Eris or Sol, the Sun
[Authorial note: This is very much a work in progress. There is a lot of LaTeX code here. Details on the remainder of the things to be covered are at the bottom].
The Solar myth of the Republic:
Eris, the Sun, is the central figure in creation myths of the Republic. They believe that Eris was firstborn among all gods, coalescing and giving birth to himself amidst the swirling dust of the Empty. All light and life comes from Eris. By his decree was Mother Earth born, and of her the siblings Sky and Ocean.
[To be continued]
[Possibly add the Solar myth of the Dominion here]
The hard science:
First we need to consider what types of stars can give rise to Earth-like biologies on appropriate geologic timescales. Once that is accomplished, we will then look at what sorts of galaxies can give rise to these stars, and where in those galaxies they reside. This will inform the density of stars seen in the night sky, hence the pattern of constellations.
This task was commenced by downloading the star catalogue hygdata_v3.csv from http://www.astronexus.com/hyg with the fields described at https://github.com/astronexus/HYG-Database/blob/master/README.md
To obtain the temperature of each star from the colour index, the formula of Ballesteros was used as described (https://en.wikipedia.org/wiki/Color_index, Ballesteros, F. J. (2012). "New insights into black bodies". EPL 97 (2012) 34008. arXiv:1201.1809, https://arxiv.org/abs/1201.1809).
That is, the temperature in degrees Kelvin is given by:
T = 4600\left[ {\frac{1}0.92\left( {B - V} \right) + 1.7 + \frac{1}0.92\left( {B - V} \right) + 0.62} \right]
Assuming that a star behaves as a black-body radiator, the temperature then determines the emission spectrum as a function of wavelength, via Planck's Law:
B\left( {\lambda ,T} \right) = \frac2{c^2}h{\lambda ^5}\left[ {\exp \left( {\fracchkT\lambda } \right) - 1} \right]
To determine how much of a star's wavelength emission spectrum occurs within the range of visible light, necessary for life on earth, we need to integrate Planck's Law.
A\left( \lambda \right) = \int {\frac2{c^2}h{\lambda ^5}\left[ {\exp \left( {\fracchkT\lambda } \right) - 1} \right]d\lambda }
Multiply this integrand by
\frac\exp \left( { - \fracch{\lambda ^{ - 1kT} \right)\exp \left( { - \fracch{\lambda ^{ - 1kT} \right)
which results in
\[A\left( \lambda \right) = 2{c^2}h\int {\frac{\lambda ^{ - 5\exp \left( { - \fracch{\lambda ^{ - 1kT} \right)\left[ {1 - \exp \left( { - \fracch{\lambda ^{ - 1kT} \right)} \right]d\lambda } \]
Now observe that for \({r^2} < 1\), \(\frac{1}1 - r = \sum\limits_{n = 0}^\infty r^n \) and that in this case \(r = \exp \left( { - \fracch{\lambda ^{ - 1kT} \right)\)\\
Consequently \(A\left( \lambda \right) = 2{c^2}h\int \lambda ^{ - 5\sum\limits_{n = 1}^\infty {\exp \left( { - \fracch{\lambda ^{ - 1kTn} \right)} d\lambda } \)\\
Since both \(\lambda >0 \) and \(T > 0\) the integrand is continuous so we can interchange the order of integration and summation.
