Compute Tidal Locking Radius: Planet/Moon Area

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rieman zeta
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How does one compute the tidal locking radius of say a planet on a putative moon its area?

Is there a formula?
rieman zeta
 
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There's a formula for how fast tidal lock occurs, see for instance

http://groups.google.com/group/rec.arts.sf.science/msg/e05283a619187a8f?dmode=source&hl=en

Both you and Geoffrey have rightly commented on my poorly-defined
variables, so let me re-state this. BTW, I'm cribbing this formula (very
slightly modified) from somewhere else, namely Burn's chapter in
"Satellites" (U of Az Press), edited by Burns & Mathews:

T = 16 rho omega a^6 (Q/k2) / ( 45 G M^2 )
rho = density of body being despun [kg/m^3]
omega = inital rotation rate of body being despun [rad/s]
= 2 pi / P, where P is the inital rotation rate
a = semi-major axis of orbit [m]
Q/k2 = dissipation function divided by the 2nd order Love #
M = mass of body doing the despinning [kg]

I hope this is clearer - I've taken the formula for the despinning
timescale out of Burn's chapter and modified it very slightly.


The very rough justification for this formula goes like this:

tidal height is proportional to (M/a)^3
stored tidal energy is proportioanl to tidal height squared
some fraction of the stored tidal energy gets disapated every cycle. A cycle occurs every time the planet rotates.

This gives a M^2/a^6 dependence on the tidal braking torque, or an a^6/M^2 dependence on the "time constant". This addresses only the dependence on mass and distance, but those are the main variables of interest.

'a' here is the semi-major axis of the orbit of the body being locked around the more massive body, what you would be calling (I think) the "tidal locking radius" of the more massive body.

Note that you have to specify the time allowed for the lock occurred - theoretically, anything will lock up given enough time.

For more details, see the quoted textbook source. I really don't know much more than what I've quoted (plus the comments I've added about the a^6 dependency) - specifically I don't have much insight into the numerical values of Q and k2 (though Brian Davis probably does, I don't think he's on this board).
 
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thanks

Although I am light years ahead of where I was before your reply, I would invite others to continue to edify me.

Thanks
Riemann Zeta