The Martini Hypothesis

by Daniel Brockman, 20060604

Contents


The Martini Hypothesis
Boundary Conditions
  Boundary Condition I: Vertical
  Boundary Condition II: Horizontal
Mathematical Analysis
  L ~ 2/3
  the seas will rise 70m
  First Calculation
  Second Calculation
  Third Calculation
Conclusion: at least 46m
Exercises
Appendix

The Martini Hypothesis

When I told my good friends Brian and Doug that the seas would rise 250 feet when the polar ice melts, both raised doubts about that estimate. Brian said the number was too big, because the water would spread out over the land, so the depth would be shallower than if it were to somehow rise straight up.

Doug said essentially the same thing, while pointing out that the fluid in a martini glass rises more with the first olive than with the second.

Because I know both Doug and Brian are astute in general and have especially good sense about geophysical conditions and fluids, I owed the Martini Hypothesis some serious consideration.

Boundary Conditions

I thought it a hard problem to figure out the precise slope of the "martini glass" in this case, but I felt I could establish some boundary conditions, namely vertical and horizontal, between which the true answer must lie.

Boundary Condition I: Vertical (BC1): The water will rise most if it spreads out minimally. We can restate the Martini Hypothesis as asserting the 70m estimate occurs when the water doesn't spread out at all, and the 2nd olive has as much effect as the 1st. We will assert for BC1 that the water doesn't spread out at all. Minimal spreadout would mean the water rises only over places already wet, which is less than the whole surface of the earth. This would occur if the land rose straight up at the shoreline, so that the water would never spread inland. With BC1, the martini glass has vertical sides like a tumbler.

Boundary Condition II: Horizontal (BC2): The water will rise least with maximal spreadout. We can restate the Martini Hypothesis as asserting the water rises much less than 70m, because it spreads out, and the 2nd olive has little effect compared with the 1st, and even the 1st may not have much effect. We will assert for BC2 that the water spreads out as far as it possibly could. That would occur if the land were flat everywhere with an altitude of 1cm. Under BC2, the entire earth would flood. With BC2, the martini glass is completely shallow and flat like a plate.

We know the shoreline isn't vertical as in BC1, and we know the land isn't flat as in BC2. The true situation lies somewhere between these. Let us proceed with a mathematical analysis.

Mathematical Analysis

If you should spot a typographical or other error in this exposition, then I shall be grateful for your informing me so I can correct it.

L = fraction of the earth's surface which is seawater (before the ice melts)
L ~ 2/3.

I've been told all my life that 2/3 of the earth's surface is seawater, and I suppose that's about right. This turns out to be the crucial variable. Note: 46.67 is 2/3 of 70.

The most credible estimate I have found of the depth after the melt comes from the NSIDC (National Snow and Ice Data Center) which has published the estimate that the seas will rise 70m (about 230 feet) when the polar ice melts.

First Calculation

Volume is given by

 V = S * h

where
V = volume
S = surface area
h = height

Applying BC1, if we cover a fraction L of Se, the surface of the earth, with meltwater, it rises to a height of 70m. So

Vm = L * Se * 70m

Under the Martini Hypothesis, this depth will never be reached, because the water will spread out. With BC2, this same volume Vm of water must spread out over the whole earth to a depth of at least h, and we must solve for h. The volume of water must be the same, regardless of how it eventually spreads out, because it is the same melted ice from the same and only two poles of the earth. With BC2,

Vm = Se * h
L * Se * 70m = Se * h
L * 70m = h
46.67m = h

So the water must rise at least to the minimum level of 46.67m to which BC2 constrains it.

Second Calculation

Some readers may object that one can't simply call Se the area of the surface of the earth, because the earth is a sphere, and the surface area is a function of the radius and π, and the calculation must take these into account. This is sensible and I owe you an explanation.

The surface of a sphere is given by

S = 4 * π * r2

If we let

re = pre-melt radius of the earth (a subject to which we shall return)

Then the surface of the earth we call a sphere is

Se = 4 * π * re2

If the radius of the earth re doesn't vary with the depth of the water that may cover it, nor with the part of the surface that's covered with water, and if π doesn't vary, and 4 doesn't vary, then Se doesn't vary with the depth of the water that covers it. Since Se doesn't vary, we get the same conclusion.

Third Calculation

At this point, some readers may object that the radius does indeed vary when the ice melts. Under BC1, the radius will be 70m greater where there is water, and under BC2, the radius will be enlarged by some lesser amount, which amount is what we want to determine. Since the radius varies, the surface will vary. Since the surface varies, the volume will vary. Since the volume will vary, the height of the water will vary. All this is sensible reasoning, and I hear your cry.