\[A\left( \lambda \right) = 2{c^2}h\sum\limits_{n = 1}^\infty {\int \lambda ^{ - 5\exp \left( { - \fracchnkT\lambda } \right)d\lambda } } \]
Now substitute \(u = \frac{1}{\lambda }\) so that \(\fracdud\lambda = - \frac{1}{\lambda ^2}\) and \(d\lambda = {-\lambda ^2}du\) \[A( u ) = - 2{c^2}h\sum\limits_{n = 1}^\infty {\int u^3}\exp \left( { - \fracchnukT} \right)du} } \]
Next substitute \(v = - \frac{u}kT\) so that \(\fracdvdu = - \frac{1}kT\) and \(du = - kTdv\)
\[A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\int v^3}\exp \left( {chnv} \right)dv} } \] Now apply the rule for integration by parts, \(\int {fg'} = fg - \int {f'g}\), letting \(f = {v^3}\) so that \(f' = 3{v^2}\) and letting \(g' = \exp \left( {chnv} \right)\) so that \(g = \frac\exp \left( {chnv} \right)chn\)
\[A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac{3}chn\int v^2}\exp \left( {chnv} \right)dv} } \right]} \]
Apply the rule for integration by parts, \(\int {fg'} = fg - \int {f'g}\), a second time. Let \(f = {v^2}\) so that \(f' = 2v\) and let \(g' = \exp \left( {chnv} \right)\) so that \(g = \frac\exp \left( {chnv} \right)chn\)
\[\begin{gathered} A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac{3}chn\left[ {\frac{v^2}\exp \left( {chnv} \right)chn - \frac{2}chn\int {v\exp \left( {chnv} \right)dv} } \right]} \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac{6}{c^2}{h^2}{n^2}\int {v\exp \left( {chnv} \right)dv} } \right]} \hfill \\ \end{gathered} \]
A third time integrate by parts so that \(\int {fg'} = fg - \int {f'g} \). Let \(f = v\) so that \(f' = 1\) and let \(g' = \exp \left( {chnv} \right)\) so that \(g = \frac\exp \left( {chnv} \right)chn\).
\[\begin{gathered} A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ \begin{gathered} \frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} \hfill \\ + \frac{6}{c^2}{h^2}{n^2}\left[ {\fracv\exp \left( {chnv} \right)chn - \frac{1}chn\int {\exp \left( {chnv} \right)dv} } \right] \hfill \\ \end{gathered} \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac6v\exp \left( {chnv} \rightc^3}{h^3}{n^3} - \frac{6}{c^3}{h^3}{n^3}\int {\exp \left( {chnv} \right)dv} } \right]} \hfill \\ \end{gathered} \]
Substitute \(w = chnv\) so that \(\fracdwdv = chn\) and \(dv = \frac{1}chndw\)
\[\begin{gathered} A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ \begin{gathered} \frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac6v\exp \left( {chnv} \rightc^3}{h^3}{n^3} \hfill \\ - \frac{6}{c^3}{h^3}{n^3}\left[ {\frac{1}chn\int {\exp \left( w \right)dw} } \right] \hfill \\ \end{gathered} \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac6v\exp \left( {chnv} \rightc^3}{h^3}{n^3} - \frac{6}{c^4}{h^4}{n^4}\int {\exp \left( w \right)dw} } \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac6v\exp \left( {chnv} \rightc^3}{h^3}{n^3} - \frac6\exp \left( {chnv} \rightc^4}{h^4}{n^4 \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\left( {\frac{v^3}chn - \frac3{v^2}{c^2}{h^2}{n^2} + \frac6v{c^3}{h^3}{n^3} - \frac{6}{c^4}{h^4}{n^4 \right)\exp \left( {chnv} \right)} \right]} \hfill \\ \end{gathered} \]
Proceding to undo the substitutions:
\[\begin{gathered} A\left( u \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\left( { - \frac{u^3}ch{k^3}{T^3}n - \frac3{u^2}{c^2}{h^2}{k^2}{T^2}{n^2} - \frac6u{c^3}{h^3}kT{n^3} - \frac{6}{c^4}{h^4}{n^4 \right)\exp \left( { - \fracchnukT} \right)} \right]} \hfill \\ = - 2{c^2}h\sum\limits_{n = 1}^\infty {\left[ {\left( { - \frackT{u^3}chn - \frac3{k^2}{T^2}{u^2}{c^2}{h^2}{n^2} - \frac6{k^3}{T^3}u{c^3}{h^3}{n^3} - \frac6{k^4}{T^4}{c^4}{h^4}{n^4 \right)\exp \left( { - \fracchnukT} \right)} \right]} \hfill \\ \end{gathered} \]
Which leads to the final result
