Problem Set 4 Solutions AY 7b Spring 2012 Problem 1 For the very simple model of the MW where Θ() = 200 km s 1, we know that where Ω = Θ/. Since Θ const, we can rewrite this as v r (l) = (Ω Ω 0 ) sin l (1) v r (l) = Θ ( 1 1 ) sin l (2) We define the angle l as shown in Figure 1. Because we can see that for <, the galactic longitude l will not cover the full angle ranges, to avoid having to define the limiting l, we instead use a galactocentric azimuthal angle φ, as shown in Fig 1. We use this figure to find an expression for sin l in terms of φ,, and. GC φ d Figure 1: Galactocentric coordinate diagram l The Law of Cosines gives d 2 = 2 + 0 2 2 cos φ d = 2 + 0 2 2 cos φ (3) The Law of Sines gives sin l = d sin φ sin l = d sin φ (4) So we have use Eqs 3, 4 to write l and v r (l) in terms of and φ: [ ] l = arcsin d sin φ v r (l) = Θ ( 1 1 ) [ d sin φ ] (5) (6) 1
Using Eqs 5, 6 and Θ() = 200 km s 1, = 8 kpc, we plot the orbits with = 4, 6, 10 and 12 kpc in Figure 2 below: 100 V r (l) vs l, using Θ() =200 km s 1 and =8 kpc 50 V r (l) (km s 1 ) 0 50 100 = 4 kpc 6 kpc 10 kpc 12 kpc 50 0 50 Galactic longitude l (degrees) Figure 2: V r (l) for various orbits in the MW Problem 2: C&O 24.3 For IAU C0923-545, the apparent visual magnitude is V = m V = +13.0, and the absolute visual magnitude is M V = 4.15. We are given that d = 9.0 kpc is its distance from Earth. (a) The estimated amount of interstellar extinction between the GC and Earth is found by using equation 24.1, giving that m V M V A V + 5 = 5 log 10 d (7) So the amount of interstellar extinction is A V = m V M V + 5 5 log 10 d = +13.0 ( 4.15) + 5 5 log 10 (9000 pc) (b) The amount of interstellar extinction per kiloparsec is A V = 2.38 (8) A V /9.0 kpc = 0.26 mag kpc 1 (9) Problem 3: C&O 24.9 Data in Table 24.1, and T H,ISM = 15 K. Assume that the disk of the MW has radius 8 kpc and height 160 pc. 2
Useful constants: M = 2 10 33 g m H = 1.67 10 24 g 1 pc = 3 10 18 cm k B = 1.38 10 16 erg K 1 (a) Estimate the average thermal energy density of hydrogen gas in the disk of the MW. We assume that the internal energy of the MW hydrogen gas is thermal (hence using the average temperature). For each H atom, E thermal = 3 2 kt, so the total energy in H gas will be E tot = EN = 3 2 knt, where N is the number of H atoms in the MW. We estimate N using N = M H gas/m H. Finally we write the energy density as u = E tot /V = EN/V = (EN)/(π 2 h) = 3 kt N 2 π 2 h (10) From Table 24.1, M gas = 0.5 10 10 M = 10 43 g, so N = M gas /m H = 6 10 66 atoms. E thermal = 3/2k B T = 3.105 10 15 erg, and V = π 2 h = π(8000 pc) 2 (160 pc)(3 10 18 cm) 3 = 8.686 10 65 cm 3 So u = EN/V = (3.105 10 15 erg)(6 10 66 )/(8.686 10 65 cm 3 ) u thermal = 2.1 10 14 erg cm 3 (11) (b) From pg. 897 of C&O, the average MW magnetic field in the ISM is B 0.4 nt = 4 µg. Eq. (11.9) gives u m = B2 2µ 0 (using SI), where µ 0 = 4π 10 7 N A 2. So u m = 6.4 10 14 J m 3 = 6.4 10 13 erg cm 3 (12) Using this rough estimate, u m 30 u thermal, which implies that it the energy density stored in the magnetic field of the MW is significant, so the ISM magnetic fields play a major role in the structure of the MW. Problem 4: C&O 24.15 (a) Assuming that Oort s high-velocity stars had v v esc 300 km s 1 (from pg. 908), and assuming that the velocity vectors of these stars are pointing radially away from the Galactic center, our rough order-of-magnitude mass estimate for M( ) is found using 2GM( ) v esc =, (13) where v esc = 300 km s 1, = 8 kpc. We solve to find M( ) = v 2 esc 2G = 1.6 1044 g M( ) = 8.1 10 10 M (14) The estimate from Ex. 24.3.1 gives M = 8.8 10 10 M, which is fairly close to our estimate. 3
(b) The extremely high-velocity stars discussed on pg. 908 have v 500 km s 1, so using this as the value of v esc and repeating the calculation gives M( ) = 4.5 10 44 g This extra mass could be due to the dark matter halo. M( ) = 2.25 10 11 M (15) (c) A problem with using the motions of stars in the solar neighborhood to determine the true mass of the galaxy is that much of the mass of the MW is located at r >, and in fact appears to be spherically distributed, this mass external to does not affect the dynamics of stars in the solar neighborhood. Problem 5: C&O 24.18 (a) Kepler s third law (Eq. 2.37) gives P = 2π [G(m 1 + m 2 )] 1/2 3/2 (16) We also know that P = 2π/Θ. So we can combine these expressions to get Θ(): (b) The Oort constants are defined in Eq. (24.39) and (24.40) as [ A 1 ] 2 d Θ 0 0 [ B 1 ] 2 d + Θ 0 0 [ ] 1/2 G(m1 + m 2 ) Θ = (17) First we find that d = 1 [G(m 1 + m 2 )] 1/2 = 1 2 3/2 2 Now using the equations for the Oort constants, we find Θ (18) A = 3 Θ 0 4 B = 1 Θ 0 4 (19) (20) (c) Assuming Θ 0 = 220 km s 1 and = 8 kpc, the Keplerian expressions give A = 20.6 km s 1 kpc 1 (21) B = 6.88 km s 1 kpc 1 (22) (d) The Hipparcos astrometry values are A = 14.8 ± 0.8 km s 1 kpc 1, B = 12.4 ± 0.6 km s 1 kpc 1. These values definitely don t agree with our Keplerian-derived values. The reason why these values differ is because in our derivation, when we found /d, we assumed that m 1 and m 2 are constant, which is bad because there is an dependence in M galaxy. Assuming that dm 2 /d = 0 is basically the same as assuming that the entire mass of the MW lies within the solar circle (i.e. inside ). 4
Problem 6: C&O 24.27 (a) Since the scale height of the MW disk is much less than the disk radius, and we re modelling the disk as essentially being an infinite plane, so the gravitational acceleration is always in the ẑ direction (i.e. g ẑ 0, g ˆx = g ŷ = 0), the appropriate Gaussian surface is a cylinder. Center the cylinder at the z = 0 plane, and have the ends be parallel to the midplane. The cylinder has ends of area A, and is a total height 2z, so each end is a distance z away from the midplane. Adding in the assumption that the disk has constant density ρ, Eq. 24.56 gives So the form of the gravitational acceleration g is 2Ag = 4πGM in = 4πGρA(2z) (23) g = 4πGρz (24) (b) Newton s second law: F = m a = m z, and since in this case we have a = g, we can write g = 4πgρz = z (25) earranging this gives d 2 z + kz = 0, dt2 k = 4πGρ (26) (c) The time-dependent equations for position and velocity for this simple harmonic oscillator are z(t) = z 0 cos(ωt + δ), w(t) = ż = ωz 0 sin(ωt + δ) (27) (28) where ω = k = 4πGρ and δ is a phase constant. (d) Assume ρ = 0.15 M pc 3. The oscillation period is P = 1/f = 2π/ω. Using M = 2 10 33 g, G = 6.67 10 8 cm 3 s 2 g 1, 1pc = 10 18 cm and 1 yr π 10 7 sec, we get P = 6.8 10 7 yr. (e) Currently, z = 30 pc, and w = 7.2 km s 1. To use our equations from part (c), we must choose an initial condition to give δ. Choose it so that the Sun passes through z = 0 at t = 0, so δ = π/2. So we can write the Sun s current position and velocity as z = z 0 sin(ωt) and w = ωz 0 cos(ωt). Combining the two equations gives ( ωt = arctan ω z ) (29) w and plugging in the current values gives ωt = 0.36 rad. So we can estimate z 0 = 85 pc (30) The quoted scale height for the thin disk, as given in Table 24.1, is h thin = 350 pc. So even though our rough estimate is off by a factor of 4, to order of magnitude our answers compare fairly well. (f) The Sun s orbital period is found using P orbit = 2π /Θ 0 for = 8 kpc and Θ 0 = 220 km s 1, giving P orbit = 2.2 10 8 yr. So we find the ratio P orbit /P osc = 3.2. 5