What is the minimum size for the Sun?

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How small could our Sun be and still "burn" with nuclear fusion and emit the same spectrum of light and other radiation as the real Sun does?



Edit:



The goal is to have a small sun inside a huge vessel, such as an O'Neill cylinder.










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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.









  • 3




    Does it need to be a star or can it be a fusion generator powering a light?
    – Nick T
    Nov 20 at 21:58






  • 1




    @NickT This question is about a star. But the answers suggest that I will need another source of "sun light", which I will ask about in a follow up question.
    – user57423
    Nov 21 at 6:30










  • Well, I'm starting to question whether this belongs here; actually it's hypothetical but probably better suits on science.stackexchange.com ?
    – LMD
    Nov 21 at 13:52











  • I think that your requirement of "same wavelengths" is going to limit you a lot. I don't even understand exactly what you mean, but I have a feeling that you're not so sure either. Everything will look white to a human anyway, after a pretty short time.
    – pipe
    Nov 21 at 15:18






  • 1




    @user57423 Oh! Quick response! Yes, Earth temperature isn't uniform which is why I worded my comment the way I did: the energy differences that produce our temperature differences are relatively minute. If, for example, you shrink our sun to the size of earth, the goldilocks zone would shrink down to a few meters, maybe less. Areas closer might melt the ship, areas further away might get Pluto-like conditions. This is a direct consequence of the "Inverse-square law" (see wikipedia) and limit how evenly things are irradiated. So a long, thin sun or spherical ship might be your best options.
    – stux
    Nov 22 at 17:57














up vote
6
down vote

favorite












How small could our Sun be and still "burn" with nuclear fusion and emit the same spectrum of light and other radiation as the real Sun does?



Edit:



The goal is to have a small sun inside a huge vessel, such as an O'Neill cylinder.










share|improve this question









New contributor




user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.









  • 3




    Does it need to be a star or can it be a fusion generator powering a light?
    – Nick T
    Nov 20 at 21:58






  • 1




    @NickT This question is about a star. But the answers suggest that I will need another source of "sun light", which I will ask about in a follow up question.
    – user57423
    Nov 21 at 6:30










  • Well, I'm starting to question whether this belongs here; actually it's hypothetical but probably better suits on science.stackexchange.com ?
    – LMD
    Nov 21 at 13:52











  • I think that your requirement of "same wavelengths" is going to limit you a lot. I don't even understand exactly what you mean, but I have a feeling that you're not so sure either. Everything will look white to a human anyway, after a pretty short time.
    – pipe
    Nov 21 at 15:18






  • 1




    @user57423 Oh! Quick response! Yes, Earth temperature isn't uniform which is why I worded my comment the way I did: the energy differences that produce our temperature differences are relatively minute. If, for example, you shrink our sun to the size of earth, the goldilocks zone would shrink down to a few meters, maybe less. Areas closer might melt the ship, areas further away might get Pluto-like conditions. This is a direct consequence of the "Inverse-square law" (see wikipedia) and limit how evenly things are irradiated. So a long, thin sun or spherical ship might be your best options.
    – stux
    Nov 22 at 17:57












up vote
6
down vote

favorite









up vote
6
down vote

favorite











How small could our Sun be and still "burn" with nuclear fusion and emit the same spectrum of light and other radiation as the real Sun does?



Edit:



The goal is to have a small sun inside a huge vessel, such as an O'Neill cylinder.










share|improve this question









New contributor




user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











How small could our Sun be and still "burn" with nuclear fusion and emit the same spectrum of light and other radiation as the real Sun does?



Edit:



The goal is to have a small sun inside a huge vessel, such as an O'Neill cylinder.







hard-science stars astrophysics






share|improve this question









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user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited Nov 21 at 16:15





















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asked Nov 20 at 20:15









user57423

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user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.




This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.








  • 3




    Does it need to be a star or can it be a fusion generator powering a light?
    – Nick T
    Nov 20 at 21:58






  • 1




    @NickT This question is about a star. But the answers suggest that I will need another source of "sun light", which I will ask about in a follow up question.
    – user57423
    Nov 21 at 6:30










  • Well, I'm starting to question whether this belongs here; actually it's hypothetical but probably better suits on science.stackexchange.com ?
    – LMD
    Nov 21 at 13:52











  • I think that your requirement of "same wavelengths" is going to limit you a lot. I don't even understand exactly what you mean, but I have a feeling that you're not so sure either. Everything will look white to a human anyway, after a pretty short time.
    – pipe
    Nov 21 at 15:18






  • 1




    @user57423 Oh! Quick response! Yes, Earth temperature isn't uniform which is why I worded my comment the way I did: the energy differences that produce our temperature differences are relatively minute. If, for example, you shrink our sun to the size of earth, the goldilocks zone would shrink down to a few meters, maybe less. Areas closer might melt the ship, areas further away might get Pluto-like conditions. This is a direct consequence of the "Inverse-square law" (see wikipedia) and limit how evenly things are irradiated. So a long, thin sun or spherical ship might be your best options.
    – stux
    Nov 22 at 17:57












  • 3




    Does it need to be a star or can it be a fusion generator powering a light?
    – Nick T
    Nov 20 at 21:58






  • 1




    @NickT This question is about a star. But the answers suggest that I will need another source of "sun light", which I will ask about in a follow up question.
    – user57423
    Nov 21 at 6:30










  • Well, I'm starting to question whether this belongs here; actually it's hypothetical but probably better suits on science.stackexchange.com ?
    – LMD
    Nov 21 at 13:52











  • I think that your requirement of "same wavelengths" is going to limit you a lot. I don't even understand exactly what you mean, but I have a feeling that you're not so sure either. Everything will look white to a human anyway, after a pretty short time.
    – pipe
    Nov 21 at 15:18






  • 1




    @user57423 Oh! Quick response! Yes, Earth temperature isn't uniform which is why I worded my comment the way I did: the energy differences that produce our temperature differences are relatively minute. If, for example, you shrink our sun to the size of earth, the goldilocks zone would shrink down to a few meters, maybe less. Areas closer might melt the ship, areas further away might get Pluto-like conditions. This is a direct consequence of the "Inverse-square law" (see wikipedia) and limit how evenly things are irradiated. So a long, thin sun or spherical ship might be your best options.
    – stux
    Nov 22 at 17:57







3




3




Does it need to be a star or can it be a fusion generator powering a light?
– Nick T
Nov 20 at 21:58




Does it need to be a star or can it be a fusion generator powering a light?
– Nick T
Nov 20 at 21:58




1




1




@NickT This question is about a star. But the answers suggest that I will need another source of "sun light", which I will ask about in a follow up question.
– user57423
Nov 21 at 6:30




@NickT This question is about a star. But the answers suggest that I will need another source of "sun light", which I will ask about in a follow up question.
– user57423
Nov 21 at 6:30












Well, I'm starting to question whether this belongs here; actually it's hypothetical but probably better suits on science.stackexchange.com ?
– LMD
Nov 21 at 13:52





Well, I'm starting to question whether this belongs here; actually it's hypothetical but probably better suits on science.stackexchange.com ?
– LMD
Nov 21 at 13:52













I think that your requirement of "same wavelengths" is going to limit you a lot. I don't even understand exactly what you mean, but I have a feeling that you're not so sure either. Everything will look white to a human anyway, after a pretty short time.
– pipe
Nov 21 at 15:18




I think that your requirement of "same wavelengths" is going to limit you a lot. I don't even understand exactly what you mean, but I have a feeling that you're not so sure either. Everything will look white to a human anyway, after a pretty short time.
– pipe
Nov 21 at 15:18




1




1




@user57423 Oh! Quick response! Yes, Earth temperature isn't uniform which is why I worded my comment the way I did: the energy differences that produce our temperature differences are relatively minute. If, for example, you shrink our sun to the size of earth, the goldilocks zone would shrink down to a few meters, maybe less. Areas closer might melt the ship, areas further away might get Pluto-like conditions. This is a direct consequence of the "Inverse-square law" (see wikipedia) and limit how evenly things are irradiated. So a long, thin sun or spherical ship might be your best options.
– stux
Nov 22 at 17:57




@user57423 Oh! Quick response! Yes, Earth temperature isn't uniform which is why I worded my comment the way I did: the energy differences that produce our temperature differences are relatively minute. If, for example, you shrink our sun to the size of earth, the goldilocks zone would shrink down to a few meters, maybe less. Areas closer might melt the ship, areas further away might get Pluto-like conditions. This is a direct consequence of the "Inverse-square law" (see wikipedia) and limit how evenly things are irradiated. So a long, thin sun or spherical ship might be your best options.
– stux
Nov 22 at 17:57










2 Answers
2






active

oldest

votes

















up vote
20
down vote



accepted










Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^4/7,quad Tpropto M^4/7$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:




  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_odot$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_odot$ (Boyaijan et al. 2008).

Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.






share|improve this answer






















  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    Nov 20 at 21:34










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    Nov 20 at 22:09











  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    Nov 20 at 22:55










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    Nov 20 at 22:57










  • @Mark A natural brown dwarf, probably not; something constructed on purpose? Perhaps. You'd have to do the math on the equilibrium conditions. Getting the resources to build something like this is a bit different to the scale of our constructions, of course.
    – Luaan
    Nov 21 at 11:42

















up vote
13
down vote













The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation






share|improve this answer




















  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    Nov 20 at 20:47










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    Nov 20 at 20:49










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    Nov 20 at 20:56






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    Nov 20 at 21:23






  • 2




    It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    Nov 21 at 0:12










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
20
down vote



accepted










Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^4/7,quad Tpropto M^4/7$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:




  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_odot$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_odot$ (Boyaijan et al. 2008).

Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.






share|improve this answer






















  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    Nov 20 at 21:34










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    Nov 20 at 22:09











  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    Nov 20 at 22:55










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    Nov 20 at 22:57










  • @Mark A natural brown dwarf, probably not; something constructed on purpose? Perhaps. You'd have to do the math on the equilibrium conditions. Getting the resources to build something like this is a bit different to the scale of our constructions, of course.
    – Luaan
    Nov 21 at 11:42














up vote
20
down vote



accepted










Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^4/7,quad Tpropto M^4/7$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:




  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_odot$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_odot$ (Boyaijan et al. 2008).

Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.






share|improve this answer






















  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    Nov 20 at 21:34










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    Nov 20 at 22:09











  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    Nov 20 at 22:55










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    Nov 20 at 22:57










  • @Mark A natural brown dwarf, probably not; something constructed on purpose? Perhaps. You'd have to do the math on the equilibrium conditions. Getting the resources to build something like this is a bit different to the scale of our constructions, of course.
    – Luaan
    Nov 21 at 11:42












up vote
20
down vote



accepted







up vote
20
down vote



accepted






Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^4/7,quad Tpropto M^4/7$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:




  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_odot$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_odot$ (Boyaijan et al. 2008).

Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.






share|improve this answer














Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^4/7,quad Tpropto M^4/7$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:




  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_odot$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_odot$ (Boyaijan et al. 2008).

Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 21 at 1:24

























answered Nov 20 at 20:26









HDE 226868

62k12215399




62k12215399











  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    Nov 20 at 21:34










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    Nov 20 at 22:09











  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    Nov 20 at 22:55










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    Nov 20 at 22:57










  • @Mark A natural brown dwarf, probably not; something constructed on purpose? Perhaps. You'd have to do the math on the equilibrium conditions. Getting the resources to build something like this is a bit different to the scale of our constructions, of course.
    – Luaan
    Nov 21 at 11:42
















  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    Nov 20 at 21:34










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    Nov 20 at 22:09











  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    Nov 20 at 22:55










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    Nov 20 at 22:57










  • @Mark A natural brown dwarf, probably not; something constructed on purpose? Perhaps. You'd have to do the math on the equilibrium conditions. Getting the resources to build something like this is a bit different to the scale of our constructions, of course.
    – Luaan
    Nov 21 at 11:42















Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
– Mark
Nov 20 at 21:34




Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
– Mark
Nov 20 at 21:34












so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
– theRiley
Nov 20 at 22:09





so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
– theRiley
Nov 20 at 22:09













@Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
– HDE 226868
Nov 20 at 22:55




@Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
– HDE 226868
Nov 20 at 22:55












@theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
– HDE 226868
Nov 20 at 22:57




@theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
– HDE 226868
Nov 20 at 22:57












@Mark A natural brown dwarf, probably not; something constructed on purpose? Perhaps. You'd have to do the math on the equilibrium conditions. Getting the resources to build something like this is a bit different to the scale of our constructions, of course.
– Luaan
Nov 21 at 11:42




@Mark A natural brown dwarf, probably not; something constructed on purpose? Perhaps. You'd have to do the math on the equilibrium conditions. Getting the resources to build something like this is a bit different to the scale of our constructions, of course.
– Luaan
Nov 21 at 11:42










up vote
13
down vote













The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation






share|improve this answer




















  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    Nov 20 at 20:47










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    Nov 20 at 20:49










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    Nov 20 at 20:56






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    Nov 20 at 21:23






  • 2




    It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    Nov 21 at 0:12














up vote
13
down vote













The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation






share|improve this answer




















  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    Nov 20 at 20:47










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    Nov 20 at 20:49










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    Nov 20 at 20:56






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    Nov 20 at 21:23






  • 2




    It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    Nov 21 at 0:12












up vote
13
down vote










up vote
13
down vote









The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation






share|improve this answer












The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 20 at 20:21









Arkenstein XII

1,630219




1,630219











  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    Nov 20 at 20:47










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    Nov 20 at 20:49










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    Nov 20 at 20:56






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    Nov 20 at 21:23






  • 2




    It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    Nov 21 at 0:12
















  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    Nov 20 at 20:47










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    Nov 20 at 20:49










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    Nov 20 at 20:56






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    Nov 20 at 21:23






  • 2




    It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    Nov 21 at 0:12















This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
– HDE 226868
Nov 20 at 20:47




This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
– HDE 226868
Nov 20 at 20:47












@HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
– user57423
Nov 20 at 20:49




@HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
– user57423
Nov 20 at 20:49












@user57423 Looks like I misread what you were saying; my bad. Thanks!
– HDE 226868
Nov 20 at 20:56




@user57423 Looks like I misread what you were saying; my bad. Thanks!
– HDE 226868
Nov 20 at 20:56




1




1




You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
– Artemijs Danilovs
Nov 20 at 21:23




You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
– Artemijs Danilovs
Nov 20 at 21:23




2




2




It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
– SJuan76
Nov 21 at 0:12




It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
– SJuan76
Nov 21 at 0:12










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