\[\begin{gathered} A\left( {\lambda ,T} \right) = \sum\limits_{n = 1}^\infty {\left[ {\left( {2ck\frac{T}n{\lambda ^3} + \frac6{k^2}{h}\frac{T^2}{n^2}{\lambda ^2} + \frac12{k^3}c{h^2}\frac{T^3}{n^3}\lambda + \frac12{k^4}{c^2}{h^3}\frac{T^4}{n^4 \right)\exp \left( { - \fracch{k}\frac{n}T\lambda } \right)} \right]} \hfill \\ \end{gathered} \]
This can be re-expressed as a definite integral with various constants
\[\left. A_a}\left( {T,\lambda } \right)} \right|_{\lambda = {x_1^{\lambda = {x_2 = \sum\limits_{n = 1}^\infty {\left[ {\left( C_1}\frac{T}n{\lambda ^3} + {C_2}\frac{T^2}{n^2}{\lambda ^2} + {C_3}\frac{T^3}{n^3}\lambda + {C_4}\frac{T^4}{n^4 \right)\exp \left( C_5}\frac{n}T\lambda } \right)} \right]} _{\lambda = {x_1^{\lambda = {x_2\]
where:
\[\begin{gathered} {C_1} = ck = 4.1390815735 \times {\text{1\text{0^\text{ - 15 \hfill \\ {C_2} = \frac6{k^2}{h} = 1.7260834414 \times {\text{1\text{0^\text{ - 12 \hfill \\ {C_3} = \frac12{k^3}c{h^2} = 2.3993761206 \times {\text{1\text{0^\text{ - 10 \hfill \\ {C_4} = \frac12{k^4}{c^2}{h^3} = 1.6676499033 \times {\text{1\text{0^\text{ - 08 \hfill \\ {C_5} = - \fracch{k} = - 1.4387768775 \times {\text{1\text{0^\text{ - 02 \hfill \\ \end{gathered} \]
Obviously this needs significant additional work, not the least part being obtaining pictures with which to replace my LaTeX mathematical code.
Having written up my derivation, I then need to write Matlab (or other) code to apply this integral to generate wavelength emission spectra for every star in the catalogue.
Once the spectra are obtained, I can compare those of other stars to that of our own Sun, thus identifying what other stars might support life as we know it.
This is effectively generating a modifed Hertzprung-Russel diagram for just those stars of interest, including an additional axis for emission wavelength.
I may then look at mathematical modelling of how the Planck emission spectrum for this fictional Sun differs from the electromagnetic spectrum on the planet surface.
The Solar myth of the Republic:
Eris, the Sun, is the central figure in creation myths of the Republic. They believe that Eris was firstborn among all gods, coalescing and giving birth to himself amidst the swirling dust of the Empty. All light and life comes from Eris. By his decree was Mother Earth born, and of her the siblings Sky and Ocean.
[To be continued]
[Possibly add the Solar myth of the Dominion here]
The hard science:
First we need to consider what types of stars can give rise to Earth-like biologies on appropriate geologic timescales. Once that is accomplished, we will then look at what sorts of galaxies can give rise to these stars, and where in those galaxies they reside. This will inform the density of stars seen in the night sky, hence the pattern of constellations.
This task was commenced by downloading the star catalogue hygdata_v3.csv from http://www.astronexus.com/hyg with the fields described at https://github.com/astronexus/HYG-Database/blob/master/README.md
To obtain the temperature of each star from the colour index, the formula of Ballesteros was used as described (https://en.wikipedia.org/wiki/Color_index, Ballesteros, F. J. (2012). "New insights into black bodies". EPL 97 (2012) 34008. arXiv:1201.1809, https://arxiv.org/abs/1201.1809).
That is, the temperature in degrees Kelvin is given by:
T = 4600\left[ {\frac{1}0.92\left( {B - V} \right) + 1.7 + \frac{1}0.92\left( {B - V} \right) + 0.62} \right]
Assuming that a star behaves as a black-body radiator, the temperature then determines the emission spectrum as a function of wavelength, via Planck's Law:
B\left( {\lambda ,T} \right) = \frac2{c^2}h{\lambda ^5}\left[ {\exp \left( {\fracchkT\lambda } \right) - 1} \right]
To determine how much of a star's wavelength emission spectrum occurs within the range of visible light, necessary for life on earth, we need to integrate Planck's Law.
