AST111 PROBLEM SET 6 SOLUTIONS
Homework problems
1. Ideal gases in pressure balance
Consider a parcel of molecular hydrogen at a temperature of 100 K in proximity
to a parcel of ionized hydrogen at a temperature of 104 K. Assume the two parcels
are in pressure balance. That means both parcels have the same pressure.
How much denser is the molecular hydrogen?
Answer:
Molecular hydrogen has a molecular weight of about 2 atomic mass units per
particle. Ionized hydrogen has a molecular weight of about 1/2 atomic mass units
per particle where we count the free electron. Using P = ρkT
µ
PH2 =
PH =
ρH2
k100
2mH
ρH
k104
0.5mH
Set the two equal and solve for the density ratio
104 2
ρH2
=
= 400
ρH
100 0.5
2. Jeans escape
The timescale for Jean’s escape depends on a unitless factor eY /Y and a timescale
h/v0 which is of order a minute for most
pplanets in the Solar system. Here h is
the atmospheric scale height and v0 = 2kT /m depends on particle mass and
temperature
is a characteristic thermal velocity. The parameter Y ≡ ve2 /v02 where
p
ve = 2GM/R is the escape velocity.
1
2
AST111 PROBLEM SET 6 SOLUTIONS
(a) Consider a molecular species with mass m that gives Y = 20. What is the
timescale for thermal escape of this molecule? Should it be present in the
planet’s atmosphere?
(b) How much more massive a molecule would allow it to remain in the atmosphere for 4 billion years?
(c) What is the ratio of the escape velocity of Jupiter compared to that of the
Earth?
(d) Taking into account only the ratio of semi-major axis, what is the ratio of
the equilibrium temperature of Jupiter compared to that of the Earth? (Ignore
differences in albedo, emissivity and atmospheric opacity).
(e) A molecule on the Earth that has Y = 20 has approximately what value on
Jupiter?
A light molecule that escapes from Earth can last forever in Jupiter’s atmosphere.
Answers:
a) e20 /20 = 2 × 107 and when we multiply by 60 seconds and divide by the
number of seconds in a year (3 × 107 ) we find 50 years. This is so short that the
molecule should not be present in the atmosphere.
q
2kT
b) The thermal velocity v0 =
∝ m−1/2 whereas the escape velocity is
m
independent of particle mass m. The parameter Y ≡ ve2 /v02 ∝ m.
By trial and error I found Y = 39 gives an escape time of 4 billion years. This
is about twice as large as Y = 20 which means the molecule mass is only twice as
large! The lifetime is extremely sensitive to the mass!
c) The escape velocity
1
1
ve ∝ M 2 R− 2
Jupiter is 318 times
p more massive than the Earth and its equatorial radius is
11.2 times larger.
318/11.2 = 5.3. The ratio of the escape velocity of Jupiter
compared to that of the Earth is 5.3.
1
d) Equilibrium temperature depends on Teq ∝ d− 2 and Jupiter has a semimajor axis of 5.2 AU so its equilibrium temperature is about 1/2 that of the
Earth.
AST111 PROBLEM SET 6 SOLUTIONS
3
e)
Y =
GM
ve2
=
∝ M (T R)−1
2
v0
kT R
Temperature is 1/2 times smaller, mass is 318 times larger, radius is 11.2 times
larger. Y is larger by a factor 318/(0.5 × 11.2) = 5.32 /0.5 = 57. On Jupiter
Y = 20 × 57 = 1140. The molecule escapes from the Earth but not from Jupiter.
3. Thermal tides on Mars
On Mars, the fractional change in atmosphere temperature between noon and
midnight is about 40%. This temperature difference (sometimes called a thermal
tide) can be estimated from the total heat absorbed by the entire atmosphere.
The solar heat input into a planetary atmosphere per day is
L∗
tday
4πd2
where Ab is the bond albedo, R is the planet radius, d is the distance from planet
to star, tday is the time length of a day (a full day, like 24 hours).
Qabs ∼ πR2 (1 − Ab )
(a) Show that the mass of the entire atmosphere per unit area is approximately
P0 /g where g is the gravitational acceleration and P0 is the pressure at the
base of the atmosphere.
(b) Show that the total heat Qdt required to raise the temperature of the entire
atmosphere by a temperature difference ∆T is
Qdt = cP
Po
4πR2 ∆T.
g
(c) Show that the fractional increase in temperature between day and night is
approximately
∆T
L∗ (1 − Ab )gtday
∼
.
T
16πd2 P0 cP T
(Hint: set Qabs = Qdt ).
The bond albedo for Mars is 0.16, the equilibrium temperature is 222K, the
mean heliocentric distance is 1.524 AU, the mass of the planet is 6.42 × 1023 kg,
the mean radius is 3390 km, the sidereal rotation period 24.6hours. The specific
heat for CO2 is 0.205 cal/g/C. The surface gravity is 370 cm s−2 . The surface
pressure is 0.0056bar . The gravitational acceleration g = GM/R2 = 370 cm s−2 .
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AST111 PROBLEM SET 6 SOLUTIONS
The specific heat cP = 0.25cal/g/C × 4.18 × 107 erg/cal = 9.6 × 106 erg/C. tday =
24.6 × 60 × 60s. Using the above equation, we get ∆T /T = 0.40.
