Atmospheres AST 111 Lecture 6
This Lecture!
❑Vertical structure of atmospheres ❑Thermodyamics (ideal gas law, adiabatic) ❑Thermal structure of atmospheres ❑Convection ❑Jean’s escape
Hydrostatic Equilibrium• In equilibrium there is a relation
between pressure, density and gravity.
• Consider a slab with thickness dz, density ρ.
• This slab exerts a force on the slab below it due to its mass and gravity.
• Per unit area this force becomes a pressure P.
• Because of the mass of the slab there is a change in pressure across it.
dz
Pressure P is force per unit area A Force is mass M times acceleration gMg
A= ⇢gdz
�P = �⇢gdz
dP
dz= �⇢g
g gravity acceleration
g =GM
R2
Ideal Gas Law
P = nkBT n number of particles/volume kB Boltzmann constant
PV = NkBT N is number of particles
PV= n’RgasT n’ number of moles Rgas constant = kB NA NA Avogadro constant
Hydrostatic Equilibrium and Scale height
Note: neglecting variation of temperature with z
dP
dz= �⇢g rP = �⇢r�equivalent to
P = nkTEquation for a gas n number density of particles k Boltzmann constant T temperatureAssume a pressure and density profile
P(z), 𝜌(z) n~𝜌/m with m mean mass of particle
P (z) ⇠ ⇢(z)kT/m
d⇢(z)
dz
kT
m= �g⇢(z) Solution is an exponential
⇢ / e�z/h
from equation of state
insert into hydrostatic eqn
is a scale heighth ⇠ kT
gm
Pressure and density as a function of height (isothermal)
P = nkT Gas law
n=𝜌/m number of particles per unit volume n related to average mass of a particle m
m = mHµ 𝜇 mass of particle in atomic mass units mH atomic mass unit
(approximately the mass of a hydrogen atom)
P / ⇢ / nSince P, ⇢, n / e�z/h
Scale heights for planetary atmospheres
~
100~ ~ 1/ 3300
earth
jupiter
kThgm
gg
g is proportional to M/R2 The mass of the earth is about 1/300th of that of Jupiter The radius of the earth is about 1/10th of that of Jupiter. T is not that much different The ratio of accelerations is about 1/3
Scale heights for planetary atmospheres
Oddly, the scale heights of the atmospheres for terrestrial and giant planets are not hugely different in size, even though densities, pressures, temperatures and masses differ by many orders of magnitude. ❑ Typical values are in the tens of
km, which indeed is small enough that the plane-parallel approximation is a good one over a few scale heights.
❑ Exceptions are the tenuous atmospheres of Mercury, Pluto and various moons which have larger scale heights.
PlanetIsothermal scale height (km)
Venus 15.9Earth 8.5Mars 11.1Jupiter 27Saturn 59.5Uranus 27.7Neptune 20Pluto 60
Vertical structure• Note that the scale height estimate ignored the possibility
that the temperature could depend on height; T(z). • To get a better model of atmospheric structure we must also
consider the thermal structure. P(z)=n(z) k T(z)
• Even though the atmospheric scale heights of most of the planets are similar, their densities and so pressures are NOT similar.
• Chemical processes and phases are dependent upon pressure as well as temperature. So planets have a wide variety of chemical structure and composition.
Processes which affect the structure of atmospheres
1. The top of the atmosphere is irradiated by the Sun. Radiation is absorbed and scattered. Includes electromagnetic heating processes.
2. The bottom. Energy from internal heat sources, and reradiation of absorbed sunlight by the planet’s surface
3. Chemical reactions in the atmosphere. This leads to changes in opacity and so thermal structure.
4. Chemical interactions between the air and crust 5. Clouds, dust and photochemically produced haze, change
the albedo and opacity and also release energy. 6. Volcanoes — gas and dust 7. Biological processes can change the composition.
Energy transportPreviously: energy transport by radiation and conduction.
Convection is the transport of energy due to density differences and circulating flows associated with them.