A\left( \lambda \right) = \int {\frac2{c^2}h{\lambda ^5}\left[ {\exp \left( {\fracchkT\lambda } \right) - 1} \right]d\lambda }
Multiply this integrand by
\frac\exp \left( { - \fracch{\lambda ^{ - 1kT} \right)\exp \left( { - \fracch{\lambda ^{ - 1kT} \right)
which results in
\[A\left( \lambda \right) = 2{c^2}h\int {\frac{\lambda ^{ - 5\exp \left( { - \fracch{\lambda ^{ - 1kT} \right)\left[ {1 - \exp \left( { - \fracch{\lambda ^{ - 1kT} \right)} \right]d\lambda } \]
Now observe that for \({r^2} < 1\), \(\frac{1}1 - r = \sum\limits_{n = 0}^\infty r^n \) and that in this case \(r = \exp \left( { - \fracch{\lambda ^{ - 1kT} \right)\)\\
Consequently \(A\left( \lambda \right) = 2{c^2}h\int \lambda ^{ - 5\sum\limits_{n = 1}^\infty {\exp \left( { - \fracch{\lambda ^{ - 1kTn} \right)} d\lambda } \)\\
Since both \(\lambda >0 \) and \(T > 0\) the integrand is continuous so we can interchange the order of integration and summation.
\[A\left( \lambda \right) = 2{c^2}h\sum\limits_{n = 1}^\infty {\int \lambda ^{ - 5\exp \left( { - \fracchnkT\lambda } \right)d\lambda } } \]
Now substitute \(u = \frac{1}{\lambda }\) so that \(\fracdud\lambda = - \frac{1}{\lambda ^2}\) and \(d\lambda = {-\lambda ^2}du\) \[A( u ) = - 2{c^2}h\sum\limits_{n = 1}^\infty {\int u^3}\exp \left( { - \fracchnukT} \right)du} } \]
Next substitute \(v = - \frac{u}kT\) so that \(\fracdvdu = - \frac{1}kT\) and \(du = - kTdv\)
\[A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\int v^3}\exp \left( {chnv} \right)dv} } \] Now apply the rule for integration by parts, \(\int {fg'} = fg - \int {f'g}\), letting \(f = {v^3}\) so that \(f' = 3{v^2}\) and letting \(g' = \exp \left( {chnv} \right)\) so that \(g = \frac\exp \left( {chnv} \right)chn\)
\[A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac{3}chn\int v^2}\exp \left( {chnv} \right)dv} } \right]} \]
Apply the rule for integration by parts, \(\int {fg'} = fg - \int {f'g}\), a second time. Let \(f = {v^2}\) so that \(f' = 2v\) and let \(g' = \exp \left( {chnv} \right)\) so that \(g = \frac\exp \left( {chnv} \right)chn\)
\[\begin{gathered} A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac{3}chn\left[ {\frac{v^2}\exp \left( {chnv} \right)chn - \frac{2}chn\int {v\exp \left( {chnv} \right)dv} } \right]} \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac{6}{c^2}{h^2}{n^2}\int {v\exp \left( {chnv} \right)dv} } \right]} \hfill \\ \end{gathered} \]
A third time integrate by parts so that \(\int {fg'} = fg - \int {f'g} \). Let \(f = v\) so that \(f' = 1\) and let \(g' = \exp \left( {chnv} \right)\) so that \(g = \frac\exp \left( {chnv} \right)chn\).
