Is there any way to avoid the tidal locking of a planet orbiting a red dwarf in the habitable zone?
For example, could a planet with a 90° obliquity and large moon avoid such a situation?
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Yes: It has a companion planet or an excessively large moon, with the two bodies orbiting their common center of mass (much like the Earth and the Moon). They could be tidally-locked to each other, but they cannot be tidally-locked to their star.
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2$\begingroup$ Can you provide some evidence for why this would work? Why couldn't it be that each body becomes tidally locked with the star first, thus preventing the planets from being tidally locked with one another? $\endgroup$ – HDE 226868♦ Nov 13 '16 at 19:57
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5$\begingroup$ I don't believe this answer is deserving of being the best answer. It does not attempt to validate or cite references for its statements. After all, anyone can say anything on the internet. I would prefer more proof for the statements above. $\endgroup$ – zephyr Nov 14 '16 at 14:17
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$\begingroup$ I thought anyone can edit; it just goes to the review queue. $\endgroup$ – JDługosz Feb 4 '17 at 5:11
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Leconte et al. (2015) suggested that the presence of an atmosphere could prevent or at least slow tidal locking. The star should exert two separate torques: one on the atmosphere and one on the solid body of the planet: $$T_a=-\frac{3}{2}K_ab_a(2\omega-2n),\quad T_g=-\frac{3}{2}K_gb_g(2\omega-2n)$$ where $$K_a\equiv\frac{3M_*R_p^3}{5\bar{\rho}a^3},\quad K_g\equiv\frac{GM_*R_p^5}{a^6}$$ for stellar mass $M_*$, planetary radius $R_p$, mean density $\rho$, semi-major axis $a$, mean motion $n$, rotation rate $\omega$, and response to torques $b_a$ and $b_g$. The two torques could be equal, and assuming that the atmosphere transfers some angular momentum to the surface of the planet, this could prevent tidal locking. There are several equilibria at which this could occur:
answered Nov 12 '16 at 16:38
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$\begingroup$ Would I be wrong to say this is very possibly occurring on Venus? $\endgroup$ – userLTK Feb 3 '17 at 17:25
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The more likely case is actually a spin-orbit resonance that is not 1:1 but a half odd multiple, like the 3:2 case of our own Mercury. Having eccentricity in the orbit encourages this situation.
I’ve been meaning to write this up on the Worldbuilding.SE but I have not re-found enough references. But see this video.
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$\begingroup$ very exciting video. U should have posted that information on WB, what if I didn't looked here? $\endgroup$ – MolbOrg Nov 19 '16 at 12:23
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$\begingroup$ Mercury is a great example, and around a red-dwarf with a shorter orbital period (like the 11 day period of proxima centauri b), with a mercury orbit that would be 264 hours of sun, 264 hours of night. Not ideal, but a happy medium between Earth's 12 hours of sunlight and Mercury's several months. $\endgroup$ – userLTK Feb 3 '17 at 17:33
