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Q.

The distance of the Earth from the Sun is 4 times that of the planet Mercury from the Sun. The temperature of the Earth in radiative equilibrium with the Sun is  290 K . Find the radiative equilibrium temperature of the Mercury (in Kelvin). Assume all three bodies to be the black bodies.

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answer is 580.00.

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Detailed Solution

Preceived=πRp2Psun4πrs2​Pemitted=σe4πRp2Tp4In equilibrium, Pr=Pe​⇒Tp2 ∝1rs​⇒TearthTmercury=rmercuryrearth⇒290Tmercury=14​⇒Tmercury=580 K
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