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)

  • 1
    $\begingroup$ "But I think the effect is more complicated than that, affecting the isosurface of equipotential" -- honestly, altering the equipotential surface is all tides directly do. On Earth's oceans, the presence of the continents together with the thinness of the ocean hampers the response to this forcing, resulting in complex patterns as detailed here, but I can't see this mattering for Jupiter. $\endgroup$
    – user10851
    Apr 17, 2015 at 21:58

1 Answer 1


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:

  1. The tidal effect of a body on its primary is made larger by increasing the primary's radius, decreasing the body's orbital radius, or increasing the body's mass.
  2. In the case of increasing the primary's radius, the increase in tidal effect quickly diminishes in comparison to the corresponding increase in the primary's mass which accompanies it. Thus, even though the tidal effect of Io on Jupiter is much stronger than the effect of the Moon on the Earth, it matters much much less since Jupiter is so much more massive. This is such a strong effect that it is true even though Jupiter is $\approx \frac{1}{4}$ as dense.
  3. The effect of Io on Jupiter's tides is on the order of $10^{-4.5} \frac{m}{s^2}$

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.

  • $\begingroup$ This answer is incorrect. This is not how the tides work. $\endgroup$ Apr 18, 2015 at 10:24
  • $\begingroup$ @DavidHammen The tides are an extremely complicated function of the tidal forcing function, which itself is a function of the above difference in force. Since the OP was asking for intuition (see also the WorldBuilding Thread), this clarifies the effect. $\endgroup$
    – geometrian
    Apr 18, 2015 at 17:55
  • $\begingroup$ But the intuition you gave is wrong. You noted another answer elsewhere, which is correct. This answer is wrong. $\endgroup$ Apr 18, 2015 at 18:04
  • $\begingroup$ @DavidHammen The answer on the linked page models the tidal forcing function, which as I literally just said is not what I am reporting here. If you fancy this answer incomplete, there's a wonderful little button labeled "edit" that I encourage you to use. $\endgroup$
    – geometrian
    Apr 18, 2015 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.