## Related questions with answers

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 $5.00 \times 10^{24}$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?

Solution

VerifiedSteady state situation means that rate of energy received (power) $P_{\text{in}}$ is the same as the planet emits $P_{\text{out}}$, i.e.

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

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

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

where $A$ 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

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

where $\sigma = 5.67 \cdot 10^{-8} \: \mathrm{W m^{-2} K^{-4}}$, $T$ is temperature of the planet and $A*$ 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

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

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

$\begin{align*} e \frac{ P }{ 4 \pi R^2 } \pi r^2 &= e \sigma T^4 4 \pi r^2 /: ( e \pi r^2 )\\ \frac{ P }{ 4 \pi R^2 } &=4 \sigma T^4 \\ \frac{ 4 \pi R^2 }{ P } &= \frac{ 1 }{4\sigma T^4 } \\ R^2 &= \frac{ P }{16\pi \sigma T^4 } \\ R^2 &= \frac{ 5 \cdot 10^{24} \: \mathrm{W} }{ 16 \cdot \pi \cdot 5.67 \cdot 10^{-8} \: \mathrm{W m^{-2} K^{-4}} \cdot ( 287 \: \mathrm{K} )^4 } / \sqrt{} \\ R &= \boxed{ 1.61 \cdot 10^{10} \: \mathrm{m} }. \tag{5} \end{align*}$

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