Complete six of the following eight problems. You can consult your textbook, notes, calculator but not your neighbours.

Problem 1

Suppose that the gravitational potential near the Sun were dominated by a spherical distribution of matter of uniform density. What would be the values of the Oort constants $A$ and $B$ in terms of the distance from the Sun to the centre of the sphere and the density of the matter?

First let's start with the definitions of the Oort constants. We have

$A = - \frac{1}{2} R [\frac{V}{R}] = \frac{1}{2} [\frac{V}{R} - \frac{d V}{d R}],$ $B = - \frac{1}{2} \frac{1}{R} [V R] = - \frac{1}{2} [\frac{V}{R} + \frac{d V}{d R}] .$

Therefore, we need to write the circular velocity near the Sun as a function of radius and we are in business. The gravitational force is

$F = - \frac{G M_{r} m}{r^{2}} = - \frac{4 π G ρ m r}{3} = - m \frac{v^{2}}{r}$

so

$\frac{v^{2}}{r^{2}} = \frac{4 π G ρ}{3} .$

The value of $v / r$ is constant so $A = 0$ . The value of

$B = - \frac{1}{2} \frac{1}{R} \sqrt{\frac{4 π G ρ}{3}} [R^{2}] = - \sqrt{\frac{4 π G ρ}{3}} .$

Problem 2

If you assume that once a hydrogen atom collides with a dust grain, it sticks and immediately finds a partner to form H₂, how long does it take a hydrogen atom to join in a molecule in terms of the radius and density of the dust grains and the thermal velocity of the hydrogen atoms? What is the timescale for v_th=0.1 km s^-1, r_dust=0.1 μm and n_dust=10^-12 cm^-3?

The cross section for the hydrogen atom and a grain is $σ = π r^{2}$ where $r$ is the radius of the grain. The timescale is given by

$τ = \frac{1}{σ n_{d u s t} v} = \frac{1}{π r^{2} n_{d u s t} v_{t h}} \approx 1 0^{10} y e a r$

Problem 3

Suppose the gravitational potential of a spherical distribution of matter is given by

$Φ = - \frac{G M}{π} \frac{\sin (r / a)}{r}$

Derive the density distribution and the circular velocity as a function of radius $r$ . Up to what radius does the density make sense? What is the total mass within this radius? Believe it or not, this is approximately the density distribution of a neutron star.

We can write that

$\nabla^{2} Φ = \frac{1}{r} \frac{d^{2}}{d r^{2}} (r Φ (r)) = 4 π G ρ$

so

$\frac{G M}{π a^{2}} \frac{\sin (r / a)}{r} = 4 π G ρ$

and

$ρ = \frac{M}{4 π^{2} a^{2}} \frac{\sin (r / a)}{r} .$

For the density to make sense, it must be positive, so $r / a < π$ and the maximum radius is $r = π a$ . The mass enclosed is $M$ , but we will prove this by looking at the force. The force is given by

$F = - \frac{d Φ}{d r} = \frac{G M}{π} [\frac{\cos (r / a)}{a r} - \frac{\sin (r / a)}{r^{2}}]$

so the mass enclosed is

$M_{r} = - \frac{F r^{2}}{G} = - \frac{M}{π} [\frac{r \cos (r / a)}{a} - \sin (r / a)] = M$

where the final equality holds at $r / a = π$ where the sine vanishes and the cosine equals -1. For the circular velocity we have

$- \frac{v^{2}}{r} = \frac{G M}{π} [\frac{\cos (r / a)}{a r} - \frac{\sin (r / a)}{r^{2}}]$

so

$v^{2} = \frac{G M}{π} [\frac{\sin (r / a)}{r} - \frac{\cos (r / a)}{a}]$

Problem 4

The nuclear star cluster of M33 has a core radius $r_{c} \approx 0.4$ pc and a measured velocity dispersion σ=24 km s^-1 and a luminosity of $L_{V} \approx 2.5 \times 1 0^{6} L_{s u n}$ . Estimate the total mass of the star cluster and the mass-to-light ratio.

The globular cluster M4 has a core radius of $r_{c} \approx 0.5$ pc and a measured velocity dispersion σ=6 km s^-1 and a luminosity of $L_{V} \approx 6 \times 1 0^{4} L_{s u n}$ . Estimate the total mass of the star cluster and the mass-to-light ratio. Explain why the mass-to-light ratios differ by so much.

Here we have to pick a model for the star clusters and stick with it to do both problems. You could use a uniform density sphere, a Plummer sphere, a Hernquist model or whatever. Let's use a uniform sphere. First let's write the potential energy of the sphere

$V = - \frac{3 G M^{2}}{5 r}$

where $M$ is the mass of sphere and $r$ is the radius. The core radius is not the total radius but the projected radius where the light has fallen by a factor of two from the centre. For the uniform sphere, this is the radius at which the total projected distance through the sphere is equal to radius of the sphere.

