I am trying to find a precise orbit for a moon that would allow it to always be in an eclipse. So, basically, I need help finding a formula for the semi-major axis and mass ratio that would get the exact values I need.

If you want exact numbers, Planetary mass is 0.8955 E, density 5.821 g/cm3, and the orbit is 54,879,809 seconds. So I need help finding a moon with the right mass and semi-major axis to have an orbit of 54879809 seconds when orbiting a planet of said mass and density. I also don't want it to be outside of the Hill Sphere, which is another issue I have found.

I tried myself, but it is a lot to deal with and I am having a lot of trouble finding it. As the numbers I am getting tend to fall past the roche limit. Which is an issue as I don't want the moon being torn apart by its planet's gravity.

Your help is much appreciated. As I am doing a relatively hard sci-fi, and I know that this configuration is theoretically possible, but I want to have exact numbers, and don't want to ditch it due to my own inability to come across values that make it work.

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    $\begingroup$ You want a moon that is never sunward of its primary? I believe the only way is to put it at Lagrange point 2, which is unstable. $\endgroup$ Oct 8, 2023 at 20:07
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    $\begingroup$ I agree with @AntonSherwood, although I read it the opposite way (it was always sunward), but that would put it in L1, which is also unstable. Please clarify on which side of the planet you want the moon. $\endgroup$
    – JBH
    Oct 8, 2023 at 21:48
  • $\begingroup$ I wanted it to always be an ecclipse, so, sunward. I don't feel it would necessarily need to be at a Lagrangian, it just needs to have the right Period to where its orbital period around the planet is equal to the orbital period around the star. Doesn't feel that it would necessitate a Lagrangian, though using one would definitely be an answer. $\endgroup$ Oct 9, 2023 at 2:35
  • $\begingroup$ Last time I checked yearly orbits, they always end up away from Hill sphere of the planet, regardless of star/planet mass ratio. So I think what you want is really impossible. $\endgroup$
    – Vesper
    Oct 9, 2023 at 6:37
  • $\begingroup$ "I know that this configuration is theoretically possible". How do you know that? It is, to me, categorically impossible from theory alone (see answer below) $\endgroup$ Oct 9, 2023 at 7:46

2 Answers 2


Mathematically impossible (under reasonable assumptions)

Ok, here are the simplifying assumptions that I will make for the mathematical discussion of what you want to obtain:

  1. Circular orbits (you don't mention the eccentricity of your orbits anyway)
  2. The mass of the star is very large compared to the mass of your planet. The mass of the planet is very large compared to the mass of the staellite.

Those assumptions are here to make the mathematical treatment significantly more pleasing to the eyes, but shouldn't affect the qualitative conclusions. Let's get started

Your requirements:

  • same period $T$ for your planet around your star and your moon around your planet
  • distance from the moon to the planet $d$ smaller than the Hill radius $R_H$: $d<R_H$

We will show that under our assumptions, those two requirements are mathematically incompatible

for a circular orbit of a planet around a much more massive star, the Hill radius simplifies to: $$R_H = a (\frac{m_p}{3m_s})^{1/3}$$ with $m_p$ the mass of the planet, $m_s$ mass of star and $a$ the orbital distance.

Let's now use Kepler's 3rd law to reformulate this in terms of orbital period. Under our assumptions, the law for the orbit of the planet around the star can be written: $$\frac{a^3}{T^2}=\frac{Gm_s}{4\pi^2}$$

So we have $$R_H^3 = \frac{Gm_s}{4\pi^2}T^2 * \frac{m_p}{3m_s} = \frac{Gm_p}{12\pi^2}T^2$$

According to the same law, the orbital distance of the moon around the planet is related to its period by: $$d^3 = \frac{Gm_p}{4\pi^2}T^2$$ Where $T$ is the same period by hypothesis. The condition $d<R_H$ (which implies $d^3<R_H^3$) then leads to: $$ \frac{Gm_p}{4\pi^2}T^2 < \frac{Gm_p}{12\pi^2}T^2$$ $$ \frac{1}{4} < \frac{1}{12}$$

Which is a contradiction.

