Deriving temperature time evolution from radiative cooling equation

[fasterthanjoao]

First year astronomy question, goes as follows:

"A closely exact description of the cooling described above is to consider the differential equation which says that rate of charge of thermal energy equals the rate of radiative
output. That is:

$$\frac{d}{dt} \frac{\left(3MkT\right)}{m_p} = -4*pi*R²*T^4$$ -eqtn1

where M, R are constant. Show that the resulting time evolution of the temperature is then given by:

T(t)```````````````1
--- = ---------------
T(0) ``` (1+3t/t_cool)^(1/3)

(kept getting problems in my latex for that part, so i just typed it out..., i'd like to get this question sorted before i have to go to math)

anyway, its also said that:

````````````3Mk
t_cool = -----------
``````````4*pi*R²*T³*sigma

everything has its usual meaning, sigma is the Stefan-Boltzmann constant, k is the Boltzmann constant.

------------

i'm not really that sure what to do, I tried rearranging eqtn one after taking dT/dt out, "multiplying" each side by dt then integrating. after that, I'm really not much closer to the point. any guides on where to go appreciated. thanks.

 

[SpaceTiger]

I don't see the problem. This equation is separable:

$$\frac{dT}{T^4}=\frac{-4\pi m_pR^2\sigma}{3Mk}dt$$

 

[fasterthanjoao]

yeah i see that, I've tried to take it some steps further but I'm getting in a bit of a mess. could you try the next step or two, not sure what it is I'm having the problem with.

 

[SpaceTiger]

$$\frac{1}{3}(T_0^{-3}-T^{-3})=\frac{-4\pi m_pR^2\sigma t}{3Mk}$$

$$T^{-3}=T_0^{-3}+\frac{4\pi m_pR^2\sigma t}{Mk}$$

$$\frac{T_0^3}{T^3}=1+\frac{4\pi m_pR^2T_0^3\sigma t}{Mk}$$

That help?

 

[fasterthanjoao]

very much so. thanks. glad to lay that to rest