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PROBLEM SET #7 SOLUTIONS
AST142
1. Jeans Mass
A spherical cloud is made of pure molecular hydrogen, and has uniform number
density 106 cm3 , uniform temperature 100 K, and mass 1M .
(a) Show that this cloud is not gravitationally unstable.
(b) To what temperature would it have to cool in order to become gravitationally
unstable?
(c) In the early universe the interstellar medium contains very little metal. One consequence of this is that the cooling of gas is much less efficient and gas might be
at much higher temperatures. Do you expect stars formed in the early universe to
be more or less massive than stars formed today? Suppose the gas does not cool
below 104 K. What is the mass of a typical star?
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Solution
(a) The Jeans mass for this density, composition (molecular hydrogen: m = 3.347 ×
10−24 g) and temperature is
kT 3/2
3 1/2
MJ =
= 4.1 × 1033 g = 2.1M
mG
4πρ
As this is above 1M the cloud will not collapse (unless things happen such as
shocks, magnetic fields ....).
(b) The Jeans mass depends on temperature to the 3/2 power. Inverting the relation
2/3
T ∝ MJ
2.12/3 = 1.639 and T = 100K/1.639 = 61K.
(c) As the Jeans mass MJ ∝ T 3/2 and T might be higher, stars formed in the early
universe might be more massive than stars formed today. Above we have 2 solar
masses for T = 100 K. If the temperature is 100 times higher the Jeans mass is
1000 times higher or 2000 solar masses. The typical ‘star’ would be much more
massive.
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PROBLEM SET #7 SOLUTIONS
AST142
2. Stromgren Sphere
The number of UV photons per second above 13.6 eV (that can ionize atomic hydrogen) emitted by a massive young star covers a wide range. B2: 4×104 4s O8.5: 2×1048 s,
O4, 9 × 1049 s. In diffuse media HII regions range from 1 to 30 pc large. There are other
classes of objects called compact and ultra compact HII regions. Compact HII regions
have sizes from 0.005 pc to 0.5 pc and electron densities of 2 × 103 to 105 cm−3 . The
ultra compact HII regions (seen in radio continuum) have sizes below 0.01 pc. Assume
a recombination coefficient of α = 2.6 × 10−13 cm3 s−1 .
Assuming an ultra compact HII region is caused by an O4 star, give a limit on the
electron density of the ISM.
3 Q∗
Rs =
4π αn2e
1/3
———————————Solution
1/3 Q∗
ne −2/3
5 × 1048 s−1
10cm−3
Taking equation 16.20 from Ryden and Peterson. We invert this
1/2
n
R −3/2 Q∗
s
e
=
10cm−3
10pc
5 × 1048 s−1
Rs = 10pc
The UV flux is 18 times larger than the value above giving a factor of 4.24. The
radius is 103 times smaller giving a factor of 3.2e4.
181/2 × (10−3 )−1.5 = 4.24 × 3.2 × 104 = 1.3 × 105
giving a density of
ne > 1.3 × 106 cm−3
3. N-body units
Problem #3 of Workshop 7 http://astro.pas.rochester.edu/~aquillen/ast142/
Work/work7.pdf
4. The power law disk
Problem #4 of Workshop 7 http://astro.pas.rochester.edu/~aquillen/ast142/
Work/work7.pdf
PROBLEM SET #7 SOLUTIONS
AST142
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5. Disk Rotation
Problem #5 of Workshop 7 http://astro.pas.rochester.edu/~aquillen/ast142/
Work/work7.pdf
—————————–
Solutions
N-body units
N-body units at a = 1 for G = 1 and m = 1 the rotational velocity is 1, the rotation
rate Ω = 1 and the orbital period is P = 2π/Ω = 2π. At a = 2 the orbital period is
2π × 23/2 .
To get things in years divide N-body time by 2π.
A hundred years you need to integrate to 100 × 2π.
Power law disk
First let us integrate a power law
Z amax
a1−α − a1−α
min
a−α da = max
1−α
amin
To normalize the probability distribution
A=
1−α
− a1−α
min
1−α
amax
To make sure that the mass integrates to Md
B = Md A = Md
so that
Z
amax
Md =
1−α
1−α
− amin
a1−α
max
a−α Bda.
amin
To get a surface density we need to correct by 2π and radius.
Md
1−α
1−α
Σ(a) =
1−α
1−α a
2π amax
− amin
Integrating the mass out out from amin to a we find (including normalization)
M (a) = Md
1−α
a1−α − amin
1−α
a1−α
max − amin
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PROBLEM SET #7 SOLUTIONS
AST142
Disk rotation
Rotation has v ⊥ r.
The rotation velocity
r
vkep =
Gµ
a
with
µ = M∗ + M (a)
vkep should be approximately correct as it includes the total mass within the radius of
the particle and this is the sum of the disk mass out to a (which is M (a)) and that of
the central star.
This is not exact because the mass distribution is not spherically symmetric. In fact
the disk outside the location of the particle does exert a force on the particle because
the mass distribution is not spherical.