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.