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 S_{e}, the surface of the
earth, with meltwater, it rises to a height of 70m. So
V_{m} = L * S_{e} * 70m
Under the Martini Hypothesis, this depth will never be reached, because
the water will spread out. With BC2, this same volume V_{m}
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,
V_{m} = S_{e} * h
L * S_{e} * 70m = S_{e} * 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 S_{e}
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 * r^{2}
If we let
r_{e} = premelt radius of the earth (a subject to which we shall return)
Then the surface of the earth we call a sphere is
S_{e} = 4 * π * r_{e}^{2}
If the radius of the earth r_{e} 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 S_{e} doesn't vary
with the depth of the water that covers it.
Since S_{e} 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 * r^{3}
Under BC1, the volume of water will be a fraction L of the
difference in volumes of a sphere with a postmelt
radius ( r_{e} + 70m ) and a sphere
with a premelt radius r_{e}
That is, the volume of water V_{m}, which is melted ice,
is given by
V_{m} = L * [ (4/3) * π * ( r_{e} + 70m )^{3}  (4/3) * π * r_{e}^{3} ]
Under BC2, this same volume of water spreads out over the
entire earth, so the seas don't rise so much, and the volume V_{m}
is given by
V_{m} = (4/3) * π * ( r_{e} + h )^{3}  (4/3) * π * r_{e}^{3}
Reconciling these two formulae for the same volume V_{m},
and solving for h, we have
(4/3) * π * ( r_{e} + h )^{3}  (4/3) * π * r_{e}^{3} = L * [ (4/3) * π * ( r_{e} + 70m )^{3}  (4/3) * π * r_{e}^{3} ]
( r_{e} + h )^{3}  r_{e}^{3} = L * [ ( r_{e} + 70m )^{3}  r_{e}^{3} ]
( r_{e} + h )^{3} = L * [ ( r_{e} + 70m )^{3}  r_{e}^{3} + (1/L) * r_{e}^{3} ]
( r_{e} + h )^{3} = [ L * ( r_{e} + 70m )^{3} + ( 1  L ) * r_{e}^{3} ]
r_{e} + h = [ L * ( r_{e} + 70m )^{3} + ( 1  L ) * r_{e}^{3} ] ^{(1/3)}
h = [ L * ( r_{e} + 70m )^{3} + ( 1  L ) * r_{e}^{3} ] ^{(1/3)}  r_{e}
As detailed in the appendix, we know that
6,380,000m > r_{e} > 6,350,000m
and we have assumed that
L = 2/3 .
Substituting these values into our derived formula for h, we find that
if r_{e} = 6,380,000m, then h = 46.66m ~ L * 70m ,
and
if r_{e} = 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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