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 --- 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" ------ 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)