• As a liquid or gas is heated it expands and becomes less dense and therefore lighter. If a cooler denser material is above the hotter layer the warmer material will rise through the cooler material to the surface.
• The rising material will dissipate its heat (energy) into the surrounding environment, become more dense (cooler), and will sink to start the process over.
Convection energy transport
• Consider a small blob of gas. • Raise it without removing heat. • Let it expand so that pressure is equalized. • If the result is lighter (hotter) than the ambient then the blob
is buoyant and convection will occur. • Convection causes eddies or convective cells. Energy is
transported as a result of fluid motions. • If convection occurs then it usually dominates over other
types of energy transport
Convection in Planetary interiors
Convection in the Sun Granulation, Convection cells
Convection in our atmosphere
Variation of temperature within an atmosphere
we have used the approximation of constant temperature to derive the dependence of pressure on height z. ❑ But if we know that convection often
dominates vertical heat transport ❑ To good approximation, convective
heat transport is adiabatic and this sets dT/dz
dP gdz
ρ= −
Simulation by A. Malagoli and F. Cattaneo, U. Chicago
Hydrostatic equilibrium,
https://youtu.be/hfZw6_1X0PA
Adiabatic processesThe first law of thermodynamicsdQ = dE + PdV dQ heat absorbed by the parcel from its surroundings dE the change in internal energy PdV the work done on the parcel by its surroundings
If a change in energy of the system is made without any heat flow dQ = 0 Adiabatically and so
dE = -PdVIf the system is an ideal gas, N molecules PV = NkBT PdV + VdP = NkB dT
Here N is number of molecules
Adiabatic processes (continued)
Heat capacity (=specific heat) at constant volume
Heat capacity at constant pressure
CV ⌘✓@Q
@T
◆
V
CV =
✓@E
@T
◆
V
CP ⌘✓@Q
@T
◆
P
Because dQ = dE + PdVCP =
✓@E
@T
◆
P
+ P
✓@V
@T
◆
P
Because dQ = dE + PdV
For an adiabatic process dQ=0 and dE=-PdVCV dT = �PdV
nRgas or NkB or kB depending on form of ideal gas law used CV
cP = cV + kB per molecule
Specific heats
CP Amount of heat to raise 1 mole by 1 degree C cP Amount of heat to raise 1 gram by 1 degree C
CP = cP m (grams per mole)
cP sometimes used to describe the amount of heat to raise one particle by 1 degree C
� ⌘ CP
CV
Specific-heat ratio, a.k.a. adiabatic index
The specific-heat ratio, γ, for atoms and molecules
❑ Atomic gas (e.g. He): j = 3 (translation in 3-D), γ = 5/3. ❑ Cold diatomic molecular gas (e.g. N2): j = 5, including
rotation (spherical angles θ,φ); γ = 7/5. ❑ Warm diatomic molecular gas: j = 6, including vibration; γ = 4/3.
❑ Large specific heat means many degrees of freedom
The specific heats in pure, ideal gases:
cV = jk/2 j number of mechanical degrees of freedom that can be excited at temperature T
Because cP = cV + kB � =j + 2
j
Adiabatic processes (continued)
1
V
V
V V
NkPdV VdP PdVC
C NkNkVdP PdV PdV PdVC C
dP dVP V
γ
γ
+ = −
# $ # $+− = + = =% & % &
' ( ' (
− =
� lnP = � lnV + constant
ln(PV �) = constant
PV �= constant
ideal gas law
Integrate:
d(PV ) = NkB dT
PdV + V dP = �NkBCV
PdV if adiabatic
Convection zones in atmospheres
Differentiate to find
How does T depend on P in an atmosphere, if heat transfer is adiabatic?
PV �= constant
V =NkBT
P
when adiabatic
Use the ideal gas law to eliminate V
PT �P��= constant
PT �/(1��)= constant
dP
dT=
�
� � 1
P
T
Convection zones (continued)We can get a relation between T and the altitude z by using the chain rule and hydrostatic equilibrium…
using the ideal gas law to eliminate P and ρ:
Integrate this to get T(z)
dT
dz= �� � 1
�
⇢Tg
P= �� � 1
�
NkT
PV
µg
k
dP
dz=
dP
dT
dT
dz=
�
� � 1
P
T
dT
dz= �⇢g
dT
dz= �� � 1
�
µg
kThis is known as the dry lapse rate!
