For a moon orbiting a gas giant which is in turn orbiting a star, I want to know how long the eclipse would last for an observer standing on the moon when the planet passes between the moon and the star.

The actual values I intend to use in the end are likely to vary from these, somewhat, but I like simplicity so here's the parameters for this simplified version:

The star = our Sun.

The planet = Jupiter radius, orbits the star at a distance of 1 AU (Earth's distance from the Sun) giving it an orbital year equivalent to an Earth year (or close enough to be a negligible difference, for simplicity)

The moon = orbits the planet at a distance of 4 million KM, which should give it an orbital time around the planet of about 42 days. It is not tidally locked and rotates on its axis at a rate that will make the same point on the surface face the star exactly once every 24 hours (again, for simplicity)

For simplicity (seeing the trend here?), assume all orbits and equators of all 3 bodies share a plane, and are circular, no eccentric or inclined orbits or tilted axes, etc., and the observer is standing on a point on the equator of the moon. Assume the eclipse "starts" when the star is directly over head (noon) of the observer, and the observer says in the same location until the eclipse ends.

Please let me know if I left out any necessary variables, I probably have them available and just forgot to include them or didn't realize they were necessary.

How long before the observer sees the star/sun again? (rounded or approximate answers would work, but the more precise the better, and an explanation of how the numbers were determined is also appreciated)

  • $\begingroup$ Related: Earth-like Moon around the Gas Giant. Eclipse length? - not a full duplicate only because your numbers are different. Also, selected answer there is very well laid out, but unfortunately contains a math error which I pointed out in my comment. $\endgroup$
    – Alexander
    Commented Jun 6, 2019 at 18:05
  • $\begingroup$ @Alexander Certainly related, yes. But it's not just a change in the actual numbers of the known variables, it's also a change in which variables are known, and approached from the opposite end of the problem. I did review that question and answer, before asking mine, but could not work out how to reach the answer I need, going backward through that math, and then trying to go back forward through it with my adjusted numbers. I might re-attempt it, if this question doesn't yield the results, but I'm no more hopeful with my math skills than the person that asked that other question. $\endgroup$
    – Harthag
    Commented Jun 6, 2019 at 18:26
  • $\begingroup$ Well, you do know the star's size. You do know the gas giant's orbit and size. And you do know the moon's distance from the giant. That's all that we need here. Going through the math in the other question's answer can be a little challenging, I know. My ballpark estimate (without doing the actual math) is 6 hours, but this is very approximate. $\endgroup$
    – Alexander
    Commented Jun 6, 2019 at 19:19
  • $\begingroup$ @Alexander in my answer below, which uses a very different approach, I came up with about 6.83... hours, so a 6 hour estimate matches up pretty well with that for not having done any of the math at all on it. $\endgroup$
    – Harthag
    Commented Jun 6, 2019 at 19:35

2 Answers 2


Let's establish some notation:

  • The star is labelled $S$, the giant planet is $P$ and the moon is $M$.

  • The diameter of the star is $d_S$, the diameter of the planet is $d_P$ and the diameter of the moon is $d_M$.

  • The radius of the orbit of the planet is $r_P$ and the radius of the orbit of the moon is $r_M$.

  • At a distance equal to the radius $r_M$ of the orbit of the moon $M$, the planet $P$ casts a shadow (which astronomers like to call the umbra, using the Latin word for shadow) of diameter $d_U$.

See the following diagram showing the position of the star $S$, the planet $P$ and the moon $M$ when the totality phase of the eclipse begins for an observer sitting on the surface of the moon; we consider that the planet and the moon orbit in the plane of the picture, and we are looking at them from the north celestial pole.

The eclipse begins for an observer

The position of the three bodies when the totality phase of the eclipse begins for an observer. $S$ is the star with diameter $d_S$. $P$ is the giant planet with diameter $d_P$ and revolving around the star on an orbit with radius $r_P$. $M$ is the moon with diameter $d_M$ revolving around the giant planet on an orbit with radius $r_M$. $d_U$ is the diameter of the shadow ("umbra") of the giant planet at a distance equal to the radius $r_M$ of the orbit of the moon $M$. Own work, available on Flickr under the CC BY 2.0 license. The diagram is absolutely not to scale.

