3 ms·

> Doesn't sound like a lot but it's still 524 times higher than the effect of the moon's gravity which creates the tides. Contrary to common belief, tides are

> Doesn't sound like a lot but it's still 524 times higher than the effect of the moon's gravity which creates the tides. 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 https://www.youtube.com/watch?v=pwChk4S99i4 for a pretty good explanation

pge 6y ago
Just to add to this answer and perhaps save you the click, what he is referring to is that the force pulling the water upward at high tide is not the direct gravitational pull of the moon. It is that the gravitational force from the moon at the edge of the earth is greater than the force at the center (because it is closer to the moon). Similarly the force on the opposite side of the earth is less (because that side is further from the moon than the earth's center). So the water molecules are drawn away from the center of the earth (near side because they are being pulled slightly harder than the center of the earth, and far side because they are being pulled slightly less hard than the center of the earth). Hence the high tides on both sides of the earth, not just the side closer to moon.
- mcguire 6y ago
  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 of the Earth. 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 is too fast. The former wants to into a lower orbit while the latter wants to go into a higher orbit.
  - mcguire 6y ago
    In a further translation from gibberish: 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.
- dwaltrip 6y ago
  If I correctly understood the video linked above, the primary cause is actually due to the tidal acceleration (e.g. moon's gravity) of objects on sides 1 and 3 (see diagram below), relative towards the earth's surface, being mostly radially inward [1]. The majority of the ocean water along the sides of Earth is being pulled in very slightly, and in aggregate across the massive surface of the ocean, this results in enough pressure to push up the water at sides 2 and 4. Tides are pushed up due to pressure, not pulled up. 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&t=285 https://www.youtube.com/watch?v=pwChk4S99i4&feature=youtu.be... --- Diagram: 1 4 E 2 M 3 E = Earth M = Moon Numbers = 4 "sides" of the Earth, relative to the Earth-Moon line
vilhelm_s 6y ago
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 vs - 2 * (Gravitational constant) * (1e19 kg) / (1000 km)^3= -1.335×10^-9 reciprocal seconds squared
- curiousgeorgio 6y ago
  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).
  - coliveira 6y ago
    Nothing of this disproves the effect of the polar ice. The forces still apply, they're just not shifting daily.
    - curiousgeorgio 6y ago
      > 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.

pge 6y ago

Just to add to this answer and perhaps save you the click, what he is referring to is that the force pulling the water upward at high tide is not the direct gravitational pull of the moon. It is that the gravitational force from the moon at the edge of the earth is greater than the force at the center (because it is closer to the moon). Similarly the force on the opposite side of the earth is less (because that side is further from the moon than the earth's center). So the water molecules are drawn away from the center of the earth (near side because they are being pulled slightly harder than the center of the earth, and far side because they are being pulled slightly less hard than the center of the earth). Hence the high tides on both sides of the earth, not just the side closer to moon.

mcguire 6y ago
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 of the Earth. 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 is too fast. The former wants to into a lower orbit while the latter wants to go into a higher orbit.
- mcguire 6y ago
  In a further translation from gibberish: 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.
dwaltrip 6y ago
If I correctly understood the video linked above, the primary cause is actually due to the tidal acceleration (e.g. moon's gravity) of objects on sides 1 and 3 (see diagram below), relative towards the earth's surface, being mostly radially inward [1]. The majority of the ocean water along the sides of Earth is being pulled in very slightly, and in aggregate across the massive surface of the ocean, this results in enough pressure to push up the water at sides 2 and 4. Tides are pushed up due to pressure, not pulled up. 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&t=285 https://www.youtube.com/watch?v=pwChk4S99i4&feature=youtu.be... --- Diagram: 1 4 E 2 M 3 E = Earth M = Moon Numbers = 4 "sides" of the Earth, relative to the Earth-Moon line

vilhelm_s 6y ago

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

curiousgeorgio 6y ago
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).
- coliveira 6y ago
  Nothing of this disproves the effect of the polar ice. The forces still apply, they're just not shifting daily.
  - curiousgeorgio 6y ago
    > 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.