text: # 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.
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