Astronomy 142 Recitation #8

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1 Astronomy 14 Recitation #8 March 13 Formulas to remember Free-fall time: tff = 3π 3Gρ Pressure from random motions of stars: P = ρv r, where ρ is the mass density of stars (mass of stars per unit Galactic volume including the space between the stars, not the mass density of the stars themselves) and v r the random component of motion of a typical star, separate from any systematic (e.g. orbital) motion. One-D hydrostatic equilibrium: dp = ρ gz Mass per unit area in a hydrostatically-supported galactic disk like the Milky Way s: = gz G = vz GH µ π π Relaxation time: 3 v 1 R N tc = = 4πG mnln ( Rr) v N 4ln Virial theorem: In thermal equilibrium or steady state, d I dt = K+ U = K+ E, where I, K, U, and E are the moment of inertia, total kinetic energy, total potential energy and total mechanical energy of a closed system of particles whose interactions can be characterized by a scalar potential. It is often the case that d I dt =, for which it simplifies to K = U. Rotation curves Abbreviations: Workshop problems vr ( ) = GM r Point mass: Keplerian rotation vr ( ) = r 4πGρ 3 Constant density: solid-body rotation v( r) = 4 πgρ r = constant 1 r density distribution: flat rotation curve kpc = 1 pc; Myr = 1 years; Gyr = 1 years. Warning! The workshop problems you will do in groups in Recitation are a crucial part of the process of building up your command of the concepts important in AST 14 and subsequent courses. Do not, therefore, do your work on scratch paper and discard it. Better for each of you to keep your own account of each problem, in some sort of bound notebook. 13 University of Rochester 1 All rights reserved

2 Astronomy 14, Spring A constant-density, spherical molecular cloud undergoes free-fall collapse. Consider molecules at two particular initial radii: those at the initial cloud radius r, and those halfway between center and edge, at r. Which molecules reach the center first? Explain in detail.. Suppose that, in the Sun s neighborhood of the Galactic disk, the density can be thought of as an infinite planar distribution, with a plane of surface density µ (mass per unit disk area) volume density ρ (mass per unit volume) that decreases sharply with increasing elevation z above or below z =. a. Explain why, under these assumptions, the equation of hydrostatic equilibrium can be written to good approximation as dp = πgρµ. b. Solve this equation for the density as a function of z, and show that the density scale height thus obtained is the same as obtained in class on Tuesday. 3. The number of pairs among N objects. This result comes up in so many contexts in physics that it s worth carrying the derivation in your head. a. In how many distinctly different ways can one arrange N distinguishable objects? (If you prefer a concrete situation: suppose you have N books to put on a shelf. In how many different ways can you place them on the shelf?) b. Suppose the N objects were indistinguishable from one another. In how many distinctly different ways can one arrange these objects. (For example, N identical books on a shelf.) c. In how many distinctly different ways can one arrange N objects, of which n and N n are of two distinguishable types, but of which the n objects of the first type are indistinguishable from one another, and similarly the N n are indistinguishable from one another? (For example, n identical blue books and N n identical green books on a shelf.) d. So how many pairs are there among N identical objects? 4. What is the total gravitational potential energy of a cluster of N stars that have typical mass m and typical separation r? 5. Use the virial theorem to show that a cluster of N stars with typical random velocity v and typical separation r has total mass M = rv G 6. A typical diameter for both open and globular clusters is 6 pc. Open clusters typically have about 6 1 members, compared to 1 for a globular cluster. Random velocities of their members are 1 around 1 km sec 1 for open clusters and 1 km sec for globular clusters. Calculate the relaxation times for typical open and globular clusters. Then look up the range of ages thought to apply to each of these sorts of clusters. Are the clusters thermalized or not? Learn your way around the sky, lesson 7. (An exclusive feature of AST 14 recitations.) Use the lab s celestial globes, TheSky running on the lab computers, and the SIMBAD database, 13 University of Rochester All rights reserved

