# Homework Help: Temperature of Earth

1. Jun 2, 2007

### Coren

1. The problem statement, all variables and given/known data

Trying to answer the question of "what would happen if Sirius take the place of the Sun", I've begin to try to calculate the average temperature of Earth in function of time for any incident flux (and therefore, the average temperature today), knowing the real average temperature of Earth and that 4'6Gy later it has reached it from zero, considering the star like a black body.

2. Relevant equations

Luminosity is given by:

L=a·4·pi·sigma·R^2·T^4 W

Where "R" is radius of the body, "T" its temperature, "sigma" the Stephan-Boltzmann constant and "a" is the absortivity of the body.

Flux is given by:

F=L/(4·pi·d^2) W/m^2

With "d" the distance where you want to calculate the flux.

Temperature in a body is:

T=c·Q/M K

Where "c" is the calorific capability of the body, "Q" is the total energy of the body, excluding mechanic ("heat" I think is how you call in english), and M is the mass of the body.

3. The attempt at a solution

In this problem, we have that Q is the absorbed energy minus the emited energy:

Q=Qabs-Qem

If Sun (or the star) is a black body, we have that (subindex s and e means "star" and "Earth"):

Qabs=Fs·pi·Re^2·a·t ; t=time

Fs=Ls/4·pi·d^2 ; Ls=4·pi·sigma·Rs^2·Ts^4

Qem=Le·t ; Le=4·pi·sigma·Re^2·Te^4

We will consider flux of the star and product with the effective section of the Earth and with absortibity like a constant, having:

Qabs=Fabs·t

Because of there are constants too in Qem, we will write it like:

Qem=A·Te^4·t where A=4·pi·sigma·Re^2

Now, with the hipotesis of calorific capability constant throught time, we have:

Te=c·(Qabs-Qem)/Me

So then:

dTe/dt=c·(Fabs-4·A·Te^3·(dTe/dt)·t-A·Te^4)/M

Making some algebra:

dTe/dt=c·(Fa-ATe^4)/(Me+c·4·A·Te^3·t)

Leads to a differential equation that I'm unable to solve. In adition I don't know if my logic is correct.

¿Can you help me?

Thanks.

Last edited: Jun 2, 2007
2. Jun 2, 2007

### Coren

I've been thinking, and I've found that the original problem is much easier.

If we have that:

Pabs=Fs·pi·Re^2·(1-A) where A is the Albedo Bond, and 1-A=absortibity

Pabs=Ls·Re^2·(1-A)/(4·d^2)

Pem=4·pi·sigma·Re^2·Te^4

In equilibrium: Pabs=Pem so..

Te=[Ls·(1-A)/(sigma·pi·16·d^2)]^0'25

Knowing that, with Ls the luminosity of Sun, Te is 253 K we can have A...

Lsun=3'65·10^26 W

sigma=5'67·10^-8 W/(m^2·K^4)

d=1'49598·10^11 m

We have:

A=0'28.

With that, now we could have the temperature if Sirius was here.

Lsirius=1'4359·10^28

Then, Te=634 K (381 degrees more...).

But I still want to know what is the solution of the other equation, and if it was well done.

3. Jun 4, 2007

### aiy1tm

I'm a bit confused about the phrasing of the question. It is something like.. plug in the luminosity of Sirius in place of the Sun... how much does the equilibrium temperature of Earth change (to an approximation)?

If so, you seem to have done the right thing. One thing to be careful about is that the flux from the star only strikes half the of the surface of the Earth, while the earth will emit radiation (blackbody assumption) from its whole surface area. These are just factors of two though...

If I follow your logic correctly, you're just leaving the time dependence in there. And it looks ok to me (not completely checking the details). One promising detail is that if you take the long time limit on your differential equation, the temperature does level off somewhere (so you do have an equilibrium situation, eventually, according to your equations). It doesn't look pleasant to solve completely though. Could perhaps do it numerically -- but I don't know what you'd insert for the specific heat of Earth.

4. Jun 4, 2007

### Coren

Yes, that was the original idea. But I thought it would be interesting too the fact of solve the "time for equillibrium".

I did, the surface I took for absortion was Sa=pi·Re^2 (effective section, like a disk). And the surface for emision Se=4·pi·Re^2.

I tought that knowing the temperature for the flux of Sun (Te today), there would be enought data to solve the specific heat. Perhaps with numerical methods as you just suggested.