Let us perform an alternate calculation. The volume of a sphere as a function of its radius is

V = (4/3) * π * r3

Under BC1, the volume of water will be a fraction L of the difference in volumes of a sphere with a post-melt radius ( re + 70m ) and a sphere with a pre-melt radius re

That is, the volume of water Vm, which is melted ice, is given by

Vm = L * [ (4/3) * π * ( re + 70m )3 - (4/3) * π * re3 ]

Under BC2, this same volume of water spreads out over the entire earth, so the seas don't rise so much, and the volume Vm is given by

Vm = (4/3) * π * ( re + h )3 - (4/3) * π * re3

Reconciling these two formulae for the same volume Vm, and solving for h, we have

(4/3) * π * ( re + h )3 - (4/3) * π * re3 = L * [ (4/3) * π * ( re + 70m )3 - (4/3) * π * re3 ]
( re + h )3 - re3 = L * [ ( re + 70m )3 - re3 ]
( re + h )3 = L * [ ( re + 70m )3 - re3 + (1/L) * re3 ]
( re + h )3 = [ L * ( re + 70m )3 + ( 1 - L ) * re3 ]
re + h = [ L * ( re + 70m )3 + ( 1 - L ) * re3 ] (1/3)
h = [ L * ( re + 70m )3 + ( 1 - L ) * re3 ] (1/3) - re

As detailed in the appendix, we know that

6,380,000m > re > 6,350,000m

and we have assumed that

L = 2/3 .

Substituting these values into our derived formula for h, we find that

if re = 6,380,000m, then h = 46.66m ~ L * 70m ,

and

if re = 6,350,000m, then h = 46.66m ~ L * 70m .

We attribute to rounding errors the departure of h from 46.67 by a few centimeters. But what we really see here, of course, is that the radius of the earth is so immensely large, relative to the difference in sea levels, that the change in radius when the ice melts has insignificant implication for the change in depth of the water.

Conclusion: at least 46m

In conclusion, we have presented several mathematical analyses of the change in sea level due to the melting of the polar ice, and of modifying the foreseen change with respect to the Martini Hypothesis. We find that, if the Martini Hypothesis applies, then the sea level will rise at least 46m and likely more than that.

Exercises

We leave the following as exercises for the reader:
1. Prove the result that h = L * 70m, without using decimal points and without using any number larger than 9, except 70.
2. Explain why h ~ L * 70m, even if the earth is an oblate spheroid or ellipsoid, rather than a sphere.
3. Determine h if L = 1/2.
4. Determine h if L = 3/4.

Appendix

Different sources offer different values for the radius of the earth, with a value of about 6,378,137m for the equatorial radius being widely accepted. Some sources offer a value of f ~ 1/298.26 for the flattening factor, which is the fraction by which the equatorial radius differs from the polar radius.

1. 6,378,137m, f=1/298.257223563, World Geodetic System 1984. http://aa.usno.navy.mil/publications/docs/Circular_179.pdf
2. 6,378,137.0m, IERS (International Earth Rotation Service) Conventions 2003. http://tai.bipm.org/iers/conv2003/conv2003.html
3. 6,378,137.d0, f inv = 298.257222101d0. ftp://tai.bipm.org/iers/conv2003/chapter4/gconv.f
4. 6,378,136.6+/-0.1m, f 298.25642. http://maia.usno.navy.mil/conv2000/chapter1/tn32_c1.pdf
5. 6378km, 6378.127km, 6370.09km to 6371.03km. http://www.agu.org/reference/gephys/4_yoder.pdf
6. Equatorial radius 6378.137km, 6378.136km. http://www.agu.org/reference/gephys/5_cazenave.pdf
7. 6371.0km. http://www.agu.org/reference/gephys/8_masters.pdf
8. Equatorial radius 6378.1363km. http://www.agu.org/reference/gephys/24_dickey.pdf
9. 3939mi. http://www.karlscalculus.org/measureearth.html
10. 6400km. http://apollo.lsc.vsc.edu/classes/met130/notes/chapter1/thin_env.html
11. Equatorial radius 6478.14km. http://solarsystem.nasa.gov/planets/profile.cfm?Object=Earth
12. Equatorial 6378km, polar 6357km. http://www.nasm.si.edu/ceps/etp/ss/ss_planetdata.html

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