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Comment by vilhelm_s | Hacker News Reader

It seems this would made the effect from the icebergs bigger rather than smaller, because the gradient decays by r^3 instead of r^2, so the distance is more important?

Like, the gravitational acceleration is a = GM/r^2 while the gradient is da/dr = -2GM/r^3

So for moon vs glacier at 1000km you'd get

- 2 * (Gravitational constant) * (mass of moon) / (391184 km)^3 = - 1.638×10^-13 reciprocal seconds squared

- 2 * (Gravitational constant) * (1e19 kg) / (1000 km)^3= -1.335×10^-9 reciprocal seconds squared

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It's the gradient applied over the whole ocean that causes the tides. In other words, the forces are not enough to create any kind of local change in the depth of the ocean. It's not a gravitational "pulling" as we're accustomed to think about, but more of a global squeezing of water from that gradient applied to the entire ocean. I'll have to give some more thought to the idea of a gravity source on the surface of the earth, but I doubt it would work the same way (we're comparing a relatively small mass in a concentrated location on the earth with that same mass distributed mostly evenly over the earth).

Nothing of this disproves the effect of the polar ice. The forces still apply, they're just not shifting daily.

> The forces still apply, they're just not shifting daily.

No, the forces are completely different. If we have an object on the surface of the earth that has enough mass to roughly produce the same nearby gravitational acceleration as that felt by the moon (which is minuscule and undetectable by most instruments), that object would not produce changes in ocean levels as we see with the moon. Again, the oceans are not rising/falling due to the moon's gravity pulling on them. It only happens because the moon is far enough away that its tiny gravitational acceleration on the earth is (1) felt everywhere on earth, and (2) felt everywhere on earth in slightly different amounts.

For a smaller, closer object (even with similar nearby gravitational acceleration), the tidal forces will not be the same because that gravitational acceleration will fall off to near zero in a very short distance.

You are mixing tidal forces with local attraction forces due to ice. Even though the forces due to ice are not felt everywhere in the planet, they do have an effect that can be calculated, and it is stronger in the body of water closer to the poles. The result of this force causes water attraction in the polar regions. So the effect that is felt in other parts of the planet is not due to the gravity force, but due to the displacement of water happening in polar regions - after all the water needs to go somewhere.

I was originally responding to the claim that "it's still 524 times higher than the effect of the moon's gravity which creates the tides" - that's why I mentioned tidal forces. If the ice creates that much direct, local gravitational attraction[1], then my point is we shouldn't be comparing it to the moon at all (because the forces involved are not the same as those involved with the moon and tides).

[1] Even the claim about the ice sheet (and its melting) contributing significantly (via gravity) to global sea level change seems dubious since, as noted elsewhere in this discussion, the Earth Gravitational Model appears to be affected much more by factors other than ice sheet thickness or surface features.

As you said, it's the gradient which matters. The gradient determines how much the sea level changes over a distance. But, the sea level change is distinct from the tides. The difference in water level between two connected locations at the same moment in time is the same for the ocean and the lake.

The video you linked to compares lakes and oceans because the lunar tides vary with time. The lake level difference between Cleveland and Buffalo at 6 will be the same as the sea level difference between New York and Providence at ~5:30. You need to compare your sea level to the sea level a quarter of the way around the world to understand why your local sea level changes from 6:00 to 12:00.

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