Download AST 341 - Homework IV - Solutions

AST 341 - Homework IV - Solutions
TA: Marina von Steinkirch, [email protected]
State University of New York at Stony Brook
November 12, 2010
1
(2 points) Show that the equation for hydrostatic equilibrium,
Mr ρ
dP
= −G 2 = −ρg
dr
r
can also be written in terms of the optical
depth, τ , as
dP
g
= .
dτ
κ
The differential optical depth can be written as dτ = −κρds, where ds is the
measured distance of the photon’s path and we can write ds = dr in this case,
1 dP
1 −ρg
dP
g
×
=
× −ρg →
→
= .
dτ
dr
dτ
−κρdr
dτ
κ
1
Mass Sun
Mass H
1.98843 × 1030 kg
1.00794
6.022143×1023
= 1.67372 × 10−27 kg
10 eV
1.602 × 10−18 J
Luminosity Sun
Age of Sun
3.846 × 1026 W
1.44 × 1017 s or 4.57 × 109 years
Table 1: Data for the Problem
2
(2 points) Assuming that 10 eV could be released by every atom in the Sun through chemical reactions, estimate how long the Sun could
shine at its current rate through chemical process alone. For simplicity, assume that Sun is
composed entirely of Hydrogen. Is it possible that the Sun’s energy is entirely chemical?
Why?
Looking to table 1, we see that if the mass of Sun was only composed of hydrogen, we would have a total of
Msun
∼ 1.19 × 1057
MH
atoms. The total energy released would be
1.9 × 1057 atoms × 1.602 × 10−18 J = 3.0428 × 1039 J
. The time of consume of this energy would then be
T =
3.0428 × 1039 J
= 7.9 × 1012 s ∼ 2.5 × 105 years.
3.846 × 1026 J/s
Clearly we see that it is not possible that the time scale for entirely chemical
composition is too short for the age of the Sun.
3
(2 points) Prove that equation
P = K 0 T γ/(γ−1)
2
follows from
P V γ = K.
Starting from the ideal gas law P V = nRT → V = nRT
P
nRT = k,
P
P
P 1−γ (nRT )γ = k
1/1−γ
k
P =
(nRT )γ
4
=
k 1/1−γ
T γ/γ−1
(nR)γ/1−γ
=
k 0 T γ/γ−1 .
(2 points) Describe the density structure associated with an n=0 polytrope.
Polytropes are the solution of the Lane-Emden equation in which pressure depends on the density in the form P = Kργ . From the book, we see that the
density in terms of a scaling factor is given by
ρ(r) = ρc [Dn (r)]n , where 0 ≥ Dn ≥ 1,
and ρc the central density of the polytropic stellar model. Making n = 0, we
then have P (r) = ρc , a model of a star with constant density at all depths
inside.
5
5.1
(2 points)
What can you conclude about the concentration of
density with radius for increasing polytropic index?
The density with R increases as the polytropic index increases, i.e. the star
becomes less centrally concentrated.
5.2
From the trend that you observe for the analytic solutions to the Lane-Emden equation, what would you
expect regarding the density concentration of an adiabatically convective stellar model, compared to a model
in radiative equilibrium?
We should observer that the adiabatically convective star would have a more
spread-out distribution of density in relation to its radius, and the star in radiative equilibrium with a more centrally-concentred distribution of density, i.e., a
3
radiative star (which has generally higher density)) should have higher density
to the radius to a convective star (which has generally lower density).
5.3
Explain your conclusion in part c in terms of physical
process of convection and radiation.
Higher density in the core than in the outer layers means that there is less
material in the outer layer than in a star of identical mass. If the distribution
is more spread-out, the material is supported only by the radiation. The basic
difference between the two forms of energy transfers: Convection means a large
temperature gradient, raising opacity, slowing the rate of the radiation transfer.
Radiation has temperature gradient low, low opacity. In conclusion, since the
density decreases with increasing radius, convection occurs, rising material to
near the surface.
4