It is therefore impossible for a satellite to have an orbital distance within the Hill sphere that gives you the same orbital period as that of the planet around the star.

The orbital distance required for that would be $3^{1/3}$ times larger than the Hill radius.

The longest period you can get to be exactly at the Hill radius would be $T_{satellite}=\frac{T_{planet}}{\sqrt{3}}$ (this would probably still be an unstable orbit). Note that the mass ratio is irrelevant to the discussion (if and only if assumption 2. is verified)

N.B: The use of Kepler's law for the satellite near the Hill radius is not justified because, by definition, the influence of the sun at that point is not negligible. The problem then turns into a 3-body problem where Kepler has no business making laws. That means that the quantitative results above are not valid near this limit.

However, this does not invalidate the general reasoning ab absurdo. Indeed, we make the assumption that we have a stable orbit, meaning that the influence of the earth dominates significantly (2-body problem). We then proceed with the reasoning above, and we can use Kepler's law under this hypothesis. We obtain the aforementioned contradiction, which invalidates our set of assumptions (namely that we can have a stable 2-body orbit with such a period)

  • $\begingroup$ I see, dissapointing, but thank you for informing me, that would explain why all of my calculations wound up with it not being in a stable orbit. $\endgroup$ Oct 9, 2023 at 13:56

Short Answer:

The most believable way to have a moon in such an orbit would be if a super advanced civilization put it there quite recently on a geological or astronomical time scale.

Long answer:

If you want exact numbers, Planetary mass is 0.8955 E, density 5.821 g/cm3, and the orbit is 54,879,809 seconds.

An Earth year is over 31,000,000 seconds. A Julian calendar year of 365.25 Earth days is 31,557,600 seconds long.

So the orbital periods of your planet and your moon should both be about 1.739 Earth years or about 635.1829 Earth days long.

Unless the planet is tidally locked to its star, it is unlikely to have a day that long. So the planet will probably rotate with a period of a few Earth hours or Earth days, and the spot where there is an eclipse will move over the surface of the planet as the planet rotates.

If you want the eternal eclipse to always eclipse the same area of the planet, the planet must be tidally locked to its star.

Around a planet with a mass of 0.8955 Earths, a moon with a mass of 0.01 Earth, about that of Earth's moon, would have a period of 1.739 Earth years if it orbited at a distance of 3,019,596 kilometers or 1,876,290 miles.

Assuming that your planet orbits its star with a period of 1.739 Earth years or 635.1839 Earth days, and orbits at the distance to receive exactly as much heat and light from its star as Earth receives from The Sun, the spectral class of the star can be calculated.

Here is a link to a question, one of the answers having a table with the details of different types of stars.


According to the table, a planet receiving as much radiation from an F8V type star - with 1.18 the mass of the Sun and 2.031 the luminosity of the Sun - as Earth gets from the Sun, would orbit at a distance of 1.425 Astronomical units (AU) and have an orbital period of 572.18 Earth days.

A planet receiving as much radiation from an F5V type star - with 1.33 the mass of the Sun and 3.434 the luminosity of the Sun - as Earth gets from the Sun, would orbit at a distance of 1.853 AU and have an orbital period of 799.11 Earth days.

According to the table at:


An F7V class star would have a mass of 1.21 Sun and a luminosity of 2.45 Sun. A planet orbiting that star at the distance to receive the same amount of radiation as Earth gets from the Sun would orbit at about 1.565 Au. According to the orbital period calculator it would have a period of 1.7797 Earth years or 650 Earth days.

A habitable planet would not have to orbit at the distance where it receives exactly as much radiation from its star as Earth gets from the Sun. It could orbit a little closer or a little rather and be a little warmer or a little colder than the Earth.

