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Specifically in this case, how large would Jupiter be seen in the Io's sky in comparison to the moon in the Earth's sky? How to calculate this?

Also, if somebody knows any online software which lets you input moons and planets to see how their sky looks like to also fill in the answer, it's appreciated.

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    $\begingroup$ You can calculate the angle subtended by an object if you know the size of the distant object and the distance from the observer to the distant object. From there it's easy to make an illustration of what the number means in practice. Compare How large does a spacecraft need to be to be visible from the surface of the Earth at 400 km altitude?. $\endgroup$
    – user
    Apr 15, 2016 at 12:10
  • $\begingroup$ Doesnt the atmosphere composition and density also goes into account since in Earth its atmosphere works as a magnifying glass? $\endgroup$
    – Pablo
    Apr 15, 2016 at 12:14
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    $\begingroup$ @Pablo That's not really a significant effect, except perhaps in very unusual circumstances $\endgroup$
    – Mike Scott
    Apr 15, 2016 at 12:16
  • $\begingroup$ @Pablo - please provide a source for your belief that the earth's atmosphere acts as a magnifying glass. For all intents and purposes, it doesn't. $\endgroup$
    – WhatRoughBeast
    Apr 15, 2016 at 15:57
  • $\begingroup$ @whatroughbeast may be I'm missunderstanding, after reading your question I've found several sources aparently stating it does not , though check this link in the optical atmospheric difraction section, last paragraph " The effect is to make an object visibly larger while being more indistinct as the dust distorts the image." en.wikipedia.org/wiki/Atmospheric_diffraction $\endgroup$
    – Pablo
    Apr 15, 2016 at 17:43

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By a straightforward bit of trigonometry, if the distance from the observer to the (centre of) the planet is x km, then the radius, r km, of the planet subtends $\arcsin(r/x)$, and so the angular size of the planet is twice this: $2\arcsin(r/x)$

For Io x = 420,000 km, and the radius of Jupiter is r = 70,000 km, so the angular size is $2\arcsin(1/6)=19^\circ$. For comparison, the moon from the Earth has an angular size of about half a degree.

Stellarium can be set up to view from other worlds of the solar system

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