21ST CENT.AST.W/WKBK+SMARTWORK >BI<
6th Edition
ISBN: 9780393415216
Author: Kay
Publisher: NORTON
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Chapter 20, Problem 16QP
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A galaxy's rotation curve is a measure of the orbital speed of stars as a function of distance
from the galaxy's centre. The fact that rotation curves are primarily flat at large galactocen-
tric distances (vrot(r) ~ constant) is the most common example of why astronomer's believe
dark matter exists. Let's work out why!
Assuming that each star in a given galaxy has a circular orbit, we know that the accelera-
tion due to gravity felt by each star is due to the mass enclosed within its orbital radius r and
equal to v?/r. Here, ve is the circular orbit velocity of the star. (a) Show that the expected
relationship between ve and r due to the stellar halo (p(r) xr-3.5) does not produce a flat
rotation curve. (b) Show that a p(r) ∞ r¯² density profile successfully produces a flat ro-
tation curve and must therefore be the general profile that dark matter follows in our galaxy.
How astronomers determine the distance of a galaxy? Explain.
Chapter 20 Solutions
21ST CENT.AST.W/WKBK+SMARTWORK >BI<
Ch. 20.1 - Prob. 20.1CYUCh. 20.2 - Prob. 20.2CYUCh. 20.3 - Prob. 20.3CYUCh. 20.4 - Prob. 20.4CYUCh. 20 - Prob. 1QPCh. 20 - Prob. 2QPCh. 20 - Prob. 3QPCh. 20 - Prob. 4QPCh. 20 - Prob. 5QPCh. 20 - Prob. 6QP
Ch. 20 - Prob. 7QPCh. 20 - Prob. 8QPCh. 20 - Prob. 9QPCh. 20 - Prob. 10QPCh. 20 - Prob. 11QPCh. 20 - Prob. 12QPCh. 20 - Prob. 13QPCh. 20 - Prob. 14QPCh. 20 - Prob. 15QPCh. 20 - Prob. 16QPCh. 20 - Prob. 17QPCh. 20 - Prob. 18QPCh. 20 - Prob. 19QPCh. 20 - Prob. 20QPCh. 20 - Prob. 21QPCh. 20 - Prob. 22QPCh. 20 - Prob. 23QPCh. 20 - Prob. 24QPCh. 20 - Prob. 25QPCh. 20 - Prob. 26QPCh. 20 - Prob. 27QPCh. 20 - Prob. 28QPCh. 20 - Prob. 29QPCh. 20 - Prob. 30QPCh. 20 - Prob. 31QPCh. 20 - Prob. 32QPCh. 20 - Prob. 33QPCh. 20 - Prob. 34QPCh. 20 - Prob. 35QPCh. 20 - Prob. 36QPCh. 20 - Prob. 37QPCh. 20 - Prob. 38QPCh. 20 - Prob. 39QPCh. 20 - Prob. 40QPCh. 20 - Prob. 41QPCh. 20 - Prob. 42QPCh. 20 - Prob. 43QPCh. 20 - Prob. 44QPCh. 20 - Prob. 45QP
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- I answer is not 100, I also tried 21. I need help! Thank you!arrow_forwardProblem 1: ( Compute the Oort constants A and B for Keplerian rotation: 0(R) = O, (R/Ro) terms of O, and Ro. If Keplerian rotation described the rotation of the Milky Way near the Sun, what would the numerical value of A and B (in kms-1 kpc-1) be and how does this -0.5 in compare to the observed values?arrow_forwardAnother commonly calculated velocity in galactic dynamics is the escape velocity vesc, that is the minimum velocity a star must have in order to escape the gravitational field of the galaxy. (a) Starting from the work required to move a body over a distance dr against f show that the escape velocity from a point mass galaxy is vse = 2GM/r where r is your initial distance. (b) Since we know galaxies aren't actually point-masses, also show that vesc from r for a galaxy with a p(r) x r-² density profile is vse = 2v²(1+ ln(R/r)). Here you must assume that R is a cutoff radius at which the mass density is zero. (c) The largest velocity measured for any star in the solar neighbourhood, at r=8 kpc, is 440 km/s. Assuming that this star is still bound to the galaxy, find the lower limit (in kiloparsecs), to the cutoff radius R and a lower limit (in solar units) to the mass of the galaxy. Note the solar rotation velocity is 220 km/s.arrow_forward
- The goal of the project is to find as many of the valuable images of galaxy collisions as possible while minimising the time invested in classification by the experts — they really don’t want to go through all of the “test negative” images to find a significant number of missed collisions. In order to better understand the capabilities of several citizen scientists for future projects, an expert classified all of the images in the library. From this information, the sensitivity and specificity of citizen scientists Beta, Gamma, and Delta in identifying images of galaxy collisions are shown in the table. Who do you think has the best performance given the goals of the project? Justify your answer.arrow_forward1. The current (critical) density of our universe is pe = 10-26kg/m³. Assume the universe is filled with cubes with equal size that each contain one person of m = 100kg. What would the length of the side of such a cube have to be in order to give the correct critical density? How many hydrogen atoms would you need in a box of 1 m³ to reach the critical density? The matter we know, which consists mostly of hydrogen, constitutes only 4.8% of the current critical energy density of our universe. So how many hydrogen atoms are actually in a box of 1 m3 in our universe? Deep space is very empty and a much better vacuum than we can obtain on earth in a laboratory.arrow_forwardAnother commonly calculated velocity in galactic dynamics is the escape velocity vesc, that is the minimum velocity a star must have in order to escape the gravitational field of the galaxy. (a) Starting from the work required to move a body over a distance dr against f show that the escape velocity from a point mass galaxy is vsc = 2GM/r where r is your initial distance. (b) Since we know galaxies aren't actually point-masses, also show that vesc from r for a galaxy with a p(r) xr¯² density profile is vese that R is a cutoff radius at which the mass density is zero. = 2v(1+ ln(R/r)). Here you must assume (c) The largest velocity measured for any star in the solar neighbourhood, at r=8 kpc, is 440 km/s. Assuming that this star is still bound to the galaxy, find the lower limit (in kiloparsecs), to the cutoff radius R and a lower limit (in solar units) to the mass of the galaxy. Note the solar rotation velocity is 220 km/s.arrow_forward
- I need the answer as soon as possiblearrow_forwardThe Tully-Fischer method relies on being able to relate the mass of a galaxy to its rotation velocity. Stars in the outer-most regions of the Milky Way galaxy, located at a distance of 50 kpc from the galactic centre, are observed to orbit at a speed vrot determine the mass in the Milky Way that lies interior to 50 kpc. Express your answer in units of the Solar mass. 250 km s-1. Using Kepler's 3rd Law,arrow_forwardThe Kormendy relation for ellipticals can be written as He = 20.2+ 3.0 log R. where R. is the half-light radius (in kpc) and 4e is the surface brightness (in magnitudes per square arc second) at R.. An elliptical galaxy obeying this relation will have a total luminosity Lo R for some index 7. What is the correct value of n? O a. n=-6/5 O b. n= 4/5 T23D Oc n= 16/5 O d. n cannot be determined with the information we have.arrow_forward
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