[Mathematica 11.1]
control= Volume of ocean
Quantity[1.332*10^9, ("Kilometers")^3]
control= seawater density
Quantity[Interval[{1020, 1035}], ("Kilograms")/("Meters")^3]
%%*%
Quantity[Interval[{1.35864*10^21, 1.37862*10^21}], "Kilograms"]
control= seawater specific heat
Quantity[Interval[{3.926, 4.022}], ("Joules")/("Grams" "KelvinsDifference")]
%%*%
Quantity[Interval[{5.33402064*10^24, 5.54480964*10^24}],
("Joules")/("KelvinsDifference")]
In[104]:= UnitConvert[Quantity[1.332*10^9, ("Kilometers")^3], "mi"]
Out[104]= Quantity[3.19563794427132*10^8, ("Miles")^3]
control= mass of atmosphere
Quantity[5.1441*10^18, "Kilograms"]
But I couldn't get it to tell me the specific heat of air.
--rwg
Date: 2017-01-12 05:47
From: "Cordwell, William R" <wrcordw(a)sandia.gov>
To: math-fun <math-fun(a)mailman.xmission.com>
Reply-To: math-fun <math-fun(a)mailman.xmission.com>
>From NOAA:
According to the U.S. Geological Survey, there are over 332,519,000 cubic
miles of water on the planet. A cubic mile is the volume of a cube
measuring one mile on each side. Of this vast volume of water, NOAA's
National Geophysical Data Center estimates that 321,003,271 cubic miles is
in the ocean.
R_e ~ 6371 km. V_ocean ~ 1.33799*10^9 km^3.
V / (4*pi*r^2) ~ 2.62318 km. (barring arithmetical errors). That is small
enough so that correcting for the earth being curved will not substantially
change the answer.
The Challenger Deep is ~ 11 km deep.
BC
-----Original Message-----
From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of
Marc LeBrun
Sent: Wednesday, January 11, 2017 11:18 PM
To: math-fun <math-fun(a)mailman.xmission.com>
Subject: [EXTERNAL] [math-fun] terrestrial estimation question
Were the Earth smoothed, preserving volume, how deep would the ocean be?
