Worldbuilder in dire need!

I'm trying to figure out the eclipse length of a habitable Earth-like moon that is rotating around a gas giant. The story that I work on is centered on the Earth-like moon, but math was never my strongest suit and I'm in dire need of some mathematicians, astronomers or science enthusiasts.

I wanted the Earth-like moon to be exactly the same as our own Earth. Well, almost.

Basic Info:

  • Year (one full orbit of the gas giant around the Sun) has 256 days.
  • A day (1 full rotation around its own axis) of the Earth-like moon is 24 hours.
  • One full orbit of the Earth-like moon around the Gas Giant = I wanted it to be precisely 8 days (192 hours)

My idea was to introduce 8th day in a week, one that people would call "Longnight" which would basically be a whole day without Sun due to the eclipse from the gas giant.

  • Size of the gas giant and distance of the two bodies aren't specified. (Since I have very limited knowledge of the astrophysics. Feel free to adjust.) The eclipse (8th day) probably won't take as much as 24 hours, but I'll appreciate anything that can give at least a bit of "long-night, a day full of darkness" feel to it, even if it takes another, 9th day a week.

(Optional: It's a world full of magic and divine beings, so if the distances or other aspects don't correlate with the real physics, we can ignore some laws and say "It's magic. Gods are keeping the moon in orbit/atmosphere together." or something like that.)

I'm just really curious about the eclipse length and the ways it could be done possible. Thank you for all the ideas.

  • 5
    $\begingroup$ For answering the question you'll get a full prize of a youtube link that redirects you to Eight Days a Week song by Beatles ! :D $\endgroup$
    – Jotunn
    Feb 13, 2017 at 4:51
  • $\begingroup$ I can't work out the math specifically right now, but it is certainly possible given that the volume of the gas giant and the length of the orbit of the earth-moon is at our disposal. $\endgroup$
    – Jaich
    Feb 13, 2017 at 5:10
  • $\begingroup$ About the day length. Is it the sidereal day or the perceived day? You define it as the former but it seems like you want the latter. $\endgroup$
    – PatJ
    Feb 13, 2017 at 5:11
  • $\begingroup$ How big is your gas giant (compared to Earth-like moon)? $\endgroup$ Feb 13, 2017 at 8:28

4 Answers 4


For the TL;DR, see the bottom of this answer.

Okay, so first of all, the orbital period of the gas giant around its star is $256 \times 24$ hours, and I'd like to establish the distance from the planet to its star. Since you haven't specified anything about the star itself, I'll go with our Sun for simplicity's sake. Also for simplicity's sake (or to retain everybody's sanity, including my own) I will approach this as two two-body problems rather than a three-body problem. This reduces the attainable precision, but greatly simplifies the math. For a representative gas giant, I'll use Jupiter.

As an approximation for the planet's orbit around its star, we can use the formula for a small body orbiting a central body:

$$ r = \sqrt[3]{\frac{\mu T^2}{4 \pi ^2}} $$


  • $r$ is the orbit's semi-major axis in meters (note: this is not the same thing as the orbital altitude, but can be approximated as the orbital radius)
  • $\mu$ is the standard gravitational parameter, $\mu = GM$
    • $G$ is the gravitational constant, in units relevant here $ 6.67408 \times 10^{-11} ~\text{m}^3 \text{kg}^{-1} \text{s}^{-2} $
    • $M$ is the mass of the central body (in this case, the star) in kg
    • $\mu_\text{Sun} \approx 1.327 \times 10^{20} ~\text{m}^3 ~\text{s}^{-2}$
  • $T$ is the orbital period in seconds

We know that the desired $T = 256 \times 24 \times 60 \times 60 = 22\,118\,400$ seconds. Let's plug all those values in and see what comes out:

$$ r = \sqrt[3]{\frac{1.327 \times 10^{20} \times 22\,118\,400^2}{4 \pi ^2}} \approx \sqrt[3]{1.644442 \times 10^{33}} \approx 1.1803375 \times 10^{11} $$

