Thre is an upwards force on a thin horizontal slab of air in the atmosphere due to the pressure gradient across the slab – the pressure below is slightly larger than the pressure above, leading to a net force.br /br /This force is usually almost exactly canceled by the downwards gravitational force acting on the slab:br /br /blockquote-A dp/dz delta z = g rho A delta zbr /br /dp/dz = -grho/blockquotebr /br /since p(infty) = 0;br /blockquotebr /p(z) = int_z^infty grho dz/blockquotebr /br /2.2.1 Geopotentialbr /br /geopotential, Phi, at a point is defined as the work against gravity required to raise a 1kg mass from sea level (or some other reference altitude) to that point.br /br /We can then define a geopotential height:br /br /blockquoteZ = Phi(z)/g_0 = (1/g_0)*int_0^z g dz/blockquotebr /br /where g_0 is the standard gravitational acceleration (~9.8 m/s/s). Geopotential height can be used as a vertical coordinate, most usefully when energy plays an important role in the system.br /br /It is generally inconvenient to deal with density, as it is difficult to measure in the field – so, eliminating rho from the hydrostatic equation:br /br /blockquotedp/dz = -pg/R_dT_v/blockquotebr /br /integrating between two levels (1,2) leads to:br /blockquotebr /Z_2-Z_1 = (R_d/g_0)*int_p2^p1 T_v (dp/p)/blockquotebr /br /2.2.2 Scale height and hypsometric equationbr /br /For an isothermal atmosphere, the above equation equation becomes:br /blockquotebr /Z_2 – Z_1 = H ln(p2/p1)/blockquotebr /where H = R_dT_v/g_0 = 29.3T_v is the scale height of the atmosphere.br /br /Below the turbopause, the atmosphere is (chemically) well mixed: the pressures and densities of the various species fall off uniformly, with a scale height of ~8.5 kmbr /br /Above the turbopause, the veritcal distribution is more strongly controlled by diffusion and each species has its own scale height which is inversely proportional to the molecular weight of that species.br /br /Example 2.2: If the ratio of O to H at Z=200km is 10^5, what is the ratio at Z=1400km, assuming that the atmosphere is isothermal in this region with a temperature of 2000K?br /span style=”font-style: italic;”At these heights, because the atmosphere is very thin, the radiative cooling is very inefficient and the atmosphere becomes very “hot” to maintain radiative balance./spanbr /br /Call the ratio RRbr /blockquotebr /RR(1400km) = p_O(200km)*exp(-1200km/H_O) / p_H(200km)*exp(-1200km/H_H)br /br /RR(1400km) = RR(200km) * exp(-1200km*(1/H_O – 1/H_H))br /br /RR(1400km) = 2.47/blockquotebr /br /As the temperature usually varies with height, we can define a mean virtual temperature bar{T_v} w.r.t. ln pbr /br /2.2.3 Thickness and constant pressure surfaces.br /br /Z_2-Z_1 is the thickness of the intervening layer, and is proportional to the mean T_v of the layer. Roughly speaking, if you imagine the air between two levels as being unable to cross the level boundaries, upon heating, the air expands, pushing the levels apart.br /br /Example 2.3: Calculate the thickness of the 1000mb – 500mb layer in a) the tropics where bar{T_v} is 9^oC and b) the polar region, where bar{T_v} = -40^oCbr /br /blockquoteDeltaZ = (R_dbar{T_v}/g_0)ln(1000/500) = 20.3bar{T_v}br /br /a) bar{T_v} = 282K, DeltaZ = 5725mbr /b) bar{T_v} = 233K, DeltaZ = 4730m/blockquotebr /br /Some more examples:br /br /ulliFrom the surface up to the tropopause, a hurricane’s core is warmer than the surroundings – consequently, the pressure surfaces are pushed further and further down as one descends from the tropopause. The intensity, as measured by the drop of pressure from average pressure for each altitude decreases with height – warm core lows have the greatest intensity near the ground and diminish with altitude./liliSome upper air lows do not extend all the way down to the ground. The downwards deflection aloft must be countered by an upwards deflection of isobaric surfaces at lower altitudes, which requires a cold core. A strong enough cold core, combined with a weak enough (or absent) upper warm anomaly can lead to surface high pressures.br //liliMost mid-latitude disturbances have a cold core disturbance (low pressure aloft, high at sea level) to the west of a warm core disturbance. The position of the greatest pressure depression slopes westward with height./li/ulbr /br /br /Example 2.4br /Calculate the geopotential height of the 1000mb surface when the pressure at sea level is 1014mb. Use H_0 = 8kmbr /br /blockquoteZ_2 – 0 = H_0ln(p_2/p_1) = H_0 ln (1.014) ~H_0*(0.014) = ~112m/blockquotebr /br /Example 2.5 br /Derive the relationship for Z(p), in terms of p_0 and T_0, given a constant lapse rate Gammabr /blockquotebr /T = T_0 – Gamma z/blockquotebr /br /Hydrostatic equation:br /blockquotebr /dp/p = -g/R dz/Tbr /dp/p = -g/R dz/(T_0 – Gamma z)/blockquotebr /br /integrating from sea level up:br /blockquotebr /ln(p/p0) = -g/R int_0^z dz/(T_0 – Gamma z)br /br /ln(p/p0) = -(g/RGamma) ln[T_0/(T_0 – Gamma z)]br /br /p = p0*-(T_0 -Gamma z)/T_0]^(g/RGamma)br /br /z = (T_0/Gamma)[1-(p/p_0)^(RGamma/g)]/blockquote

# Wallace and Hobbs 2.2 – The Hydrostatic Equation

Previous post: Wallace and Hobbs 2.1 – The Gas Laws

Next post: Wallace and Hobbs 2.3 – The First Law of Thermodynamics