Geopotential Altitude⚠ unverified

Understanding the Parameters

• Geometric altitude $h$ — the honest, physical height above mean sea level: what a tape measure, radar altimeter, or GPS reports. This is the real distance.

• Geopotential altitude $H$ — a potential-energy coordinate wearing the costume of a length. It is defined so that lifting a unit mass to geopotential altitude $H$ against a constant $g_0$ costs the same energy as lifting it to geometric altitude $h$ against the true, weakening $g(h)$. It is not a distance anything actually travels.

• $H < h$, always — because gravity weakens with height, lifting to a real height $h$ costs less energy than a constant-$g_0$ world would charge. The equivalent constant-$g$ height is therefore lower. The formula confirms it: $R/(R+h) < 1$ for any $h > 0$.

• How small is the correction? — tiny, and this is the point. At $11\ \text{km}$, $H = 10{,}981\ \text{m}$ — a deficit of only $19\ \text{m}$ (0.17 %). At $32\ \text{km}$, $H = 31{,}840\ \text{m}$, a deficit of $160\ \text{m}$ (0.5 %). Below a few km the two are interchangeable for most engineering. The correction matters for rigour and standards conformance, not usually for the answer.

• Earth radius $R$ — a hard-coded constant. Note the Earth is not a sphere: the equatorial radius is $6378\ \text{km}$ and the polar $6357\ \text{km}$, a 21 km spread that dwarfs the $r_0$-vs-mean-radius quibble above. The single-$R$ model is a deliberate standardisation, not an oversight — the ISA wants one number everywhere, not the local truth.

• When to bother — if your altitude comes from a pressure altimeter it is already a pressure altitude (see Pressure Altitude) derived from the ISA equations, so no conversion is needed. Convert when you have a genuinely geometric height (GPS, radar, survey) and want to feed the ISA formulas properly.

Derivation (Approaching a Proof)

The definition is an energy equivalence. Define geopotential $\Phi$ as the work per unit mass done against gravity in climbing from sea level to geometric altitude $h$:

$$\Phi(h) = \int_0^h g(z)\,dz.$$

Now define geopotential altitude $H$ as the height that would produce that same potential under a constant $g_0$:

$$g_0 H \equiv \Phi(h) = \int_0^h g(z)\,dz \quad\Longrightarrow\quad H = \frac{1}{g_0}\int_0^h g(z)\,dz.$$

This is the whole idea; the rest is evaluating the integral. By Newton's law of gravitation (see Gravitational Force), outside the Earth the acceleration falls with the inverse square of distance from the centre. Writing it in terms of altitude above the surface,

$$g(z) = g_0\left(\frac{R}{R+z}\right)^2.$$

Substitute and pull the constants out:

$$H = \frac{1}{g_0}\int_0^h g_0 \frac{R^2}{(R+z)^2}\,dz = R^2\int_0^h \frac{dz}{(R+z)^2}.$$

$g_0$ cancels — the result depends only on geometry, not on the strength of gravity. The integral is elementary; with $u = R+z$, $\int u^{-2}du = -u^{-1}$:

$$H = R^2\left[-\frac{1}{R+z}\right]_0^h = R^2\left(\frac{1}{R} - \frac{1}{R+h}\right) = R^2\cdot\frac{(R+h) - R}{R(R+h)} = R^2\cdot\frac{h}{R(R+h)}.$$

Cancelling one factor of $R$:

$$H = \frac{R h}{R + h}. \qquad\blacksquare$$

Sanity checks. As $h \to 0$, $H \to 0$ ✓. For $h \ll R$ (all of aeronautics), expand:

$$H = \frac{h}{1 + h/R} \approx h\left(1 - \frac{h}{R}\right) = h - \frac{h^2}{R},$$

so the deficit grows quadratically: at $h = 11\ \text{km}$, $h^2/R = 11000^2/6.371\times10^6 = 19.0\ \text{m}$ — matching the exact value above. As $h \to \infty$, $H \to R$: even an infinitely distant mass has finite potential energy relative to the surface, capped at $g_0 R$ — the same finiteness that makes Escape Velocity a finite number rather than infinite.

Dimensional check. $Rh/(R+h) = (\text{m}\times\text{m})/\text{m} = \text{m}$ ✓ — a length, as required, even though what it really encodes is energy per unit weight.

History and Development

• The inverse-square law (1687). Isaac Newton's Principia established $g \propto 1/r^2$, which is the only physical input to this derivation. Everything else is calculus.

• Geopotential in meteorology (1900s–1920s). As upper-air balloon soundings proliferated after Teisserenc de Bort's discovery of the stratosphere (1902), meteorologists needed height coordinates in which the thermodynamic equations stayed simple. Working in geopotential lets the hydrostatic equation keep a constant $g_0$, so that pressure-height relations hold exactly rather than approximately. The geopotential metre and later the dynamic metre entered standard practice; Vilhelm Bjerknes and the Bergen School's quantitative weather-prediction programme leaned on such energy-consistent coordinates.

• Codification (1952–1976). The ICAO Standard Atmosphere (1952) and the U.S. Standard Atmosphere 1976 both define their layer boundaries — 11 km, 20 km, 32 km — in geopotential, not geometric, height, and specify the conversion radius $r_0 = 6{,}356{,}766\ \text{m}$. This value is not the mean Earth radius; it is an effective radius chosen so that the conversion reproduces the standard's adopted $g$ variation at about 45° latitude, where $g_0 = 9.80665\ \text{m}/\text{s}^2$ is itself defined. Meteorologists still report upper-air data on geopotential height surfaces (the "500 hPa height" of every weather chart) for exactly the reason this page exists: it makes the energy bookkeeping honest while keeping $g$ constant.

Related Concepts: ISA Temperature, ISA Pressure, ISA Density, Stratospheric Temperature, Pressure Altitude, Density Altitude, Gravitational Force, Escape Velocity

Notes: $R$ hard-coded at $6{,}371{,}000\ \text{m}$ (mean radius); the ISA standard specifies the effective $r_0 = 6{,}356{,}766\ \text{m}$ — difference is 0.04 m at 11 km, negligible. $H < h$ always; deficit grows as $h^2/R$ (19 m at 11 km, 160 m at 32 km). ISA layer boundaries (11/20/32 km) are defined in geopotential. $g_0$ cancels in the derivation — the result is pure geometry. Default $h = 1.0\ \text{m}$ is a placeholder.