So your planet orbits at a distance of about $1.2 \times 10^8$ km, or 120 million km, to its star. This is comparable to Venus' orbit around the Sun (Venus' semi-major axis is about $1.08 \times 10^8$ km, with an orbital period of $224.7 \times 24$ hours). That's awfully close for a gas giant in anything resembling our solar system, but it's the only way to get the planet orbital period you ask for while keeping the star Sun-like. You could twiddle the knob for star mass ($M = M_\text{Sun}$, influencing $\mu_\text{Sun}$ above) until you are happy with the outcome; for inspiration, look no further than to Wikipedia's list of main sequence star example parameters which gives the stars' mass in terms of solar masses, from which you can calculate the corresponding value for $\mu$.

The arc length of a circle sector is given by $ L = \theta \times r $ where $\theta$ is the angle subtended. We know the approximate arc length (the diameter of the Sun: twice its radius of $695\,700$ km) and distance ($1.2 \times 10^8$ km) and want angle subtended, so we get $$ 2 \times 695\,700~\text{km} = \theta \times 1.2 \times 10^8~\text{km} \Rightarrow \theta = \frac{2 \times 695\,700}{1.2 \times 10^8} = 0.011595 $$

Because $\theta$ comes out in radians, we multiply by 57.296° to get the angle subtended in degrees, which turns out to be 39.86 arcminutes or 0.664 degrees. A quick check against Wikipedia gives the Sun's angle subtended from Earth (at an orbital radius of $1.5 \times 10^8$ km) as 31.6-32.7 arcminutes, so while possibly not perfect, this is well within the ballpark. The same calculation for an orbital radius of $1.5 \times 10^8$ km gives 31.9 arcminutes, squarely in the range given.

You specified the orbital period of the moon around the gas giant to be 192 hours, or $192 \times 3\,600 = 691\,200$ seconds. We can use the vis-viva equation to calculate the corresponding orbital radius. We have $$ v^2 = \mu \left( \frac{2}{r} - \frac{1}{a} \right) $$

For a circular orbit, $r = a$ (orbital radius is equal to the semi-major axis of the orbit) and thus $$ \left( \frac{2 \pi r}{T} \right)^2 = \mu \left( \frac{2}{r} - \frac{1}{r} \right) $$

We have $\mu_\text{Jupiter} \approx 1.267 \times 10^{17} ~\text{m}^3 ~\text{s}^{-2}$ and $T = 691\,200 ~\text{s}$. By rearranging, we get $$ r = \sqrt[3]{\mu \left(\frac{T}{2\pi}\right)^2} \approx 1\,153\,080 ~\text{km} $$

So the Earth-like moon orbits the gas giant at an orbital radius of about 1.15 million km, because that's the orbital radius (for a perfectly circular orbit, one with eccentricity $e = 0$ or semi-major axis equal to semi-minor axis) that corresponds with the desired orbital period. This happens to be very similar to the radius of the orbit of Ganymede (which is 1.07M km, and an eccentricity of about 0.0013 in case you wondered), providing a nice sanity check for the result; Ganymede orbits Jupiter in 171 hours, only slightly less than your desired 192 hours, so at least to a first order approximation this checks out.

The formula for computing the length of the umbra (central shadow) of an eclipse is $$ L = \frac{r \times R_o}{R_s - R_o} $$ where $r$ is the distance from the star to the occulting object (in our case, the gas giant), $R_o$ is the radius of the occulting object, and $R_s$ is the radius of the star. We thus have a shadow cone of length (where all distance and size values are in kilometers) $$ L = \frac{1.2 \times 10^8 \times 71\,492}{695\,700 - 71\,492} \approx \frac{8.58 \times 10^{12}}{624\,208} \approx 13.74 \times 10^6 $$

Because $ 1.15 \times 10^6 \lt 13.74 \times 10^6 $, the moon passes through the shadow cone cast by the planet, so we have a full eclipse (the moon passes through the umbra cast by the planet). Now, for how long does the eclipse last?

