Habitable moon of a gas giant: working out the size of the tide

I'm creating a fictional planet that orbits a gas giant (a lot like Jupiter). I'm working out all the issues listed here and here. But what I'm really having trouble with is the size of the tide. In a moon with 115% of Earth's mass with an orbital period of 100 days (periapsis distance: 6052380km; apoapsis distance: 6444500km) around Jupiter-like gas giant, how big would the tide be?

• This really belongs in the physics S.E. – iAdjunct Jun 22 '16 at 2:12
• sorry, i'm new here. still trying to understand everything – Victor Jun 22 '16 at 2:14
• Welcome! There are lots of StackExchange sites. You'll probably get an answer to that here, but the physics S.E. would probably be better. – iAdjunct Jun 22 '16 at 2:15
• It looks on-topic here. It might also be on-topic there, but I'm pretty confident that we have people here who can answer it. Welcome to Worldbuilding! Check out our short tour for more about our scope. – Monica Cellio Jun 22 '16 at 3:37

Short answer, expect tides about 6.05 times greater the lunar tides on earth.

Tidal height is essentially impossible to calculate in detail because it depends on so many possible factors. This is why lunar tides on earth vary by such a large amount as different factors result in different results at various shores around the earth.

However, we can compare average tidal accelerations fairly easily, and assume that the tides on your earth-like moon will be proportional, i.e., that there will be a similar range of actual tidal factors (ocean floor geometry, winds, etc).

Tidal acceleration (ta) is 2 * G * M * r / (R^3)

G is the gravitational constant
M is the mass of the body causing the tidal force
r is the distance from the center of the body having tides
T is the center-center distance of between the 2 bodies

Of course, you can plug in actual numbers (with consistent units, MKS strongly recommended for most cases), or you could just realize that you don't care about the actual tidal acceleration, but just the ratios (mostly: see next paragraph)

E2 (Earth # 2) is 15% heavier. If we assume overall density is the same, the planet will have a radius that is 4.77% greater the earth, and a surface gravity that is also 4.77% greater than earth. The heavier surface gravity will mean that idealized tidal height is reduced by the same percentage.

Earth / Lunar system
M = 1 (lunar mass)
r = 1 (earth radius)
R = 1 (earth moon distance)
How convenient, everything is 1, therefore ta1 = 1
In more common units ta1 is about 1.1e-7 g

Jupiter / E2 system
M = 2.59e4 (lunar masses)
r = 1.0477 (earth radii)
R = 16.229 (earth moon distances)
ta2 = 2.59e4 * 1.0477 / (16.229 ^ 3) = 6.34

But since the surface gravity of E2 is 4.77% heavier, the nominal tidal height will be only 6.34/1.0477 = 6.05 times the nominal height of lunar tides on earth.

The earth lunar tide component varies from none to about 10 meters (Bay of Fundy has tides that can exceed 15 meters, but the sun tidal force is obviously included in that). I found different values for the smallest tides online, but Tahiti specifically mentioned as having no lunar tidal component.. Personally, I would expect the more extreme tidal multipliers to be clearly non-linear, but the multiple could either be smaller or larger depending on the details and you would need very exacting and detailed computation based on ocean floor geometries, yada, yada.

Perhaps this is good enough for your tidal estimates. The Bay of Fundy equivalent may be too much for you to allow, if so, just assume that it does not occur.

Edit: I made an error in surface gravity of E2, have corrected the answer.