\[\begin{gathered} A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ \begin{gathered} \frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} \hfill \\ + \frac{6}{c^2}{h^2}{n^2}\left[ {\fracv\exp \left( {chnv} \right)chn - \frac{1}chn\int {\exp \left( {chnv} \right)dv} } \right] \hfill \\ \end{gathered} \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac6v\exp \left( {chnv} \rightc^3}{h^3}{n^3} - \frac{6}{c^3}{h^3}{n^3}\int {\exp \left( {chnv} \right)dv} } \right]} \hfill \\ \end{gathered} \]
Substitute \(w = chnv\) so that \(\fracdwdv = chn\) and \(dv = \frac{1}chndw\)
\[\begin{gathered} A\left( v \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ \begin{gathered} \frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac6v\exp \left( {chnv} \rightc^3}{h^3}{n^3} \hfill \\ - \frac{6}{c^3}{h^3}{n^3}\left[ {\frac{1}chn\int {\exp \left( w \right)dw} } \right] \hfill \\ \end{gathered} \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac6v\exp \left( {chnv} \rightc^3}{h^3}{n^3} - \frac{6}{c^4}{h^4}{n^4}\int {\exp \left( w \right)dw} } \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\frac{v^3}\exp \left( {chnv} \right)chn - \frac3{v^2}\exp \left( {chnv} \rightc^2}{h^2}{n^2} + \frac6v\exp \left( {chnv} \rightc^3}{h^3}{n^3} - \frac6\exp \left( {chnv} \rightc^4}{h^4}{n^4 \right]} \hfill \\ = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\left( {\frac{v^3}chn - \frac3{v^2}{c^2}{h^2}{n^2} + \frac6v{c^3}{h^3}{n^3} - \frac{6}{c^4}{h^4}{n^4 \right)\exp \left( {chnv} \right)} \right]} \hfill \\ \end{gathered} \]
Proceding to undo the substitutions:
\[\begin{gathered} A\left( u \right) = - 2{c^2}h{k^4}{T^4}\sum\limits_{n = 1}^\infty {\left[ {\left( { - \frac{u^3}ch{k^3}{T^3}n - \frac3{u^2}{c^2}{h^2}{k^2}{T^2}{n^2} - \frac6u{c^3}{h^3}kT{n^3} - \frac{6}{c^4}{h^4}{n^4 \right)\exp \left( { - \fracchnukT} \right)} \right]} \hfill \\ = - 2{c^2}h\sum\limits_{n = 1}^\infty {\left[ {\left( { - \frackT{u^3}chn - \frac3{k^2}{T^2}{u^2}{c^2}{h^2}{n^2} - \frac6{k^3}{T^3}u{c^3}{h^3}{n^3} - \frac6{k^4}{T^4}{c^4}{h^4}{n^4 \right)\exp \left( { - \fracchnukT} \right)} \right]} \hfill \\ \end{gathered} \]
Which leads to the final result
\[\begin{gathered} A\left( {\lambda ,T} \right) = \sum\limits_{n = 1}^\infty {\left[ {\left( {2ck\frac{T}n{\lambda ^3} + \frac6{k^2}{h}\frac{T^2}{n^2}{\lambda ^2} + \frac12{k^3}c{h^2}\frac{T^3}{n^3}\lambda + \frac12{k^4}{c^2}{h^3}\frac{T^4}{n^4 \right)\exp \left( { - \fracch{k}\frac{n}T\lambda } \right)} \right]} \hfill \\ \end{gathered} \]
This can be re-expressed as a definite integral with various constants
\[\left. A_a}\left( {T,\lambda } \right)} \right|_{\lambda = {x_1^{\lambda = {x_2 = \sum\limits_{n = 1}^\infty {\left[ {\left( C_1}\frac{T}n{\lambda ^3} + {C_2}\frac{T^2}{n^2}{\lambda ^2} + {C_3}\frac{T^3}{n^3}\lambda + {C_4}\frac{T^4}{n^4 \right)\exp \left( C_5}\frac{n}T\lambda } \right)} \right]} _{\lambda = {x_1^{\lambda = {x_2\]
where:
\[\begin{gathered} {C_1} = ck = 4.1390815735 \times {\text{1\text{0^\text{ - 15 \hfill \\ {C_2} = \frac6{k^2}{h} = 1.7260834414 \times {\text{1\text{0^\text{ - 12 \hfill \\ {C_3} = \frac12{k^3}c{h^2} = 2.3993761206 \times {\text{1\text{0^\text{ - 10 \hfill \\ {C_4} = \frac12{k^4}{c^2}{h^3} = 1.6676499033 \times {\text{1\text{0^\text{ - 08 \hfill \\ {C_5} = - \fracch{k} = - 1.4387768775 \times {\text{1\text{0^\text{ - 02 \hfill \\ \end{gathered} \]
Obviously this needs significant additional work, not the least part being obtaining pictures with which to replace my LaTeX mathematical code.
Having written up my derivation, I then need to write Matlab (or other) code to apply this integral to generate wavelength emission spectra for every star in the catalogue.
Once the spectra are obtained, I can compare those of other stars to that of our own Sun, thus identifying what other stars might support life as we know it.
This is effectively generating a modifed Hertzprung-Russel diagram for just those stars of interest, including an additional axis for emission wavelength.
I may then look at mathematical modelling of how the Planck emission spectrum for this fictional Sun differs from the electromagnetic spectrum on the planet surface.
Type
Star
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