6
$\begingroup$

I am a scriptwriter working on a science fiction story. As best as I can I like to reconcile my imaginary places to within the constraints of reality. Here is my problem:

Kettrah enter image description here

Kettrah is by most means a tidally locked planet which mean that it rotates on one axis in the same time it takes to complete its orbit. However this planet is different in that it also rotates while it is tidally locked. Thus the north pole is locked to the star and the planet spins around that axis while the locked pole moves on its own axis to always face the star.

If you're having trouble picturing this: Imagine Uranus, but with one pole always facing the Sun.

enter image description here Source

My questions are:

Can a planet have two rotational axes?

If so is there a way to lock one pole to a star?

$\endgroup$
  • $\begingroup$ Is this related to this answer? $\endgroup$ – Samuel Jul 30 '15 at 22:00
  • 2
    $\begingroup$ See physics.stackexchange.com/questions/19201/…. Specifically, "Euler's rotation theorem guarantees that any rotation of a rigid object can be expressed as a rotation around a single axis." However, also look at math.stackexchange.com/questions/44696/…. $\endgroup$ – HDE 226868 Jul 30 '15 at 22:15
  • 1
    $\begingroup$ A good description for what you want might be the planet Uranus, but with North (or South) always pointing at the sun. $\endgroup$ – Samuel Jul 30 '15 at 22:26
  • 2
    $\begingroup$ @JoshBelmont Well consider the Uranus diagram. The only way for the pole to continue pointing at the Sun is for something to pull it around. I'm not sure how large or fast it would need to be going, but I think an orbiting moon could pull the planet around. A moon in a polar orbit is not something that forms naturally though, it'll have to be captured by the planet. $\endgroup$ – Samuel Jul 30 '15 at 22:44
  • 1
    $\begingroup$ @JoshBelmont Again, I'm not sure about it, but the lopsided core might help. The moon will be orbiting the center of mass and will have increased tidal forces on the South side as is passes more closely. $\endgroup$ – Samuel Jul 30 '15 at 23:17
3
$\begingroup$

Yes, this is possible; the rotation that keeps the pole pointed toward the star would be considered to be precession. If this planet was in a close orbit around a red dwarf star, it would be close enough that it would be tidally locked, and gravity would provide the necessary torque, while remaining in the star's habitable zone.

$\endgroup$
  • $\begingroup$ Thanks Monty Wild and especially @Samuel for helping me figure this one out! I will take the extremely sparse answers section to mean that this topic can be debated both ways. But debated means plausible, and that is a very good thing. Lopsided core with a polar orbiting large moon around a red dwarf star so that it precesses. Now that's what I am talking about! Consider this one answered. $\endgroup$ – Josh Belmont Aug 1 '15 at 6:59
  • 4
    $\begingroup$ I disagree. The precessing axes won't follow the motion of the sun, but will be much slower, taking many orbits to turn around. This will not be stable: the planet will podhole and the axis will shift until it is a normal tidal face locking. $\endgroup$ – JDługosz Jun 19 '16 at 18:10
  • $\begingroup$ @JDługosz Precession doesn't have to happen at a specific speed - it could, in principle, happen in exactly one orbit. A lucky strike by another planetary body, earlier in the system's history, could have set it spinning just so... I think it's unlikely, but not impossible. youtube.com/watch?v=ty9QSiVC2g0 $\endgroup$ – Joanna Marietti Sep 20 '16 at 15:56
  • 3
    $\begingroup$ @JoannaMarietti the precession does happen at a specific speed determined by the torque applied. It is not “set spinning” by some previous event, but requires continued application of force. This is normally due to tidal effects against the equator’s bulge, and it involves a change of angular momentum. Your proposal is off by 3 or 4 orders of magnitude and is not a reasonable value, and would never last with a high value. $\endgroup$ – JDługosz Sep 20 '16 at 23:24
  • $\begingroup$ The video: yea, freshman physics class demo of a bicycle wheel. I know what precession is so what is that supposed to add? You’ll notice in the demo that the off-center gravity is the driving force in this case, and driving it makes the wheel drop over time, and (harder to tell) causes friction agaist the axle bearings. It runs down as this is a dissapatory process. Given the wheel demo, how do you think that a good whack would make it precess at a different rate? Do you know what determines the rate you see? … $\endgroup$ – JDługosz Sep 20 '16 at 23:30
3
$\begingroup$

No

An object rotating without any outside forces on it will have one stationary angular momentum vector. In order to have a precession of this angular momentum vector, torque must be applied. This torque added (integrated) over time will add to the current angular momentum vector to give the new angular momentum vector. In order for the angular momentum vector to always point towards (or away from) the sun the torque applied would have to be strong enough in magnitude that it could stop the planet rotation in quarter year. That might not be so bad except the direction the torque would be rather difficult to achieve. If you consider the axis of motion of the planet around the sun as up, then the torque would have to be pushing the bottom of the planet and pulling on the top. Unfortunately, since the planet is rotating, those halves are switching twice a day, so any gravitational pull would net do nothing, unless your day was really long (like if a day was equal to a year and thus the planet was tidally locked)

But there is another type of precession, a torque free precession where the axis of instantaneous rotation changes while the axis of angular momentum remains stationary. However, this could also not produce the desired movement as the two axis cannot point in opposite directions (or even be perpendicular) but in your described movement the axis of rotation points in completely the opposite direction every half year.

$\endgroup$
0
$\begingroup$

I don't think it is possible to achieve this for a planet orbiting a star. But you could in effect achieve this with a moon that's tidally locked to a planet, which it orbits in a sun-synchronous orbit. This requires a fast-rotating planet (so it has a strong equatorial bulge). For Earth it's not possible (all sun-synchronous orbits lie within the Roche limit), but for a Jupiter-like planet it might in principle work. The mass of a gas giant would easily allow for an Earth-sized moon.

Mind, a moon in such a specific, nearly polar orbit is pretty unlikely... also, the planet would probably have massive volcanic activity because of the planet's tidal influence.

$\endgroup$

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.