Stars & galaxies, weekly problems

Problem 1.

1. Answer a: The distance r = 300 kpc is beyond the extent of the dark halo (which has radius 200 kpc), hence equate v_{c}^{2} = GM/r, where M = 1.0 × 10^{12}M_{o}dot is the total mass of the halo, to find v_{c} = 170 km s^{−1}

2. Answer b: The 1/r^{2} density profile gives rise to a flat rotation curve inside the halo. Therefore the circular velocity at r = 100 kpc is the same as at r = 200 kpc where it is v_{h}^{2} = GM/r, with M = 2.0 × 10^{12}M_{o}dot, hence v_{h} = 207km s^{−1}.

3. Answer c: The circular velocity is due to dark halo and buldge combined, so v_{c}^{2} = GM_{halo}(< r)/r+GM_{bulge}(< r)/r = v_{h}^{2}+GM_{bulge}(< r)/r. Since the bulge is uniform density, M_{bulge}(< r) = (r/r_{b})^{3} M_{b}, hence M_{b}(r < 2kpc) = (2/4)^{3} M_{b}, hence v_{c} = 220 km s^{−1}

4. Answer b: The expression for Oort’s constant A is given by Eq. (5.5) in the notes,

A = −\frac{1}{2} (\frac{dV_{c}}{dr} − \frac{v_{c}}{r})
= −\frac{1}{2} (\frac{GM_{b}}{r_{b}^{3}} \frac{r}{v_{c}} − \frac{v_{c}}{r})
= 48.9 km s^{−1} ,

with v_{c} = 220 km s^{−1}from above

Problem 2.

1. Answer b: The 3D velocity dispersion σ_{3D}^{2} = σ_{1}^{2} + σ_{2}^{2} + σ_{3}^{2} which, if the orbits or isotropic, σ_{3D}^{2} = 3σ_{1D}^{2}, hence σ_{3D} = 1386km s^{−1}.

2. Answer b: The virial mass is given by Eq. (9.4) in the notes,

M = \frac{Rσ_{3D}^{2}}{G} = 2.2 × 10^{14}M_{⊙} .  (1)

3. Answer a: The virial temperature is given by Eq. (9.7) in the notes,

T = \frac{2μm_{p}}{3k_{B}} \frac{GM}{R} = 1.7 × 10^{8} K ,  (2)

where the mean molecular weight per particle, μ = 0.5, if we assume fully ionised hydrogen. At this temperature, the typical kinetic energy of a particle, k_{B}T ∼ 15 keV, is much large than the binding energy of Hydrogen, 13.6 eV, which is why the gas is indeed highly ionised.

4. Answer b: Given mass and radius, we can compute the 3D, and hence 1D velocity dispersion of the cluster, σ_{1D} = σ_{3D}/\sqrt{3}, with σ_{3D}^{2} = GM/R. Then the crossing time is

t_{cr} = \frac{R}{σ_{1D}} = 820 Myr.  (3)