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

  1. Jan 3, 2015 #1
    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?
  2. jcsd
  3. Jan 3, 2015 #2

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    You would have libration - a rocking back and forth. The moon does this.
  4. Jan 5, 2015 #3
    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 ωl is the libration angular frequency. I say direction(s) because for a planet's libration, both 0d and 180d are stable directions.
  5. Jan 5, 2015 #4
    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.
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