# Impact of tidal forces to a liquid in a half-full tank

The german Wikipedia article about tidal forces contains a sentence which mentions, that tidal forces have an impact to an half-full fuel tank:

Wo keine mechanisch feste Verbindung besteht, können Gezeitenkräfte spürbar werden: [... zum Beispiel] bei Flüssigkeit in einem halbvollen Tank.

Translation:

Where there is no mechanically firm connection, tidal forces can be felt: [... for example] when liquid is in a half-full tank.

Although the article mentions that tidal forces impact liquid fuel inside a half-full tank, there's no detail about what the impact on the fuel is.

After reading the article I now have the following question:

What are possible impacts caused by tidal forces to a liquid in a half-full tank?

• Depends on your proximity to and the mass of an astronomical body. Obviously the tidal forces havent been a problem on Earth since we figured out how to fly before we ever cared about tidal forces on fuel tanks.
– anon
Jul 23 '18 at 12:38
• Don't be so quick to accept an answer. It discourages others from writing an answer. Jul 23 '18 at 13:39
• It looks like the German version of wikipedia is on par with the English one regarding quality and clarity, which is a bit sad. Jul 23 '18 at 14:19
• @Uwe - You are perhaps overstating things, by an order of magnitude. Lake Superior can be viewed as a very, very, very large tank. It is over 500 km long and is occasionally subject (in theory) to 5 cm spring tides. That's smaller than a tenth of a millimeter per kilometer. We see large tides in the oceans because the oceans are a whole lot larger than is Lake Superior. Jul 23 '18 at 16:15
• @David Hammen: Yes it is smaller than a tenth of a millimeter per kilometer. The largest tides in the oceans appear at the coasts due to resonances. The ocean tides far away from any coast are much smaller, about 40 cm.
– Uwe
Jul 23 '18 at 17:14

What are possible impacts caused by tidal forces to a liquid in a half-full tank?

For all practical purposes, none.

The tidal along the radial axis from the center of the planet to the spacecraft's center of mass is approximately $2g\frac r R$, where $g$ is the gravitational acceleration at the spacecraft's center of mass, $r$ is the distance between the spacecraft's center of mass and some other point on the spacecraft, and $R$ is the distance between the spacecraft's center of mass and the center of the planet. It's half that for separations perpendicular to the radial axis.

The spacecraft center of mass and the fuel tank would have to be separated by kilometers for tidal forces to be a significant effect. Even then, whether 10-3 g0 (a thousandth of Earth surface gravity) qualifies as "significant" is dubious. Typical values are in the micro g level, which means that surface tension completely dominates if the spacecraft is not rotating and in free fall, that centrifugal force completely dominates if the spacecraft is in free fall but is rotating, and that thrust completely dominates if the spacecraft is under thrust.

A half-full tank represents a potential controllability / stability issue with regard to slosh; half-full is when slosh is at its worst. Nobody that I know of considers tidal forces when modeling slosh. A half-full (or less) tank in free fall also represents a potential issue with regard to getting fuel to the thruster. This is why a huge number of patents on propellant management devices exist. I've yet to read about a PMD that takes advantage of tidal forces.

Tidal forces in a tank are a non-issue, for now at least. They might be when humanity builds a kilometers long multi-generation starship. That is a problem for some future generation.

• Thoughtful answer! Nothing, not even a General Products hull can protect you from tidal forces in orbit around things much more massive than Earth though.
– uhoh
Jul 23 '18 at 23:16

Assume you have a spherical tank 1 meter in radius in a 400 km (altitude) circular orbit, much like the ISS. The magnitude of the radial gravity gradient (and hence the tidal acceleration) is given by $2\frac{GM}{r^{3}}$∆r, where ∆r is the radial distance from the tank's center of mass. Substituting GM for Earth and 6778 km for the distance from Earth's center yields $\left(2.56 \times {10^{-6}}\right)$∆r $\frac{m}{s^{2}}$. So one liter of water (used because the mass density is very nearly 1 kg/liter) at the top side of the tank (the point farthest away from Earth) or the bottom side would feel a tidal force of only $2.56 \times {10^{-6}}$ Newtons.

This is about one quarter of the force solar light pressure exerts on a perflectly reflecting surface of one square meter, at 1 AU from the sun!

In a theoretical world, where you could postulate the absence of any forces such as surface tension, light pressure, atmospheric drag, etc., then the liquids in a half-filled tank would migrate (slowly!) to the tank's top and bottom, potentially starving a drain, if it is at the middle of the tank instead of at the top or bottom ends. If that drain is a feed to an engine, or especially to a turbopump, this is really bad. But—as @DavidHammen says—these other forces dominate, so tidal forces would not determine where the liquids go in the tank.

PMDs easily overcome those other forces so they would have no trouble at all handling tidal forces in any solar system scenario.

• How similar is a spherical tank to a spherical cow? Jul 24 '18 at 0:01
• @RonJohn Quite similar, actually, except that 1) the tank is not made of flesh, and 2) the tank doesn't emit milk equally over 4$\pi$ steradians! ;-) Jul 24 '18 at 0:05

A roughly spherical shaped object in space (like earth) transforms into a more egg-shaped spheroid with two buldges (high tides) due to differences in graviational influence. One on the side of the attractor (i.e. a moon) and the other on the opposite side.

Gravitational force will decrease exponential to the distance from the objects barycenter. Therefore, liquids on the side of the moon will accelerate quicker towards the moon while liquids on the opposite site will 'fall behind'.

The answer to your question is different depending on the location of the fuel tank.

On earth, the fuel tank is mechanically connected to earth and therefore, the fuel will generally experience the same behaviour than water in oceans. However, it is limited in its movement by the sidewalls of the tank. If the tank is in a high tide area, the fuels volume will increase and the pressure will decrease.

In space, the entire crafts trajectory will always be partially influenced by the moon (assuming you're between VLEO and HEO for the sake of simplicity). If you want to read more about that, search for 'n body problem'. The fuel inside the tank experiences tidal forces as well, depending on the location of the fuel relative to the barycenter of the overall craft/fuel tank. They will (if not influenced by stronger forces) 'fall behind' or get ahead relative to the moon.

Edit: Tidal forces on a fluid increase, as the distance of a liquid from the barycenter increases. This is due to the difference in gravitational influence from the external attractor. Therefore, tidal forces in fuel tanks in space are probably diminishable (If someone bothers to do the math, please go ahead).

• The 'mun'... :P? Jul 23 '18 at 13:24
• @MagicOctopusUrn I played too much KSP... :o Jul 23 '18 at 13:25
• Re Gravitational force will decrease exponential to the distance from the objects barycenter. Gravitation decreases as $1/R^2$, which is a far cry from an exponential ($\exp(-R)$) decrease. Jul 23 '18 at 13:47
• "If the tank is in a high tide area, the fuels volume will increase and the pressure will decrease." - Um, what? You don't magically gain more fuel in your tank at high tide. Further, even if you somehow did, that should mean that your pressure also increased (assuming the tank is a sealed container - if it's open, the pressure would stay constant as any extra air was expelled). If the tank was not completely full, the fuel might become more distributed, but this wouldn't change the actual volume of the fuel at all. Jul 23 '18 at 15:50
• On earth, the fuel tank is mechanically connected to earth and therefore, the fuel will generally experience the same behaviour than water in oceans. Um, no. Look at Lake Superior as if it was a very large fuel tank. Theoretically, it is subject to tides. In practice, the centimeter scale tides that should result are all but impossible to measure because (1) they're very tiny, and (2) they're completely overwhelmed by winds and seasonal effects. Jul 23 '18 at 16:08