3 Astronomy 14, Spring 13 and any other resources you would like to use, to answer these questions about the celestial sphere and the constellations. 7. Literary astronomy. It has been safe in recent years to assume that all high-school graduates have read the Harry Potter books, whence came the literary-astronomy examples we used in AST 111. But for several generations it has been safe to assume that all high-school graduates have read Lord of the Rings. So... Early in The Fellowship of the Ring, Tolkien tells us that Frodo, Sam and Pippin set off on their journey from Hobbiton to Crickhollow on Frodo's birthday, September. Next afternoon, the hobbits encounter a group of elves wandering through the woods. They are all going the same direction, so hobbits and elves walk together until very late that night, reaching a camp at which they dine and spend the night. A while after midnight Frodo looks at the eastern sky: Away high in the east swung Remmirath, the Netted Stars, and slowly above the mists red Borgil rose, glowing like a jewel of fire. Then by some shift of airs all the mist was drawn away like a veil, and there leaned up, as he climbed over the rim of the world, the Swordsman of the Sky, Menelvagor with his shining belt. a. What constellation does Tolkien mean by Menelvagor? b. So what does he mean by Borgil, and by the Netted Stars? c. Did Tolkien get it right? Can one see these objects recently risen at such an hour, on 4 September? 8. First, review your solution to the "Learn your way " problems 5-7 on February 13, in which you computed the effect of the elliptical shape of the Earth s orbit on local sidereal time, under the assumption that the orbit, though elliptical, lies in the Earth s equatorial plane. (It would help if you would also retrieve the digital version of Excel plot you made in problem 7.) Of course the real orbit (path of the Sun) doesn t lie in the equatorial plane at all; the Sun travels through the sky along the ecliptic instead of the celestial equator. These paths are not the same angular length; yet both the real and mean Sun get from equinox to solstice (6 hours of sidereal time) in a quarter of a year. Thus the angular speed ω o of the real Sun varies along its track because of the tilt of the ecliptic a. For simplicity, suppose the Earth s orbit is circular, but that the ecliptic and the celestial equator are tipped by ψ = 3.5, and that they intersect at the vernal and autumnal equinoctes. Also suppose for simplicity that the departure of the real Sun (following the ecliptic) from the mean Sun (following the equator). Calculate the difference between the angular speeds of mean and real Sun, ωo = ω ωo, at the vernal equinox. b. Argue that ωo has the same value at the vernal and autumnal equinoctes, and the opposite of these values at the solstices, and that these are the maximum tilt-related differences between angular speed of real and mean Sun. c. Propose a simple form for ω o ( t ) angular difference θ o ( t ) that satisfies the results of part b, and integrate it to obtain the between real and mean Sun. 13 University of Rochester 3 All rights reserved

4 Astronomy 14, Spring 13 d. Plot θ o ( t ) difference ( t) over the course of a year, and compare the result to that obtained for the angular θ ε due to orbital eccentricity. Plot also the net angular difference between real and mean Sun, by adding these two results. This combined result is called the equation of time, and relates sidereal time (or, if you like, sundial time) to clock time (mean solar time). 13 University of Rochester 4 All rights reserved

5 Astronomy 14, Spring 13 Solutions 1. They all reach the center at the same time. If the density is uniform, the free-fall time is independent of position. This reflects the fact that the outer components are subjected to greater gravitational acceleration from the interior material, and it catches up with everything else at the center.. a. Officially, what we should mean by µ is µ = ρ ( z) If we are interested in solving the hydrostatic equilibrium equation for positive z, and not very high above z =, then the dominant mass is the plane at z <, for which the surface density is µ. Thus dp µ = ρ gz = ρ π G πgρµ =. b. We use ρ v z for P, assume the vertical component of random velocity, v z, to be independent of z, and proceed to separate and integrate: dρ vz = π Gρµ ρ( z) z dρ π Gµ = ρ ρ( ) vz ( ) ln ρ z π Gµ z ( ), v = z z = z = ρ v z ; z π Gµ zz ρ z = ρ e. ( ) ( ) Note that z = H, as we defined H on Tuesday as the scale height above and below the plane, so this is the same result as obtained on Tuesday (pages 19- of the lecture notes), but obtained a little less inelegantly. 3. a. Select the first object; there are N different choices. Select the second: there are N 1 different N N 1. choices for that, so the number of different outcomes so far is the product of these, ( ) There are N next choices, raising the total to N( N 1)( N ). And so on: after all N objects are selected and arranged, the number of different ways it could have been done is ( )( ) ( )( ) N N 1 N 1 = N! b. If the objects are indistinguishable, and different arrangements cannot be told apart, then there s only one distinct arrangement possible. Apparently indistinguishability among the objects reduces the number of distinct, different arrangements. 13 University of Rochester 5 All rights reserved

