How Big Would Jupiter Look From Europa?

I was talking with Tanya the other day and the conversation, as it does, steered me towards eventually saying "imagine a Jupiter-rise on Europa." Tanya chipped in that moon-rises would be pretty snazzy to watch too, and it got us thinking: just how snazzy would Jupiter- and moon-rises be if we were to set up camp on Europa?

I did what any self-respecting geek would do in such a situation: some maths.

I'll stick to the results in this post, but if you want the specifics of the maths (it's accessible to anyone with so much as a GCSE* in the subject), you can check that out at the sister-post over at MathsQS.

The first thing we need to sort out is "how can we talk about how big something looks in the sky?" The answer to this is fairly easy: we imagine standing still and drawing a line from our eye to one side of the object, and another from our eye to the opposite side of the object in the sky. We measure the angle between these two lines, and quote this as its 'angular size'. For a lot of objects we might want to look at, this value is pretty small and we use different units than we're probably used to to talk about those. For this question, though, our old faithful 'degrees' are fine.

For an example and for use as a comparison, when you look at our Moon in the sky above Earth its angular size is about half (0.5) a degree.

Now let's set up camp on Europa. 5 AU from the Sun orbits giant Jupiter, with a whole host of satellites in its family. The largest of these, in order of distance from Jupiter, are Io, Europa, Ganymede and Callisto**. Europa's an ice moon, and it came up in conversation as it's one of my favourite moons, being one of the solar system's most promising places for finding life beyond Earth.

We don't need to go there to find out how big Io, Ganymede and Jupiter would look in its sky. Using some GCSE level maths we can work out their angular sizes as long as we know two things: the radius of what we're looking at and how far away it is. With Jupiter that second value isn't too hard to consider- we just pretend Europa's orbit's a perfect circle and we've simplified the model so we can deal with it. With Io and Ganymede, however, it's a different story: all three bodies are orbiting Europa with different periods, so their mutual distances vary wildly. I decided to look at the points at which they're closest to each other, as that, in my opinion, would be the most exciting point. Take a look at my post on MathsQS if you want the nitty-gritty, but to cut a medium-length story short we come out with the following values:

With a radius of 2631.2 km, and being around 399,300 km away at closest approach***, Ganymede would appear, to us at our campsite on Europa, to have an angular size of 0.755 degrees. That's a bit more than 3/4 of a degree, compared to our Moon's 1/2 a degree: it'd look roughly one and a half times as big as our Moon does.

Slightly smaller than Ganymede, with a radius of 1830 km, but also slightly closer, at 249,300 km****, Io would appear to take up 0.841 degrees of the sky, so it'd appear a bit bigger than Ganymede.

At 671,100 km away, but measuring a whopping 61,911 km in radius, Jupiter would dominate the sky, taking up nearly 12 degrees of the sky when full. That's almost twenty-four times larger than the Moon appears in our sky!

As an extra thing to think about, we also discussed the fact that the sizes of Io and Ganymede as viewed from our base camp on Europa would cycle as our orbits took us further away and then closer again. Given that the orbits of Io, Europa and Ganymede are in resonance, these cycles would be very regular. As it happens, Ganymede's cycle would repeat every 7 (Earth days), meaning that you could potentially tell what day it is from Ganymede's apparent size!

* Or O'Level, whatever that is...
** I Eat Green Caterpillars. Clever, eh?
*** I subtracted Ganymede's average orbital distance from Europa's to get this. This actual distance may never happen, but it gives us an idea.
**** See *** for the general gist.


  1. How big would Callisto using or subtracting moon sizes be from the point-of-view from the surface of Europa?

    I would be interested in knowing what that would look like,

  2. If you go to the post referenced at the end of the third paragraph you'll get an idea of how to work out the answers to questions like this. I'd have a go, but I'm a bit busy at the mo!

    You need to take into account the fact that both Callisto and Europa are orbiting Jupiter so the distance between them (and therefore their apparent sizes in the sky as observed from one another) changes constantly and, I'd imagine, quite wildly.

    If you don't want to have a go yourself, you could always head over to the forum (link in the bar at the top of the page) and post it for others to see, discuss and comment on!


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