6
$\begingroup$

Could 2 moons that orbit same terrestrial planet never see each other if they orbit the planet at same time?

Moons have different mass and gravity.

|improve this question
$\endgroup$
  • 3
    $\begingroup$ If I recall correctly, for some time people seriously thought about the possibility of there being a planet invisible to us due to being on the other side of the sun. EDIT: A quick google search yields this: en.wikipedia.org/wiki/Counter-Earth. It seems that contrary to what I remembered nobody took the idea really seriously. Still the physics to make it work would be similar to what you are looking for. $\endgroup$ – Vincent Aug 19 at 15:59
  • 1
    $\begingroup$ If we consider the rocks making up a Saturn-like ring and call them moons, I suppose that some opposite rocks do not see each other (or at least not for quite a long time) - but of course the presence of gazillion other rocks srongly hints to the existence of more rocks behind the planet ... $\endgroup$ – Hagen von Eitzen Aug 19 at 19:16
  • $\begingroup$ Such an orbit is theoretically possible, but in practical reality it's an unstable orbit due to tidal forces. You would need multiple moons in resonance to be able to create an orbit that keeps two moons opposite each other. $\endgroup$ – stix Aug 19 at 20:40
20
$\begingroup$

In theory if the two moons were in the exact same orbit on opposite sides of the planet then yes. Having the moons closer to the planet and smaller also makes that easier. For example geostationary satellites over opposite sides of earth will never have direct line of sight to each other.

In practice though that would be a very unstable arrangement (even if there were no other moons to disrupt things) and would also be very unlikely to form naturally.

So it would be very unlikely to form naturally and if it did form it would be unstable ... so realistically the answer is "no" but if you can explain away the improbabilities somehow then "yes".

The moons having different masses doesn't change their behavior in this case. If they are in the same orbit they are in the same orbit.

|improve this answer
$\endgroup$
  • 2
    $\begingroup$ There is alsa some kind of magic in the world so i could just say "A WIZARD DID IT" $\endgroup$ – AIwithstick Aug 19 at 8:40
  • 4
    $\begingroup$ Powerful wizard. Moons are BIG. $\endgroup$ – Tim B♦ Aug 19 at 8:42
  • 2
    $\begingroup$ There's something to be said for human timescale and geological/universal timescale. It's true it would be unstable over 10's of thousands of years, but it could work for at least a couple hundred years, or even a millennium or two. The civilization could have myths about a 2nd moon that people don't believe until it starts to crest the planet, causing religious and scientific disturbances. $\endgroup$ – computercarguy Aug 19 at 16:22
  • 2
    $\begingroup$ Would you mind explaining why synchronous orbits would be unstable? If they're given the same exact parameters I don't get why it wouldn't be stable $\endgroup$ – Ferdz Aug 19 at 17:22
  • 4
    $\begingroup$ @asgallant I'm not sure what you mean? I clearly say that the orbit will be unstable. $\endgroup$ – Tim B♦ Aug 19 at 18:53
8
$\begingroup$

Yes, this is possible.

A large moon and a smaller moon can share the same orbit if one is 60 degrees ahead of the other. In such an orbit, the smaller moon would be at one of the stable Lagrangian points L4 and L5. If the orbital radius is less than $\frac{1}{\cos (30^{\circ})} = \frac{2}{\sqrt{3}}R_M \approx 1.15 R_M$ (where $R_M$ is the radius of the planet), then the planet will block the line of sight between the two moons. That is, each moon will be beyond the horizon as seen from the other moon.

Of course, such orbits would be very close to the planet. Would the moons break apart due to tidal forces? The answer to that is given by the Roche limit, which for a rigid satellite is

$$ d = R_M \left( 2\frac{\rho_M}{\rho_m} \right)^{1/3} $$

where $\rho_M$ and $\rho_m$ are the densities of the planet and the moon respectively. If the moons orbit outside this radius, they will survive. If they are inside the radius, they will break apart. For our scenario, we need the Roche limit to be less than $1.15 R_M$, so the density of the moons must be at least 30% larger ($\frac{3^{3/2}}{2^2}$) than the density of the planet.

Summary of requirements

  • The moons and planet should form an equilateral triangle (the Lagrangian point).
  • The moons share an orbit that is less than 15.5% larger than the radius of the planet (so that the line of sight is blocked).
  • One moon must be at least 24.96 times larger than the other (to allow a stable orbit).
  • Both moons at least 29.9% denser than the planet (to avoid destruction by tidal forces). For example, perhaps the planet is rocky while the moons are primarily composed of metals like iron.
  • The moons must have rigid solid interiors (otherwise they will deform due to tidal forces and eventually break apart).
|improve this answer
$\endgroup$
  • 2
    $\begingroup$ Nice answer. Maybe add a summary at the end of the conditions required. $\endgroup$ – Apollys supports Monica Aug 20 at 0:53
  • 1
    $\begingroup$ Beautiful answer! $\endgroup$ – Vincent Aug 20 at 8:50
0
$\begingroup$

Orbits are elliptical, normally quite eccentric - our moon's almost circular orbit is unusual. For two moons not to see each other, both their orbits would have to be extremely circular and almost exactly in the same plane.

The system would be unstable. If one moon lead the other by a tiny fraction it would be accelerated by the lagging moon and the lagging moon would be dragged by the leading one. This would rapidly cause the system to collapse.

However, it is not impossible.

|improve this answer
$\endgroup$
  • 2
    $\begingroup$ They wouldn't have to be circular necessarily. Two elliptical orbits of the same eccentricity but opposite longitude of the AN and opposite inclination would never see each other. That, however, is incredibly unlikely and unstable. $\endgroup$ – Hearth Aug 19 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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