I' trying to solve for the surface temperature Ts=((n+1)So(1-A)4)1/4, and thus So=L/4𝞹a2 but I cannot seem to find the value for solar flux or its luminosity.

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    $\begingroup$ Your equations seem a little messed up, and I don't trust a blind MathJax conversion. Are there some exponents in there that you left out? $\endgroup$ Mar 7, 2018 at 3:09
  • $\begingroup$ @NathanTuggy I agree. It's probably something like one of these, but I don't know what the (n+1) is en.wikipedia.org/wiki/… See also cow: space.stackexchange.com/q/25671/12102 $\endgroup$
    – uhoh
    Mar 7, 2018 at 3:21

1 Answer 1


The solar flux at 1 AU (Earth's distance from the sun) is ~1361 W/m2. Flux falls off as the square of distance. Saturn, on average -- and thus Titan, on average -- is about 9.58 AU from the sun, so the falloff factor is 1 / 9.582 or 0.0109.

Titan should average about 1361 x 0.0109 = 14.8 W/m2 when not shadowed by the gas giant.

When Saturn is at its furthest from the sun, 10.12 AU, the flux would be closer to 13.3 W/m2.


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