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Consider a system in which a central star is orbited by a planet with liquid water oceans, which is itself orbited by a moon.

Given the masses and distances between these three objects, is there some formula that outputs the minimum and maximum tide heights the planet's oceans cycle through for every orbit of the moon?

For simplicity, the effects of local topography on the tides are being ignored.

Also, if there is such a formula, could it be applied to solar systems in which there is more than one central star and/or more than one moon orbiting the planet?

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Wikipeida (quoting Icarus ) gives

$$\frac{15}{8}\frac{mA^4}{Mr^3}$$

Where $m$ and $M$ are the masses of the moon and planet, respectively; $r$ is the orbit radius of the moon and $A$ is the radius of the planet. For Earth this is a little less than a metre.

Systems with multiple moons (or moon and sun, like on Earth) will have multiple bulges which can add up.

This tells you very little about the sizes of actual tides, which are strongly magnified by topography. It predicts tides of 0.7 m, whereas real tides have ranges of between 0 and 16 m.

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  • $\begingroup$ Calculating tides without considering topography is like working out the sound of a violin only considering the strings and ignoring the hollow body. Nearly all the sound of a violin comes not from the strings but from resonance in the body. Likewise, real tides are a resonance effect driven by the moon, not a tidal bulge. $\endgroup$ – James K Mar 30 at 9:35

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