Question

Estimate the global temperature of our planet (in Celsius) if we consider it as a black-body in thermal equilibrium if Temperature of the Sun is 5700 K, Radius of the Sun is 7 × 10^5 km and distance between earth and sun is 150 × 10^6 km. (two decimal places only)​

Answer 1

Answer:

$\boxed{\mathfrak{Global \ temperature \ of \ our \ planet = 2.18 \degree C}}$

Given:

Temperature of sun ($\sf T_s$) = 5700 K

Radius of sun ($\sf R_s$) = $\sf 7 \times 10^5$ km

Distance between earth and sun (r) = $\sf 150 \times 10^6$ km

Explanation:

Power radiated by sun's surface:

$\rm P = \sigma A_s {T_s}^{4} \\ \\ \rm P = \sigma 4\pi{R_s}^{2}{T_s}^{4}$

$\rm \sigma \longrightarrow Stefan's \: constant$

Intensity at earth's surface:

$\rm I = \dfrac{P}{4\pi r^2} \\ \\ \rm I = \dfrac{ \sigma \cancel{4\pi}{R_s}^{2}{T_s}^{4}}{ \cancel{4\pi} {r}^{2} } \\ \\ \rm I =\dfrac{ \sigma {R_s}^{2}{T_s}^{4}}{ {r}^{2} }$

Power absorbed by earth:

$\rm P_{absorbed} = I \times \pi {R_e}^2 \\ \\ \rm P_{absorbed} = \frac{ \sigma \pi{R_s}^{2}{R_e}^2{T_s}^{4}}{ {r}^{2} }$

Power emitted by earth:

$\rm P_{emitted} = \sigma \times 4\pi {R_e}^{2} \times {T_e}^{4}$

$\rm R_e \longrightarrow Radius \ of \ earth$

$\rm T_e \longrightarrow Temperature \ of \ earth$

At Equilibrium,

$\rm \implies P_{absorbed} = P_{emitted} \\ \\ \rm \implies \frac{ \cancel{ \sigma} \cancel{\pi}{R_s}^{2} \cancel{{R_e}^2}{T_s}^{4}}{ {r}^{2} } = \cancel{ \sigma} \times 4 \cancel{\pi{R_e}^{2}} \times {T_e}^{4} \\ \\ \rm \implies \frac{{R_s}^2 {T_s}^4}{r^2} = 4 {T_e}^4 \\ \\ \rm \implies{T_e}^4 = \frac{{R_s}^2 {T_s}^4}{4r^2} \\ \\ \rm \implies T_e = {(\frac{{R_s}^2 {T_s}^4}{4r^2} )}^{ \frac{1}{4} } \\ \\ \rm \implies T_e = \frac{T_s \sqrt{R_s} }{ \sqrt{2r} } \\ \\ \rm \implies T_e = T_s \sqrt{ \dfrac{R_s}{2r} }$

By substituting value we get:

$\rm \implies T_e = 5700 \times \sqrt{ \dfrac{7 \times {10}^{5} }{2 \times 150 \times {10}^{6} } } \\ \\ \rm \implies T_e \approx 275.33 \: K \\ \\ \rm \implies T_e \approx 2.18 \degree C$