.MCAD 304020000 1 74 213 0
.CMD PLOTFORMAT
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0 0 1 1 1 0 0 1 1 
0 1 0 0 1 1 NO-TRACE-STRING
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0 1 1 0 1 1 NO-TRACE-STRING
0 2 2 0 1 1 NO-TRACE-STRING
0 3 3 0 1 1 NO-TRACE-STRING
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0 1 5 0 1 1 NO-TRACE-STRING
0 2 6 0 1 1 NO-TRACE-STRING
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.CMD FORMAT  rd=d ct=10 im=i et=3 zt=15 pr=3 mass length time charge temperature tr=0 vm=0
.CMD SET ORIGIN 0
.CMD SET TOL 0.001000000000000
.CMD SET PRNCOLWIDTH 8
.CMD SET PRNPRECISION 4
.CMD PRINT_SETUP 1.200000 1.218750 1.200000 1.200000 0
.CMD HEADER_FOOTER 1 1 *empty* *empty* *empty* 0 1 *empty* *empty* *empty*
.CMD HEADER_FOOTER_FONT fontID=14 family=Arial points=10 bold=0 italic=0 underline=0 colrid=902837712
.CMD HEADER_FOOTER_FONT fontID=15 family=Arial points=10 bold=0 italic=0 underline=0 colrid=902837712
.CMD DEFAULT_TEXT_PARPROPS 0 0 0
.CMD DEFINE_FONTSTYLE_NAME fontID=0 name=Variables
.CMD DEFINE_FONTSTYLE_NAME fontID=1 name=Constants
.CMD DEFINE_FONTSTYLE_NAME fontID=2 name=Text
.CMD DEFINE_FONTSTYLE_NAME fontID=4 name=User^1
.CMD DEFINE_FONTSTYLE_NAME fontID=5 name=User^2
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.CMD DEFINE_FONTSTYLE_NAME fontID=10 name=User^7
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.CMD UNITS U=1
.CMD DIMENSIONS_ANALYSIS 0 0
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.CMD COLORTAB_ENTRY 128 0 0
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.CMD COLORTAB_ENTRY 128 128 128
.CMD COLORTAB_ENTRY 192 192 192
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.TXT 2 2 35 0 0
Cg a72.000000,72.000000,28
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard {\fs20 COPYRIGHT MIRON 
KAUFMAN 1995}}
.TXT 7 16 39 0 0
Cg a34.625000,34.625000,39
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}{\f1\fcharset0\froman Bookman Old Style;}}
\plain\cf1\fs28 \pard {\f1\fs36\b Thermal Physics\par Computer Project
\par Lab #1}}
.TXT 15 -17 27 0 0
Cg b73.000000,73.000000,498
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}{\f1\fcharset0\froman Times New Roman;}{\f2
\fcharset2\fnil Symbol;}}\plain\cf1\fs28 \pard {\f1 We estimate the 
mass of the earth atmosphere by assuming the temperature to be 
uniform. This of course is just a first-order approximation of 
reality.  First we get the variation of density with altitude by 
integrating the barometric formula dp/dh = -  g}{\f2 r}{\f1  by using 
the equation of state of ideal gases }{\f2  r}{\f1  }{\f1 = pM/RT.  So 
dp/dh = -(gM/RT)p. Separate the variables: dp/p = -dh/L, where L = 
RT/gM. Finally integrate on both sides: ln(p/p}{\f1\dn 0}{\f1 ) = -h/L 
or p = p}{\f1\dn 0}{\f1 exp(-h/L), and                }{\f2 r} = {\f2 r}{
\dn 0}exp(-h/L).  }
.TXT 21 27 132 0 0
Cg a46.000000,46.000000,25
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard This is the gas constant.}
.EQN 1 -23 106 0 0
{0:R}NAME:8.3145*({0:joule}NAME)/({0:K}NAME)
.EQN 5 0 133 0 0
{0:T}NAME:273*{0:K}NAME
.EQN 4 0 137 0 0
{0:g}NAME={0}?_n_u_l_l_
.EQN 4 0 146 0 0
{0:M}NAME:28.8*(10)^(-3)*{0:kg}NAME
.TXT 0 22 148 0 0
Cg a47.000000,47.000000,90
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard This is the molar mass 
of air. Since air is mainly N{\dn 2 }the molar mass is close to 28g/mole.}
.EQN 7 -23 155 0 0
{0:R.E}NAME:6.37*(10)^(6)*{0:m}NAME
.TXT 0 23 157 0 0
Cg a47.000000,47.000000,25
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard This is the earth radius.}
.EQN 5 -23 156 0 0
{0:p.0}NAME:1.013*(10)^(5)*{0:Pa}NAME
.TXT 0 24 158 0 0
Cg a46.000000,46.000000,32
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard This is the atmospheic pressure.
