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Dust along the ecliptic scatters light, causing “light pollution” throughout the orbital plane of the solar system. If an object like, say, the sun's corona, had its brightness measured from above the ecliptic, would the measurement reflect that?

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  • $\begingroup$ Just to be clear, are you assuming the dust lies primarily only in the ecliptic? $\endgroup$
    – Dave
    Oct 29 '18 at 20:09
  • $\begingroup$ no assumptions, data, the orbital plane has lots of dust, see zodiacal lights etc $\endgroup$
    – user24677
    Oct 29 '18 at 20:15
  • $\begingroup$ Unclear what you mean by "would the measurement reflect that". $\endgroup$
    – ProfRob
    Oct 29 '18 at 21:02
  • $\begingroup$ had its brightness measured from above the ecliptic - in contrast to what? $\endgroup$
    – user1569
    Oct 29 '18 at 22:05
  • $\begingroup$ just in contrast to ambient light, like always when measuring brightness. re: would measurement reflect that, similar to measuring brightness of stars in day time or night time $\endgroup$
    – user24677
    Oct 29 '18 at 23:20
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The answer is yes: dust is highly constrained to the ecliptic plane and so viewing from above the plane would result in less light that needs to be subtracted from observations of the solar corona. The zodiacal light is quite strong in the direction of the sun because the dust strongly scatters in the forward direction. Near the Sun, its surface brightness is somewhere around 21 - 23 mag/arcsec$^2$ from our viewpoint in the ecliptic, and this would be an additive contamination to any measurement of the Sun's corona. From the ecliptic pole, this light foreground would be much less. Of course, a careful observer would know to also measure the zodiacal light contribution and to subtract it off, so both observers would get the same brightness for the corona roughly. The observer at the pole would have lower noise for a comparable instrument and observing time.

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  • $\begingroup$ Zodiacal light can also be subtracted because it has different polarisation properties to the Thomson scattering in the Sun's k-corona. It is also not time-varying like the Sun's corona. $\endgroup$
    – ProfRob
    Oct 30 '18 at 18:05
  • $\begingroup$ @RobJeffries your proposition contradicts the answer. since collaborative Q&As are for selecting the accurate answer, if you have propositions that falsify this one, could you contribute an answer, so that the selection process can do its magic? $\endgroup$
    – user24677
    Nov 1 '18 at 22:40
  • $\begingroup$ @juli I have upvoted this answer. It seems a perfectly reasonable and clear set of statements. I haven't a clue whether it answers you question, but the answerer is blameless in that regard $\endgroup$
    – ProfRob
    Nov 1 '18 at 23:56

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