6 Astronomy 14, Spring 13 c. Of the N! different arrangements, a factor of n! don t count because different arrangements of the first objects are indistinguishable. Similarly a factor of ( N n)! don t count because the second type of objects aren t distinguishable either. Thus the number of distinctly different arrangements needs to be reduced by both of these factors, to N! n! N n! ( ) d. And thus for n = : the number of pairs among N objects is 4. Each pair of objects has potential energy N! N N =!! ( N ) u Gm r ( 1) =, and there are ( ) ( 1) ( 1) un N Gm N N Gm N U = = r r for any cluster with enough members to be worthy of the name. 5. In spherical symmetry N N 1 pairs, so d I dt and its averages can safely be taken to be zero, so 6. This comes out to t c 13 University of Rochester 6 All rights reserved N U Gm N K = mv = = 4r 1 4r rv M = Nm = v = G G R N = 4 Myr (open cluster), Gyr (globular cluster). v = N 4 ln Ages must be much greater than relaxation time for thermal equilibrium. Open clusters range from extremely young (< 1 Myr) to moderate (Pleiades 1 Myr, Hyades 66 Myr), rarely very old (M67, 4 Gyr). The youngest ones aren t thermalized but by the time one gets to the Hyades they probably are. Globular clusters are all about 1 Gyr old. So they are mostly close to thermalized but not by very much. 7. a. A swordsman with a shining belt is a giveaway: Menelvagor = Orion. Elsewhere in his books and notes Tolkien identifies the constellation with Oromë, one of the godlike Valar who run the world on which Middle-Earth is situated. The two names are similar enough that Tolkien must have named his divine hunter after our constellation Orion. b. As Orion rises in the east, we indeed find a bright red star almost directly above the belt -- which itself runs almost straight up and down as the constellation rises, as seen from our latitude -- and a nebulous group of "netted" stars directly above that. So Remmirath = the Pleiades, and Borgil = Aldebaran, α Tauri, brightest of the Hyades.

7 Astronomy 14, Spring 13 (I note that there's a bunch of web pages by Tolkien zealots in which these objects are identified, and all the ones I've seen identify Borgil with Betelgeuse, α Ori. To do so ignores Tolkien's detailed description -- if the mists cover the Belt, they will cover Betelgeuse too (see below) -- and the fact that another bright red star fits the description much better.) c. Here's a chart of the eastern sky at 1AM, standard time, on 4 September, at latitude 43 N. (The Shire is probably supposed to be a bit further north.) Pleiades ("Remmirath") Aldebaran ("Borgil") ("Menelvagor") So Tolkien got it precisely right. 8. a. Since the real Sun, following the ecliptic, is taking a longer path toward the next equinox than the mean Sun, its angular speed is smaller by projection between the two paths: ( ) cos ω o VE = ω ψ, so 8 1 ( VE) ( VE) ωo = ω ωo = rad sec. 13 University of Rochester 7 All rights reserved

8 Astronomy 14, Spring 13 b. The geometry of real and mean Solar motion is the same at both equinoctes, so the same value of ω o applies. If this value applied all through the year, though, real time would not catch up with sidereal time every vernal equinox: they would be different by h 4 h ω o ( VE ) 1 year =, π rad so in other parts of the path the angular speed difference must offset this. We suppose the offset must be symmetrical, and so that the speed differences at the solistices are equal and opposite to those at the equinoctes. c. The simplest symmetrical form is sinusoidal. Taking the time of vernal equinox to be t 1, and noting that the frequency of the oscillation must be ω in order that it have two peaks and two troughs each year, we get whence ( 1 ) ( t) ( VE) ( t t ) ωo = ωo cos ω, ( VE) ωo θo ( t) = sin ( ω( t t1 )). ω d. I did this in the same Excel document used last time: 1 Mean Sun - Sun (minutes) Obliquity Mean Sun - Sun (minutes) Combined effects -8-1 Eccentricity Time (days) Time (days) 13 University of Rochester 8 All rights reserved