Answers:
Force on surface is due to weight of gravity on the atmosphere. The total mass
of atmosphere in a column with area A is m. The total weight is mg. The pressure
on the surface is force divided by area or P0 = mg/A. The mass per unit area is
m/A = P0 /g.
The mass per unit area is P0 /g so the total mass of the atmosphere is this times
the surface area or 4πR2 P0 /g. The heat required to raise this by ∆T depends on
the specific heat or
Po
Qdt = cP 4πR2 ∆T
g
Heat absorbed is
L∗
tday
4πd2
We set this to be equal to that required to raise the temperature Qabs = Qa .
Po
L∗
πR2 (1 − Ab )
t
=
c
4πR2 ∆T
day
P
2
4πd
g
and solve for ∆T
gL∗ (1 − Ab )tday
∆T =
cP P0 16πd2
Divide this by T to get the expression
Qabs ∼ πR2 (1 − Ab )
∆T
L∗ (1 − Ab )gtday
∼
T
16πd2 P0 cP T
This expression assumes that all the energy during the day is going into heating
up the atmosphere and none is going out as radiation during the daytime. A better
calculation would take into account the energy emitted during the daytime.
—————————————————–
Workshop problems
1. Isothermal gas
Explain why γ = 1 gives a gas that is isothermal under compression. In other
you can compress it without changing its temperature.
AST111 PROBLEM SET 6 SOLUTIONS
5
Answer:
Inverting the pressure/temperature relation
dT
γ−1T
=
=0
dP
γ P
That means that there is no change in temperature for a change in pressure.
Likewise P V γ = P V = constant and P V = N kT together imply that T is a
constant.
Interstellar medium is often considered to be isothermal because its temperature
is set by absorbing and emitting radiation.
2. The Dry Lapse Rate
Consider a convecting atmosphere.
(a) Show that the factor that the temperature changes per atmospheric scale
height is equal to (1 − γ)/γ. In other words what is ∆T /T across a scale
height h?
(b) Show that the factor that the temperature changes per atmospheric scale
height is usually less than 1.
(c) Compare the temperature as a function of opacity for an optically thick atmosphere undergoing radiative energy transport to the temperature change
across an atmospheric scale height for a conductive and adiabatic atmosphere
computed in the previous problem. A dense atmosphere can have extremely
high opacity. Explain why when there is a strong green house effect (and the
planet’s surface temperature is much higher than the effective temperature)
that convective heat transport is unlikely.
(d) Compute ∆T /T across a scale height h for monatomic, diatomic and polyatomic gases.
Answers:
a,b ) As the atmosphere is convective the temperature drops with radius with
the dry lapse rate. h = kT /(gµ) and
dT
γ − 1 µg
=−
dz
γ k
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AST111 PROBLEM SET 6 SOLUTIONS
divide
dT
dz
by h giving
dT
γ−1
h=−
T
dz
γ
Across a scale height
∆T
γ−1
=−
T
γ
The adiabatic index is positive and usually greater than 1 and this means that
the factor that the temperature changes in a scale height is less than 1. We used
here the scale height computed with a fixed temperature. This is a reasonable
approximation since | ∆T
|<1
T
c) The dry lapse rate (giving the temperature gradient during convection) is
just too shallow to give a large temperature gradient. When a large temperature
gradient is present likely energy transport is radiative.
d) For monotomic ∆T /T = 0.4 across a scale height, diatomic, 0.28, and for
polyatomic, 0.25, using γ = 5/3, 7/5, and 4/3 respectively. For atmospheres with
lots of molecules, the temperature gradient is very shallow.
3. Optically thick radiative transfer
The absorption coefficient of gas in a circumstellar disk might be of order α ∼
1g−1 cm2 . The optical depth through the disk τ = sαρ where s is the path length
and ρ the density. Note that the product sρ is a mass per unit area. We can
integrate ρdz through the disk to estimate a disk mass surface density. Here z
is in the Rdirection normal to the disk midplane. Integrating from the midplane
∞
upwards 0 ρdz = Σ/2 where the factor of two arises because we only integrate
for positive z and Σ is the the mass per unit area in the disk.
Taking into account only the integrated column
R ∞ through half of the disk (from
the midplane upwards) the optical depth τ = 0 ραdz = αΣ/2.
(a) If the disk has a mass surface density of Σ = 20g cm2 what is the optical
depth through half of it? (using the above given absorption coefficient).
(b) If turbulence in the disk and associated accretion heats the gas in the midplane
it is hotter than the disk surface. If the disk is optically thick then energy
through the disk may be transported by radiation, through absorption and reemission of photons. Both the top and bottom of the disk can radiate. What
is the ratio of midplane to effective disk temperature?
AST111 PROBLEM SET 6 SOLUTIONS
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Answers:
The optical depth is τ = αΣ/2 = 1 × 20/2 = 10. Which equation relating temperature to opacity should we use? We want effective temperature of radiation
near the top related to the midplane temperature. Probably we should use something like the ground temperature equation, and as a quick approximation ignore
that radiation from the ground only goes in one direction. Using that equation
3/4τ + 1 = 8.5. The temperature ratio goes as the 1/4th power of this or 1.7.
The midplane is about twice as hot as the disk surface.
There are many ways that reality is more complex that this. For example the
absorption coefficient would be dependent on wavelength and temperature (and
metalicity and ionization state etc etc ...)