How much closer or how much farther could a habitable planet orbit? Here is a link to a list of about a dozen estimates of the inner and outer edges of the Sun's habitable zone.


Some are very different from others. I advise writers who are certain that they want one and only one habitable planet in the fictional planetary system in their story to put that planet within about 1 or 2 percent of the distance at which it will received exactly as much radiation from the star as Earth Gets from the Sun.

As far as I know it is impossible for a moon to have a long term stable orbit if its orbital period around the planet is the same length as the planet's orbital period around the star.

A planet will have a "Hill sphere" around it where stable orbits are possible. The radius of the Hill sphere is calculated from the mass of the planet, the mass of the star, and the distance between them. The closer a moon is to its planet, the shorter its orbital period will be, and the farther a moon is from its planet, the longer the orbital period will be. And you want your moon to have a very long orbital period and thus be very far from its planet and very probably far outside the Hill sphere of the planet.

And in fact the area of true stability will be smaller, no more than 0.333 or 0.5 of the Hill radius. A moon outside the area of true stability will be lost into interplanetary space rapidly by astronomical standards.


According to this Hill sphere calculator:


A planet with a mass of 0.8955 Earth orbiting a star with a mass of 1.21 Sun at a distance of 1.565 AU and with a orbital eccentricity of 0.01 would have a Hill radius of 2,097,298.7777 kilometers.

And the region of true stability would be only about 699,099 or 1,048,649 kilometers from the planet.

A satellite of a planet is likely to have a rotation period or day equal to its orbital period around the planet.

On page 3 of this article:


The longest possible length of a satellite’s day compatible with Hill stability has been shown to be about P∗p/9, P∗p being the planet’s orbital period about the star (Kipping 2009a).

This means that the orbital period of the planet around the star has to be at least 9 times the orbital period of the satellite/moon around the planet, or that the orbital period of the moon has to be 0.111111 or less of the orbital period of the planet.

And here is a link to the article cited as the source for that statement.


So your moon would not be able to remain in orbit for very long with an orbital period equal to the orbital period of the planet around its star.

Not very long by geological or astronomical standards. But perhaps long enough for the purposes of the story.

So possibly the planet developed for billions of years and developed an oxygen rich atmosphere and multicellular life. And then by some freak accident another object in that star system was captured by the planet and took up an orbit with the exact same period as the orbital period of the planet around the star.

Such an orbit would be very unstable and short lived by astronomical standards, but might last for the millions or thousands of years necessary for your story.

Or maybe the moon was not captured by chance but put into that very precise orbit by an advanced civilization for some reason. Possibly to affect any intelligent life forms on that planet.

And they could greatly reduce the enormous energy costs by making an artificial moon and putting it in orbit around the planet.

I ham thinking of an ultra lightweight inflatable object like the Echo satellites but with at least thousands of times the diameter and thus at last billions of times the volume.


And the advanced aliens might have to regularly adjust the orbit of their artificial moon because of perturbations and the stellar wind.

And of course if you have a really great story idea which requires that eternal eclipse on a (probably small) part of the planet then you probably won't have to bother much with explaining how it is possible.


  • $\begingroup$ nice answer! I didn't know it was impossible before reading this thread. I posted a complementary answer with a short mathematical derivation of the impossibility of having such an orbital period within the Hill sphere $\endgroup$ Oct 9, 2023 at 7:26
  • $\begingroup$ @Barbaud Julien It is not exactly totally impossible but almost totally impossible. The most plausible way I could think of was that an extremely advanced civilization built an artificial moon and put it into a very precise orbit, perhaps to see how it would affect the natives. People who find that idea too crazy should realize that a moon in such an orbit happening naturally is much more improbable. $\endgroup$ Oct 9, 2023 at 7:39
  • $\begingroup$ It is totally impossible to get a stable orbit verifying the conditions OP is looking for. It might be possible to create a trajectory that looks like such an orbit on a human timeframe, but seems quite unlikely $\endgroup$ Oct 9, 2023 at 7:42

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