5 ms·

At a distance of 100 km, the gravitational acceleration caused by the greenland ice sheet is .019 m/s^2 or ~0.2% g. Doesn't sound like a lot but it's still 524

by jjk166 6y ago

At a distance of 100 km, the gravitational acceleration caused by the greenland ice sheet is .019 m/s^2 or ~0.2% g. 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. The ice sheet has the same gravitational effect as the moon at a distance of approximately 2300 km.

jaspax 6y ago
This is the answer which makes the most sense to me.
pbhjpbhj 6y ago
Hmm, but a water molecule 100 km from an Arctic ice sheet is also experiencing the pull of an "equal" weight of water 100km the other direction, albeit ever so slightly offset angularly. 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.
- mehrdadn 6y ago
  That pull would be there regardless though.
- rovolo 6y ago
  The interviewee is talking about ice on land (Greenland and Antarctica), not about ice in the ocean. If the ice on land melts, then those water molecules migrate away from Greenland because they drop to sea level. That loss of mass would happen regardless of the gravitational pull of water 200 km away and would change the gravitational gradient of the water 100 km away. 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.
curiousgeorgio 6y ago
> 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
deleted 6y ago
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by jjk166 6y ago

jaspax 6y ago

This is the answer which makes the most sense to me.

pbhjpbhj 6y ago

Hmm, but a water molecule 100 km from an Arctic ice sheet is also experiencing the pull of an "equal" weight of water 100km the other direction, albeit ever so slightly offset angularly. 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.

mehrdadn 6y ago
That pull would be there regardless though.
rovolo 6y ago
The interviewee is talking about ice on land (Greenland and Antarctica), not about ice in the ocean. If the ice on land melts, then those water molecules migrate away from Greenland because they drop to sea level. That loss of mass would happen regardless of the gravitational pull of water 200 km away and would change the gravitational gradient of the water 100 km away. 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.

curiousgeorgio 6y ago

> 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

deleted 6y ago

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