How the last days look like just before a tidal lock?

[Czcibor]

Would a planet before getting a tidal lock to its star have for some transitory period days which would be centuries long? Or is there a threshold where it would just get from long days suddenly to tidal lock with everything levelling out?

 

[Vanadium 50]

You would have libration - a rocking back and forth. The moon does this.

 

[lpetrich]

That libration would only happen if the planet was rigid enough to have a triaxial distribution of mass. Here's the equilibrium configuration:

• The long axis: the direction to the star
  
• The intermediate axis: the direction of motion in the planet's orbit
  
• The short axis: the orbit's pole

But it would not happen if the planet was mostly fluid, like a gas giant. Its lack of rigidity would mean a lack of "handles" for libration.

A good analogy to the late stages of spindown is an unbalanced wheel. Its equations of motion also closely resemble the equations of motion for the planet's longitude libration, and have much the same behavior. As the wheel spins down, it changes from circulation (complete rotations) to libration in pendulum fashion.
$$ \frac{d^2 \theta}{dt^2} = - (\omega_l)^2 \sin \theta ,\ \frac{d^2 \theta}{dt^2} = - \frac12 (\omega_l)^2 \sin (2\theta) $$
The unbalanced-wheel equation and the planet's libration equation, where θ is the orientation angle around the stable direction(s) and ω_l is the libration angular frequency. I say direction(s) because for a planet's libration, both 0d and 180d are stable directions.

 

[lpetrich]

I will solve the second equation using Jacobi elliptic functions.

Libration: $\theta = \arcsin ( \sqrt{m} \text{sn} (\omega_l t, m) )$
Transition (m = 1): $\theta = \arcsin ( \tanh (\omega_l t) )$
Circulation: $\theta = \arcsin ( \text{sn} ((\omega_l / \sqrt{m}) t, m) )$

They have limits:
Libration: $\theta = \sqrt{m} \sin (\omega_l t)$
Circulation: $\theta = (\omega_l / \sqrt{m}) t$

Here, m is the elliptic-function parameter. A common alternative is to use its square root as the parameter.

 

[pmacias]

The last days before a planet becomes tidally locked to its star would likely involve a gradual increase in the length of its days. As the planet's rotation slows down due to the gravitational forces of its star, the days would become longer and longer until they eventually match the planet's orbital period around the star. This process could take centuries or even millennia, as it depends on the specific characteristics of the planet and its star.

It is possible that there could be a transitional period where the days become extremely long, potentially lasting for centuries. However, it is more likely that the planet would reach a threshold where it becomes tidally locked with its star, with the days suddenly becoming equal to the planet's orbital period. This is because once the planet's rotation slows down enough, the gravitational forces from the star would be strong enough to keep the same side of the planet facing the star at all times.

Ultimately, the exact timeline and process of a planet becoming tidally locked would vary depending on the specific conditions of the planet and its star. But it is certain that as a planet approaches its tidal lock, the days would gradually become longer until they reach a point of equilibrium with the planet's orbital period.