3.5 Permanent tide system

Make sure that your spherical harmonic coefficients are in the right permanent tide system. Generally, satellite altimetry products use the mean-tide system. Conversion between different permanent tide systems involves either modifying one spherical harmonic coefficient or adding a zonally uniform correction to the geoid undulations. To convert zero-tide coefficients to mean-tide, use:

$$\bar{C}_{2,0}^{\mathrm{(mean tide)}} - \bar{C}_{2,0}^{\mathrm{(zero tide)}} = 1 \times \frac{(-0.198\text{ m})r^{3}g}{a^{2}GM\sqrt{5}}$$ (16)

$$= -1.39\times10^{-8},$$
to convert tide-free coefficients to mean-tide, use

$$\bar{C}_{2,0}^{\mathrm{(mean tide)}} - \bar{C}_{2,0}^{\mathrm{(tide free)}} = (1+k) \times \frac{(-0.198\text{ m})r^{3}g}{a^{2}GM\sqrt{5}}$$ (17)

$$= (1+k)\times(-1.39)\times10^{-8},$$

where $k$ s the (fundamentally unknowable) zero frequency Love number, which must be adopted. (For example, for EGM96, $k$=0.3 was adopted). $g$ is the mean gravity.

Alternatively, you could add the permanent tide correction in the space domain, after computing the geoid height $N$ (Rapp, 1989):

$$N^{\mathrm{(mean tide)}} - N^{\mathrm{(zero tide)}} =$$

$$1 \times (-0.198 ) \times \left(\frac{3}{2}\sin^{2}\bar{\varphi} - \frac{1}{2}\right)$$ (18)

$$N^{\mathrm{(mean tide)}} - N^{\mathrm{(tide free)}} =$$

$$(1 + k) \times (-0.198 ) \times \left(\frac{3}{2}\sin^{2}\bar{\varphi} - \frac{1}{2}\right).$$ (19)

This latter method appears to be the preferred one, because you can easily convert between permanent tide systems after the more expensive computation of the geoid undulation.