Homework # 5 Solutions
1. In example given in the notes describing the observations of collapsing clouds, it is indicated in some cases that there is a locus
of points for which vr is constant. vr is the radial velocity seen by
an observer in the ±z direction (see the relevant figure).
−1
For the case that
p v(r) = v0r , find equations for (r, θ), or
(x, y) where r = x2 + y 2 and θ = tan−1 (x/y), of the locus of
points for which vr = 2v0 /Ri = constant. Here, Ri is the outer
radius of the collapsing region.
Using v(r) = v0 /r, and defining the angle θ relative to the axis
perpindicular to the line-of-sight, a point along the line-of-sight R
which is a distance z from the axis satisfies sin θ = z/r. Then the
component of the velocity along the line-of-sight vr = v(r) sin θ =
(v0 /r) sin θ. Solving for r:
r = (v0 /vr ) sin θ = (Ri /2) sin θ.
This crosses the line-of-sight twice, showing there are two points
in the flow with the same observed radial velocity.
Show that in the case v(r) = v0 r there is no locus of points for
which vr is constant.
In this case, find the curve for the fixed value of vr = v0 Ri /2.
We have vr = v(r) sin θ = v0 r sin θ. Solving for r:
r = (vr /v0 ) / sin θ = (Ri /2) / sin θ.
This curve only crosses the line-of-sight once, showing there is
only one point in the flow with a given value of the observed
radial velocity.
2. In a Stromgren sphere containing H and He in solar proportions,
why can’t the ionization front for He+ ever be larger than that
for H+ ?
All photons capable of ionizing He can ionize H. Therefore,
even after all photons capable of ionizing He are used, other photons remain to ionize further H beyond that point.
3. Suppose the number density of bright F dwarfs, at the Sun’s radial
distance from the Galactic center, varied with height above the
Galactic plane as
n (z) = n0e−z/zh
where zh = 285 pc. Also suppose that the velocity dispersion of
these dwarfs varies with z as
2 (1 + z/z ) ,
σz2 = σz0
h
σz0 = 14 km s−1 .
Show that the surface mass density within 1 kpc of the plane is
Σ (< 1 kpc) ≃ 90 M⊙ pc−2 .
What factor is this larger than the surface mass density in the
Galactic disc at the Sun’s position?
We can use equation 3.94 of the text. Then
i
−1 d h
2
2πGΣ (< z) =
n (z) σz .
n (z) dz
We find
2
2
zσz0
σz0
[1 + z/zh − 1] = 2 ≃ 2.4 km2 pc−1 s−2 .
2πGΣ (< z) =
zh
zh
Then
Σ (< z) =
2.4
−2
−2
M
pc
≃
89M
pc
.
⊙
⊙
2π × 4.3 × 10−3
If, instead, one assumed that the velocity dispersion was independent of height above the Galactic plane, what would you find
for Σ(< 1 kpc)?
We find
2
σz0
2πGΣ (< z) =
[1] ≃ 0.69 km2 pc−1 s−2 .
zh
Then
Σ (< z) =
0.69
M⊙ pc−2 ≃ 26M⊙ pc−2 .
−3
2π × 4.3 × 10