Download Homework #9 - Solutions - Department of Physics and Astronomy

Astronomy 204: Homework #9-Solutions
Problem 14-4: What stellar populations would you expect to find (and why)
a)
in the nucleus of our galaxy
The nuclei of galaxies typically contain population II stars, so we would likely
expect the same for the nucleus of the Milky Way. However, there have been molecular
clouds observed near the galactic center suggesting the possibility of some population I
stars.
b)
in the spiral arms of the Andromeda Galaxy
These stars should be similar to the stars in the spiral arms of our own galaxy
which are primarily population I stars.
c)
in the Pleiades star cluster
The Pleiades star cluster is a young open star cluster located in the spiral arms,
thus we would expect population I stars.
d)
in intergalactic space (beyond the halo)
We would expect to see only old extreme population II stars (and possibly
population III stars which are predicted to exist but have not been conclusively observed).
e)
in the galactic bulge
Population II since there is little gas and dust for new star formation.
Problem 14-5: A globular cluster is in an elliptical orbit (e = 0.9) about the center of
our Galaxy, reaching apogalacticon (farthest distance from the center) at the distance of
40 kpc. What is the perigalacticon (nearest) distance, and how long will the cluster
require to complete one orbit?
a (1 − e 2 )
We can use the formula r =
letting θ=0° for perigalacticon and θ=180° for
(1 + e cos θ )
apogalacticon.
2
a ⎡1 − ( 0.9 ) ⎤
⎣
⎦ = 1.9a
40 kpc =
⎡⎣1 + ( 0.9 ) cos180°⎤⎦
a = 21.0kpc
( 21.0kpc ) ⎡⎣1 − ( 0.9 )
rperigalacticon =
⎡⎣1 + ( 0.9 ) cos 0° ⎤⎦
2
⎤
⎦ = 2.1 kpc
Now apply Kepler’s third lay for galaxies with a = 4.3 x 109 AU and a mass for the Milky
Way of 1012 M:. You may neglect the mass of the cluster.
4.3 × 109 AU )
(
a3
2
P =
=
= 8.0 × 1016 yr 2
12
M Galaxy + M Cluster
(10 M : )
3
P = 2.8 × 108 yr
Problem 15-5: Why are none of the Blamer lines seen as interstellar absorption lines,
even though hydrogen is the most abundant element in the Universe?
To absorb a Balmer photon the electron in a hydrogen atom must be in the second level
and then it jumps up as it absorbs. Since interstellar gas is so cool, almost all of the
hydrogen electrons are in the ground state.
Problem 15-18: Let’s examine again what conditions would result in the formation of the
Sun.
a)
Consider the following: A core of a giant molecular cloud has a radius of about 1
pc, a temperature between 30 and 100 K, and a number density of some 1010 /m3. Is it
likely to collapse to form a solar-type star?
We simply need to check the mass and see if it is comparable to the sun. Assume that the
cloud is molecular hydrogen.
⎛ 1010 ⎞
kg
= 3.34 ×10−17 3
3 ⎟
m
⎝ m ⎠
ρ = 2mH N = (1.67 ×10−27 kg ) ⎜
3⎛
4
4
kg ⎞
M = π R 3 ρ = π ( 3.086 × 1016 m ) ⎜ 3.34 ×10−17 3 ⎟ = 4 ×1033 kg ≈ 2000 M :
3
3
m ⎠
⎝
Thus, for this cloud to collapse to a solar-type star, it must lose 99.95% of its initial mass.
We can also use equation 15-9 to evaluate the size scale of the collapsing region for this
density and temperature range.
L ≈ 107
T
L1 = 107
30
= 9.5 ×1015 m ≈ 0.3 pc
3.34 ×10−17
L2 = 107
100
= 1.7 ×1016 m ≈ 0.55 pc
−17
3.34 × 10
ρ
thus, this cloud is too large to be likely to collapse to a solar-type star.
b)
Small knots within cores of giant molecular clouds have radii of about 0.1 pc,
temperatures between 30 and 200 K, and number densities of some 1012/m3. Is one more
likely to collapse to form a solar-type star than the case in part (a)?
3⎛
4
4
kg ⎞
M = π R 3 ρ = π ( 3.086 ×1015 m ) ⎜ 3.34 ×10−15 3 ⎟ = 4 ×1032 kg ≈ 200M :
3
3
m ⎠
⎝
For this knot to collapse to a solar-type star, it must loose 99.5% of its initial mass.
L1 = 107
30
= 9.5 × 1014 m ≈ 0.03 pc
3.34 ×10−15
L2 = 107
200
= 2.4 × 1015 m ≈ 0.08 pc
3.34 × 10−15
Certainly this cloud is more likely to collapse to form a solar-type star than the cloud in
part (a) since it must lose a smaller amount of its initial mass and its size is more
comparable to the size scale of a collapsing cloud.
c)
t ff =
For the more likely case (part b), what is the free-fall time of the collapse?
6.44 ×104
ρ
1
2
0
=
6.44 ×104
1
−15 2
( 3.34 ×10 )
= 1.12 ×1012 s = 3.5 ×104 yr
d)
For the more likely case, what is the average luminosity radiated by the protostar
during its collapse phase?
From the virial theorem, we know that the energy available to radiate during gravitational
collapse for the knot in part (b) is half the available gravitational potential energy.
Using 0.1pc = 3.086 x 1015 m for the size scale yields:
1
GM 2
Eradiation = U ≈
2
2L
2
2
⎛
−11 N − m ⎞
32
6.67
10
×
⎜
⎟ ( 4 ×10 kg )
2
kg ⎠
1
= ⎝
= 1.7 ×1039 J
15
2
( 3.086 ×10 m )
Using the timescale of collapse from part (c), the average luminosity during the collapse
is
E
1.7 × 1039 J
L = radiation =
= 1.6 × 1027 W ≈ 4 L:
12
1.1×10 s
t ff
which is a reasonable value.