If a hypothetical blackbody planet's components were solid at all temperatures, would the substellar point on the airless tidally-locked (1:1) planet eventually heat up to the surface temperature of the star it is orbiting? This material would be a very poor heat conductor.

Also, would the temperature on the opposite point approach the cosmic background temperature 2.7 K?

  • $\begingroup$ You have said this hypothetical black body is "solid at all temperatures"... So no, it obviously won't melt :-) $\endgroup$ – Rory Alsop Feb 27 '20 at 14:45
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    $\begingroup$ @RoryAlsop I thought I may have had to emphasize that. $\endgroup$ – Pyrania Feb 27 '20 at 14:53
  • $\begingroup$ If this can be simplified to a uniform solid sphere of thermal conductivity $k$ and constant emissivity $\epsilon$ with incident intensity $I$ the temperature distribution across the surface probably has a simple solution expressed in some set of polynomials. As I type this I think I'm remembering an existing answer here, I'll leave another comment if I find it. in the mean time this is a reasonable Stack Exchange question, it could be on-topic here and in Physics SE. $\endgroup$ – uhoh Feb 27 '20 at 17:07
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    $\begingroup$ FOUND IT! Why would a tidally-locked rocky planet have a first-order spherical harmonic surface temperature distribution? I'll leave a comment there under that answer to have a look here. $\endgroup$ – uhoh Feb 27 '20 at 17:08
  • $\begingroup$ astronomy.stackexchange.com/a/25661/17743 $\endgroup$ – Keith McClary Mar 4 '20 at 23:18

The planet would reach an equilibrium where the amount of heat absorbed is the same as the amount of heat radiated. If there is no way to transfer heat on the planet (no conduction, no atmosphere), then that condition must apply locally.

The flux radiated from a blackbody surface (in W/m$^2$) is given by $\sigma T^4$, where $\sigma$ is Stefan's constant and $T$ is the temperature.

If the substellar point is a distance $d$ from the star, and the luminosity of the star is given by $L \simeq 4\pi R^2 \sigma T_{*}^4$ (assuming it too is a blackbody with radius $R$ and temperature $T_*$) and assuming $d \gg R$ to avoid some unnecessary geometric unpleasantness, then the flux absorbed at the substellar point is $L/4\pi d^2$. The flux is all absorbed, since you wish to assume a blackbody.

The temperature of the substellar point at equilibrium is therefore given by $$\sigma T^4 = \frac{4\pi R^2 \sigma T_*^{4}}{4\pi d^2},$$ $$ T = T_* \left(\frac{R}{d}\right)^{1/2}$$

Since we assume that $d \gg R$ then clearly $T < T_{*}$.

At other points on the planet's surface it will receive a reduced flux from the star, simply becasue the flux from the star is incident at an angle to the exposed surface, so the equilibrium temperature will be lower.

On the unlit side of the planet there is no illumination from the star, but an almost isotropic flux from the cosmic microwave background equal to $\sigma T_{\rm CMB}^4$ over the entre surface. Therefore in the absence of any other source of heat, then that side will assume the temperature of the CMB at equilibrium.


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