JEE Advanced 2023 Paper 2 · Physics · Calorimetry and Heat Transfer · medium

A planet of negligible atmospheric heat capacity orbits a star at distance d=1.5×10^11 m. The star is a black body of radius 7×10^8 m and surface…

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A planet of negligible atmospheric heat capacity orbits a star at distance

d=1.5\times10^{11}\ \mathrm{m}

. The star is a black body of radius

7\times10^{8}\ \mathrm{m}

and surface temperature

5800\ \mathrm{K}

. The planet reflects

30\%

of the incident radiation and, being a fast rotator, radiates as a black body uniformly from its whole surface. Using Wien's constant

2.898\times10^{-3}\ \mathrm{m\,K}

, the wavelength at which the planet's emission peaks in steady state is nearest to

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Answer: (C) $11.3\ \mu\mathrm{m}$

Power absorbed by the planet (radius

r_p

): intensity at distance

d

\sigma T_s^4\,\dfrac{R_s^2}{d^2}

, intercepted over the cross-section

\pi r_p^2

, and only

(1-0.3)

is absorbed. Power radiated:

4\pi r_p^2\sigma T^4

. Balancing,

T^4=\dfrac{0.7}{4}\,T_s^4\dfrac{R_s^2}{d^2}

, so

T=T_s\sqrt{\dfrac{R_s}{2d}}\,(0.7)^{1/4}=5800\times0.0483\times0.9147\approx256\ \mathrm{K}

. Wien's law gives

\lambda_m=\dfrac{2.898\times10^{-3}}{256.3}\approx11.3\ \mu\mathrm{m}

. Without the albedo one gets

280\ \mathrm{K}

and

10.3\ \mu\mathrm{m}

; using

R_s/d

instead of

R_s/2d

gives

8.0\ \mu\mathrm{m}

; applying

0.7^{1/2}

gives

12.4\ \mu\mathrm{m}

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