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Dwarf tides

In the various discussions about the dynamics of the potentially habitable planet around Gliese 581, the issue of tide-locking comes up. How do we know that a planet like this may face this problem? It's closer to the parent star, of course, but the star is a lot less massive as well.

It turns out it's not too hard to estimate the tidal forces, at least at the order-of-magnitude level, for a clone of Earth put around another star. More details below.

The tidal force is dependent on several factors, specifically the masses of the two bodies, the distance between them, and the size of the body affected by the tides (the larger it is, the larger the gravitational gradient across it and hence the larger the tides are). Additionally, the actual time it take to tide-lock a body depends on factors like the rigidity of the body. For this calculation, let's imagine we "clone" Earth, at exactly the right distance from the other star to get exactly the same amount of light as it gets from the Sun. With this in mind, the size, mass, and composition of the planet don't matter, so the only factors that matter are the mass of the star and the planet's orbital distance.

For Gliese 581, these are quite well known. The mass, m, of Gliese 581 A is relatively easy to determine from the orbital information, and is 0.31 times that of the Sun. To get the distance, d, we need to know how bright the star is. Astronomers call the total absolute brightness of the star its luminosity, L, and often normalize it to the Sun's luminosity, so that a star twice as luminous as the Sun has L = 2. The luminosity of Gliese 581 A works out to 2.0 x 10-3. It's quite a bit fainter than the Sun. Our "clone" of Earth thus has to orbit at d = (2.0 x 10-3)0.5 = 0.0447 times as far from Gliese as the Sun is from Earth. Gliese 581 C orbits at 0.041 times as far, so you can see it's pretty close to the correct distance!

From the tidal force equations, you can see that the primary tidal force component is proportional to m, and also proportional to d-3. Thus the expected tide force is equal to (0.31) x (0.0447)-3, or about 3500 times as much as the Sun's tide on Earth. This will cause a tidal lock much sooner than the Earth would ever experience. It doesn't help that Gliese 581 C is larger than Earth, since a larger planet will experience larger tides, all else being equal, as mentioned above.

Can we generalize this to other stars? It turns out we can, at least approximately. A star's luminosity is consistently related to its mass, at least for most fairly low-mass stars like the Sun and red dwarfs like Gliese 581. If we normalize the star's mass and luminosity to multiples of the Sun's, the luminosity generally follows a power law with an exponent equal to approximately 3.5. The value is empirical, from curve-fitting data, so there's some uncertainty; I've seen values anywhere from 3 to 4 depending on the data used. For the sake of this discussion, let's represent the exponent by a. So the luminosity, L, = ma. Remember, the distance we need for our Earth clone is just L0.5. Thus the Earth-habitable distance (d) is proportional to ma/2. Finally, since the tide is proportional both to m and to d-3, the estimated tide strength for our copy of Earth at another star is m * m-3a/2 = m (1-3a/2). This exponent is negative for all a > 2/3, meaning the tidal force strength goes up as mass goes down. For all ordinary stars, a is greater than 1 (larger masses lead to larger luminosities more quickly than the mass itself goes up). Thus it's safe to say, as a general rule, tidal forces for an Earthlike planet go up as stellar mass goes down.

Assuming the validity of the mass-luminosity relationship, you can estimate the tides knowing just the luminosity, since you can estimate the mass as well. Let's assume Gliese 581 follows the mass-luminosity relationship already described with a = 3.5. With a luminosity of 2.0x10-3, m = 0.169 and the tidal force estimate, m (1-3a/2), is approximately 7000 times that of the Sun acting on Earth. This estimate of the mass is too low, but the tidal force is within a factor of 2 of a more direct calculation.

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