We want to compute $d_U$ given $d_S$, $d_P$, $r_P$ and $r_M$. (Why do we want to compute $d_U$? Because we want to find out how much time the moon $M$ will take to move across the shadow.)


  1. The following calculations are not exact, but rather decent approximations good enough for a work of fiction. Exact calculations would take more time than I have available.

  2. The calculation refers to the totality phase of the eclipse, from the moment the sun is completely covered by the giant planet to the moment the first sliver of sunlight becomes visible.

From triangle similarity we have

$$\frac {r_M}{r_P} = \frac {d_P - d_U}{d_S - d_P}$$

which means that

$$d_U = d_P - \frac {r_M}{r_P} (d_S - d_P)$$

Plugging in the numbers

$d_S$ = 1,391,400 km (diameter of the Sun)

$d_P$ = 142,984 km (diameter of Jupiter)

$r_P$ = 149,597,870 km (1 astronomical unit)

$r_M$ = 4,000,000 km (given)

we find that

$$d_U = 142{,}984 - \frac {4{,}000{,}000}{149{,}597{,}870} (1{,}391{,}400 - 142{,}984) \approx 109{,}600 \text{ km}$$

The moon travels 2 $\pi$ × 4,000,000 kilometers in 42 days, or 598,399 km per day. The 109,600 km wide umbra will then be traversed in 109,600 / 598,399 = 0.183 days, or 4 hours 24 minutes.

During this time the moon has rotated a bit, about 60 degrees, so that the observer is still in the shadow -- the moon must move on its orbit a distance of about two thirds of its radius to bring the observer into the light. Assuming that the moon has the same radius as the Earth, 6,400 km, this will take about 4,000 / 598,399 days or about 10 minutes.

So the grand total is 4 hours and 34 minutes of total eclipse.

P.S. What about the total duration from the moment the planet touches the sun to the moment the entire sun is again visible? Given that the radius of the orbit of the moon is much smaller than the radius of the orbit of the planet, the diameter of the penumbra is only a little larger than the diameter of the planet, or let's say some 160,000 km. The moon will traverse the penumbra (and the umbra) in about 6 hours 50 minutes. What this means that if the eclipse (partial + total) begins at noon chances are it will end after sunset...

Note that this post-scriptum is even more approximative than the calculation for the totality phase, but still good enough for fiction.

  • $\begingroup$ Looking at your diagram, the way it's drawn both dP-dU and dS-dP both produce 2 separate triangles, the bases of which both contribute to the distances and ratios. But RM/RP only accounts for one side, one part of the ratio. It seems like this doubling of triangles and distances wasn't taken in to consideration in the equations. Though admittedly I'm having trouble wrapping my head around exactly how it would have an effect, my gut says it should be having some effect, but I don't see it in the math. What am I missing? $\endgroup$
    – Harthag
    Commented Jun 6, 2019 at 21:24
  • $\begingroup$ @Dalila: The factors of 2 cancel out, don't they? $\frac {\frac 12 (d_S - d_P)}{\frac 12 (d_P - d_U)} = \frac {r_P}{r_M}$. One triangle has the short cathetus is $\frac 12 (d_S - d_P)$ and the long cathetus about $r_P$, etc. $\endgroup$
    – AlexP
    Commented Jun 6, 2019 at 21:27
  • $\begingroup$ THAT'S IT ! Thanks for humoring me. I knew it was a thing, it's just been too long since I've played with fractions, so I didn't realize that the thing was canceled out of the rest of the process by being itself. :P $\endgroup$
    – Harthag
    Commented Jun 6, 2019 at 21:31
  • $\begingroup$ @AlexP Could you explain your P.S. a little more? I managed to follow the formulas in the main answer (which results in a period of 21.91 hours of eclipse in the 6 day orbital period of my moon, which is exciting), but am struggling with the P.S. Is the radius of the orbit of the moon its orbital period? Same for the planet orbital radius. Is the 160000 in your answer just a randomly chosen number, so I could go with e.g. 150000 for mine from 139822? And how did you then arrive at the 6 hours 50 minutes from the umbra 4 hours 34 minutes? $\endgroup$ Commented Dec 25, 2023 at 19:06
  • $\begingroup$ @SaintDiabolus: The radius of the orbit of satellite M is the distance between the center of its primary P and the center of S. In practice, the sun S is very far away, so that the the cone of the shadow of the planet is very extended. For this reason, the difference between the diameter of the penumbra and the diameter of the planet is just about equal to the difference between the diameter of the planet and the diameter of the full shadow. $\endgroup$
    – AlexP
    Commented Dec 25, 2023 at 21:57

