8
$\begingroup$

I have read about about the habitability of red dwarf systems on Wikipedia, as well as some web articles on similar topics. The problem is, it does not explain why and how it happened. Google search comes up inconclusive. Since then, it became a riddle for me.

We all know that higher mass supergiant stars have stronger gravity than lower mass dwarf stars. Therefore, I can assume that stronger gravity results in stronger tidal forces as well. So, I can conclude:

High mass star systems should have a higher tidal lock probability than lower mass dwarf systems, right?

From that assumption that contradicts the reality, here comes another question:

Does it mean that the tidal lock probability has more to do with the size of the planet relative to the host star size than its mass or gravitational forces?

But here is my other assumption: Probably a more concentrated (almost point source) of gravity from smaller red dwarf stars locks planets more effectively than a broad giant star. (It is kind of analogous to a small speaker magnet that has stronger magnetic field compared to the Earth's magnetic field when measured).

$\endgroup$
2
  • 2
    $\begingroup$ In short, the habitable zone is much closer to the star for red dwarves, and it's the distance from the star that more prominently affects tidal forces. $\endgroup$ Apr 3, 2020 at 19:29
  • $\begingroup$ Could this be an artifact of sampling bias? The way we search for exoplanets might give us a false sense of the probabilities because it's much easier to find them around certain types of stars. $\endgroup$
    – eps
    Apr 3, 2020 at 22:38

1 Answer 1

16
$\begingroup$

A rough back-of-the-envelope way of seeing what's going on...

The tidal force is due to the difference in gravitational force, so follows an inverse cube law:

$$F_\mathrm{tide} \propto M_\ast R^{-3}$$

where $M_\ast$ is the stellar mass and $R$ is the distance. So at the same distance from a less massive star, the tidal force will indeed be weaker. But what's usually of interest is the effect on planets in the habitable zone. As a first approximation, define the habitable zone as the location where a planet receives the same amount of insolation from its star as the Earth does from the Sun. The inverse square law then gives:

$$R_\mathrm{HZ} \propto L_\ast^{1/2}$$

where $L_\ast$ is the luminosity of the star. For main sequence stars, we can approximate using a power-law mass–luminosity relationship:

$$L_\ast \propto M_\ast^{k}$$

where the exponent $k$ is traditionally given as 3.5, although it may vary between about 2 and 4 in the mass range of interest. Combining these gives:

$$F_\mathrm{tide,\ HZ} \propto M_\ast^{1-3k/2}$$

The exponent is negative for $k > \frac{2}{3}$, which is the case for the main-sequence mass–luminosity relationship. Therefore the tidal force in the habitable zone increases with decreasing stellar mass: the smaller distance to the habitable zone for these stars provides a stronger effect than the decreased mass.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .