Homework # 4 Solutions
1. In this problem, you are asked to calculate the expected extinction from a cool
cloud. Assume it is spherical and located at a distance of 100 pc from the
Earth. Assume the cloud is isothermal with a temperature of 10 K. Assume it
is embedded in a warm diffuse ISM with temperature 10,000 K and a density
of 1 cm−3 .
Use the Bonnor-Ebert critical spherical configuration to estimate the total
cloud mass. What is the mass? (You do not have to solve the differential
equations, use the known expression.)
For diffuse ISM, po = 1.38 × 1012 erg cm−3 . The Bonnor-Ebert critical
configuration has
4
1.4
No kT
po = 3 2
, M ≃ 5.1M⊙
G M
µ
We used µ = 2 because the cloud is molecular.
Assuming the density profile in the cloud has the form of the isothermal
singular sphere, compute an expression for the visual extinction as a function
of angular distance from the cloud center from stars located behind the cloud
as seen from the Earth. You may use the relation between visual extinction
and column density given in chapter 1 of the text. Make sure the mass of
the isothermal singular sphere is the same as that of the Bonner-Ebert critical
sphere. Hint: what is the radius of the cloud with this density profile?
For the singular isothermal sphere, the number density of H2 is n(r) =
no ro2 /r 2 . Let no be the density where pressure equilibrium with the ISM is
obtained
po
nISM TISM
no =
=
= 1000 cm−3
No kT
T
The mass of the cloud is equal to
M = 4π
µ
no ro3
No
so the radius is ro ∼ 0.20 pc.
The column density along a line of sight with impact parameter (offset from
the center) R through the cloud is
!
p
Z zm
Z zm
2 − R2
r
dz
ro2
o
√
ρ (r (z)) dz = 2no
NH2 =
= 2no tan−1
.
2
2
R
R
z +R
0
−zm
The extinction formula is
Av =
NH2
NH
.
2 =
21
1.8 × 10 cm
0.9 × 1021 cm−2
To get R in arcsec, you multiply by 180 ∗ 3600/(πd) using the small angle
approximation.
Now, instead, assume the density profile in the cloud has the form
i−1
h
ρ = ρo 1 + 13 (r/R)2
.
Re-compute the visual extinction as a function of angular distance from
the cloud center.
A typo here: I should have used ro in place of R in the above expression,
and I do so from here on. Note that now no refers to the central density which
is 14 times larger than no in the previous problem in order to have the same
density at the boundary ro of the cloud. The total mass now satisfies
µ
M = 4πno
No
Z
ro
0
r 2 dr
µ
2 = 4πno N
o
1 + 13 (r/ro )
ro
√
13
3 √
13 − tan−1
Therefore,
ro (new) = ro (old )
√
13 √
141/3
13 − tan−1
An integration gives the new column density
NH2 = 2 p
n o ro
13 (1 + 13R2 /ro2 )
tan−1
√ −1/3
13
≃ 0.22 pc.
s
13 (ro2 − R2 )
(ro2 + 13R2)
!
.
√
13 .
which has a maximum value of
NH2 ,max = 2no ro (13)−1/2 tan−1 (13)1/2 ≃ 7.1 × 1021 cm−2 .
Compare the two predicted extinction cuves to data for Barnard 68 (a figure
is included in the course notes) and comment.
The maximum extinction is about 8 magnitudes. Comparing to the figure
for B68 in the notes, the shape of the curve seems right but the amount of
extinction is not large enough and the radius of the cloud is too large.
However,
√
increasing po by
about
a
factor
of
10
decreases
M
by
a
factor
10,
decreases
√
√
ro by a factor 10 and increases the column density by a factor 10 as can be
analytically seen from the above equations. This gives good agreement with
the maximum extinction and the radius of the cloud.
2. Show that the gravitational potential at a distance r from a point mass m is
the same as that as that inside an isothermal singular sphere at the radius r
where the enclosed mass is M(r) = m.
I had meant to show that the gravitational forces were the same, a result
that follows easily from the discussion in the text. The potentials are not, and
that of the isothermal sphere has a logarithmic divergence unless it is bounded
by a pressure condition and the potential set to a value, such as zero, there.