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Gliese 581c is the first Earth-like extrasolar terrestrial planet discovered. Its parent star, Gliese 581, is a red dwarf that radiates electromagnetic waves with power 5.00×10245.00 \times 10^{24}W, which is only 1.30% of the power of the Sun. Assume the emissivity of the planet is equal for infrared and for visible light and the planet has a uniform surface temperature. If an average temperature of 287 K is necessary for life to exist on Gliese 581c, what should the radius of the planet’s orbit be?

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Answered 6 months ago
Answered 6 months ago
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Steady state situation means that rate of energy received (power) PinP_{\text{in}} is the same as the planet emits PoutP_{\text{out}}, i.e.

Pin=Pout.\begin{align*} P_{\text{in}} &= P_{\text{out}}. \tag{1} \end{align*}

We can also keep in mind that power PP and intensity II are connected as P=IAP = I A, where A is area that receives radiation. Therefore, we can write power received as

Pin=eIA,\begin{align*} P_{\text{in}} &= e I A, \tag{2} \end{align*}

where AA is effective area of the planet (which is calculated in part a). Now we have to find power that planet emits, for which we can assume that radiates as black-body

Pout=eσT4A,\begin{align*} P_{\text{out}} = e \sigma T^4 A*, \tag{3} \end{align*}

where σ=5.67108Wm2K4\sigma = 5.67 \cdot 10^{-8} \: \mathrm{W m^{-2} K^{-4}}, TT is temperature of the planet and AA* is surface area of the planet (calculated in b). Problem here is that we don't know the intensity of the sun, but we know it's power. Then we can write

I=P4πR2.\begin{align*} I &= \frac{ P }{ 4 \pi R^2 }. \tag{4} \end{align*}

Then we can combine result (2), (3) and (4) into (1) and use numbers form the problem to calculate the distance between Gliese 581c and Gliese 581 as

eP4πR2πr2=eσT44πr2/:(eπr2)P4πR2=4σT44πR2P=14σT4R2=P16πσT4R2=51024W16π5.67108Wm2K4(287K)4/R=1.611010m.\begin{align*} e \frac{ P }{ 4 \pi R^2 } \pi r^2 &= e \sigma T^4 4 \pi r^2 /: ( e \pi r^2 )\\ \frac{ P }{ 4 \pi R^2 } &=4 \sigma T^4 \\ \frac{ 4 \pi R^2 }{ P } &= \frac{ 1 }{4\sigma T^4 } \\ R^2 &= \frac{ P }{16\pi \sigma T^4 } \\ R^2 &= \frac{ 5 \cdot 10^{24} \: \mathrm{W} }{ 16 \cdot \pi \cdot 5.67 \cdot 10^{-8} \: \mathrm{W m^{-2} K^{-4}} \cdot ( 287 \: \mathrm{K} )^4 } / \sqrt{} \\ R &= \boxed{ 1.61 \cdot 10^{10} \: \mathrm{m} }. \tag{5} \end{align*}

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