That is the question. I know the concepts of luminosity $(L=4\pi R^2F)$ and the flow $F=\sigma· T^4$, with $T$ the temperature in its surface. But how I use that to know if the Earth would get warmer or not if $R$ increases and $T$ decreases?

  • 2
    $\begingroup$ This depends in part on the change in $R$ and $T$. $\endgroup$
    – HDE 226868
    Dec 19, 2015 at 15:09
  • $\begingroup$ Yeah but the question is: the bigger the luminosity, the most the Sun warms the Earth? Or it depends only on the flow? $\endgroup$ Dec 19, 2015 at 15:41
  • $\begingroup$ @Carlos: As HDE said it depends on the precise values. Without pluggin in numbers you cannot know, as expansion and cooling have two competing effects on the luminosity. $\endgroup$ Dec 19, 2015 at 16:02
  • $\begingroup$ Yes, I know that, but more luminosity implies more heat? That's the question $\endgroup$ Dec 19, 2015 at 20:22
  • $\begingroup$ @CarlosVázquezMonzón Yep, via the formula for effective temperature. $\endgroup$
    – HDE 226868
    Dec 19, 2015 at 23:10

1 Answer 1


The equilibrium temperature of the Earth, $T_E$, scales roughly as $L^{1/4}$, which is proportional to $R^{1/2} T$, where $L$, $R$ and $T$ are the solar luminosity, radius and temperature.

The actual approximate relationship is derived by equating the power received by the Earth, which is proportional to the solar luminosity $L$, with the power radiated by the Earth, which is proportional to $T_E^4$ for a blackbody. Hence $T_E \propto L^{1/4}$.

So the answer to your question depends on by how much you increase the radius compared with the decrease in temperature.

There will be second order effects that do depend on the spectrum of radiation from the Sun (and therefore its temperature) compared with the wavelength dependence of the albedo and emissivity of the Earth. So I will post a better question...

  • $\begingroup$ @CarlosVazquezMonzon I will add something $\endgroup$
    – ProfRob
    Dec 20, 2015 at 22:55
  • $\begingroup$ "The equilibrium temperature of the Earth scales roughly as L^1/4..." How do you know that?? Are you saying that the equilibrium temperature is the effective temperature of the Earth? $\endgroup$ Dec 20, 2015 at 22:56

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