<p>Ignoring meteorological factors and any dust or satellites, this is still an incomplete problem. You would also need to know the rotation axis. For example, Uranus rotates completely on its side (i.e. it rotates at 97.77 degrees, while earth rotates at 23 degrees, 26 minutes and 21.4119 seconds). Such things become important for factors like the Arctic Circle. You are also missing the distance between the planet and the star. That becomes important when you consider that Pluto is colder than Mercury. This is needed for the formulation of the sterdian.</p>

<p>In your question, you said to ignore atmospheric effects. However, the atmosphere also acts as a secondary source of radiation (the infamous greenhouse effect). Even if you neglect eddies, advection, etc. you also need to consider that the atmosphere can absorb radiation and emit it back to the surface, otherwise you would find the average temperature of the surface to be the same as the average temperature of the earth.</p>

<p>Edit: I found a decent approximation (approximation being the key word).</p>

<p>Let $Q_S$ is the incoming solar radiation flux
$$Q_s= S(1-\alpha)(\frac{\bar{d}}{d})^2cos(\zeta)\tau_s $$
Where $S$ is the solar constant, $\alpha$ is the albedo, $d$ is the distance of the earth from the sun, $\bar{d}$ is the mean distance of the earth from the sun and $\tau_s$ is the transmissivity of the atmosphere (including any clouds above).</p>

<p>The last factor, $cos(\zeta)$ can be computed by$$cos(\zeta)=sin(\psi)sin(\delta)+cos(\psi)cos(\delta)cos(h)$$ where $$\delta=23.45\times\frac{\pi}{180}cos(\frac{2\pi\times(d-d_{solst})}{d_{year}})$$
Where $d$ is the Julian day, $d_{solst}$ is the julian day of the solstice (173) and $d_{year}$ is the number of days in the year (365.25)</p>

<p>Additionally, $h$ is the local hour, as defined by $$h=\frac{(t_{UTC}-12)\times\pi}{12}+\frac{\lambda\pi}{180}$$ where $\lambda$ is the longitude and $t_{UTC}$ is the time in UTC (in hours).</p>

CC BY-SA 3.0
2016-10-17T15:52:10.420Z
Note that the approximation is valid for earth. The number 23.45 should be replaced with the axial tilt, and the number 12 should be adjusted for the revolution of the earth.

This is also neglecting turbulent fluxes and considering only the net solar radiation, as well as the downward longwave radiation from the atmosphere (the "greenhouse" effect).
BarocliniCplusplus
CC BY-SA 3.0
2015-08-06T02:15:06.430Z
Also, orbital eccentricities.
naught101
CC BY-SA 3.0
2015-08-06T14:59:07.713Z
Ok, assuming I had all of that, how would I calculate it? Also, the point of this is to get the temperature before accounting for the atmosphere; I'm already accounting for the atmosphere in another formula.
Kaelan Cooter
CC BY-SA 3.0
2015-08-05T22:07:15.937Z
2016-10-17T15:09:19.273Z
2016-10-17T15:09:19.273Z
BarocliniCplusplus
<p>I think that if you just want the insolation (at the top of the atmosphere), then you can just follow the argument here:<a href="https://en.wikibooks.org/wiki/Climate_Change/Science/Distribution_of_Insolation" rel="nofollow">Climate Change/Science/Distribution of Insolation (Wikimedia)</a>.</p>

<p>Then you can make some assumption about the radiative transfer through the atmosphere to get at the temperature. </p>

CC BY-SA 3.0
2015-08-15T00:37:26.537Z
If you were to provide a summary here of the information from your link you might get more up votes
Fred
CC BY-SA 3.0
2015-08-14T18:50:42.523Z
2016-10-10T06:31:20.850Z
2016-10-10T06:31:20.850Z
brian
<p>I'm trying to find equations that would help me determine the amount of solar radiation hitting a certain latitude on a certain planet given the following inputs:</p>

<ul>
<li>the degrees of latitude of the location in question</li>
<li>this hemisphere's current season (winter or summer)</li>
<li>size of the planet</li>
<li>luminosity of the star(s) the planet orbits</li>
</ul>

<p>Without taking into consideration wind, air pressure, or any atmospheric effects.</p>

<p>Ideally I would like to determine the average solar radiation of a given location in both the winter and summer. </p>

<p>My end goal is to determine the average surface temperature of a given latitude on a planet using the base solar radiation and the effects of wind, air pressure, and surface ocean currents.</p>

CC BY-SA 3.0
2015-08-03T04:14:01.380Z
so are you saying it's a cloudless planet?  you might see http://www.acs.org/content/acs/en/climatescience/energybalance/predictedplanetarytemperatures.html
f.thorpe
CC BY-SA 3.0
2015-08-03T15:12:52.297Z
@farrenthorpe That's actually what I'm using to determine the overall planetary temperature, but this question is asking the same but for a specific latitude
Kaelan Cooter
CC BY-SA 3.0
2015-08-04T05:03:07.213Z
latitude dependence requires much more than a simple energy balance equation. There are too many factors that have various regimes, so without knowing the particulars of your "planet" (e.g. surface type, atmospheric composition, etc) you are left with hundreds of parameters that need to be set before you can get into a simplified equation.  If you have an EARTH science question (which includes the study of all terrestrial worlds in our solar system), I'm sure someone on the site would try to help you.
f.thorpe
<atmosphere><atmospheric-radiation>
Equation for solar radiation at a given latitude?
Kaelan Cooter
# Equation for solar radiation at a given latitude?