By considering the shadow cone to be a triangle with the base length of the diameter of the planet and the height calculated above, we can use Pythagoras' theorem to calculate the length of the resulting hypotenuse. (This turns out to be almost identical to the height, approximately $1.37400 \times 10^7$ versus the height approximately $1.37439 \times 10^7$ km.) We can then apply the intercept theorem which states that when dividing a triangle by a line parallell to the base of the triangle, the length of the new base line is to the original base line as the hypotenuse of the part of the triangle is to the total hypotenuse length. By approximating the required inner height as the orbital radius of the moon around the gas giant, we end up with $$ \frac{DE}{2 \times 71\,492} = \frac{1\,150\,000}{13\,740\,000} \approx \frac{0.083697}{1} $$ where $DE$ is the length of the line connecting the edges of the triangle at the orbital radius of the moon. Hence the occluded path for the moon is approximately $ \frac{2 \times 71\,492}{0.083697} \approx 1.708 \times 10^6$ km. (This is really the base of a circle segment where the moon traces the circle segment, but the difference is small enough to be negligible at these levels of precision.)

The circumference of a circle of radius $1.15 \times 10^6$ km is $$ 2 \pi r = 2 \pi \times 1.15 \times 10^6 \approx 7.226 \times 10^6 ~\text{km} $$

Thus, passing through the umbra cast by the planet takes $ \frac{1.708 \times 10^6}{7.226 \times 10^6} \approx 0.2364 $ of the orbital period of the moon. Multiplying by the orbital period of 192 hours gives us a duration of 45.4 hours within the total eclipse zone (the umbra).

Note that there are three things I am actually ignoring in the calculations above. First, I'm positing that all bodies are orbiting within your solar system's ecliptic; if their orbits are inclined relative to one another, you need to take the angle at which they are orbiting (the inclination) into account. Doing so complicates the math a fair bit for no significant gain, as bodies that form naturally within a solar system are likely to orbit close to the ecliptic. I'm leaving that entirely as an exercise for the reader.

Second, I'm ignoring the planet's orbital motion around the star. When the (Earth-like) moon is orbiting around the (gas giant) planet, and the (gas giant) planet is orbiting around the star, this is going to have the effect of either making the apparent eclipse slightly shorter or slightly longer. (Which it happens to become depends on the relative direction of the orbital movement.) I'm too lazy to account for this, so I just don't, but it shouldn't be more than some trigonometry if you really care to do that part of the math yourself.

Third, I'm ignoring the fact that the Earth-sized moon is going to tug a little at the planet. The barycenter of the system won't actually be at the planet's center, but a little outside of the planet's center, which will cause the two to be joined in somewhat of an orbital dance. This is very similar to how, in our solar system, Jupiter perturbs the Sun, despite being only $\frac{1}{1\,047}$ the mass, or how Earth's moon perturbs Earth.


One set of values that match your criteria are:

  • Star mass $1.99 \times 10^{30}$ kg (1 solar mass) (by choice)
  • Star diameter $1\,391\,400$ km (1 solar diameter) (by choice)
  • Planet mass $1.8986 \times 10^{27}$ kg (1 Jupiter mass) (by choice)
  • Planet diameter $142\,984$ km (1 Jupiter diameter) (by choice)
  • Planet orbital period around star $256 \times 86\,400$ seconds (by decree)
  • Planet orbital radius around star $1.2 \times 10^8$ km
  • Moon orbital period around planet $192 \times 3\,600$ seconds (by decree)
  • Moon orbital radius around planet $1.15 \times 10^6$ km
  • Eclipse length $45.4 \times 3\,600$ seconds

There are many other sets of values that can match your criteria. If you aren't happy with the above, just pick different values for the masses and radii involved, and recompute; there is nothing magical about the sizes of Sun or Jupiter. Just don't forget to change the value of $\mu$ accordingly!

  • 1
    $\begingroup$ I sure hope I got all the math correct. If anyone spots any mistakes, please let me know so that I can correct them. Thanks! $\endgroup$
    – user
    Feb 13, 2017 at 15:56
  • $\begingroup$ This is amazing! I thank you for all the work that you've put into this, really. I also thank you for the explanation how the calculation works, though I'm afraid I'm not able to put all the values into my calculator. Eclipse of 18 hours is great! It will do. The distances that you provided help a lot for better visualization. 1,15M km from the earth-moon to the gas giant. 120M km from the gas giant to the star. This is a bit tricky and I didn't expect for the gas giant to be restricted as such. (It makes sense, of course.) $\endgroup$
    – Jotunn
    Feb 13, 2017 at 21:14
  • 1
    $\begingroup$ I've been trying to get the hang of this with the help of some online distance calculators, but the math just doesn't work for me. :D Also, instead of M for star mass, they want M for planet mass... And then the results are just wierd. Nevermind! Here's the youtube reward as promised! youtube.com/watch?v=jZzEvqDQfIA $\endgroup$
    – Jotunn
    Feb 13, 2017 at 21:49
  • 1
    $\begingroup$ I'm worried something is wrong. Ganymede has an eclipse time of 220 minutes at most. alpo-astronomy.org/jupiter/GaliInstr.pdf The length of the occluded path should be smaller than the diameter of Jupiter. This moon is slightly further out, but I'd be surprised at figure of over 4 hours. $\endgroup$
    – James K
    Feb 17, 2017 at 23:43
  • 1
    $\begingroup$ Michael's calculations went wrong when he tried to apply the intercept theorem. Instead of DE = 1.708×10E6, it should have been DE = 1.31×10E5 (nearly a diameter of the gas giant, not 13x the diameter). Further on, it results in 0.0181 of the moon's orbital period, which is just 3.48 hours. $\endgroup$
    – Alexander
    Jun 2, 2017 at 21:54

I'm scratching my head, but if I'm reading this right then the answer to your question is in the question. :)

As far as I can tell, there's only two possible answers.

First, the eclipse would last 24 hours. Since you outlined that the world has a 24 hour day and the eighth day is covered by the eclipse, the eclipse must last the entire day.

Or secondly, it will last 12 hours. Since the world does spin and one side of the planet would be nighttime anyways, the eclipse would only need to last for the daylight portion of the day and that side of the planet would still be in darkness for the entire 24 hours. The opposite side wouldn't have even known there was an eclipse this way though.

But in all honesty, since you can place the planet pretty much anywhere inside the gas giant's orbit and that you have plenty of wiggle room in the gas giant's size, there's nothing stopping the eclipse from being pretty much any amount of time your want it to be. If you want the eclipse to last longer then the planet is closer to the gas giant. If you want the eclipse to be shorter than the planet is farther away from the gas giant.

Although a good thing to keep in mind is that if you have a long eclipse then the gas giant will also be larger in your sky. The shorter the eclipse the smaller the gas giant is in your sky.

Hope I helped!

  • $\begingroup$ I was interested in the physics behind it and the distances and times that I can't calculate myself. But thank you for the answer. :) $\endgroup$
    – Jotunn
    Feb 14, 2017 at 23:37

You need to take into account two effects, given that you fixed the size of the moon: the orbital speed and the apparent size of the moon.

A long lasting eclipse is granted by a slow orbiting moon, which is achieved by placing it far from the planet. But placing it far will also make it look smaller, so less able to shield the star.

Vice versa, if it is closer to planet it appears bigger, but also move faster in the sky, shortening the duration of the eclipse.

Since you don't mention how far is the planet from the star and how big the star is, you can also play with these two extra parameters.

  • $\begingroup$ I'd very much like to play, but I don't know the math, that's the problem... :/ Thank you for the aswer, though! $\endgroup$
    – Jotunn
    Feb 14, 2017 at 23:36

Another problem: The eclipse time would change every day. The shadow of the gas giant will be in a different position as it goes around the star.

So you wouldn't get an "eclipse day" on such a regular schedule.

The only way I see it working as a fixed "every 8 days" would be if the gas giant was tidally locked to the star (is that even possible for a gas giant?) and the moon was close enough to the gas giant to be pulled along by the gas giant's rotation.

I think that to pull off those two conditions, the gas giant would have to be so close to the star and the moon so close to the planet that the moon would be very hot.


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