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What I mean is the following. Imagine two smooth massive spherical bodies ($M_1\neq{M_2}$), with equal and homogeneous mass densities. Both masses have a layer of water on them for which holds ($R_1$ and $R_2$ are the radii of the masses):

$$\frac{R_1}{R_2}=\frac{{{waterlevel}_1}}{{{waterlevel}_2}}$$

The two masses rotate around each other. Their rotations around their rotation axes are the same: $\omega (\frac{rad}{sec})$. The rotation (pseudo)vectors are perpendicular to the plane in which both bodies revolve around each other. Finally, let's assume for simplicity that their orbits are circular.

My question: Will the tides on both bodies be scale independent? Which is to say, can we see any difference in the tides if (in the very hypothetical case) our bodies had the size "fitting" to the planets, by which I mean that if we were shrunk or blown up so it would seem to us that both planets had the same size?

It's my guess that no difference would be seen as both masses of water on the planets would fall at the same rate towards the other, resulting in (relatively) the same tides. Which implies scale invariance.

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The object with less mass would experience higher tides. To "see" this better imagine a very small mass in a very low, very fast, stable circular orbit around a very massive object. If we were size proportional on each object we would barely see a bulge on the massive one, but the smaller would be ovoid shaped from tidal forces, so if we were comparatively smaller on it the tidal bulge would seem huge.

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Assuming that the ocean has no friction and no inertia - in other words, that it instantaneously matches the gravitationally “flat” surface of the planet, the range of the tides raised on a planet by its moon is $$ρθ^3$$ where

ρ is the density of the moon relative to the planet (so 1 if they are made of identical material),

θ is the angular radius of the moon as seen from the planet, in radians,

and the tidal range is measured as a fraction of the planet’s radius.

This formulation is very suitable for the kinds of thought experiments you are doing because it does not require any mass as such - only relative density - or explicit sizes as such.

Your “scale independence” amounts, if I understand it correctly, to measuring the tide height in units in which $R=1$. In that case, if the two bodies are the same density, the smaller body will have the relatively higher tides because, viewed from it, the bigger body appears larger in the sky.

In real life, oceans do have viscosity and do have inertia, so we don’t really have a pair of tidal waves running along the equator at 1,000 miles an hour. Moreover, the water tugged around by tidal forces sloshes around like soup in a soup plate. But the simple $ρθ^3$ formula does represent the strength of the driving forces and how they scale.

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  • $\begingroup$ Suppose the Sun was a solid and (locally) flat body of mass with the Earth orbiting it at a small distance. The thickness of the layer of water on the Sun is in proportion with the layer of water on the Earth. It's thicker than the layer of water on the Earth (assuming the Earth is also perfectly flat, without mountains) by a factor $\frac{R_{sun}}{R_{earth}}$. Aren't the forces the Sun and the Erth exert on each other the same? Doesn't the water on Earth gets elongated by the same relative amount because of its free fall to the Sun as the water on the Sun by its free fall to the Earth? $\endgroup$ – descheleschilder Jul 12 at 15:19

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