Change of sea level in relation to sea ground

  • Thread starter G.H.Hardy
  • Start date
  • #26
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
684
Nope. It doesn't show that, either. It is just an equipotential surface.
 
  • #27
907
2
Nope. It doesn't show that, either. It is just an equipotential surface.
What's the difference between the geoid and the sea level, then?
 
  • #28
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
684
Sorry, I misread your post. You are correct, the geoid (the EGM96 geoid; there are multiple geoid models) more or less shows mean sea level -- at least for the parts of the Earth covered by ocean. The reason I said "more or less":
  • The choice of the potential. The geoid is a geopotential surface, at least over ocean. The goal is to use the potential value that results in a best fit with sea level.

  • Exposed land. The geoid is not a geopotential surface of the underlying spherical harmonic gravity models over land. There is quite a bit of fudging done over the land-covered areas of the Earth to arrive at those geoid undulations for the exposed land portion of the Earth. The nature of this fudging is getting problematic with the EGM08 gravity model.

  • The nature of the model. The EGM96 geoid does not quite show mean sea level. Neither does the GGM02C geoid. The EGM96 geoid is a tide-free model while GGM02C is a zero tide model. Neither of them is a mean tide model. Tide-free is essentially what the Earth would look like if the Sun and Moon didn't raise any tides whatsoever. The Sun and Moon do raise tides, both in the oceans and in the Earth itself. Those tides have a zero frequency (permanent) component and time-varying components. Zero tide models result from ignoring the time-varying components but keeping the permanent tides. Finally, those time-varying components do not have a zero mean in terms of potential. Accounting for this non-zero mean results in a mean tide model.
 
  • #29
5,439
9
I like diagrams so here is one.

With the aid of this diagram and our definition of sea ground (thank you hamster) I think I understand what the teacher was proposing so I will explain my understanding of this proposal for further discussion.

The diagram show a section of varying sea floor and the water column above it. The proposal is that where the sea-floor is above mean sea-floor level the actual sea level is below mean sea level.

The reason is suggested to be an increase in gravity at local sea level, perhaps because the lump of land marked L is much denser than the equivalent volume of water.

Of course, so long as our test mass is small and outside the mean solid surface of the earth, gravity increases as we approach this mean surface. Gravity is naturally stronger at B than A anyway, since it is closer to the centre of the earth.
 

Attachments

Last edited:
  • #30
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
684
The diagram show a section of varying sea floor and the water column above it. The proposal is that where the sea-floor is above mean sea-floor level the actual sea level is below mean sea level.
That doesn't make sense. Actual sea level differs from mean sea level because of the tides, not because of some gravity anomaly. Remove those tidal effects in one way or another (three ways; see my previous post) and what will be left is some form of mean sea level. The geoid, at least over the oceans, *is* the surface corresponding to mean sea level.

The difference between the geoid and the reference ellipsoid, the diagrams linked by hamster and I, essentially reflect the fact that the Earth's surface and material close to the Earth's surface (close meaning up to 200+ km deep) is dynamic over the scale of tens of thousands of years (isostatic rebound from the last ice age) and even tens millions of years (tectonic activities such as the collision of India and Asia).
 
  • #31
5,439
9
That doesn't make sense.
I think that's a trifle harsh.
I didn't even say I agreed with the proposal, I was just trying to clarify it so we are all talking about the same thing.

In particular, unless the local surface of the sea (regardless of how it got there) is coincident with the mean level (yes I know it's the same as the geoid for practical purposes) it will experience a different gravitational acceleration ie stronger or weaker gravity, increasing from A in the direction of B in my diagram.
So if B is below mean sea level as shown then it will experience stronger gravity, simply because it is closer to the centre of the earth. It cannot be any other way.

This has nothing to do with tides or isostacy.

Similarly the sea bed below B is where it is.

My interpretation of the original proposal is that the stronger gravity at B is attributed to the extra height of mass I have noted as L in my diagram.

Again I do not say I agree with this explanation, but I think that was the original point for discussion.
 

Related Threads on Change of sea level in relation to sea ground

  • Last Post
Replies
11
Views
5K
  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
17
Views
4K
  • Last Post
2
Replies
31
Views
38K
  • Last Post
Replies
14
Views
4K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
8
Views
4K
Replies
31
Views
16K
  • Last Post
Replies
10
Views
6K
Replies
52
Views
9K
Top