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 .

Explanation

`P_(received)=(piR_(p)^(2))((P_(sun))/(4pir_(s)^(2)))`
`P_(emitted)=sigma (e)4piR_(p)^(2)T_(p)^(4)`
In Thermal Equilibrium
`P_(received)=P_(emitted)`
`implies (T_(P))^(2) prop (1)/(r_(s))`
`(T_(earth))/(T_(mercury))=sqrt((r_(mercury))/(r_(earth)))`
`T_(mercury)=(290K)((4)/(1))^(1//2)`
`=580 K`