(Asked by: username_0 on 2015-08-03)

I'm trying to find equations that would help me determine the amount of solar radiation hitting a certain latitude on a certain planet given the following inputs:

* the degrees of latitude of the location in question
* this hemisphere's current season (winter or summer)
* size of the planet
* luminosity of the star(s) the planet orbits

Without taking into consideration wind, air pressure, or any atmospheric effects.

Ideally I would like to determine the average solar radiation of a given location in both the winter and summer.

My end goal is to determine the average surface temperature of a given latitude on a planet using the base solar radiation and the effects of wind, air pressure, and surface ocean currents.

**Tags:** atmosphere, atmospheric-radiation

> **Comments:**
> **username_1**: so are you saying it's a cloudless planet?  you might see http://www.acs.org/content/acs/en/climatescience/energybalance/predictedplanetarytemperatures.html  
> **username_0**: @farrenthorpe That's actually what I'm using to determine the overall planetary temperature, but this question is asking the same but for a specific latitude  
> **username_1**: latitude dependence requires much more than a simple energy balance equation. There are too many factors that have various regimes, so without knowing the particulars of your "planet" (e.g. surface type, atmospheric composition, etc) you are left with hundreds of parameters that need to be set before you can get into a simplified equation.  If you have an EARTH science question (which includes the study of all terrestrial worlds in our solar system), I'm sure someone on the site would try to help you.  

---

## 2 answers

### Answer by username_2 on 2015-08-05. Score: 11

Ignoring meteorological factors and any dust or satellites, this is still an incomplete problem. You would also need to know the rotation axis. For example, Uranus rotates completely on its side (i.e. it rotates at 97.77 degrees, while earth rotates at 23 degrees, 26 minutes and 21.4119 seconds). Such things become important for factors like the Arctic Circle. You are also missing the distance between the planet and the star. That becomes important when you consider that Pluto is colder than Mercury. This is needed for the formulation of the sterdian.

In your question, you said to ignore atmospheric effects. However, the atmosphere also acts as a secondary source of radiation (the infamous greenhouse effect). Even if you neglect eddies, advection, etc. you also need to consider that the atmosphere can absorb radiation and emit it back to the surface, otherwise you would find the average temperature of the surface to be the same as the average temperature of the earth.

Edit: I found a decent approximation (approximation being the key word).

Let $Q\_S$ is the incoming solar radiation flux
$$Q\_s= S(1-\alpha)(\frac{\bar{d}}{d})^2cos(\zeta)\tau\_s $$
Where $S$ is the solar constant, $\alpha$ is the albedo, $d$ is the distance of the earth from the sun, $\bar{d}$ is the mean distance of the earth from the sun and $\tau\_s$ is the transmissivity of the atmosphere (including any clouds above).

The last factor, $cos(\zeta)$ can be computed by$$cos(\zeta)=sin(\psi)sin(\delta)+cos(\psi)cos(\delta)cos(h)$$ where $$\delta=23.45\times\frac{\pi}{180}cos(\frac{2\pi\times(d-d\_{solst})}{d\_{year}})$$
Where $d$ is the Julian day, $d\_{solst}$ is the julian day of the solstice (173) and $d\_{year}$ is the number of days in the year (365.25)

Additionally, $h$ is the local hour, as defined by $$h=\frac{(t\_{UTC}-12)\times\pi}{12}+\frac{\lambda\pi}{180}$$ where $\lambda$ is the longitude and $t\_{UTC}$ is the time in UTC (in hours).

> **Comments:**
> **username_3**: Also, orbital eccentricities.  
> **username_0**: Ok, assuming I had all of that, how would I calculate it? Also, the point of this is to get the temperature before accounting for the atmosphere; I'm already accounting for the atmosphere in another formula.  
> **username_2**: Note that the approximation is valid for earth. The number 23.45 should be replaced with the axial tilt, and the number 12 should be adjusted for the revolution of the earth.  This is also neglecting turbulent fluxes and considering only the net solar radiation, as well as the downward longwave radiation from the atmosphere (the "greenhouse" effect).  

---

### Answer by username_4 on 2015-08-14. Score: 0

I think that if you just want the insolation (at the top of the atmosphere), then you can just follow the argument here:[Climate Change/Science/Distribution of Insolation (Wikimedia)](https://en.wikibooks.org/wiki/Climate_Change/Science/Distribution_of_Insolation).

Then you can make some assumption about the radiative transfer through the atmosphere to get at the temperature.

> **Comments:**
> **username_5**: If you were to provide a summary here of the information from your link you might get more up votes  

---

