Contrary to common belief, tides are not caused by the direct influence of the moon's gravity (it's far too weak to have any effect)[1]. The tidal forces are caused by the gravitational gradient from the moon (and the "centrifugal" forces from our path around the earth-moon barycenter), and I don't believe you'd get the same effects from a gravity source on the surface of the earth.
Even a lot of very respectable scientists and textbooks get this wrong.
[1] See https://www.youtube.com/watch?v=pwChk4S99i4 for a pretty good explanation
Or, to put it another way, the surfaces of the Earth closest to and farthest away from the Moon are traveling at the same orbital velocity around the center of the Earth/Moon system. However, they should be in different orbits; the point closest to the Moon is too slow for the orbit it is in and the point farthest away is too fast. The former wants to into a lower orbit while the latter wants to go into a higher orbit.
The analogy they used is that tides are more like a pimple being squeezed than taffy being stretched.
[1] See timestamp 4:45 in the video: https://www.youtube.com/watch?v=pwChk4S99i4&feature=youtu.be...
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Diagram:
1
4 E 2 M
3
E = EarthM = Moon
Numbers = 4 "sides" of the Earth, relative to the Earth-Moon line
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
vs
- 2 * (Gravitational constant) * (1e19 kg) / (1000 km)^3= -1.335×10^-9 reciprocal seconds squared
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.
[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.
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.
The lunar system is seemingly much less symmetrical but we still get some amphidromic points (points of zero tidal range) and neep tides (lowest "high" tide) can be very small.
Another way to think of it: there's not an "equal" weight of water above Greenland because it's raised above sea level by the land. The weight of rock underneath the ice sheet needs to be included as well.