point) or counter-clockwise? Explain the basis of your answer. Clockwise. Since the arms are all going towards the clockwise direction the galaxy must be moving in the clockwise direction 3. Based on the colors you see in the image, what can you say about the ages of stars you would find in this galaxy? Think back to what you know about stellar evolution and what you know about galaxies. I would say that the galaxy is relatively younger 1

Figure 2 (below) shows the luminosity profile of NGC 3198: the total luminosity of all stars inside radius R, where R is the distance from the center of the galaxy. The vertical axis is in units of billions of L sun . The radius R on the horizontal axis is in units of kiloparsecs (kpc). Figure 2: Luminosity Profile of NGC 3198 4. What are the units of the vertical axis of the graph? Lumens per second Luminosity within R (billions of Sun) As you can see, the total luminosity of ALL the stars in NCG 3198 is about 9 Billion L sun . About half of that light comes from inside the radius R = 5 kpc and about half of comes from between 5 and 15 kpc. The curve stops increasing at 15 kpc because there are almost no stars beyond that radius in NGC 3198, as you can see from Figures 1 & 2. If all of the stars in NGC 3198 were exactly like the Sun, we could just multiply the total luminosity by 1 M sun / L sun (1 solar mass per unit of solar luminosity) to infer that the total mass of all the stars in the galaxy is about 9 billion M sun . However, the light actually comes from a mix of main sequence stars, red giant stars, white dwarfs, etc. with a wide range of luminosities and masses. By the time you average over all the stars and their numbers, it turns out that you need to multiply by a mass-to-light ratio of about 4 M sun / L sun to infer the stellar mass. Stellar Mass = Interior Luminosity × (4 M sun / L sun ) 2

Figure 3 (below) shows the rotation curve of NGC 3198, a plot of the disk’s rotation speed (measured from Doppler shifts) versus distance from the center of the galaxy. The rotation curve is measured from radio observations of hydrogen gas, and it extends to 30 kpc, roughly twice the radius of the galaxy in the visible image. Figure 3: Rotation Curve of NGC 3198 We can also use Newton’s laws of gravity to infer the mass of the galaxy from the velocities of the gas, just as we have used the orbital speeds of stars to infer the mass of the Galactic Center black hole in a previous lab. Here we use our familiar formula M = v 2 × R/G, but to make your lives easier, we have converted the units for you to express it in a scaled form: Total Mass = (2.3 × 10 5 M sun ) × (v in km/s) 2 × (R in kpc). When we apply this formula to the measured rotation speed at radius R, it tells us the total mass of the galaxy inside a circle of radius R. For example, if you measured a velocity of 110 km/s at a radius of 2 kpc, you would infer that the mass inside 2 kpc is Total Mass inside 2 kpc = (2.3 × 10 5 M sun ) × (110 km/s) 2 × (2 kpc) = 5.6 × 10 9 M sun 5. Using the Stellar Mass formula and the luminosity profile in Figure 2: Stellar Mass = Interior Luminosity × (4 M sun / L sun ) along with the Total Mass formula and the observed rotation curve shown in Figure 3: Total Mass = (2.3 × 10 5 M sun ) × (v in km/s) 2 × (R in kpc) 4

Compute the “stellar mass” and “total mass” of NGC 3198 interior to these 5 radii on the next page: R = 2 kpc, 5 kpc, 15 kpc, 20 kpc, and 30 kpc. Computed Masses R = 2 kpc Stellar mass interior to R, from luminosity profile: (2.7 X 10^5 MSUN) * (4 M sun / L sun ) = 8.7 * 10^5 Total mass interior to R, from rotation curve: Total mass = (2.3 × 10 5 M sun ) × (110 in km/s) 2 × (2 in kpc) = 5.6 Billion M R = 5 kpc Stellar mass interior to R, from luminosity profile: (4.38 x 10^5 Msun) * (4 Msun/Lsun) = 17.52 * 10^5 Total mass interior to R, from rotation curve: = 22.5 billion m R =15 kpc Stellar mass interior to R, from luminosity profile: 8.35 x 10^5 MSun * 4 Msun/ LSun = 33.4* 10^5 Total mass interior to R, from rotation curve: Total Mass = 2.3 X 10^5 Msun X 150 in km/s^2 x 15 in ppc = 77.6 billion m R = 20 kpc Stellar mass interior to R, from luminosity profile: 8.49 x 10^5 MSun * 4 M sun /L sun = 33.96 x 10^5 Total mass interior to R, from rotation curve: 2.3 x 10^5 MSun x 140 in k/s x 20 in kpc = 90 billion M R = 30 kpc 5

6. Is the stellar mass interior to 2 kpc sufficient to explain the observed rotation speed measured at 2 kpc? In other words, do the stars themselves have enough mass to approximately produce the gravitational force necessary to explain the speed of the gas at 2 kpc? No, the stellar mass does not have enough to produce the amount of force necessary only if the mass is increased by a substantial amount. 7

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