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Transcript of AST 111 Lecture 6 - University of aquillen/ast111/Lectures/ AST 111 Lecture 6. This

  • 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 g Mg

    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/ 3 300

    earth

    jupiter

    kTh gm

    g g

    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.

    Planet Isothermal scale height (km)

    Venus 15.9 Earth 8.5 Mars 11.1 Jupiter 27 Saturn 59.5 Uranus 27.7 Neptune 20 Pluto 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 transport Previously: 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 g dz

    ρ= −

    Simulation by A. Malagoli and F. Cattaneo, 
 U. Chicago

    Hydrostatic equilibrium,

    https://youtu.be/ hfZw6_1X0PA

  • Adiabatic processes The first law of thermodynamics dQ = 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 = -PdV If 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=-PdV CV 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 PdV C

    C NkNkVdP PdV PdV PdV C C

    dP dV P V

    γ

    γ

    + = −

    # $ # $+ − = + = =% & % &

    ' ( ' (

    − =

    � lnP = � lnV + constant

    ln(PV �) = constant

    PV � = constant

    ideal gas law

    Integrate:

    d(PV ) = NkB dT

    PdV + V dP = �NkB CV

    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