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As far as I am aware, the latest spectral types that have been assigned are around Y2, for objects like WISE 0855-0714 that have temperatures around 250 K or so. I've also seen several directly-imaged exoplanets have had their spectra classified as L- or T-type (e.g. Bonnefoy et al. 2016 who find that the giant planets HR 8799 d and e are good matches for L6–L8 dwarfs).

Jupiter is colder and less massive than the observed Y dwarfs and differs from isolated brown dwarfs because it is being illuminated by the Sun, but there is thermal emission which could be observed from the nightside of the planet. Has Jupiter's nightside spectrum (i.e. the spectrum of Jupiter excluding contributions from reflected sunlight) been observed in sufficient detail to compare it to the spectra of Y dwarfs, and if so does it fit the trends observed in spectral type Y or is it so different that it cannot be spectrally classified as a Y dwarf?

For clarity: I am asking purely in terms of spectral classification, i.e. the classification of the spectrum. I am not asking about whether Jupiter is a brown dwarf making the Sun+Jupiter a binary system. That has been asked elsewhere.

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This Wikipedia page lists Jupiter as a disputed sub-brown dwarf because its mass reportedly is sufficient to qualify for one. Distance to another star like the Sun doesn't matter, nor does the visibility in the infrared spectrum either since every planet is emitting some infrared light. The fact that Jupiter emanates more energy than it gets from the Sun can't be accounted for either, since that's also a matter of distance from the Sun, and Saturn and Neptune do so too. But, as the article states, Jupiter obviously has enough mass to possibly count as a sub-brown dwarf. Sub-brown dwarfs are still planets because they don't undergo nuclear fusion unlike brown dwarfs and hydrogen-fusing stars and never did.

Edit: There is no reason to only classify the night side just because the day side is illuminated by a(nother) star. Spectral classifications are given to the entire object. Since sub-brown dwarfs reportedly have a mass of at least one Jupiter mass, you may want to classify Jupiter a Y9V sub-brown dwarf.

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    $\begingroup$ This doesn't really answer the question, which is about the classification of Jupiter's spectrum. It isn't about whether or not Jupiter qualifies as a brown dwarf. $\endgroup$ – user24157 Jul 24 '20 at 15:51
  • $\begingroup$ @antispinwards We don't know where a lower limit of Y-type objects might be set and therefore this is hard to answer with a number. Reportedly, sub-brown dwarfs have a mass of at least a Jupiter mass, so Jupiter would qualify as a sub-brown dwarf. In my opinion, the entire Y-type sub-brown dwarf class is unnecessary since they're actually planets (gas giants) if they don't undergo nuclear fusion. $\endgroup$ – Ioannes Jul 24 '20 at 16:20
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    $\begingroup$ Once again, this is not about whether or not Jupiter is a (sub-)brown dwarf. It's about its spectrum and how it compares to objects that have been classified as spectral type Y. $\endgroup$ – user24157 Jul 24 '20 at 16:44
  • $\begingroup$ @antispinwards You already stated "Jupiter is colder and less massive than the observed Y dwarfs". So the spectrum is obviously less intensive than that of the observed ones. When looked from outer space on the objects, all of them have a pitch dark night side or are entirely like this when not illuminated by a star. No visible light is emitted on the exterior. $\endgroup$ – Ioannes Jul 24 '20 at 17:10
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    $\begingroup$ No, I'm specifically asking whether it fits into spectral class Y, which is what is in the question title and the body of the question. If I were asking about whether Jupiter is a brown dwarf or a planet (which is another question entirely), I would have asked that. Spectral classification is the classification of spectra. It is not the classification of the objects that produce those spectra. To take a different example, some late M-type dwarfs are hydrogen-burning stars, some are young brown dwarfs, but the spectra are classified as type M regardless. $\endgroup$ – user24157 Jul 24 '20 at 18:21

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