You've probably heard at some point that tides on Earth are mostly caused by the moon, along with some smaller but still noticeable effects from the sun. In other words, the two objects' tidal forces are comparably strong (rather than being many orders of magnitude different: Uranus doesn't appreciably affect our tides!). You've probably also heard (or seen, during an eclipse) that the moon and the sun appear to be about the same size in the sky, even though the sun is vastly larger (but farther away). Remarkably, it turns out that these two facts are directly related.

Here's the idea. Let's say that the distance from Earth to some distant object is D, the radius of that distant object is R, and its density is p (I won't bother typing the usual "rho"). Ignoring constant numerical factors that would be the same for every (spherical) object, the mass of that distant object is proportional to p R

In other words, the tidal force exerted on the Earth by a distant object is proportional to the density times the cube of its angular size. Since the moon and the sun have about the same angular size, it's only the sun's lower density that makes its effects on our tides less significant. And as expected, planets like Uranus have a

Neat, huh?

Here's the idea. Let's say that the distance from Earth to some distant object is D, the radius of that distant object is R, and its density is p (I won't bother typing the usual "rho"). Ignoring constant numerical factors that would be the same for every (spherical) object, the mass of that distant object is proportional to p R

^{3}. The gravitational acceleration due to that distant object is proportional to M/D^{2}= (p R^{3})/D^{2}, but if you're experienced in the math of Netwon's gravity it's fairly straightforward to show that*tidal forces*are instead proportional to M/D^{3}= (p R^{3})/D^{3}. (Tidal forces refer to the*difference*in gravitational force on opposite sides of the earth, and that extra power of 1/D essentially comes from a linear approximation of the changing force that's proportional to R_{Earth}/D.) Factoring that a little differently, that means that tidal forces are proportional to p (R/D)^{3}. But using a little trig, R/D is just the tangent of (half of) the angular size of the object in the sky, and for small angles that*equals*the angular size.In other words, the tidal force exerted on the Earth by a distant object is proportional to the density times the cube of its angular size. Since the moon and the sun have about the same angular size, it's only the sun's lower density that makes its effects on our tides less significant. And as expected, planets like Uranus have a

*much*less significant effect, since their angular size is tiny by comparison (and their density is in the same ballpark).Neat, huh?

## no subject

jon-leonard.livejournal.com## no subject

spoonless.livejournal.com## no subject

## no subject

steuard.livejournal.comSo if the sun's effect were as big as the moon's, I might guess that full and new moons would lead to tides twice as high as the real-life average (while neap tides might come close to having no tide at all, or maybe four small swells a day: I'm not sure). There are probably some islands out there that would be under water during a double-high tide (just as there are a few places today where land is only exposed at low tide), but those places are presumably already routinely submerged during big storms.