I recall reading an essay by Asimov (I think) around 1980 stating that the tides are a function of a power of the diameter of the primary, so (surprisingly) small close moons of Jupiter raise large tides.
Looking at the geometry, back of the envelope I figure it's the ratio of the squares of the distances of the far and near faces, which is not a "higher power" but still a magnifying factor. So I figure Io's tidal force on the cloud deck of Jupiter is about 1.8× the moon's force on our ocean.
Now what does the force accomplish? If it's simply lifting molecules of fluid against the surface gravity, then the 2.5g is factored in. But I think the effect is more complicated than that, affecting the isosurface of equipotential.
Can someone explain the answer? Google is distracted by ♃'s effects on Io which saturate the results, and old essays are generally not quoted online anywhere.
(This came up in Worldbuilding SE)
Best Answer
Let me expand on (and correct a minor error) in what I said in the alluded-to thread.
Unfortunately, I do not know the mentioned article (although "The Moon's Twin", published in 1989 in "The Magazine of Fantasy and Science Fiction", discussing the Jupiter/Io system, might be it). Therefore, I can't directly address it, but I think I can say three major things:
To simplify a bit, the highest tidal force the body exerts on its primary is the difference of the maximum and minimum forces. Let's work two examples: Earth/Moon, and Jupiter/Io.
For Earth/Moon, the highest force is on the near side of the Earth when the Moon is at perigee (average (it's complicated) $R_0=362~600~km$ distance), and the lowest is on the far side of the Earth when the Moon is at apogee (average $R_1=405~400~km$). In practice, the tidal force will be less than this, since apogee and perigee are generally different. Let's say the Earth and Moon are both spherical, the radius of Earth is $r_e=6~371.0~km$, and the mass of the Moon is $m_m=7.3477 \cdot 10^{22} kg$. The lowest and greatest accelerating tidal "force"s, given by the standard gravity equation, are then:$$ a_{min} = G\frac{m_m}{(R_1+r_e)^2} \approx 2.8918 \cdot 10^{-5} \frac{m}{s^2}\\ a_{max} = G\frac{m_m}{(R_0-r_e)^2} \approx 3.8638 \cdot 10^{-5} \frac{m}{s^2}\\ \Delta a \approx 9.720 \cdot 10^{-6} \frac{m}{s^2} $$(Side-note: this means that a $70 kg$ person weighs up to ~the weight of $2.70$ grams less because of the moon. This disagrees with the only mention I could find, but since this average acceleration very accurately predicts the force of the Moon on the Earth ($\approx 2 \cdot 10^{20}$), I think I am correct.)
So in this case, the tidal force varies by up to $\approx 14.39\%$ and constitutes at most about $1$ part in $290~329$ of the total force at Earth's surface.
Now let's look at Jupiter/Io. The values from before are (with $m_i=8.931938 \cdot 10^{22} kg$, $m_j=1.898 \cdot 10^{27} kg$, $R_0=420~000~km$, $R_1=423~400~km$, $r_j=69~911~km$):$$ a_{min} = G\frac{m_i}{(R_1+r_j)^2} \approx 2.4492 \cdot 10^{-5} \frac{m}{s^2}\\ a_{max} = G\frac{m_i}{(R_0-r_j)^2} \approx 4.8631 \cdot 10^{-5} \frac{m}{s^2}\\ \Delta a \approx 2.4139 \cdot 10^{-5} \frac{m}{s^2} $$In this case, we see that the tidal force varies by up to $\approx 33.01\%$ and constitutes at most about $1$ part in $509~800$. Note that, unlike in the Earth/Moon case, I assumed that the Jupiter/Io barycenter is the same as Jupiter's center of mass. I can do that since Jupiter is $21~256$ times more massive (and has other moons too), as opposed to Earth only being $81.3$ times as massive as its single natural moon.
Thus we have our answer: the tidal force produced by Io is on the same order of magnitude as that produced by Earth's moon, but since Jupiter is much larger, it varies by a larger percentage (the orbit also affects it for the same reason). However, since Jupiter is so much more massive than Earth, the proportional difference this makes to the gravity at Jupiter's surface is much, much less.