Explanation:
solar radiation is incident on the spacecraft and the spacecraft radiates heat to deep space, which is at an average temperature of 3 K.
[tex]Given \\Q_{sun} =950W/m^2\\\alpha =0.3\\T_{space} =3k\\sigma=5.67*10^-^8W/m^2k^4\\eplison=0.72[/tex]
to find
[tex]T_{surface} =?[/tex]
assume that Incident solar radiation on the spacecraft is uniform. The spacecraft radiates thermal energy to its surroundings which are at an average temperature of 3 K.
The spacecraft absorbs 30% of the incident radiation from the sun, because:
[tex]Q_{in} =\alpha Q_{sun} \\Q_{in} =285W/m^2[/tex]
The space craft radiates to deep space and deep space radiates to the spacecraft as well.
[tex]Q_{out.net} =epsilon*sigma(T_{surface}^4- T_{space}^4)[/tex]
When the temperature on the surface of the spacecraft reaches a steady-state value, there will be no net amount of heat transfer to or from the surface :
[tex]Q_{in}= Q_{out.net}[/tex]
Next, we can combine Eq 2 & 3 and solve for the surface temperature of the spacecraft.
[tex]T_{surface} =[T_{space} ^4+\frac{Q_{in} }{sigma*epsilon} ]^1^/^4[/tex]
[tex]T_{surface} =[3k^{4} +\frac{285W/m^2}{5,67*10^-^8*0.72} ]^1^/^4\\T_{surface} =289.05k=16C[/tex]
