<p>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.</p>

<p>The flux radiated from a blackbody surface (in W/m<span class="math-container">$^2$</span>) is given by <span class="math-container">$\sigma T^4$</span>, where <span class="math-container">$\sigma$</span> is Stefan's constant and <span class="math-container">$T$</span> is the temperature.</p>

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

<p>The temperature of the substellar point at equilibrium is therefore given by
<span class="math-container">$$\sigma T^4 = \frac{4\pi R^2 \sigma T_*^{4}}{4\pi d^2},$$</span>
<span class="math-container">$$ T = T_* \left(\frac{R}{d}\right)^{1/2}$$</span></p>

<p>Since we assume that <span class="math-container">$d \gg R$</span> then clearly <span class="math-container">$T &lt; T_{*}$</span>.</p>

<p>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.</p>

<p>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 <span class="math-container">$\sigma T_{\rm CMB}^4$</span> 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.</p>

CC BY-SA 4.0
2020-03-07T11:48:42.363Z
2020-03-07T11:48:42.363Z
ProfRob
<p>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.</p>

<p>Also, would the temperature on the opposite point approach the cosmic background temperature 2.7 K?</p>

CC BY-SA 4.0
2020-02-27T14:45:38.777Z
You have said this hypothetical black body is "solid at all temperatures"... So no, it obviously won't melt :-)
Rory Alsop
CC BY-SA 4.0
2020-02-27T14:53:48.683Z
@RoryAlsop I thought I may have had to emphasize that.
Pyrania
CC BY-SA 4.0
2020-02-27T17:07:40.577Z
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.
uhoh
CC BY-SA 4.0
2020-02-27T17:08:58.987Z
FOUND IT! [Why would a tidally-locked rocky planet have a first-order spherical harmonic surface temperature distribution?](https://astronomy.stackexchange.com/q/33097/7982) I'll leave a comment there under that answer to have a look here.
uhoh
CC BY-SA 4.0
2020-03-04T23:18:36.810Z
https://astronomy.stackexchange.com/a/25661/17743
Keith McClary
CC BY-SA 4.0
2020-03-05T16:40:03.160Z
https://iopscience.iop.org/article/10.3847/0004-637X/825/2/99 This is the link which will detail you....Kindly follow it
Pranay
<temperature><tidal-locking>
Temperature of a substellar point on an airless tidally-locked planet
Pyrania
# Temperature of a substellar point on an airless tidally-locked planet

(Asked by: username_0 on 2020-02-27)

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?

**Tags:** temperature, tidal-locking
> **Comments:**
> **username_1**: You have said this hypothetical black body is "solid at all temperatures"... So no, it obviously won't melt :-)  
> **username_0**: @RoryAlsop I thought I may have had to emphasize that.  
> **username_2**: 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.  
> **username_2**: FOUND IT! [Why would a tidally-locked rocky planet have a first-order spherical harmonic surface temperature distribution?](https://astronomy.stackexchange.com/q/33097/7982) I'll leave a comment there under that answer to have a look here.  
> **username_3**: https://astronomy.stackexchange.com/a/25661/17743  
> **username_4**: https://iopscience.iop.org/article/10.3847/0004-637X/825/2/99 This is the link which will detail you....Kindly follow it  

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## 1 answers

### Answer by username_5 on 2020-03-07. Score: 4 | Accepted answer

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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