# Temperature of body in space

1. Oct 30, 2005

### sid_galt

I want to determine the temperature of a body in space exposed to the Sun as it varies with time.
I tried this.
Let A be the area exposed to the Sun and 2A the total area of the body. Let 1380 W/m2 be the power of the sunlight falling on the body. Let $$\sigma$$ be the boltzmann constant, $$\epsilon$$ the emissivity of the body and T the temperature at a a particular instant of time, m the mass of the body and C the specific heat constant. Then
$$\displaystyle\frac{dT}{dt} = \displaystyle\frac{1380A - 2\sigma\epsilon A T^4}{mC}$$
$$\displaystyle\frac{mC}{A}\int\displaystyle\frac{1}{1380 - 2\sigma\epsilon T^4}dT = \int dt$$
I tried to integrate it on integrals.wolfram.com taking boltzmann constant as 5.6E-8 and emissivity as 0.7. The result was
$$\displaystyle\frac{0.156942mC}{A}(\arctan[0.00230862T]+arctanh[0.00230862T]) + C' = t$$

C' is here the integration constant

I dont know how to proceed further. Can anyone help please?
Thank you

Last edited: Oct 31, 2005
2. Oct 31, 2005

### sid_galt

3. Oct 31, 2005

### µ³

Are you sure it's T^4? My thermodynamics knowledge is limited but I thought the rate of heat transmition was proportional to deltaT, I might be wrong though. Otherwise, there is no analytic solution to the lower equation for T.
Edit - nevermind, it's blackbody radiation. Yeah, sorry, can't figure out what's wrong with either your physics or math. The last equation is not solvable for T.

Last edited: Oct 31, 2005
4. Oct 31, 2005

### sid_galt

But there must be someway to find temperature as a function of time for a body in space exposed to the Sun.

5. Oct 31, 2005

### Tide

Yes, there is. You can evaluate your integral numerically.

6. Oct 31, 2005

### sid_galt

But how do I integrate high temperatures into the equation, say if I want to evaluate it for a body with an initial temperature of 1000 K.
My Arctanh would give an unreal value for all temperatures higher than 434.78 K.

7. Oct 31, 2005

### µ³

if you notice that ArcTan[x]+ArcTanh[x] for x > 1 always give some value a - 1.5708i , well , make it so your integration constant takes out the imaginary part.
Edit- more specifically
for x>1
ArcTanh[x] = ArcTanh[1/x] -1/2pi*i

Edit: Another approximation:
for small x
ArcTanh[x] = ArcTan[x]

another nice identity:
(you might be able to solve for this actually)
$$tan^{-1}(x) +tan^{-1}(y) = tan^{-1}(\frac{x+y}{1-xy})$$

Last edit - I tried all the above, it doesn't work (not even with shoddy approximations).

Last edited: Oct 31, 2005
8. Oct 31, 2005

### sid_galt

I didn't notice that before. Thanks for the help.