}
.EQN 6 -25 152 0 0
{0:\r.0}NAME:({0:M}NAME*{0:p.0}NAME)/({0:R}NAME*{0:T}NAME)
.TXT 3 23 161 0 0
Cg a47.000000,47.000000,45
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard This is the air density 
at the earth surface.}
.EQN 3 -23 153 0 0
{0:\r.0}NAME={0}?_n_u_l_l_
.EQN 6 0 166 0 0
{0:L}NAME:({0:R}NAME*{0:T}NAME)/({0:M}NAME*{0:g}NAME)
.TXT 1 22 170 0 0
Cg a49.000000,49.000000,58
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard L is the distance scale 
for atmospheric density variation.}
.EQN 6 -23 168 0 0
{0:L}NAME={0}?_n_u_l_l_
.TXT 4 0 174 0 0
Cg a49.000000,49.000000,37
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}{\f1\fcharset0\fnil ;}{\f2\fcharset2\fnil Symbol;}}
\plain\cf1\fs28 \pard Since h = r - R{\dn E}{\f1  the density }{\f2 r}
(r) is:}
.EQN 1 38 176 0 0
{0:\r}NAME({0:r}NAME):{0:\r.0}NAME*({0:e}NAME)^(-(({0:r}NAME-{0:R.E}NAME)/({0:L}NAME)))
.TXT 8 -38 182 0 0
Cg a72.000000,72.000000,54
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard We graph the atmospheric 
density decrease with height:}
.EQN 7 -1 181 0 0
{0:r}NAME:{0:R.E}NAME,{0:R.E}NAME+{0:L}NAME*0.1;{0:R.E}NAME+2*{0:L}NAME
.EQN 2 3 177 0 0
&&(_n_u_l_l_&_n_u_l_l_)&({0:\r}NAME({0:r}NAME))/({0:\r.0}NAME)@2&0&(_n_u_l_l_&_n_u_l_l_)&({0:r}NAME-{0:R.E}NAME)/({0:L}NAME)
0 1 1 1 1 0 0 1 1 
0 1 1 1 1 0 0 1 1 
0 1 0 0 1 1 NO-TRACE-STRING
0 2 1 0 1 1 NO-TRACE-STRING
0 3 2 0 1 1 NO-TRACE-STRING
0 4 3 0 1 1 NO-TRACE-STRING
0 1 4 0 1 1 NO-TRACE-STRING
0 2 5 0 1 1 NO-TRACE-STRING
0 3 6 0 1 1 NO-TRACE-STRING
0 4 0 0 1 1 NO-TRACE-STRING
0 1 1 0 1 1 NO-TRACE-STRING
0 2 2 0 1 1 NO-TRACE-STRING
0 3 3 0 1 1 NO-TRACE-STRING
0 4 4 0 1 1 NO-TRACE-STRING
0 1 5 0 1 1 NO-TRACE-STRING
0 2 6 0 1 1 NO-TRACE-STRING
0 3 0 0 1 1 NO-TRACE-STRING
0 4 1 0 1 1 NO-TRACE-STRING
0 1 1 52 26 14 0 3 
.TXT 41 -2 183 0 0
Cg a72.000000,72.000000,43
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard To get the mass we 
integrate the density:  }
.EQN 6 3 184 0 0
{0:m}NAME:4*{0:\p}NAME*({0:R.E}NAME&{0:\¥}NAME*{0:m}NAME`{0:\r}NAME({0:r}NAME)*({0:r}NAME)^(2)&{0:r}NAME)
.TXT 12 0 185 0 0
Cg a69.000000,69.000000,180
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard the numerical 
integration did not converge.  To help Mathcad we try then to simplify 
the integral: (1) take out the constant factors; (2) change variable 
of integration to x = r/L.}
.EQN 13 0 73 0 0
{0:m}NAME:4*{0:\p}NAME*{0:\r.0}NAME*({0:e}NAME)^(({0:R.E}NAME)/({0:L}NAME))*({0:L}NAME)^(3)*(({0:R.E}NAME)/({0:L}NAME)&{0:\¥}NAME`({0:x}NAME)^(2)*({0:e}NAME)^(-{0:x}NAME)&{0:x}NAME)
.TXT 16 -4 212 0 0
Cg b73.000000,73.000000,28
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard {\fs20 COPYRIGHT MIRON 
KAUFMAN 1995}}
.TXT 7 1 206 0 0
Cg a72.000000,72.000000,195
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard We still get an error 
message due to raising e to a large power R{\dn E}/L = 792.6.  We next 
use the symbolic processor (MAPLE). Select the integral and then from 
Symbolic chose Evaluate symbolically. }
.EQN 14 2 192 0 0
{0:m}NAME:4*{0:\p}NAME*{0:\r.0}NAME*({0:e}NAME)^(({0:R.E}NAME)/({0:L}NAME))*({0:L}NAME)^(3)*((2)/({0:exp}NAME(({0:R.E}NAME)/({0:L}NAME)))+(2)/({0:exp}NAME(({0:R.E}NAME)/({0:L}NAME)))*({0:R.E}NAME)/({0:L}NAME)+(1)/({0:exp}NAME(({0:R.E}NAME)/({0:L}NAME)))*(
({0:R.E}NAME)^(2))/(({0:L}NAME)^(2)))
.EQN 1 88 204 0 0
({0:R.E}NAME)/({0:L}NAME)={0}?_n_u_l_l_
.TXT 16 -88 211 0 0
Cg a70.000000,70.000000,66
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard Finally select the 
equation above and use Simplify under Symbolic.}
.EQN 5 1 101 0 0
{0:m}NAME:4*{0:\p}NAME*{0:\r.0}NAME*{0:L}NAME*(2*({0:L}NAME)^(2)+2*{0:R.E}NAME*{0:L}NAME+({0:R.E}NAME)^(2))
.EQN 9 1 102 0 0
{0:m}NAME={0}?_n_u_l_l_
.TXT 6 -3 115 0 0
Cg a70.000000,70.000000,94
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}{\f1\fcharset0\froman Times New Roman;}}\plain
\cf1\fs28 \pard {\f1 The mass of the atmosphere is 5.3*10}{\f1\up 18}{
\f1 Kg}{\f1\up   }{\f1 which is small compared to the earth mass 6*10}{
\f1\up 24}{\f1 kg. }}
.TXT 45 -2 213 0 0
Cg b73.000000,73.000000,28
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil Arial;}}\plain\cf1\fs28 \pard {\fs20 COPYRIGHT MIRON 
KAUFMAN 1995}}