Is it possible for a planet to take just as long to orbit its star as a moon takes to orbit the planet? If we assume circular orbits, then

$\text{orbital period}\sim \sqrt{\frac{\text{radius}^{3}}{\text{parent mass}}}$

and thus what we need is

$\frac{r_{moon}^3}{m_{planet}} = \frac{r_{planet}^3}{m_{star}}$, or in other terms, $\left( \frac{r_{moon}}{r_{planet}} \right)^3 \simeq \frac{m_{planet}}{m_{star}}$

As far as I know, the only limiting factor on a moon's distance from its planet is that eventually the gravitational field strength of the star approaches that of the planet, pulling it out of orbit. This means that we need

$\frac{m_{star}}{r_{planet}^{2}} \ll \frac{m_{planet}}{r_{moon}^{2}}$, or in other terms, $\left( \frac{r_{moon}}{r_{planet}} \right)^2 \ll \frac{m_{planet}}{m_{star}}$

Unfortunately, doing the algebra on this leads us to conclude that $\frac{r_{moon}}{r_{planet}} > 1$ which cannot happen.

Is this the case, or have I overlooked something?

  • $\begingroup$ How do you find the ratio of radii is greater than 1? $m_{planet} / m_{star} << 1$, so $r_{moon} / r_{planet}$ must also be less than 1. $\endgroup$ Commented Jul 7, 2022 at 20:20
  • $\begingroup$ @CarlosGauss What limits the distance between moon and planet then? Why do we not see moons of Jupiter 100 million miles out? $\endgroup$ Commented Jul 7, 2022 at 20:38
  • $\begingroup$ @NuclearHoagie The mass ratio equals the cube of the radii ratio. The square of the radii ratio must then be less than the cube of the radii ratio. Any number less than 1 has a smaller cube than square. $\endgroup$ Commented Jul 7, 2022 at 20:39
  • $\begingroup$ @CarlosGauss That makes no sense to me. The balance of forces underlies the entire concept of a Hill Sphere, which is the region around a body where that body's gravity dominates, and is the only place where satellites of that body may be found. Your comment implies that so long as a body and its moon are in free orbit around a star, the moon can orbit at any distance at all, but orbits around a body can only exist in that body's Hill Sphere. $\endgroup$ Commented Jul 7, 2022 at 20:41
  • 1
    $\begingroup$ Actually I was wrong, it is only true if you can approximately assume that the attraction from the sun is the same on both, planet and moon, so if their distance is small compared to their joint distance to the sun. Or I am wrong again? $\endgroup$
    – user338734
    Commented Jul 7, 2022 at 20:49

3 Answers 3


This occurs at the Lagrange points, where the moon is stationary in the rotating reference frame. A moon at L1, between the planet and the star, or at L2, on the "outside" of the planet's orbit, has a sidereal period equal to the planet's year.

Unfortunately for your idea, L1 and L2 are unstable equilibria. The stable equilibria at L4 and L5 are much more like the moon and planet are co-orbiting the star than like the moon is orbiting the planet.

You might be imagining the case where the "synodic period" of the moon is equal to the planet's year. For example, imagine that the full moon only happened in June, and the new moon only happened in December. This corresponds to a resonant orbit where the sidereal period of the moon is twice the sidereal period of the planet. You can almost certainly come up with mass ratios where this resonance is allowed, and where the moon's orbit also fits inside the planet's "Hill sphere" where orbits are stable.

Your analysis of gravitational force balance is incorrect because it neglects the centrifugal force in the rotating frame. In the limit of a low-mass planet, the radius of the Hill sphere is approximately $r_\text{moon}^\text{max} = r_\text{planet}\sqrt[3]{\frac{m_\text{planet}}{3\cdot m_\text{star}}}$.

  • 1
    $\begingroup$ I was not going for the synodic period. Ultimately the purpose of this question was to help answer a question on the Worldbuilding stack, where someone wanted to know if a month-long solar eclipse was possible. Obviously, this can't happen on a planet, but I was trying to determine if it could occur on a gas giant's moon. The gas giant would need to move so slowly in the moon's sky that it takes a month to clear the star, hence the need for a moon and planet with very similar orbital periods. $\endgroup$ Commented Jul 7, 2022 at 20:53
  • $\begingroup$ An object at L2 would be in continuous eclipse until it chaotically wandered away from the equilibrium. A moon within the Hill sphere in the 2:1 resonant orbit could have an eclipse at every "full moon," whose duration might be some fixed fraction of the year. A moon somewhere between the 2:1 and 1:1 sidereal resonances might have eclipses which last a constant fraction of a longer synodic period; their orbital stability is a numerical question. $\endgroup$
    – rob
    Commented Jul 7, 2022 at 21:11
  • $\begingroup$ The JWST is at a Lagrange point for this reason, so the earth continuously eclipses the sun. $\endgroup$ Commented Jul 8, 2022 at 0:16
  • 2
    $\begingroup$ FWIW, the projection of the JWST halo orbit onto the celestial sphere is roughly elliptical, with semi-major axis ~30° and semi-minor axis ~16°. I have some 2D & 3D plots here: astronomy.stackexchange.com/a/49616/16685 $\endgroup$
    – PM 2Ring
    Commented Jul 8, 2022 at 0:40
  • 1
    $\begingroup$ There is no "synodic period" solution to this problem. A related Worldbuilding question turned up this paper, which gives a maximum synodic period of 1/9 the parent body's year. $\endgroup$
    – Mark
    Commented Jul 9, 2022 at 0:55

A Klemperer rosette is sort of way that it could happen. Like @rob's Lagrange point idea, it is unstable.

Pick a rosette with many bodies of three types: Sun, Planet, Moon. Arrange them like this. S, P, M, P, S, P, M, P, ...

Because there are many bodies, the nearest are nearly in a line. Each Moon is hidden behind two Planets from two Suns.


I want to add to the other answers that if the orbital periods of the planet and its moon are the same the imaginary civilization in that planet can't have an astronomical calendar.

You know that the duration of a month in a calendar is the synodic period of the moon.

In a planet-moon system we have: $$\frac{1}{P_p}=\frac{1}{P_m}-\frac{1}{P_s}$$

Where $P_p$ is the planet's orbital period, $P_m$ is the moon's orbital period, and $P_s$ is the moon's synodic period.

If $P_p$ and $P_m$ are the same, then $P_s$ reaches infinity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.