The geopotential is the amount of Work required to lift an air parcel of unit mass from the Earth’s surface ( $z = 0$ ) to a height $z$ .

Defined as

Φ (z) : = \int_{0}^{z} g d z

where it’s defined as $0$ at $z = 0$ , much like (usually) the gravitational potential.

Alternative formula for ideal gases

Due to the Hydrostatic Law, we have

Φ (z) = \int_{0}^{z} g d z = - \int_{P (z = 0)}^{P (z)} \frac{d p}{ρ}

Supposing it’s an unit mass parcel of air, we can write its specific volume $α = \frac{1}{ρ}$ and thus

Φ (z) = - \int_{P_{0}}^{P (z)} α d p

Using gravitational field formula

Opening up $g$ ,¹ we have

Φ (z) = \int_{0}^{z} \frac{GM}{( R + z ) ^{2}} d z

where $R$ is the Earth’s (or any planet’s) radius. It’s the same as

\int_{R}^{R + z} GM \frac{d r}{r ^{2}} = - GM [(R + z)^{- 1} - R^{- 1}] = GM \frac{z}{( R + z ) R}

When $z ≪ R$ , we have the usual result

Φ (z) \approx \frac{GM}{R ^{2}} z \equiv g z

WALLACE, John Michael; HOBBS, Peter Victor. Atmospheric science: an introductory survey. 2nd ed ed. Amsterdam Paris: Academic press, 2006.

Newton’s gravitational law for some mass $M$ : gravitational field $G (r) = \frac{GM}{r ^{2}}$ , from which $F_{g} = G m$ is the gravitational force. ↩