Convection zones (continued)
❑ If the composition (and therefore µ and γ ) changes with height, the dependence of T on z gets correspondingly more complicated.
❑ Nevertheless, when T varies linearly with z, it’s a strong indication that this part of the atmosphere is a convection zone: heat is transported predominantly by convection.
( ) ( ) 10g
T z T zkµγ
γ
−= −
❑ Meteorologists call this formula for the adiabatic temperature gradient the dry adiabatic lapse rate.
❑ For fixed composition and low altitude the integral gives:
T decreases linearly as z increases.
z ⌧ R
g = GM/R2
assume not varying
Dry lapse rate
dT
dz= �� � 1
�
µg
k
Derived assuming: 1. Adiabatic variations 2. Ideal Gas Law 3. Hydrostatic equilibrium
Vertical structure of Venus’s atmosphereThe simplest atmosphere, in which CO2 dominates until well above the clouds. ❑ For CO2, γ = 11/9 at the
temperature of Venus’s surface, declining to 9/7 by z = 50 km where T = 350 K.
❑ This leads to values of —(𝛾-1)𝜇g/𝛾k which match measurements of dT/dz well for z = 0-32 km and 51-56 km: two convection zones.0
10
20
30
40
50
60
70
0 5 10 15
Altitude (km)
K/km
−dTdz
−1γ µgγ k
Convection
Convection
Data from VIRA (Seiff et al. 1985)
λ = 16º
lapse rate follows adiabatic curve where convective
Alti
tude
(km
)
Lapse rate K/km
Vertical structure of Venus’s atmosphere (continued)
The cloud layers, of which there are three, largely lie outside the convection zones. ❑ The clouds are made of
liquid H2SO4 droplets. ❑ The lowest layer, near 50 km
altitude, is the thinnest and the most opaque. Below it lies a thin haze of droplets extending to about 32 km.
0
10
20
30
40
50
60
70
0 5 10 15
Altitude (km)
K/km
−dTdz
−1γ µgγ k
Data from VIRA (Seiff et al. 1985)
λ = 16º
Alti
tude
(km
)
clouds!
Lapse rate K/km
0
50
100
150
200
250
0 200 400 600 800
log Pressure (bar)
Altitude (km)
Temperature (K)
-8
-6
-4
-2
-0
-10
-12
Vertical structure of Venus’s atmosphere (continued)
Radiation and conduction take over the heat transport above ~60 km, resulting in a minimum in the temperature: a thermal inversion. ❑ This inversion is called the
tropopause; below lies the troposphere, above the thermosphere.
❑ The thermosphere has a substantial day/night (diurnal) temperature difference.
Data from VIRA (Seiff et al. 1985, Keating et al. 1985)
Tropopause
Noon Midnight
λ = 16º
Thermosphere
Troposphere
Cryosphere
Alti
tude
(km
)
log
Pres
sure
(ba
r)
0
50
100
150
200
250
0 200 400 600 800
log Pressure (bar)
Altitude (km)
Temperature (K)
-8
-6
-4
-2
-0
-10
-12
Vertical structure of Venus’s atmosphere (continued)
In the troposphere as the temperature decreases colder molecules can condense (clouds can form)
Data from VIRA (Seiff et al. 1985, Keating et al. 1985)
Tropopause
Noon Midnight
λ = 16º
Thermosphere
Troposphere
Cryosphere
Alti
tude
(km
)
log
Pres
sure
(ba
r)
Vertical structure of Earth’s atmosphere
❑ Convective troposphere.
❑ Above the troposphere, the temperature increase in the stratosphere is caused by an increase in ozone (O3), which strongly absorbs ultraviolet radiation.
