- #1
sid_galt
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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
[tex]
\displaystyle\frac{dT}{dt} = \displaystyle\frac{1380A - 2\sigma\epsilon A T^4}{mC}[/tex]
[tex]
\displaystyle\frac{mC}{A}\int\displaystyle\frac{1}{1380 - 2\sigma\epsilon T^4}dT = \int dt
[/tex]
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 don't know how to proceed further. Can anyone help please?
Thank you
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
[tex]
\displaystyle\frac{dT}{dt} = \displaystyle\frac{1380A - 2\sigma\epsilon A T^4}{mC}[/tex]
[tex]
\displaystyle\frac{mC}{A}\int\displaystyle\frac{1}{1380 - 2\sigma\epsilon T^4}dT = \int dt
[/tex]
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 don't know how to proceed further. Can anyone help please?
Thank you
Last edited: