3
$\begingroup$

The temperature on earth is around 14C celsius.

It is around 1370 W/m^2 from our sun. The earth has an albedo of around 0.3

On Tatooine (from Star Wars) i have been able to calculate it goes from 1066 W/m^2 to 2100 W/m^2 (depending if the stars eclipse each other or not). Tatooine is a desert planet with a albedo of around 0.4. How can i make a simplified function to calculate what the temperature ranges from on Tatooine?

I did read on effective temperature on wikipedia, but that is higher than the level in my physics class so i didnt understand.

In the movies they say the temperatures get quite cold at night and warm during the day, so i would expect it to go something like -10C to 40C.

We can just assume the atmosphere is the same as on earth since it is breathable for humans.

Considering as the albedo increases, the temperature should decrease with an increasing albedo.

Therefore my first attempt was

T=k(1-a)*P, where T is temp, k is a constant, a is albedo and P is w/m^2

Since the atmosphere is the same, k should be the same?

Therefore

287=k(1-0.3)*1370, solving for k gives us

k=0.3

therefore temperature on Tatooine should follow the function

f(x)=T=0.3*(1-0.4)*x-273 (to get it measured in celsius)

Where 1066<x<2100.

f(1066)=-81 and f(2100)=99

the temperature are not this extreme doe, so this wont work.

I am aware this is not an accurate science, but im taking my first physics course in high school now, so I dont understand the "proper" way of doing this. Is it completely incorrect to assume temperature is proportional to (1-a) and how much power the sun gives us, when its the same atmosphere (ish)?

If anyone wants how i found out the power the twin suns gives Tatooine i can share it here.

$\endgroup$
4
  • $\begingroup$ You are trying to use a linear approximation to the temperature change, which usually only works well for small changes to parameters. This might not work well here, although Tattooine is pretty Earth-like by astronomy standards. I do think you need to use the full, nonlinear formula for effective temperature. $\endgroup$ – Anders Sandberg Mar 17 at 16:02
  • $\begingroup$ Not quite sure how to help you, an equilibrium temperature is the simplest model you can do for a planetary temperature. Just that the luminosity $L$ is your related to the flux F (in $W/m^2$) via $L= F/4\pi d^2$, where d is the planet-star distance. For two stars, add luminosities. You will not be able to reproduce Earth's temperature without accounting for the greenhouse effect, for which you can add a factor $(1+\tau)^{1/4}$, where $\tau$ is the infrared optical depth, a number between 0 and $\infty$, so you should need the GHG for your Tattooine model as well. $\endgroup$ – AtmosphericPrisonEscape Mar 17 at 20:54
  • $\begingroup$ Perhaps you can take the hemisphere summer and winter T just to see the impact of the difference in flux. And than assume a linear dependance T on flux for flux not too dissimilar from the actual sun earth system. $\endgroup$ – Alchimista Mar 18 at 14:10
  • $\begingroup$ I tried doing this, if I assume the increase in temperature is proportional to how much power the planet gets, where the constant is there to adjust for atmosphere, i get y'=kx,hence y=k/2*x^2 + C, but C=0 because y(0)=0. Since y(1370)=13.86 I can solve for k. Then draw a graph of the different temperatures, With this function I get y(1066)=8,5 and y(2209)=37. I think this can be a good attempt, any way i can improve it? @AtmosphericPrisonEscape $\endgroup$ – Bob the Turtle Mar 19 at 16:56
2
$\begingroup$

Don't have high enough reputation to comment, but someone should mention the Stefan-Boltzmann Law: https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

When a planet (or anything) gets warmer, the amount of radiation it emits increases with the 4th power of temperature (measured on an absolute scale like Kelvins).

A planet's temperature reaches equilibrium when it receives about as much radiation as it emits back into space.

This should give you a better approximation than the linear model.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.