❑ Note pressure units:
6 -21 bar 10 dyne cm1 atmosphere = 1.013 bar
=0
50
100
150
200
250
0 200 400 600 800
log Pressure (bar)
Altitude (km)
Temperature (K)
-8
-6
-4
-2
-0
Mesopause
Stratopause
Tropopause
Thermosphere
Mesosphere
Stratosphere
Data from the MSIS-E-90 atmospheric model.
Alti
tude
(km
)
log
Pres
sure
(ba
r)
Vertical structure of Jupiter’s upper atmosphere
The parachute probe dropped by Galileo measured thermodynamic parameters as functions of altitude, taken to be height above the P = 1 bar level. ❑ Again, a convective
troposphere. ❑ Nearly isothermal strato-
mesosphere.
-200
0
200
400
600
800
1000
0 200 400 600 800 1000
log Pressure (bar)
Altitude (km)
Temperature (K)
-8
-6
-4
-2
-0
Data from Seiff et al. 1998.
Tropopause
Thermosphere
Troposphere
Clouds: NH3 NH4SH
Alti
tude
(km
)
log
Pres
sure
(ba
r)
Convection as a diffusive process
• To approximate energy transport as a diffusive process we need a diffusion coefficient (kd in cm2 s-1).
• Convective energy transport is often approximated with a diffusion coefficient called the Eddy diffusion coefficient.
• This is often done via an approximation known as “mixing length” theory.
• This theory is not exact, so Eddy-Diffusion coefficients are often “measured.”
Atmospheres of the Giant Planets
• In the thermosphere, the temperature increases with height as more energy from the Sun and solar wind is deposited into the atmosphere.
• The mesosphere is the intermediate region which in some cases can be considered isothermal.
Troposphere
Thermosphere
Mesosphere
hydrocarbon haze
Spectrum of an atmosphereIn
tens
ity o
f rad
iatio
n
Atmospheric Escape
• A particle can escape from an atmosphere if its kinetic energy exceeds the gravitational binding energy of the planet, and if it moves on an upward trajectory without intersecting another particle.
• The region of escape is called the exosphere. • The location and size of the exosphere can be
estimated from the mean free path of molecules.
Maxwellian velocity distribution function- continued
For a gas in thermal equilibrium, the velocity distribution of particles of mass m is given by distribution function
With spherical symmetry dvx dvy dvz = v2 dv Integrating over angle gives a factor of 4𝜋
f(v)dvx
dvy
dvz
= N⇣ m
2⇡kT
⌘ 32e�m(v2
x
+v
2y
+v
2z
)/(2kT )dvx
dvy
dvz
Kinetic energy/kT
Recall black body function with
e�h⌫kT
T temperature k Boltzmann constant
f(v)dv = N
r2
⇡
⇣ m
kT
⌘ 32v2e�
mv2
2kT dv
Gaussian function
Z 1
�1dx e
� x
2
2�2 =p2⇡�2 normalization
hx2i = 1p2⇡�2
Z 1
�1dx x
2e
� x
2
2�2 = �
2 variance
𝜎 standard deviation
Equipartition
21/ 2 3/ 2
2 /(2 )2( )
This holds for each mass atom or molecule.Because of collisions, and thermal radiation,different molecules have approximately the same temperature.This
mv kTmf v dv N v e dvkT
mπ
−# $ # $= % & % &' ( ' (
2 21 1 2 2
is known as the . Lower mass molecules have higher mean velocities.Lighter molecules are more likely to escape.
m v m v= equipartition of energy
f(v)dv = N
r2
⇡
⇣ m
kT
⌘ 32v2e�
mv2
2kT dv
This holds for each atom or molecule depending on its mass Because of collisions and radiation different molecules are described by the same temperature Equipartition of energy
m1hv21i = m2hv22i
Thermal Escape (Jean’s Escape)
21/ 2 3/ 2
2 /(2 )
0
0
2( )
Because of the exponential form, most particles havevelocities less than the root mean scale velocity ~ 2 / Compare to the escape velocity
mv kTmf v dv N v e dvkT
v kT m
v
π−# $ # $= % & % &
' ( ' (
( )
0
20 escape
2 /Particles with will tend to escape.
