I'm writing a program to generate solar systems but I'm having trouble calculating the expected temperature of a planet. I have found a formula to calculate this, but I haven't been able to get a remotely correct answer out of it as it doesn't clearly state what units your supposed to use.

This formula I found:

$$4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi R ^ 2 L_{\odot}(1 - a)}{(4 \pi d ^ 2)}$$

where $R$ is the planet's radius (not sure what units), $d$ is the distance from the Sun (it mentions AU), $a$ is the albedo, $L_{\odot}$ is the luminosity of the Sun (which I assume can be interchanged with the luminosity of any star), $T$ is the temperature of the planet (kelvin, this is what I'm trying to get), and $ơ$ is the Stefan-Boltzmann constant.

The site I found it on is notes for an astronomy college course. Here is the link:


Any help would be very much appreciated.


The formula

$$4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi R ^ 2 L_{\odot}(1 - a)}{4 \pi d ^ 2}$$

is correct, if you want to calculate the radiative equilibrium temperature. You only need to use the right units. We can further simplify the formula to

$$T ^ 4 = \frac{ L_{\odot}(1 - a)}{16 \pi d ^ 2 ơ}\;.$$

You should input the luminosity in watts, the distance to the star in meters and the Stefan-Boltzmann constant as

$$σ = 5.670373 × 10^{−8} \;\mathrm{W}\; \mathrm{m}^{−2}\; \mathrm{K}^{−4}.$$

The albedo is dimensionless. The resulting temperature will be in Kelvins. Let me make an example for Earth:

$d = 149,000,000,000 \;\mathrm{m}$

$L = 3.846×10^{26} \;\mathrm{W}$

Albedo of Earth is 0.29. (The Bond albedo should be used.) You will get

$$ T ^ 4 = \frac{ 3.846×10^{26}(1 - 0.29)}{16 \pi \times (149,000,000,000) ^ 2 \times (5.670373 × 10^{−8})}=4,315,325,985 \;\mathrm{K}^4\;. $$

After powering this number to 1/4, we obtain temperature 256 K, which is -17° C. This looks reasonable. The real average temperature on Earth is closer to 15° C, but the greenhouse effect is responsible for the difference.

  • $\begingroup$ Thank you so much, it would have taken me forever to figure out which units were right. $\endgroup$
    – Eegxeta
    Mar 13 '15 at 19:46
  • $\begingroup$ T(effective) is easy. y $\endgroup$ Jul 8 '17 at 15:25
  • $\begingroup$ Sorry, had to leave and couldn't edit in time. I wanted to say that modelling greenhouse will be trickier. I am doing something similar but not with a computer. I have found that each system will have its own "personality". A lot depends on initial abundances, star parameters (initial and present time), system evolution (migration, orbits, etc.) and many other factors including random luck. Observation tells us that if it is scientifically possible it's out there somewhere and if we find something we don't think is possible, we had better take a new look at our models. $\endgroup$ Jul 8 '17 at 15:43
  • $\begingroup$ is the above solution including the temperatures in the polar regions of the planet.. and if not how can be calculating? $\endgroup$ Aug 6 '17 at 14:14
  • 1
    $\begingroup$ The above solution is for average (over whole surface) temperature of a planet. The difference in temperature between equator and poles is more complicated matter and probably it would require global circulation model to obtain reasonable result. It will depend on the axis tilt, length of the day, and also on how dense is the atmosphere. If the atmosphere is much denser than on Earth, the differences will be very small between poles and equator. Without atmosphere or with thin atmosphere, the differences will be even much larger compared to Earth. $\endgroup$
    – Irigi
    Aug 13 '17 at 16:02

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