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It has been reported this week[0] that the following image by Kevin Gill is the shadow of Io crossing Jupiter.

Jupiter with shadow

Why is the shadow so large? My intuition is that Io is much smaller than Jupiter so should cast only a small shadow, considering the distance of the light source. Earth’s moon orbits at a similar distance as Io orbits Jupiter, and our moon does not cast such a large shadow on Earth. The shadow in the image would be several times the width of Earth.

The image also looks “off” to me - somehow different from pictures of Jupiter that I’m used to - as though an image of a small patch has been stretched over a sphere. Is this just my imagination?

[0] https://www.universetoday.com/143442/yes-this-is-actually-the-shadow-of-io-passing-across-the-surface-of-jupiter/

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    $\begingroup$ I've just answered this same question over here: space.stackexchange.com/questions/38938/… $\endgroup$
    – Swike
    Commented Sep 20, 2019 at 17:43
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    $\begingroup$ The other interesting thing to me is how little penumbra shows compared to images of the moon's shadow on Earth. It's a visceral indication of how much the smaller the Sun's angular size is from the neighborhood of Jupiter. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Sep 20, 2019 at 17:58

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The image is taken by Juno at a distance of 8000 km. That's very close. The distance to the center of Jupiter would be about 78000 km. We can use that to calculate the angular size of the planet to an observer at that distance.

$$\sin (\theta) = \frac{R}{d}$$ $$\theta = \sin ^{-1} \left(\frac{70\,000}{78\,000}\right)$$ $$\theta = 63^\circ$$

And that's the angular radius. The diameter of the planet would be over 120 degrees, similar to how earth appears to someone at an altitude of $8000 \ \text{km}\frac{R_\mathrm E}{R_\mathrm J}$ or about 730 km.

We can also calculate that instead of seeing nearly a hemisphere of Jupiter (~180 degrees), we are seeing an area that takes up $(\frac{\pi}{2} - \theta)\times 2$ or about $52^\circ$ from edge to edge.

Given that, the planet looks nothing like the more distant images from Hubble or Voyager. It must be quite distorted.

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    $\begingroup$ FWIW, assuming a spherical Jupiter with radius 70,000 km, and Juno's altitude 8,000 km, I calculate that the visible portion of Jupiter is equivalent to a flat disk of radius roughly 30,880 km. $\endgroup$
    – PM 2Ring
    Commented Sep 20, 2019 at 18:09
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It's easier to get perspective if you can see more.

Hubble image: "Hubble Clicks Images of Io Sweeping across Jupiter", April 20, 1999 12:00AM (EDT) Release ID: 1999-13

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On January 24 2015 the Hubble wide field camera 3 took a series of photographs of some of Jupiter's moons and their shadows:

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Click for much larger image.

"The Hubble image on the left (unlabeled at top left, labeled at bottom left) shows the beginning of the event, which took place on January 24, 2015. From left to right the moons Callisto and Io are above Jupiter's cloud tops. The shadows from Europa (not seen in the image), Callisto, and Io are strung out from left to right.

Near the end of the event, approximately 42 minutes later (right-side image), Europa has entered the frame at lower left. Slower-moving Callisto is above and to the right of Europa. Fastest-moving Io is approaching the eastern limb of the planet; its shadow is no longer visible on Jupiter. Europa's shadow is toward the left side of the image, and Callisto's shadow to the right. (The moons' orbital velocities are proportionally slower with increasing distance from the planet.) ... The farther away a moon is from the planet, the softer the shadow, because the shadow is more spread out across the disk.".

Here is the Tweet containing the image in question retweeted on Kevin M. Gill's feed. Here's the Flickr page and list of raws he used. Here's a raw page from "Jupiter - Perijove 22 - Mid-Northern Latitude Map" using a equirectangular projection. 2019-09-18 23:10 Universal time.

Io equirectangular shadow Click for much larger image.

Same data, different processing. Yes, you can say it's "off" or that the perspective is odd.

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It looks overly large because its a closeup. The round shape of the photo makes you think you are seeing all of Jupiter but it is not. Its more like a view through a lens or fisheye, its rather misleading.

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In addition to the other answers that correctly assume that this is a largely fish-eye-style distorted image, because of the positioning of the spot relative to the center of the planet, and because of the fact that the planet is fully lit, we know that this image is taken when Io is casting a "long shadow." This is very similar to how at sunset or sunrise our shadows are much larger than our bodies. This also explains why, in spite of how extremely close Io is to Jupiter, the shadow is "fuzzy" instead of crisp around the edges. The reason that the "long shadow" looks mostly circular instead of oblong is the same distortion+perspective identified by the top commenter.

Lastly, we must remember that Jupiter's atmosphere is very thick but still somewhat translucent, so we are also seeing more scattering and depth of the shadow from the side than we would from the top. This is not disimilar to how, on a misty day, the world can seem brighter than usual if the sun is shining through the mist, and darker than usual if the sun is covered by a cloud, even if that cloud is the only cloud in the sky

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