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 .

Answer

2

Explanation

`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)))` .