Define a parameter / ( in the book)
By integrating the high velocity tail of the Maxwellian distribution function,the
e
e
e
v GM rv v
Y v v Y λ
=>
≡ =
rate that particles escape can be estimated.
f(v)dv = N
r2
⇡
⇣ m
kT
⌘ 32v2e�
mv2
2kT dv
By integrating the high velocity tail of the Maxwellian distribution function, we can estimate the rate that particles escape
v0 =p
2kT/mTypical velocity Escape velocity Particles with v>ve will escape!
ve =p
2GM/r
To characterize the number that can escape Define a parameter
Y ⌘ (ve/v0)2
Thermal escape (continued)
21/ 2 3/ 2
2 /(2 )
-2 -1
2( )
To estimate an escape rate we consider the number of atoms that have velocities above .Their flux is ( ) Number of atoms which escape cm s
~
mv kT
z e
z
zv
mf v dv N v e dvkT
v vf v v
dv
π−# $ # $= % & % &
' ( ' (
>
2 /(2 )3 2~ (1 ) where ~ /z
escape
mv kT Yz ev e e Y Y mv kT
∞− − +∫
f(v)dv = N
r2
⇡
⇣ m
kT
⌘ 32v2e�
mv2
2kT dv
To estimate an escape rate we consider the number of atoms with vertical velocity component above the escape velocity 𝑣z > 𝑣e
Their flux is f(𝑣) 𝑣z
flux = number of atoms per unit time per unit area
Y = (ve/v0)2 ⇠ mv2e/(kT )
Timescale for reducing the atmosphere via thermal escape
The rate that particles escape via thermal evaporation
� ⇠ Nv0(1 + Y )e�Y particles cm-2 s-1
N is the number of particles per cm3 at the exobase. The exobase is the height where a particle traveling upward is unlikely to hit another particle
Total numbers of particles in the atmosphere Tp ~ NAh A area of planet, h is scale height The total number of particles lost per second is
dTp
dt= ��A = �Tp
hv0(1 + Y )e�Y
Timescale for reducing the atmosphere via thermal escape
exponential solution Total numbers of particles in atmosphere
for Y>1 the timescale to reduce atmospheric density by 1/e
tescape ⇠h
v0
eY
Y
dTp
dt= ��A = �Tp
hv0(1 + Y )e�Y
Tp / e�t/tescape
Atmospheric escape (continued)
Y =2kT
mv2e
v2e =2GM
r= 2gr
Y ⇠ h
r
m̄
m
h ⇠ kT
gm̄
For a particle of mass m If Y>>1 particles escape
Scale height depends on average mass of particles
average mass
Escape velocity Insert into Y
Jeans escape of light molecules from an atmosphere (continued)
The Earth’s exosphere: the temperature is typically about 600 K, and the top of Saturn’s atmosphere, for which the temperature is about 160 K. In years, tescape comes out to
Earth Saturn
He
Age of solar system
Age of Universe
2H
2N
× 39.7 10 × 4051.9 10
× 142.6 10 × 8123.0 10
× 1425.4 10 × 57412.9 10
× 94.6 10
× 101.4 10
Additional Processes that after atmospheric escape
• Non-thermal distributions • Photodissociation. H escape. • Interactions with high speed solar wind
particles. These additional processes are needed to
explain things like how Venus lost its water (if it ever had any).
First glimpses of extrasolar planet atmospheres
Venus transit
Hot exoplanet phase curves
Maxted et al. 2012 WASP-18
Review• Hydrostatic equilibrium • Scale height (and how to estimate it). • Dry lapse rate dT/dz • Atmospheric structure, thermosphere, troposphere,
mesosphere. • Jean’s (thermal) escape and how to estimate the
timescale for different molecules. • Energy transport processes that are approximated with
the diffusion equation: 1. Conduction, 2. Radiative transfer in an optically thick medium, 3. Convection.
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