Съдържание

IOAA 2007, Теория — Задача 2. A PLANET & ITS SURFACE TEMPERATURE

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2007

Theoretical Competition — 1st IOAA — December 5, 2007 · 5 декември 2007 г. · 10 т.

Условие

A fast rotating planet of radius $R$ with surface albedo $\alpha$ is orbiting a star of luminosity $L$ . The orbital radius is $D$ . It is assumed here that, at equilibrium, all of the energy absorbed by the planet is re-emitted as a blackbody radiation.

a.) What is the radiation flux from the star at the planet’s surface? [1,5 т.]

b.) What is the total rate of energy absorbed by the planet? [1,5 т.]

c.) What is the reflected luminosity of the planet? [2 т.]

d.) What is the average blackbody temperature of the planet’s surface? [2 т.]

e.) If we were to assume that one side of the planet is always facing the star, what would be the average surface temperature of that side? [2 т.]

f.) For the planet in problem d:

$\alpha = 0.25$ ,

$D = 1.523\ \mathrm{A.U.}$ ,

calculate its surface temperature in kelvins for the value of

$L = 3.826 \times 10^{26}\ \mathrm{W}$ . [1 т.]

Отговори

Решение

QUESTION 2 A PLANET & ITS SURFACE TEMPERATURE

SOLUTION

a.) Intensity

I = \frac{L}{4\pi D^{2}}

(1 point)

b.) Absorption rate

\mathcal{A} = \left(1-\alpha\right)\pi R^{2}I = \left(1-\alpha\right)\frac{LR^{2}}{4D^{2}}

(1 point)

c.) Light energy reflected by the planet per unit time is $\alpha\pi R^{2}I = \dfrac{\alpha LR^{2}}{4D^{2}}$ (1 point)

Hence the planet’s luminosity is

L_{\text{planet}} = \frac{\alpha LR^{2}}{4D^{2}}

(1 point)

d.) Here, we will neglect the planet’s internal source of energy.

Let $T$ be the black-body temperature of the planet’s surface in kelvins.

Since the planet is rotating fast, we may assume that its surface is being heated up uniformly to approximately the same temperature $T$ .

The total amount of black-body radiation emitted by the planet’s surface is from Stefan-Boltzmann law: $4\pi R^{2} \cdot \sigma T^{4}$ , $\sigma$ being Stefan-Boltzmann constant.

At equilibrium, that is when the temperature remains steady, this emission rate must be equal to the absorption rate in b.). (1 point)

Hence

4\pi R^{2} \cdot \sigma T^{4} = \left(1-\alpha\right)\frac{LR^{2}}{4D^{2}}

T = \left[\left(1-\alpha\right)\frac{L}{16\pi\sigma D^{2}}\right]^{\frac{1}{4}}

(1 point)

e.) In this case the emitted black-body radiation is mostly from the planet’s surface facing the star. The emitting surface area is now only $2\pi R^{2}$ and not $4\pi R^{2}$ . Hence the surface temperature is given by $T'$ , where

2\pi R^{2} \cdot \sigma\left(T'\right)^{4} = \left(1-\alpha\right)\frac{LR^{2}}{4D^{2}}

(1 point)

T' = \left[\left(1-\alpha\right)\frac{L}{8\pi\sigma D^{2}}\right]^{\frac{1}{4}} = \left(2\right)^{\frac{1}{4}} \cdot T \approx \left(1.19\right)T

(1 point)

f.)

T = \left[\left(1-\alpha\right)\frac{L}{16\pi\sigma D^{2}}\right]^{\frac{1}{4}}

T = \left[\left(1-0.25\right) \times \frac{3.826 \times 10^{26}}{16\pi \times 5.67 \times 10^{-8} \times \left(1.523 \times 1.496 \times 10^{11}\right)^{2}}\right]^{\frac{1}{4}}

= 209.8 \simeq 210\ \mathrm{K} = \text{-}63^{\circ}\mathrm{C}

(2 points)

Оригинал в Архива: exam_theory.pdf · официални решения: exam_theory_solution.pdf