<p>There is quite simple <a href="https://en.wikipedia.org/wiki/Tidal_locking#Timescale" rel="nofollow">formula</a> that will give you the tidal-locking half-time</p>

<p>$$T = C\, \frac{a^6 R \mu}{  M_s M_p^2}$$</p>

<ul>
<li>$a$ - semi-major axis, or simply the radius of the planet circular orbital trajectory in meters</li>
<li>$R$ - planet radius in meters</li>
<li>$M_s$ - planet mass in kg</li>
<li>$M_p$ - parent planet/star mass in kg</li>
<li>$\mu$ - rigidity, approximately $3×10^{10}$ for rocky objects and $4×10^9$ for icy ones.</li>
<li>$C$ - a prefactor dependent on units used. On Wikipedia, they write $C = 6 × 10^{10}$ if the time is to be in years, but this number is probably wrong. (<a href="https://en.wikipedia.org/wiki/Talk:Tidal_locking#Contradicting_formulae_in_.22Timescale.22" rel="nofollow">See the discussion</a>.)</li>
</ul>

<p>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 </p>

<p>$$ \frac{A^6}{  M_s^3} \ll \frac{a^6}{  M_s M_p^2}\;,$$</p>

<p>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. $M_s = 6\times 10^{24}\;\mathrm{kg}$ and $M_p = 2.9\times 10^{29}\;\mathrm{kg}$. We get</p>

<p>$$2\cdot 10^{-41}\ll 6.3\cdot 10^{-36}\;,$$</p>

<p>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 </p>

<p>$$2\cdot 10^{-35}\ll 6.3\cdot 10^{-36}\;,$$</p>

<p>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 <a href="http://en.wikipedia.org/wiki/Hill_sphere" rel="nofollow">Hill sphere</a>, which approximately gives the stable orbits.) </p>

<p>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.</p>

CC BY-SA 3.0
2015-03-14T11:35:15.470Z
2015-03-14T12:06:31.130Z
2015-03-14T12:06:31.130Z
Irigi
<p>Would twin planets orbiting around each other present a solution to tidal locking of habitable planets in red dwarf systems? It's an awfully easy equation to set up. I'm fibbing; I can't do it. I was hoping someone might show me when they get a moment.</p>

<orbit><tidal-forces><red-dwarf>
The tidal locking problem concerning Earth sized planets in habitable zone around Red Dwarfs
C hibbs
# The tidal locking problem concerning Earth sized planets in habitable zone around Red Dwarfs

(Asked by: username_0 on 2015-01-22)

Would twin planets orbiting around each other present a solution to tidal locking of habitable planets in red dwarf systems? It's an awfully easy equation to set up. I'm fibbing; I can't do it. I was hoping someone might show me when they get a moment.

**Tags:** orbit, tidal-forces, red-dwarf


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## 1 answers

### Answer by username_1 on 2015-03-14. Score: 2

There is quite simple [formula](https://en.wikipedia.org/wiki/Tidal_locking#Timescale) that will give you the tidal-locking half-time

$$T = C\, \frac{a^6 R \mu}{ M\_s M\_p^2}$$

* $a$ - semi-major axis, or simply the radius of the planet circular orbital trajectory in meters
* $R$ - planet radius in meters
* $M\_s$ - planet mass in kg
* $M\_p$ - parent planet/star mass in kg
* $\mu$ - rigidity, approximately $3×10^{10}$ for rocky objects and $4×10^9$ for icy ones.
* $C$ - a prefactor dependent on units used. On Wikipedia, they write $C = 6 × 10^{10}$ if the time is to be in years, but this number is probably wrong. ([See the discussion](https://en.wikipedia.org/wiki/Talk:Tidal_locking#Contradicting_formulae_in_.22Timescale.22).)

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

$$ \frac{A^6}{ M\_s^3} \ll \frac{a^6}{ M\_s M\_p^2}\;,$$

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. $M\_s = 6\times 10^{24}\;\mathrm{kg}$ and $M\_p = 2.9\times 10^{29}\;\mathrm{kg}$. We get

$$2\cdot 10^{-41}\ll 6.3\cdot 10^{-36}\;,$$

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

$$2\cdot 10^{-35}\ll 6.3\cdot 10^{-36}\;,$$

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](http://en.wikipedia.org/wiki/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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