# Common formula

## gravity

1. Newton’s gravitational law

symbol

description

unit

$$g$$

normal gravity at sea level

$$\frac{m}{s^2}$$

$$g_{h}$$

gravity at specific height

$$\frac{m}{s^2}$$

$$z$$

altitude

$$m$$

$$R$$

$$m$$

$g_{h} = g\left(\frac{R}{R + z}\right)^2$
2. Normal gravity at sea level ellipsoid

This is the WGS84 ellipsoidal gravity formula as taken from NIMA TR8350.2

symbol

name

unit

$$a$$

WGS84 semi-major axis

$$m$$

$$b$$

WGS84 semi-minor axis

$$m$$

$$e$$

eccentricity

$$m$$

$$g$$

normal gravity at sea level

$$\frac{m}{s^2}$$

$$g_{e}$$

gravity at equator

$$\frac{m}{s^2}$$

$$g_{p}$$

gravity at poles

$$\frac{m}{s^2}$$

$$\phi$$

latitude

$$degN$$

\begin{eqnarray} e^2 & = & \frac{a^2-b^2}{a^2} \\ k & = & \frac{bg_{p} - ag_{e}}{ag_{e}} \\ g & = & g_{e}\frac{1 + k {\sin}^2(\frac{\pi}{180}\phi)}{\sqrt{1 - e^2{\sin}^2(\frac{\pi}{180}\phi)}} \\ g & = & 9.7803253359 \frac{1 + 0.00193185265241{\sin}^2(\frac{\pi}{180}\phi)} {\sqrt{1 - 0.00669437999013{\sin}^2(\frac{\pi}{180}\phi)}} \end{eqnarray}
3. Gravity at specific altitude

This is the WGS84 ellipsoidal gravity formula as taken from NIMA TR8350.2

symbol

name

unit

$$a$$

WGS84 semi-major axis

$$m$$

$$b$$

WGS84 semi-minor axis

$$m$$

$$f$$

WGS84 flattening

$$m$$

$$g$$

normal gravity at sea level

$$\frac{m}{s^2}$$

$$g_{h}$$

gravity at specific height

$$\frac{m}{s^2}$$

$$GM$$

WGS84 earth’s gravitational constant

$$\frac{m^3}{s^2}$$

$$z$$

altitude

$$m$$

$$\phi$$

latitude

$$degN$$

$$\omega$$

WGS84 earth angular velocity

$$rad/s$$

The formula used is the one based on the truncated Taylor series expansion:

\begin{eqnarray} m & = & \frac{\omega^2a^2b}{GM} \\ g_{h} & = & g \left[ 1 - \frac{2}{a}\left(1+f+m-2f{\sin}^2(\frac{\pi}{180}\phi)\right)z + \frac{3}{a^2}z^2 \right] \\ \end{eqnarray}

## geopotential height

symbol

description

unit

$$g$$

normal gravity at sea level

$$\frac{m}{s^2}$$

$$g_{0}$$

mean earth gravity

$$\frac{m}{s^2}$$

$$g_{h}$$

gravity at specific height

$$\frac{m}{s^2}$$

$$p$$

pressure

$$Pa$$

$$R$$

$$m$$

$$z$$

altitude

$$m$$

$$z_{g}$$

geopotential height

$$m$$

$$\phi$$

latitude

$$degN$$

$$\rho$$

mass density

$$\frac{kg}{m^3}$$

The geopotential height allows the gravity in the hydrostatic equation

$dp = - \rho g_{h} dz$

to be replaced by a constant gravity

$dp = - \rho g_{0} dz_{g}$

providing

$dz_{g} = \frac{g_{h}}{g_{0}}dz$

With Newton’s gravitational law this becomes

$dz_{g} = \frac{g}{g_{0}}\left(\frac{R}{R + z}\right)^2dz$

And integrating this, considering that $$z=0$$ and $$z_{g}=0$$ at sea level, results in

$z_{g} = \frac{g}{g_{0}}\frac{Rz}{R + z}$
$z = \frac{g_{0}Rz_{g}}{gR-g_{0}z_{g}}$

## gas constant

symbol

name

unit

$$k$$

Boltzmann constant

$$\frac{kg m^2}{K s^2}$$

$$N_A$$

$$\frac{1}{mol}$$

$$R$$

universal gas constant

$$\frac{kg m^2}{K mol s^2}$$

Relation between Boltzmann constant, universal gas constant, and Avogadro constant:

$k = \frac{R}{N_A}$

## ideal gas law

symbol

name

unit

$$k$$

Boltzmann constant

$$\frac{kg m^2}{K s^2}$$

$$N$$

amount of substance

$$molec$$

$$p$$

pressure

$$Pa$$

$$R$$

universal gas constant

$$\frac{kg m^2}{K mol s^2}$$

$$T$$

temperature

$$K$$

$$V$$

volume

$$m^3$$

$pV = \frac{NRT}{N_{A}} = NkT$

## barometric formula

symbol

name

unit

$$g_{0}$$

mean earth gravity

$$\frac{m}{s^2}$$

$$g_{h}$$

gravity at specific height

$$\frac{m}{s^2}$$

$$k$$

Boltzmann constant

$$\frac{kg m^2}{K s^2}$$

$$M_{air}$$

molar mass of total air

$$\frac{g}{mol}$$

$$N$$

amount of substance

$$molec$$

$$N_A$$

$$\frac{1}{mol}$$

$$p$$

pressure

$$Pa$$

$$R$$

universal gas constant

$$\frac{kg m^2}{K mol s^2}$$

$$T$$

temperature

$$K$$

$$V$$

volume

$$m^3$$

$$z$$

altitude

$$m$$

$$z_{g}$$

geopotential height

$$m$$

$$\phi$$

latitude

$$degN$$

$$\rho$$

mass density

$$\frac{kg}{m^3}$$

From the ideal gas law we have:

$p = \frac{NkT}{V} = \frac{10^{-3}NM_{air}}{VN_{a}}\frac{kTN_{a}}{10^{-3}M_{air}} = \rho\frac{RT}{10^{-3}M_{air}}$

And from the hydrostatic assumption we get:

$dp = - \rho g_{h} dz$

Dividing $$dp$$ by p we get:

$\frac{dp}{p} = -\frac{10^{-3}M_{air}\rho g_{h} dz}{\rho RT} = -\frac{10^{-3}M_{air}g_{h}dz}{RT}$

Integrating this expression from one pressure level to the next we get:

$p(i+1) = p(i)e^{-\int^{z(i+1)}_{z(i)}\frac{10^{-3}M_{air}g_{h}}{RT}dz}$

We can approximate this further by using an average value of the height dependent quantities $$M_{air}$$, $$g_{h}$$ and $$T$$ for the integration over the range $$[z(i),z(i+1)]$$. This gives:

\begin{eqnarray} g' & = & g_{h}(\phi,\frac{z(i)+z(i+1)}{2}) \\ p(i+1) & = & p(i)e^{-10^{-3}\frac{M_{air}(i)+M_{air}(i+1)}{2}\frac{2}{T(i)+T(i+1)}\frac{g'}{R}\left(z(i+1)-z(i)\right)} \\ & = & p(i)e^{-10^{-3}\frac{M_{air}(i)+M_{air}(i+1)}{T(i)+T(i+1)}\frac{g'}{R}\left(z(i+1)-z(i)\right)} \end{eqnarray}

When using geopotential height the formula is the same except that $$g=g_{0}$$ at all levels:

$p(i+1) = p(i)e^{-10^{-3}\frac{M_{air}(i)+M_{air}(i+1)}{T(i)+T(i+1)}\frac{g_{0}}{R}\left(z_{g}(i+1)-z_{g}(i)\right)}$

## mass density

symbol

name

unit

$$N$$

amount of substance

$$molec$$

$$N_A$$

$$\frac{1}{mol}$$

$$M_{air}$$

molar mass of total air

$$\frac{g}{mol}$$

$$V$$

volume

$$m^3$$

$$\rho$$

mass density

$$\frac{kg}{m^3}$$

$\rho = \frac{10^{-3}NM_{air}}{VN_{a}}$

## number density

symbol

name

unit

$$n$$

number density

$$\frac{molec}{m^3}$$

$$N$$

amount of substance

$$molec$$

$$V$$

volume

$$m^3$$

$n = \frac{N}{V}$

## dry air vs. total air

symbol

name

unit

$$n$$

number density of total air

$$\frac{molec}{m^3}$$

$$n_{dry\_air}$$

number density of dry air

$$\frac{molec}{m^3}$$

$$n_{H_{2}O}$$

number density of H2O

$$\frac{molec}{m^3}$$

$$M_{air}$$

molar mass of total air

$$\frac{g}{mol}$$

$$M_{dry\_air}$$

molar mass of dry air

$$\frac{g}{mol}$$

$$M_{H_{2}O}$$

molar mass of H2O

$$\frac{g}{mol}$$

$$\rho$$

mass density of total air

$$\frac{kg}{m^3}$$

$$\rho_{dry\_air}$$

mass density of dry air

$$\frac{kg}{m^3}$$

$$\rho_{H_{2}O}$$

mass density of H2O

$$\frac{kg}{m^3}$$

\begin{eqnarray} n & = & n_{dry\_air} + n_{H_{2}O} \\ M_{air}n & = & M_{dry\_air}n_{dry\_air} + M_{H_{2}O}n_{H_{2}O} \\ \rho & = & \rho_{dry\_air} + \rho_{H_{2}O} \\ \end{eqnarray}

