http://www.pdas.com/atmosTable2SI.html below 10 km; fits
pressure vs altitude (10km)
An analysis of data submitted by computer: on 15-SEP-2018 at 11:54 indicates that a function of the form:
--Exponential-- y=A exp(Bx)
can fit the 22 data points with a reduced chi-squared of 0.11E+07
FIT
PARAMETER VALUE ERROR
A = 0.1034E+06 0.62E+03
B = -0.1331 0.11E-02
NO x-errors
NO y-errors
scale height =-1/B = 7.513 km
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temperature vs altitude (10km)
An analysis of data submitted by computer: on 15-SEP-2018 at 11:56 indicates that a function of the form:
--Linear-- y=A+Bx
can fit the 22 data points with a reduced chi-squared of 0.11E-02
FIT
PARAMETER VALUE ERROR
A = 288.1 0.13E-01
B = -6.490 0.22E-02
Lapse Rate = 6.5 K/km
---
pressure vs altitude (86km)
An analysis of data submitted by computer: on 15-SEP-2018 at 12:00 indicates that a function of the form:
--Exponential-- y=A exp(Bx)
can fit the 45 data points with a reduced chi-squared of 0.31E+07
FIT
PARAMETER VALUE ERROR
A = 0.9887E+05 0.31E+04
B = -0.1426 0.64E-03
scale height = -1/B = 7.013 km
---
to me all of these plots look "pretty good"
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Swihart chapter 5 #8
law of Dulong & Petit says molar specific heat of solids is about 3 R
assume a rock density of about 3 g/cm^3 = 3000 kg/m^3
the energy loss seems to be assuming sigma T^4 per area from the ground,
but more realistically the atmosphere would also heat the ground
this heat loss is made up with a Delta T=10 of some moles or mass of ground
rate of energy loss per area: sigma T^4 for 12*3600 seconds =
* ? sigma*295^4*12*3600
18551668.66420149 (J/m^2 = 10^3 erg/cm^2, energy lost over 12 hours per m^2)
Q=n C (Delta T) = n 3 R 10 = above
* ? res/(3*R*10)
74375.12066666526 !(moles)
* ? 74375.*.025 !(mole weight= 25 g)
1859.375000000000 !(kg of rock)
* ? res/3000 !(density of rock =3000 kg/m^3)
.6197916666666666 !(m)