# Barometric formula

The barometric formula, sometimes called the exponential atmosphere or isothermal atmosphere, is a formula used to model how the pressure (or density) of the air changes with altitude.

## Pressure equations

Pressure as a function of the height above the sea level

There are two different equations for computing pressure at various height regimes below 86 km (or 278,400 feet). The first equation is used when the value of Standard Temperature Lapse Rate is not equal to zero; the second equation is used when standard temperature lapse rate equals zero.

Equation 1:

${P}=P_b \cdot \left[\frac{T_b}{T_b + L_b\cdot(h-h_b)}\right]^{\textstyle \frac{g_0 \cdot M}{R^* \cdot L_b}}$

Equation 2:

$P=P_b \cdot \exp \left[\frac{-g_0 \cdot M \cdot (h-h_b)}{R^* \cdot T_b}\right]$

where

$P_b$ = Static pressure (pascals)
$T_b$ = Standard temperature (K)
$L_b$ = Standard temperature lapse rate -0.0065 (K/m) in ISA
$h$ = Height above sea level (meters)
$h_b$ = Height at bottom of layer b (meters; e.g., $h_1$ = 11,000 meters)
$R^*$ = Universal gas constant for air: 8.31432 N·m /(mol·K)
$g_0$ = Gravitational acceleration (9.80665 m/s2)
$M$ = Molar mass of Earth's air (0.0289644 kg/mol)

Or converted to Imperial units:[1]

where

$P_b$ = Static pressure (inches of mercury, inHg)
$T_b$ = Standard temperature (K)
$L_b$ = Standard temperature lapse rate (K/ft)
$h$ = Height above sea level (ft)
$h_b$ = Height at bottom of layer b (feet; e.g., $h_1$ = 36,089 ft)
$R^*$ = Universal gas constant; using feet, kelvins, and (SI) moles: 8.9494596×104 lb·ft2/(lbmol·K·s2)
$g_0$ = Gravitational acceleration (32.17405 ft/s2)
$M$ = Molar mass of Earth's air (28.9644 lb/lbmol)

The value of subscript b ranges from 0 to 6 in accordance with each of seven successive layers of the atmosphere shown in the table below. In these equations, g0, M and R* are each single-valued constants, while P, L, T, and h are multivalued constants in accordance with the table below. The values used for M, g0, and $R^*$ are in accordance with the U.S. Standard Atmosphere, 1976, and the value for $R^*$ in particular does not agree with standard values for this constant.[2] The reference value for Pb for b = 0 is the defined sea level value, P0 = 101325 pascals or 29.92126 inHg. Values of Pb of b = 1 through b = 6 are obtained from the application of the appropriate member of the pair equations 1 and 2 for the case when $h = h_{b+1}$.:[2]

Subscript b Height above sea level Static pressure Standard temperature
(K)
Temperature lapse rate
(m) (ft) (pascals) (inHg) (K/m) (K/ft)
0 0 0 101325.00 29.92126 288.15 -0.0065 -0.0019812
1 11,000 36,089 22632.10 6.683245 216.65 0.0 0.0
2 20,000 65,617 5474.89 1.616734 216.65 0.001 0.0003048
3 32,000 104,987 868.02 0.2563258 228.65 0.0028 0.00085344
4 47,000 154,199 110.91 0.0327506 270.65 0.0 0.0
5 51,000 167,323 66.94 0.01976704 270.65 -0.0028 -0.00085344
6 71,000 232,940 3.96 0.00116833 214.65 -0.002 -0.0006096

## Density equations

The expressions for calculating density are nearly identical to calculating pressure. The only difference is the exponent in Equation 1.

There are two different equations for computing density at various height regimes below 86 geometric km (84,852 geopotential meters or 278,385.8 geopotential feet). The first equation is used when the value of Standard Temperature Lapse rate is not equal to zero; the second equation is used when Standard Temperature Lapse rate equals zero.

Equation 1:

${\rho}=\rho_b \cdot \left[\frac{T_b + L_b\cdot(h-h_b)}{T_b}\right]^{\left(-\frac{g_0 \cdot M}{R^* \cdot L_b}\right)-1}$

Equation 2:

${\rho}=\rho_b \cdot \exp\left[\frac{-g_0 \cdot M \cdot (h-h_b)}{R^* \cdot T_b}\right]$

where

${\rho}$ = Mass density (kg/m3)
$T$ = Standard temperature (K)
$L$ = Standard temperature lapse rate (see table below) (K/m) in ISA
$h$ = Height above sea level (geopotential meters)
$R^*$ = Universal gas constant for air: 8.31432 N·m/(mol·K)
$g_0$ = Gravitational acceleration (9.80665 m/s2)
$M$ = Molar mass of Earth's air (0.0289644 kg/mol)

Or converted to English gravitational foot-pound-second units:[1]

Where

${\rho}$ = Mass density (slug/ft3)
${T}$ = Standard temperature (kelvins)
${L}$ = Standard temperature lapse rate (degrees Celsius per foot)
${h}$ = Height above sea level (geopotential feet)
${R^*}$ = Universal gas constant (8.9494596×104 ft2/(s·°C))
${g_0}$ = Gravitational acceleration (32.17405 ft/s2)
${M}$ = Molar mass of Earth's air (0.0289644 kg/mol)

The value of subscript b ranges from 0 to 6 in accordance with each of seven successive layers of the atmosphere shown in the table below. The reference value for $\rho_b$ for b = 0 is the defined sea level value, $\rho_o$ = 1.2250 kg/m3 or 0.0023768908 slug/ft3. Values of $\rho_b$ of b = 1 through b = 6 are obtained from the application of the appropriate member of the pair equations 1 and 2 for the case when $h = h_{b+1}$ [2]

In these equations, g0, M and R* are each single-valued constants, while $\rho$, L, T and h are multi-valued constants in accordance with the table below. The values used for M, g0 and R* are in accordance with the U.S. Standard Atmosphere, 1976, and that the value for R* in particular does not agree with standard values for this constant.[2]

Subscript b Height Above Sea Level (h) Mass Density ($\rho$) Standard Temperature (T')
(K)
Temperature Lapse Rate (L)
(m) (ft) (kg/m3) (slugs/ft3) (K/m) (K/ft)
0 0 0 1.2250 2.3768908 x 10−3 288.15 -0.0065 -0.0019812
1 11,000 36,089.24 0.36391 7.0611703 x 10−4 216.65 0.0 0.0
2 20,000 65,616.79 0.08803 1.7081572 x 10−4 216.65 0.001 0.0003048
3 32,000 104,986.87 0.01322 2.5660735 x 10−5 228.65 0.0028 0.00085344
4 47,000 154,199.48 0.00143 2.7698702 x 10−6 270.65 0.0 0.0
5 51,000 167,322.83 0.00086 1.6717895 x 10−6 270.65 -0.0028 -0.00085344
6 71,000 232,939.63 0.000064 1.2458989 x 10−7 214.65 -0.002 -0.0006096

## Derivation

The barometric formula can be derived fairly easily using the ideal gas law:

$\rho = \frac{M \cdot P}{R^* \cdot T}$

When density is known:

$P = \frac{\rho \cdot {R^*} \cdot T}{M}$

And assuming that all pressure is hydrostatic:

$dP = - \rho g\,dz\,$

Dividing the $dP$ by the $P$ expression we get:

$\frac{dP}{P} = - \frac{M g\,dz}{R^*T}$

Integrating this expression from the surface to the altitude z we get:

$P = P_0 e^{-\int_{0}^{z}{M g dz/R^*T}}\,$

Assuming constant temperature, molar mass, and gravitational acceleration, we get the barometric formula:

$P = P_0 e^{-M g z/R^*T}\,$

In this formulation, $R^*$ is the gas constant, and the term $R^*T/M g$ gives the scale height (approximately equal to 8.4 km for the troposphere).

(For exact results, it should be remembered that atmospheres containing water do not behave as an ideal gas. See real gas or perfect gas or gas for further understanding)

## Estimating the temperature

Assuming that the only energy source is from the sun, and the albedo is constant throughout the planet, we can get an estimate of a constant temperature. The solar energy flux at a distance $R_\mathrm{AU}$ can be estimated as:

$f_\odot= \frac{L_\odot}{4\pi R^2_\mathrm{AU}},$

where $L_\odot$ is the solar luminosity. The actual incoming energy can be estimated as

$E_{in}=f_\odot (1-\alpha)\pi R^2_\oplus$

where $\alpha$ is the albedo of the planet. $R_\oplus$ is the radius of the planet and $R_\mathrm{AU}$ is the distance to the Sun in astronomical units. The outgoing energy can be estimated using the Stefan-Boltzmann's law

$E_{out}=4\pi R^2_\oplus \sigma T^4_{eq}$

where $\sigma$ is the Stefan–Boltzmann constant and $T_{eq}$ is the temperature at equilibrium. Solving the equation:

$E_{in}=E_{out} \,$

leads to the following estimate of a planet's temperature

$T_{eq}=\sqrt[4]{\frac{L_\odot (1-\alpha)}{16 \pi \sigma R^2_\mathrm{AU}}}$

which for Earth is about 255 K or −18 °C