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Temperature of body in space

  1. Oct 30, 2005 #1
    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 [tex]\sigma[/tex] be the boltzmann constant, [tex]\epsilon[/tex] 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}[/tex]
    \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
    [tex]\displaystyle\frac{0.156942mC}{A}(\arctan[0.00230862T]+arctanh[0.00230862T]) + C' = t[/tex]

    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. jcsd
  3. Oct 31, 2005 #2
    Can anyone help please?
  4. Oct 31, 2005 #3
    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
  5. Oct 31, 2005 #4
    But there must be someway to find temperature as a function of time for a body in space exposed to the Sun.
  6. Oct 31, 2005 #5


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    Yes, there is. You can evaluate your integral numerically.
  7. Oct 31, 2005 #6
    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.
  8. Oct 31, 2005 #7
    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)
    [tex]tan^{-1}(x) +tan^{-1}(y) = tan^{-1}(\frac{x+y}{1-xy})[/tex]

    Last edit - I tried all the above, it doesn't work (not even with shoddy approximations).
    Last edited: Oct 31, 2005
  9. Oct 31, 2005 #8
    I didn't notice that before. Thanks for the help.
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