Since the tidal bulge is always in the same place, how would that affect ocean tides? Would they change throughout an elliptical orbit, due to changing distance from the star? How exactly would they behave? And would tides on a tidally locked moon behave the same way?


2 Answers 2


Since the tidal bulge is always in the same place, how would that affect ocean tides?

The concept of tidal bulge is a useful fiction, but fiction nonetheless. For an object in an eccentric orbit, the object's rotation rate and the object's orbital rate are rarely equal. There would be tidal effects. We see this on the Earth's Moon, where eccentricity results in moonquakes.

We also see this on Jupiter's moon Io. Io is caught in a 1:2:4 resonance with Ganymede and Europa. These resonance effects tend to make Io's orbit more eccentric. This increased eccentricity in turn increases the tidal effects by Jupiter on Io. (Io's tides would be frozen if Io was in a circular orbit.) The increase in tidal effects in turn causes internal heating in Io. This in turn results in factors that decrease Io's eccentricity. The end result is a nice hysteresis loop. Io's core is coolish and hard when it is in a fairly circular orbit. Ganymede and Europa pull it out this circular orbit. Tidal heating due to increased eccentricity makes Io's core get warmer and not quite so hard, which pulls it back to more circular orbit. And so on.

  • $\begingroup$ Thank you! This is super helpful! So, on Europa, for example, there would be high tides twice a day from the pull of Io and every other day from the pull of Ganymede? And the tides from Jupiter would increase and decrease with the fluctuating eccentricity? I understand the concept of why the tides from Jupiter would increase and decrease with changing eccentricity, but do you know on what timescale this change in eccentricity would occur? And how that would affect daily tides induced by Jupiter? $\endgroup$
    – Elhammo
    Commented Sep 28, 2022 at 18:21
  • $\begingroup$ You say that if Io's core is not quite so hard then this pulls it back to more circular orbit. How does this work? $\endgroup$ Commented Sep 28, 2022 at 19:42
  • $\begingroup$ @AdamChalcraft Tidal energy dissipation is not uniform over time. It is greatest when the rotation rate and the instantaneous orbital rate differ by the most. Averaged over time, tidal energy dissipation is proportional to $k_2/Q$ (and also proportional to the square of eccentricity). Compared with a planet or moon with a solid interior, a planet or a moon with a partially molten interior has a higher second order tidal Love number $k_2$ and a lower tidal quality factor $Q$, so $k_2/Q$ is considerably larger for a partially molten object compared to a rock-solid object. $\endgroup$ Commented Sep 28, 2022 at 21:48
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    $\begingroup$ @AdamChalcraft Another way to look at it is that the tidal forces at aphelion act to raise perihelion while the tidal forces at perihelion act to lower aphelion. The end result is circularization. $\endgroup$ Commented Sep 28, 2022 at 21:55
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    $\begingroup$ @Karl Solid bodies aren't perfectly rigid, so tides also affect them, see en.wikipedia.org/wiki/Earth_tide $\endgroup$
    – PM 2Ring
    Commented Sep 29, 2022 at 4:06

With an elliptical orbit that is "locked" to the parent the orbital velocity will change--as seen from the child the parent will move backwards during the close part of the orbit and forward during the far part of the orbit. Thus you will have tides, but they'll be small and go back and forth, not around.

Note that even with a circular orbit you can still have wobble causing the same sort of thing.

Eventually the tides will circularize the elliptical orbit and damp the wobble of the circular orbit but that takes time. It very well might not have happened by the time the star dies.


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