## virtual temperature

symbol

name

unit

$$k$$

Boltzmann constant

$$\frac{kg m^2}{K s^2}$$

$$M_{air}$$

molar mass of total air

$$\frac{g}{mol}$$

$$M_{dry\_air}$$

molar mass of dry air

$$\frac{g}{mol}$$

$$M_{H_{2}O}$$

molar mass of H2O

$$\frac{g}{mol}$$

$$N$$

amount of substance

$$molec$$

$$N_A$$

$$\frac{1}{mol}$$

$$p$$

pressure

$$Pa$$

$$p_{dry\_air}$$

dry air partial pressure

$$Pa$$

$$p_{H_{2}O}$$

H2O partial pressure

$$Pa$$

$$R$$

universal gas constant

$$\frac{kg m^2}{K mol s^2}$$

$$T$$

temperature

$$K$$

$$T_{v}$$

virtual temperature

$$K$$

$$V$$

volume

$$m^3$$

From the ideal gas law we have:

$p = \frac{NkT}{V} = \frac{10^{-3}NM_{air}}{VN_{a}}\frac{kTN_{a}}{10^{-3}M_{air}} = \rho \frac{RT}{10^{-3}M_{air}}$

The virtual temperature allows us to use the dry air molar mass in this equation:

$p = \rho\frac{RT_{v}}{10^{-3}M_{dry\_air}}$

This gives:

$T_{v} = \frac{M_{dry\_air}}{M_{air}}T$

## volume mixing ratio

symbol

name

unit

$$n$$

number density of total air

$$\frac{molec}{m^3}$$

$$n_{dry\_air}$$

number density of dry air

$$\frac{molec}{m^3}$$

$$n_{H_{2}O}$$

number density of H2O

$$\frac{molec}{m^3}$$

$$n_{x}$$

number density of quantity x

$$\frac{molec}{m^3}$$

$$\nu_{x}$$

volume mixing ratio of quantity x with regard to total air

$$ppv$$

$$\bar{\nu}_{x}$$

volume mixing ratio of quantity x with regard to dry air

$$ppv$$

\begin{eqnarray} \nu_{x} & = & \frac{n_{x}}{n} \\ \bar{\nu}_{x} & = & \frac{n_{x}}{n_{dry\_air}} \\ \nu_{dry\_air} & = & \frac{n_{dry\_air}}{n} = \frac{n - n_{H_{2}O}}{n} = 1 - \nu_{H_{2}O} \\ \nu_{air} & = & \frac{n}{n} = 1 \\ \bar{\nu}_{dry\_air} & = & \frac{n_{dry\_air}}{n_{dry\_air}} = 1 \\ \bar{\nu}_{H_{2}O} & = & \frac{n_{H_{2}O}}{n_{dry\_air}} = \frac{\nu_{H_{2}O}}{\nu_{dry\_air}} = \frac{\nu_{H_{2}O}}{1 - \nu_{H_{2}O}} \\ \nu_{H_{2}O} & = & \frac{\bar{\nu}_{H_{2}O}}{1 + \bar{\nu}_{H_{2}O}} \end{eqnarray}

