Geopotential

By 0x7df, Sun 30 August 2015, in category Atmospheric science

atmospheric science, meteorology, physics

The geopotential:

$$ \phi = gz $$

is the gravitational potential - i.e. the gravitational potential energy per unit mass - at a location in the earth's atmosphere. It is often used as a vertical coordinate instead of height above sea level, \( z\).

The geopotential at height \( z\) is the difference between the gravitational potential energy at some reference height, \( z_0\), usually taken to be zero (sea level), and that at \( z\). This difference is numerically equal to the work done in raising the parcel from \( z_0\) to \( z\):

$$ \phi(z) = \int_{z_0}^z g(z') dz' $$

The simplification to:

$$ \phi = gz $$

assumes that \( g\) is constant with height. If the variation is to be taken into account, a suitable profile for \( g(z)\) is:

$$ g(z) = \frac{g_0}{\left[ 1 + \left( z/E\right)\right]^2} $$

where \( g_0\) is the value at sea level and \( E\) is the earth's mean radius. Substituting into the definition of geopotential:

$$ \phi(z) = \int_{z_0}^z g(z') dz' $$

gives:

$$ \phi(z) = \int_{z_0}^z \frac{g_0}{\left[ 1 + z' / E \right]^2} dz' $$
$$ \phi(z) = g_0 E^2 \int_{z_0}^z \frac{1}{\left[ E+z' \right]^2} dz' $$
$$ \phi(z) = g_0 \frac{z-z_0}{\left( 1+z/E \right) \left( 1+z_0/E \right)} $$

or, if we take \( z_0\) to be zero:

$$ \phi(z) = g_0 \frac{z}{1+z/E} $$

The assumption that gravity is constant is equivalent to assuming that \( z \ll E\) so that \( z/E \ll 1\), and therefore that the denominator on the right-hand side of the above equation is unity.

In terms of differentials:

$$ d\phi = g dz $$

and taking into consideration the hydrostatic equation:

$$ dp = -\rho g dz $$

we obtain:

$$ d\phi = -\frac{dp}{\rho} = -\alpha dp $$

where \( \alpha = 1 / \rho\) is the specific volume (volume per unit mass).

The geopotential is larger than the geometric height by a factor of \( g\) - e.g. the geopotential in \( \mathrm{J/kg}\) is about 10 times the magnitude of the geometric altitude in metres. To make the geopotential have numerical magnitude more nearly equal to the geometric height, it is often expressed in units a factor of \( g_0\) smaller; these units are usually called geopotential metres, or dynamic metres, or geodynamic metres. That is, the geopotential in these units, is:

$$ \mathrm{gpm}(z) = \frac{\phi(z)}{g_0} $$
$$ \mathrm{gpm}(z) = \frac{\int_{z_0}^z g(z') dz'}{g_0} $$
$$ \mathrm{gpm}(z) = \frac{z}{1+z/E} $$

Rogers and Yau Problem 3.2b

Show that the geopotential at pressure level \( p\) of an atmosphere in hydrostatic equilibium is given by:

$$ \phi(p) = R\bar T\ln{\left(p_0/p\right)} $$

where \( \phi(p_0) = 0\).

The differential expression for hydrostatic equilibrium is:

$$ dp = -g\rho dz $$

From the definition of geopotential:

$$ \phi(z) = g \int_{z_0}^z dz' $$

we have:

$$ d\phi = g dz $$

Hence:

$$ dp = -\rho d\phi $$

Introducing the equation of state gives:

$$ dp = \frac{p}{RT} d\phi $$
$$ \int \frac{dp}{p} = -\frac{1}{R} \int \frac{d\phi}{T} $$

Performing the integration gives:

$$ \ln(p) = -\frac{1}{R\bar T} \phi(p) + C $$

where \( \bar T\) is the average temperature through the layer from \( p_0\) to \( p\). Since \( \phi(p_0) = 0\):

$$ C = \ln{p_0} $$

Hence:

$$ \phi(p) = R\bar T\ln{(p_0/p)} $$

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