Saturday, 15 October 2011

orbit - The tidal locking problem concerning Earth sized planets in habitable zone around Red Dwarfs

There is quite simple formula that will give you the tidal-locking half-time



T=C,fraca6RmuMsMp2



  • a - semi-major axis, or simply the radius of the planet circular orbital trajectory in meters

  • R - planet radius in meters

  • Ms - planet mass in kg

  • Mp - parent planet/star mass in kg

  • mu - rigidity, approximately 3×1010 for rocky objects and 4×109 for icy ones.

  • C - a prefactor dependent on units used. On Wikipedia, they write C=6×1010 if the time is to be in years, but this number is probably wrong. (See the discussion.)

Solving the problem of interaction between the parent star and between the planets will be difficult. Generally, if tidal-locking half-time between the planets is much shorter than the tidal-locking half-time between the star and the planet, the planets will tidal-lock to each other rather than to the star. So we ask, whether



fracA6Ms3llfraca6MsMp2;,



where A is the semi-major axis of the orbit of the double-planet. For typical planet in the habitable zone of the red dwarf star, a = 9,000,000,000 m (0.06 AU), while the distance of the double-planet could be for example A = 30,000,000 m. Ms=6times1024;mathrmkg and Mp=2.9times1029;mathrmkg. We get



2cdot1041ll6.3cdot1036;,



which is well-satisfied and the planets will tidal lock to each other rather than to the parent star. Were the distance between the planets A = 300,000,000 m, the inequality would yield



2cdot1035ll6.3cdot1036;,



which is not satisfied and the tidal forces of the star would probably disrupt the double-planet. (The planets would not even be in the Hill sphere, which approximately gives the stable orbits.)



So the answer is: a very close double planets can tidal-lock to each other rather than to the parent red-dwarf star, but they would have to be much closer than the Earth to the Moon.

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