## mass mixing ratio

symbol

name

unit

$$M_{air}$$

molar mass of total air

$$\frac{g}{mol}$$

$$M_{dry\_air}$$

molar mass of dry air

$$\frac{g}{mol}$$

$$M_{x}$$

molar mass of quantity x

$$\frac{g}{mol}$$

$$n$$

number density of total air

$$\frac{molec}{m^3}$$

$$n_{dry\_air}$$

number density of dry air

$$\frac{molec}{m^3}$$

$$n_{H_{2}O}$$

number density of H2O

$$\frac{molec}{m^3}$$

$$n_{x}$$

number density of quantity x

$$\frac{molec}{m^3}$$

$$q_{x}$$

mass mixing ratio of quantity x with regard to total air

$$\frac{kg}{kg}$$

$$\bar{q}_{x}$$

mass mixing ratio of quantity x with regard to dry air

$$\frac{kg}{kg}$$

$$\nu_{x}$$

volume mixing ratio of quantity x with regard to total air

$$ppv$$

$$\bar{\nu}_{x}$$

volume mixing ratio of quantity x with regard to dry air

$$ppv$$

\begin{eqnarray} q_{x} & = & \frac{n_{x}M_{x}}{nM_{air}} = \nu_{x}\frac{M_{x}}{M_{air}} \\ \bar{q}_{x} & = & \frac{n_{x}M_{x}}{n_{dry\_air}M_{dry\_air}} = \bar{\nu}_{x}\frac{M_{x}}{M_{dry\_air}} \\ q_{dry\_air} & = & \frac{n_{dry\_air}M_{dry\_air}}{nM_{air}} = \frac{nM_{air} - n_{H_{2}O}M_{H_{2}O}}{nM_{air}} = 1 - q_{H_{2}O} \\ q_{air} & = & \frac{nM_{air}}{nM_{air}} = 1 \\ \bar{q}_{dry\_air} & = & \frac{n_{dry\_air}M_{dry\_air}}{n_{dry\_air}M_{dry\_air}} = 1 \\ \bar{q}_{H_{2}O} & = & \frac{n_{H_{2}O}M_{H_{2}O}}{n_{dry\_air}M_{dry\_air}} = \frac{q_{H_{2}O}}{q_{dry\_air}} = \frac{q_{H_{2}O}}{1 - q_{H_{2}O}} \\ q_{H_{2}O} & = & \frac{\bar{q}_{H_{2}O}}{1 + \bar{q}_{H_{2}O}} \end{eqnarray}

## molar mass of total air

1. molar mass of total air from H2O volume mixing ratio

symbol

name

unit

$$M_{air}$$

molar mass of total air

$$\frac{g}{mol}$$

$$M_{dry\_air}$$

molar mass of dry air

$$\frac{g}{mol}$$

$$M_{H_{2}O}$$

molar mass of H2O

$$\frac{g}{mol}$$

$$n$$

number density of total air

$$\frac{molec}{m^3}$$

$$n_{dry\_air}$$

number density of dry air

$$\frac{molec}{m^3}$$

$$n_{H_{2}O}$$

number density of H2O

$$\frac{molec}{m^3}$$

$$\nu_{H_{2}O}$$

volume mixing ratio of H2O

$$ppv$$

\begin{eqnarray} M_{air} & = & \frac{M_{dry\_air}n_{dry\_air} + M_{H_{2}O}n_{H_{2}O}}{n} \\ & = & M_{dry\_air}\left(1 - \nu_{H_{2}O}\right) + M_{H_{2}O}\nu_{H_{2}O} \end{eqnarray}
2. molar mass of total air from H2O mass mixing ratio

symbol

name

unit

$$M_{air}$$

molar mass of total air

$$\frac{g}{mol}$$

$$M_{dry\_air}$$

molar mass of dry air

$$\frac{g}{mol}$$

$$M_{H_{2}O}$$

molar mass of H2O

$$\frac{g}{mol}$$

$$n$$

number density of total air

$$\frac{molec}{m^3}$$

$$n_{dry\_air}$$

number density of dry air

$$\frac{molec}{m^3}$$

$$n_{H_{2}O}$$

number density of H2O

$$\frac{molec}{m^3}$$

$$q_{H_{2}O}$$

mass mixing ratio of H2O

$$\frac{kg}{kg}$$

$$\nu_{H_{2}O}$$

volume mixing ratio of H2O

$$\frac{kg}{kg}$$

\begin{eqnarray} M_{air} & = & M_{dry\_air}\left(1 - \nu_{H_{2}O}\right) + M_{H_{2}O}\nu_{H_{2}O} \\ & = & M_{dry\_air}\left(1 - \frac{M_{air}}{M_{H_{2}O}}q_{H_{2}O}\right) + M_{air}q_{H_{2}O} \\ & = & \frac{M_{dry\_air}}{1 + \frac{M_{dry\_air}}{M_{H_{2}O}}q_{H_{2}O} - q_{H_{2}O}} \\ & = & \frac{M_{H_{2}O}M_{dry\_air}}{M_{H_{2}O} + M_{dry\_air}q_{H_{2}O} - M_{H_{2}O}q_{H_{2}O}} \\ & = & \frac{M_{H_{2}O}M_{dry\_air}}{\left(1-q_{H_{2}O}\right)M_{H_{2}O} + q_{H_{2}O}M_{dry\_air}} \\ \end{eqnarray}

## partial pressure

symbol

name

unit

$$p$$

total pressure

$$Pa$$

$$p_{x}$$

partial pressure of quantity

$$Pa$$

$$\nu_{x}$$

volume mixing ratio of quantity x with regard to total air

$$ppv$$

$$\bar{\nu}_{x}$$

volume mixing ratio of quantity x with regard to dry air

$$ppv$$

\begin{eqnarray} p_{x} & = & \nu_{x}p \\ p_{x} & = & \bar{\nu}_{x}p_{dry\_air} \\ p_{x} & = & N_{x}kT \end{eqnarray}

## saturated water vapor pressure

symbol

name

unit

$$e_{w}$$

saturated water vapor pressure

$$Pa$$

$$T$$

temperature

$$K$$

This is the August-Roche-Magnus formula for the saturated water vapour pressure

$e_{w} = 610.94e^{\frac{17.625(T-273.15)}{(T-273.15)+243.04}}$