Using a completely comparative approach, rather than a direct calculation approach, here's my best answer:

Totality of an eclipse involving the actual Sun, Earth, and Moon can last up to 7 minutes and 31 seconds.

The orbital period of this moon around its planet takes about 1 1/3 as much time as it takes for the moon to orbit the Earth, so it seems reasonable to assume that it's apparent speed across the sky, and therefore across the front of the star, takes 1 1/3 as long. This increases the time of totality from 7.5 minutes to 10.

EDIT: The apparent diameter of the Sun is about 30 arc-minutes, the apparent diameter of the Moon is about 31 arc-minutes, and the apparent diameter of this planet as seen from its moon would be about 71 arc-minutes. As pointed out by Mike Scott in a comment (many thanks for the heads up) While that planets apparent size is about 2.13 times the apparent size of our Moon, the key change this makes is actually the difference in the number of arc-minutes between the sizes of the star and the sizes of the obscuring objects. So it's actually a difference of only a single arc-minute between the Sun and Moon, but it's a difference of 41 arc-minutes between the Sun and the planet. A factor of 41 times, not just 2.13 as I'd originally assumed. This increases the time from 10 minutes to 410 minutes of totality

This is long enough for the Sun to have set, from the observer's location, before totality is complete, so the observer wouldn't see the Sun again until it rose the following morning.

  • 2
    $\begingroup$ No, that’s not right. The period of totality is very short on Earth because the angular size of the Moon and sun are nearly the same. If the sun subtends 30’ and the Moon subtends 31’, the period of totality is based on that 1’ of difference between them where the Moon can completely cover the sun. In this case we’re talking about 30’ and 71’ so the difference is 41’, forty times as big rather than a bit over twice the size. So maybe 400 minutes of totality. $\endgroup$
    – Mike Scott
    Commented Jun 6, 2019 at 19:11
  • 1
    $\begingroup$ @MikeScott Thanks for pointing that out. I was using the math as if the numbers were from start to end of the entire eclipse, not just the totality portion, even though I had stated that I was referring to the totality portion (which is what I really intended). I'll edit the answer accordingly. $\endgroup$
    – Harthag
    Commented Jun 6, 2019 at 19:18
  • $\begingroup$ In addition to @MikeScott's correction (though I calculated the angular diameter to be 2°1', for a different of 91')... the gas giant is also traveling slower through the zodiac compared to how our moon moves. Its umbra will be on the planet about 1.7 times as long. $\endgroup$
    – Ghedipunk
    Commented Jun 6, 2019 at 19:50
  • $\begingroup$ Also, I think a great answer should mention the penumbra, and not just the time that an observer is in the umbra (totality), as the penumbra is an interesting part of any solar eclipse already. Since the size of the penumbra is (mostly) set by the angular diameter of the sun, it would just be our own penumbra's duration times 1.7, given the gas giant's slower travel through the zodiac. $\endgroup$
    – Ghedipunk
    Commented Jun 6, 2019 at 19:54
  • 1
    $\begingroup$ @Ghedipunk: Each solar eclipse is very different on Earth because the angular diameter of the Moon is so close to the angular diameter of the sun, so that small variations in the distance Earth-Moon and Earth-Sun cause vast changes in the size of the dark spot. In fact, sometimes it happens that the Moon cannot cover the sun completely and we get an annular eclipse (when the Moon is at its most distant point when it comes in line with the Sun, and the Earth is at its nearest point from the Sun). $\endgroup$
    – AlexP
    Commented Jun 6, 2019 at 21:40

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