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If the Terrestrial Poles Were to Melt...Problem: Solution:
First, let us estimate the total volume of the south polar ice. We begin by noting that, on the Earth, 1 degree of latitude corresponds to a distance of 69.2 mi. Now, using a well marked map, Antarctica may be estimated as having an angular radius of 18 degrees of latitude, or a radius in miles of r_{A} = 69.2 mi/deg x 18 deg = 1.25 x 10^{3} mi. Its estimated surface area is then r_{A}^{2} = 4.91 x 10^{6} mi^{2}. If we assume the average thickness of the Antarctic ice to be 1 mi, then the total ice volume is V_{Ice}
= 4.91 x 10^{6} mi^{3}.
If this ice melts, the resulting volume of liquid water is 4/5 of this value or V_{Water}
= 3.93 x 10^{6} mi^{3}.
Now, let this volume of water be spread in a uniform, thin shell over a uniform Earth (for the moment, we will ignore the continents). The total surface area of the Earth is A_{E} = 4r_{E}^{2} = 1.97 x 10^{8} mi^{2}, where r_{E} = 3.96 x 10^{3} mi. Since the thickness of this shell, r_{E}, may be assumed to be << the Earth's radius, r_{E} , we may write 3.93
x 10^{6} mi^{3} A_{E}r_{E}^{1}
or, r_{E} = 1.99 x 10^{-2} mi = 105 ft. If we now return the continents to their proper place, it is easy to see that the value of r_{E} must become larger. Since only 2/3 of the earth's surface area is ocean, the thickness of the water shell must be increased by a factor of about 3/2: r_{E}
= (3/2) x 105 ft = 158 ft. ^{2}
Thus, we have estimated that if the polar ice were to melt, the Earth's oceans would rise by about 150 ft. Reference to a map showing land elevations allows us to determine that coastal areas and major river valleys (such as the Mississippi or the Amazon river valleys) would be flooded. The land distribution would appear somewhat differently than it does on the globe today. But the majority of land would still be above sea level.
Actually
the volume of the water shell is V_{Water} = (4/3)(r_{2}^{3
}- r_{1}^{3}), where r_{1} is the radius
of the Earth prior to flooding (3.96 x 10^{3} mi), and r_{2}
is the radius of the Earth after flooding (r_{2} = r_{1}
+ r_{E}).
We may simplify the above expression for V_{Water} to read V_{Water}
= (4/3)(r_{E}^{3}),
where (r_{E}^{3})
= (r_{1} + r_{E})^{3}
- r_{1}^{3}= 3r^{2}r_{E}
+ 3r_{1}(r_{E})^{2}
+ (r_{E})^{3}
3r^{2}r_{E},
since r_{E}
<< r_{1}. Thus V_{Water}
4r^{2}r_{E}
= A_{E}r_{E}.
This last expression is the one given in the text.
^{2} The actual volume of Antarctic ice is 7 x 10^{6} mi^{3}. Using this value rather than the value 4.91 x 10^{6} mi^{3} previously calculated, we find r_{E} = 225 ft., close to the values given in text books. |
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