$r^{2} = r_{c}^{2} + {(\frac{r}{2})}^{2} = r_{c}^{2} + \frac{1}{4} r^{2}$

so

$r_{c} = \frac{\sqrt{3}}{2} r$

Now let's use the virial theorem to relate the potential and kinetic energy

$- V = 2 T, \frac{3 G M^{2}}{5 r} = M σ^{2}$

so

$M = \frac{10}{3 \sqrt{3}} \frac{r_{c} σ^{2}}{G}$

where the dimensionless constant depends on the particular model that we chose. For a Plummer sphere we have

$V = - \frac{3 π}{16} \frac{G M^{2}}{a} \approx 0.59 \frac{G M^{2}}{a},$

very close to the result for the uniform sphere. Here we have

$r_{c} = \sqrt{\sqrt{2} - 1} a = 0.64 a,$

somewhat smaller than for the uniform sphere $0.866 r$ . Putting it together yields

$M = \frac{16}{3 π \sqrt{\sqrt{2} - 1}} \frac{r_{c} σ^{2}}{G} .$

For the potential in problem 3,

$V = - \frac{3}{2 π} \frac{G M^{2}}{a} = - \frac{3}{2} \frac{G M^{2}}{r}$

where $r$ is the radius of the sphere. Using the constants for the uniform sphere, we get $M = 2 \times 1 0^{35} k g = 1 0^{5} M_{s u n}$ for M33 and $M = 8 \times 1 0^{3} M_{s u n}$ for M4. In solar units we have $M / L = 0.04$ for M33 and $M / L = 0.13$ for M4. Regardless of the models that you chose you should get a larger ratio for M4 by a factor of three. This is because the M33 cluster is composed of bright young stars and the M4 globular cluster is old stars.

Problem 5

Show that if the pitch angle of spiral arm is constant, the centre of the spiral arm can be described by the following equation in polar coordinates

$r = A e x p (θ t a n i)$

wherei is the pitch angle.

The pitch angle is the angle between the arm and the tangent to a circle at the radius, so we have

$\frac{1}{r} \frac{d r}{d θ} = \tan i$

or

$\frac{d \ln r}{d θ} = \tan i$

so

$\ln r = θ \tan i + C, r = A e x p (θ \tan i)$

Problem 6

What is the distribution of the apparent ellipticities of spiral disks? You may assume that a spiral disk is infinitely thin.

The apparent ellipticity of a thin disk is given by

$e = 1 - \frac{b}{a} = 1 - \frac{a \cos i}{a} = 1 - \cos i$

where $i$ is the angle between plane of the sky and the plane of the disk. $i$ is also the angle between the line of sight and a vector perpendicular to the disk. We can figure out the distribution of $i$ by looking at spherical coordinates. The distribution of $i$ is proportional to the area of the sphere with an angle $i$ away from the $z -$ axis. We have

$d A \propto \sin i d i = - d (\cos i)$

and

$d (e) = - d (\cos i)$

so

$d (e) \propto d A$

and the distribution of ellipticities is uniform. We can do it more slowly too

$\frac{d e}{d i} = \sin i$

and divide by $d A / d i$ to get

$\frac{d e}{d A} = \frac{\sin i}{\sin i} = 1$

Problem 7

Show that, if the distance to the source is fixed, then area $π θ_{E}^{2} d_{L e n s}^{2}$ within the Einstein radius of the point-mass lens is maximized when the lens is midway between the source and us.

Let's write out the equation for the Einstein radius

$θ_{E}^{2} = \frac{4 G M}{c^{2}} \frac{d_{L S}}{d_{L} d_{S}}$

so the area is

$A = π θ_{E}^{2} d_{L}^{2} = \frac{4 G M}{c^{2}} \frac{d_{L S} d_{L}}{d_{S}} = \frac{4 G M}{c^{2} d_{S}} (d_{S} - d_{L}) d_{L} .$

Taking the derivative yields

$\frac{d A}{d d_{L}} = \frac{4 G M}{c^{2} d_{S}} [- d_{L} + (d_{S} - d_{L})] = 0$

and

$d_{S} = 2 d_{L} .$

Problem 8

Assume that an AGN is a black hole emitting at the Eddington luminosity as blackbody from a sphere located at the radius of the last stable orbit. Derive a relationship between the temperature of the blackbody radiation and the mass of the black hole.

The radius of the last stable orbit is $R = \frac{3 G M}{c^{2}}$ and the Eddington luminosity is

$\frac{L}{4 π r^{2} c} σ_{T} = \frac{G M m_{p}}{r^{2}}$

and

$L_{E d d} = \frac{4 π c G M m_{p}}{σ_{T}} = σ T^{4} 4 π {(\frac{3 G M}{c^{2}})}^{2}$

so

$T^{4} = \frac{m_{p} c^{5}}{9 G M σ σ_{T}}$