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IOAA 2021, Теория — Задача 2. Temperature of the Earth (10 points).

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Всички материали от IOAA 2021

Theory — Q2-1 — Temperature of the Earth (10 points). · 10 т.

Standard Table of Constants — values used in this problem

Stefan-Boltzmann constant: $\sigma=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ .

Standard Table of Constants — values used in this problem

1 astronomical unit (au): $1\ au=1.496\times10^{11}m$ . Effective temperature of Sun: $5778K$ .

Standard Table of Constants — values used in this problem

Sun: mean radius $695\ 500\ km$ . Earth: mean radius $6\ 378\ km$ ; semi-major axis $1.000000\ au$ .

Условие

For at least the last few million years, the Earth has been in roughly thermal equilibrium with the radiation from the Sun at the Earth's orbital distance.

2.1 Assuming our planet to be an ideal black body, calculate what the Earth's equilibrium temperature (in Celsius) would be. [4 т.]

2.2 The Earth's albedo is approximately 30%. Calculate the Earth's surface temperature (in Celsius) considering its albedo. [2 т.]

2.3 The Earth's absorbed radiation is reemitted as black body radiation from its surface, but its atmosphere re-absorbs 58% of that energy, causing a greenhouse effect. Considering this effect, calculate the Earth's surface temperature (which will be the same as the temperature of the lower atmosphere). Give your answer in Celsius.
For simplicity, consider the reabsorption effect as happening only once, and do not consider the atmosphere as a separate black body. [4 т.]

Отговори

Решение

2.1 The solar radiation intensity ( $I_T$ ) received on Earth is:

I_T=\frac{P_s}{4\pi r_{s-t}^2}=\sigma T_s^4\cdot\left(\frac{R_s}{r_{s-t}}\right)^2

[1 point]

with $P_s=4\pi\sigma R_s^2T_s^4$ .

Also, Earth would absorb energy at this rate:

P_{abs}=I_T\pi R_t^2=\pi\sigma T_s^4\cdot\left(\frac{R_s\cdot R_t}{r_{s-t}}\right)^2

With $R_t=6.4\times10^6m$ as the radius of the planet “disk”. [1 point]

Then, by thermal equilibrium, the absorbed radiation would be radiated over the planet’s surface:

P_{abs}=P_{rad}

With

P_{rad}=4\pi\sigma R_t^2T_t^4

\pi\sigma T_s^4\cdot\left(\frac{R_s\cdot R_t}{r_{s-t}}\right)^2=4\pi\sigma R_t^2T_t^4

[1 point]

T_t=T_s\cdot\left(\frac{R_s}{2r_{s-t}}\right)^{\frac12}=278.58\ K=5.43\ ^\circ C

[1 point]

This would be very cold but still viable to harbor life.

2.2 The Earth’s absorbed radiation, considering the albedo, is:

P'_{abs}=0.7\cdot\pi\sigma T_s^4\cdot\left(\frac{R_s\cdot R_t}{r_{s-t}}\right)^2

[1 point]

T'_T=T_s\ (0.7)^{\frac14}\cdot\left(\frac{R_s}{2r_{s-t}}\right)^{\frac12}=254.81\ K=-18.34\ ^\circ C

[1 point]

2.3 If Earth reabsorbs 58% of the 70% re emitted energy, then:

T''_T=T_s\ [0.7+(0.58\cdot0.7)]^{\frac14}\cdot\left(\frac{R_s}{2r_{s-t}}\right)^{\frac12}=285.68\ K=12.53\ ^\circ C

[4 points]

Оригинал в Архива: TQ-2-Q.pdf · официални решения: TQ2.pdf · допълнителни данни: TQ-2-A.pdf · допълнителни данни: Standard-Table-of-constants.docx.pdf · допълнителни данни: TQ-G.pdf · допълнителни данни: data.pdf