Since the tides on the moon have an antieffect on the other side of earth, (still don't quite understand how that works), shouldn't the gas giants deform when their moons move around in their orbits? At least for Saturn this should hold true since most bigger moons are close together.

Is the ice layer on Uranus too thick to crack under the tides of its moon?

  • $\begingroup$ The gas giants probably won't deform - do you mean will the moons deform instead? (Io has quite heavy tidal forces.) $\endgroup$
    – Andy
    Jul 12, 2016 at 11:57
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    $\begingroup$ Both, since it is a mass relation and not a moon planet relation =) Why would a Gasgiant deform, atleast a little? Isnt gas easier to move then water? $\endgroup$
    – Git
    Jul 12, 2016 at 11:59
  • $\begingroup$ Oh I see your point - yes Jupiter will be (very slightly) distorted by Io, in the same way the Earth is affected by the moon. I was wrong earlier when I claimed it "won't deform". $\endgroup$
    – Andy
    Jul 12, 2016 at 13:07
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    $\begingroup$ This would be more on topic for astronomy.stackexchange.com $\endgroup$
    – user687
    Jul 12, 2016 at 14:32
  • $\begingroup$ Uranus is not frozen. It is called an ice giant because it is composed mainly of elements heavier than hydrogen and helium (referred to as "ices"). It is still largely gaseous. $\endgroup$
    – called2voyage
    Jul 12, 2016 at 15:58

1 Answer 1


Short answer: yes, the gas giant will be deformed by its orbiting moon, but the tide raised will be vanishingly small..

The reason tides happen is because the force of gravity varies with distance. Newton's Law states that $F_g=\frac{G*M_1*M_2}{R^2}$, and has an inverse square dependence on distance. When the moon is directly overhead, you are being pulled more by it than when it is on the other side of the Earth. This means that the material on the Earth is pulled outwards, creating a bulge. The material on the opposite side of the Earth, however, does not feel the pull as strongly, so it relaxes, creating another bulge. Here is an image showing how the force of gravity varies across a planet, creating two tidal bulges. For more information on how that works, check this website.

A vector plot showing the differential gravitational forces on a body due to a satellite.

We can estimate the tidal acceleration on the surface of Neptune by Neptune's largest moon, Triton, using the formula given here: $$ a = G*M_{moon}*\frac{2*D*R_{planet}-{R_{planet}}^2}{D^2*(D-R_{planet})^2} $$ Plugging in numbers for each of those values gives us an approximate acceleration of $6.56*10^{-29}$ m/s^2. The Moon's tidal acceleration on the Earth is $1.129*10^{-6}$ m/s^2. Thus, Triton's effect on Neptune is $10^{13}$ times smaller than the Moon's effect on Earth. I have chosen the Triton/Neptune system because Triton is over twice the mass of all the Uranian satellites combined.

Given all this, Triton does not have a significant effect on Neptune.

However, Neptune has a much bigger effect on Triton. Reversing the situation, we find that Neptune's tidal acceleration on Triton is $3.36*10^{-4}$m/s^2. Whether or not that is strong enough to break the frozen nitrogen on Triton's surface depends on the tensile strength. If you want to see what kinds of stresses result from tidal forces on icy satellites, check out this program we're working on (right now it only works on Mac).

Though, actually, it turns out that one key factor which determines the strength of stressing factors on a moon is its orbital eccentricity. Triton (despite being in a retrograde orbit), only has an orbital eccentricity of 0.000016. This results in stresses on the order of 0.05 kPa, not nearly enough to break the ice (stresses due to its orbital obliquity, however, are several orders of magnitude larger, but that is beyond the scope of this question).

  • $\begingroup$ The earth does not bulge on the other side because it relaxes. Earth and moon rotate around their common barycenter, causing a centrifugal force at the opposite side. $\endgroup$
    – sweber
    Jul 12, 2016 at 16:58
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    $\begingroup$ @sweber That's not really the case, though many textbooks teach it that way. The real cause is the gravitational differential. This website has a good explanation of how the rear tidal bulge actually works. I'll go ahead and put the diagram in my answer, though, to clear up confusion. $\endgroup$
    – Phiteros
    Jul 12, 2016 at 17:06
  • $\begingroup$ Thanks for that detailed answer. Especially your calculation for Triton! $\endgroup$
    – Git
    Jul 13, 2016 at 13:50

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