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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 `290K` The radiative euilibrium temperature of the Mercury is `5.80 xx 10^(n)` Find the value of n Assume all three bodies to be black body . |
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Answer» Correct Answer - 2 `P_("recieved")=(piR_(p)^(2))((P_(sun))/(4pir_(s)^(2)))` `P_("emitted")=sigma(e)4piR_(p)^(2)T_(p)^(4)` In equilibrium `P_(r) = P_(e)` `rArr (T_(p))^(2) alpha (1)/(r_(s))` `T_(earth)/(T_(mercury))=sqrt(r_(mercury)/(r_(earth)))` . |
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