7
$\begingroup$

Imagine an earth-type world orbiting a Jupiter-like gas giant roughly within the habitable zone. If the plane on which the moon orbits the planet is tilted at an angle relative to the Solar System's ecliptic plane would you get typical seasons like that on Earth that are determined by the year of the (gas giant) planet?

I sketched a rough diagram below for reference.

Planet's location and orientation in both summer and winter

$\endgroup$
1
  • 1
    $\begingroup$ BTW, I like the idea. To start your research, consider this video. $\endgroup$
    – JBH
    Jan 31 at 23:57

1 Answer 1

7
$\begingroup$

The answer is a yes! We know moons that have orbits like that and should experience seasons, and we know what needs to happen for them to form.

Take a look at the moons of Jupiter -- and in particular, their orbital inclinations. Many fall into groups that share common characteristics and orbits. For example, consider the Himalia group. Among other things, they all have orbital inclinations of approximately 27-30 degrees relative to the plane of the Solar System, which is close to Earth's axial tilt. This is because the group of moons are likely the remains of a smaller body that was captured by Jupiter in a chaotic multi-body interaction, and subsequently fragmented.

The same is true for other groups of moons that were captured by planets: gravitational capture can force them into orbits with significant inclinations. On the other hand, Jovian moons that were not captured by Jupiter -- for example, the Galilean moons -- have much lower orbital inclinations, since they were likely formed from the circumplanetary disk Jupiter accumulated from the Solar System's protoplanetary disk, which spanned the ecliptic plane. Absent axial tilt, they would not experience seasons.

So the recipe you want is for the gas giant to gravitationally capture a roughly Earth-sized protoplanet, and for it to fall into an orbit not too close for the gas giant's tidal forces to tear it apart.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .