User:GabrielVelasquez/Solar Constant Formula

Some non-original-research using known values and basic algebra:

I disagree that the figure for the Earth's Solar constant at Solar Constant are accurate,
I've read various different quotations of the value in texts and on the internet. A little accuracy wouldn't hurt:
R= 6.955e8 m (Sun's radius)
T= 5778 °K (Sun's photosphere or Effective temperature)
a= 5.67051e-8 (Stefan-Boltzmann Constant)
d= 149597876600 meters (Earth's average distance, Mariner 10 data), 1 AU
f= flux or Insolation.
L= 4pi·R²aT⁴
L = 4pi·d²f
Since both Luminosity formulas equal 3.84181e26 Watts for the Sun and Earth,
then 4pi·R²aT⁴ = 4pi·d²f
Therefore, f=(R²aT⁴) / d²
Then ((6.955e8 m)² (5.67051e-8) (5778°K)⁴) / (149597876600)² = 1366.0785 W/m²
(Which is only off by 0.1333% the so called satellite measured solar constant.)
This is the average. If you factor in the Earths's eccentricity (0.016710219),
then the range is 1321.5430 W/m² to 1412.9039 W/m²

So far added chart to:

109 Piscium b

code for Gliese 581 c at Perihelion:

$f_{p}={\frac {((0.38\times 6.955\times 10^{8})^{2})\times (5.67051\times 10^{-8})\times (3480^{4})}{((0.073-(0.073\times 0.16))\times 149597876600)^{2}}}$

solar mass $M_{\odot }=1.98892\times 10^{30}{\hbox{ kg}}$
Solar radius $1\,R_{\odot }=6.960\times 10^{8}\,{\hbox{m}}=0.004652\,{\hbox{AU}}$ (Astronomical unit)
astronomical unit $AU_{\odot }=149,597,870.691{\hbox{ km}}$
Earth mass M_⊕ = 5.9736 × 10²⁴ kg
Gravitational constant $G=\left(6.67428\pm 0.00067\right)\times 10^{-11}\ {\mbox{m}}^{3}\ {\mbox{kg}}^{-1}\ {\mbox{s}}^{-2}.$
Sidereal year $P_{\odot }=365.25636042{\hbox{ days}}$

(((695500000)^2)*(0.0000000567051)*(5778^4))/ ((1-(1*0))*149597870691 )^2 = 1366.07868576308

Formula for planet temperature

$T_{p}={\frac {(5.67051E-8)*(3840^{4})}{(4*pi*(0.0613*1AU)^{2})}}*\ {{4*pi*(0.29*{R_{\odot }})^{2}}{(4*pi*(R)^{2}}}{pi*(R)^{2}*(1-0.296){}0.0000000567051)^{0}.25}$

=(((((0.0000000567051)*(3840^4))/(4*PI()*((0.0613*149597876600)^2))) * ((4*PI()*((0.29*695500000)^2))/(4*PI()*((11162)^2)))*((PI()*((11162)^2)*(1-0.64))))/0.0000000567051)^0.25

{R_\odot}

\sigma \cdot T_{\mathrm {p} }^{4}={\frac {\sigma \cdot T_{\ast }^{4}}{4\pi \cdot d^{2}}}\cdot {\frac {4\pi \cdot R_{\ast }^{2}}{4\pi \cdot R_{\mathrm {p} }^{2}}}\cdot \pi \cdot R_{\mathrm {p} }^{2}(1-A_{\mathrm {p} })

However I don't see why you are trying to use this equation, as many of the terms in it cancel each other out, making it unnecessarily cumbersome to work with. In fact it is rearranged on the next line as follows:

T_{\mathrm {p} }=T_{\ast }\left[\left({\frac {R_{\ast }}{d}}\right)^{2}{\frac {1-A_{\mathrm {p} }}{4}}\right]^{\frac {1}{4}}

$T_{p}={\frac {(5.67051E-8)*(3840^{4})}{(4*pi*(0.0613*1AU)^{2})}}*\ {{4*pi*(0.29*{R_{\odot }})^{2}}{(4*pi*(R)^{2}}}{pi*(R)^{2}*(1-0.296){}0.0000000567051)^{0}.25}$

Selsis et al made serious error! 97°C not 40°C

"FUNDAMENTALS OF ASTRONOMY" page 318 (QB43.3.B37 2006) by Dr.Cesare Barbieri, professor of Astronomy at the university of Pauda, Italy has a different planet Equilibium Temperature formula than the Selsis et al paper (very different) and is more accurate. It works out for Gliese 581 c to over 337°K/64°C (368.3°K /95°C at perihelion) with out the GHG effect (Earth is +33°C), with the Earth's increase added it would put it at 97°C! (130°C at perihelion). The full formula is T=(((((0.0000000567051)*(3840^4))/(4*π *((0.073*149597876600)^2))) * ((4*PI()*((0.29*695500000)^2))/(4*π *((11162)^2)))*((π *((11162)^2)*(1-0.3))))/0.0000000567051)^0.25 Using the same Albedos as they did for Earth, but nowhere near habitable. Tp=(((((5.67051E-8)*(Ts^4))/(4*π *((d*1au)^2))) * ((4*π *((Rs)^2))/(4*PI()*((Rp)^2)))*((PI()*((Rp)^2)*(1-A))))/5.67051E-8)^0.25