Transit timing analysis of the exoplanets TrES-1 and TrES-2

Sedna wrote:Sorry if you think that I am stupid after reading this, but I cannot understand the term of tidal dissipation, even though I try to get it.

No not at all. I found it difficult to grasp tidal forces too. Tidal dissipation is how easy it is for a planet to get rid of tidal energy without it affecting the orbit of the moon.

As a moon orbits around a planet, it raises a tide on that planet like the Moon does on Earth. Earth's rotation will carry this bulge ahead of the sub-lunar point (the place on Earth directly below the moon).


Since that bulge on the primary is ahead of the sub-lunar point, the bulge's gravity will pull on the moon ever so slightly. Since the bulge is ahead of the moon, the moon will be gravitationally pulled in the direction it is already orbiting around Earth. This causes it to speed up a little, and an increase in speed translates to an increase in altitude (orbital dynamics).

Earth is pretty solid being a terrestrial planet. The tide raised on it by the moon won't go away too easily. What really helps dissipate this tidal bulge on Earth is the oceans, because they're easily movable and can flow. The more efficiently a planet can get rid of this bulge, the less the bulge will be able to affect the moon.

Considering a terrestrial planet without oceans, such a planet will have more difficulty getting rid of this tidal bulge because the solid nature of the planet doesn't allow the tidal bulge to be dissipated as easily as it would if the planet were made of gas. As such, gas planets are much more efficient at getting rid of the tidal bulges caused by their moons than terrestrial planets.

In the case of a hot Jupiter, which will certainly be tidally locked to the star, any stable satellite orbital periods will be shorter than the rotation period of the planet. This will cause the tidal bulge to drag behind the moon, lowering its orbit instead. Now tidal dissipation here determines how quickly the moon will fall into the planet, instead of how long it will take to escape (like at the Earth/moon system).

The precise reason for why, (and by extension, how to measure Q_p) is a topic of continuing research. Quoting from the paper Lazarus linked to earlier,

The physical origin of the tidal Q_p in gas giant planets has been an outstanding question since at least the 1970's. The Q_p for Jupiter is constrained to be Q_p ~ 10⁵ - 10⁶ in order that Io's orbit expanded into the Laplace resonance. Early theoretical work by Hubbard (1974) showed that the turbulent viscosity generated by convective eddies required for the outward transport of heat would give rise to Q_p ~ 10⁵, perhaps explaining the observed value. Goldreich & Nicholson (1979) then pointed out that turbulent eddies in the planet have long turnover rates, compared to the forcing periods of interest, severely decreasing the turbulent viscosity used in Hubbard's calculation. Goldreich & Nicolson (1977) estimated Q_p ~ 10¹³ for "equilibrium-tide" flow in Jupiter, underpredicting the observed tidal dissipation rate by a factor of 10⁷ - 10⁸. Wu (2005) revisited Goldreich and Nicholson's calculation, revising Q_p downward to ~ 10¹²

Tidal quality factor Q is the ratio of the energy stored in the tidal bulges to the energy dissipated per tidal cycle (from here. So higher Q means the planet doesn't dissipate the energy so quickly, hence longer survival times around planets with higher Q.

Does a stable zone exist? The entirety of the Hill sphere is not stable. Using the result from here that puts the outermost stable orbit of a prograde satellite at 0.4895 times the Hill sphere (for zero eccentricity planetary orbits). For TrES-2, the outermost stable prograde orbit is at 2.09 planetary radii. Retrograde satellites are stable out to 0.9309 times the Hill sphere, or almost 4 planetary radii, slightly smaller than the orbit of our moon and comparable to Triton's orbit around Neptune. The Roche limit for an Earth-density satellite is at about 1.23 planetary radii, so the stable zone is not zero.

Wait, it seems I'm confused. Higher Q_p means the planet is worse at dissipating tidal energy? Wouldn't that cause the bulge to persist and drag the moon in quicker?

For the orbit to decay, the energy has to go somewhere... your choices are basically rotation of the planet, rotation of the satellite, heat...

I assumed that in getting rid of the tidal bulge, the energy would go into heating of the planet (with a rocky planet having lower Q and thus gets heated more than a gas planet with higher Q). Is that incorrect?
And what do I need to correct in my post up there?

Thank you very much Sirius_Alpha for your comment, that helps me out. So, if I'm right: the greater the tidal dissipation parameter is, the bigger the potential moons are (feel free to correct if it's wrong).
About Earth and Moon and going back to the time when the Moon was still a "baby". At that time, Earth had no oceans and the Moon was very close to it (approx. 30,000 km). The Q must have been very high to avoid the Moon to crash on it (or maybe the lava oceans replaced the water oceans). If it's true, I think that rocky planets in close orbits could host moons so that the ratio could be small, like the confusing-I-think Corot-7 b.

Bye

Sedna