Eötvös effect

In the early 1900s a German team from the Institute of Geodesy in Potsdam carried out gravity measurements on moving ships in the Atlantic, Indian and Pacific Oceans. While studying their results the Hungarian nobleman and physicist Lorand Roland Eötvös (1848-1919) noticed that the readings were lower when the boat moved eastwards, higher when it moved westward. He identified this as primarily a consequence of the rotation of the earth. In 1908 new measurements were made in the Black Sea on two ships, one moving eastward and one westward. The results substantiated Eötvös' claim. Since then geodesists use the following formula to correct for the Eötvös effect

$a_{r}=2\Omega u\cos \phi +{\frac {u^{2}+v^{2}}{R}}.$

Here,

$u$ is the velocity in latitudinal direction (east-west)

$v$ is the velocity in longitudinal direction (north-south)

$\Omega$ is the rotation rate of the Earth

$R$ is the Radius of the Earth

$\phi$ is the latitude where the measurements are taken.

Physical explanation

The Eötvös effect changes how much upward force must be exerted for neutral buoyancy.

Motion along the equator

To focus on the buoyancy, the example of zeppelins will be discussed. There are plans for huge cargo-lifting zeppelins, the weight of the zeppelin plus the cargo will be around 500 tons. The cruising velocity of these cargolifters will be in the order of 25 meters per second.

To calculate what it takes for the cargolifter to be neutrally buoyant when it is stationary with respect to the Earth the fact that the Earth rotates must be taken into account. At the equator the velocity of Earth's surface is about 465 meters per second. The amount of centripetal force required to cause a mass to move along a circular path with a radius of 6370 kilometer (the Earth's radius), at 465 m/s, is about 0,034 newton per kilogram of mass. For the 500 ton cargolifter that amounts to about 17000 newton. The amount of buoyancy force required is the mass of the cargolifter (multiplied with the gravitaional acceleration), minus those 17000 newton. In other words: any object co-rotating with the Earth at the Equator has its weight reduced by 0.34 percent, thanks to the Earth's rotation.

When cruising at 25 m/s due East the total velocity becomes 465 + 25 = 490 m/s, which requires a centripetal force of about 19000 Newton. Cruising at 25 m/s due West the total velocity is 465 - 25 = 440 m/s, requiring about 15000 Newton. So if the cargolifter is neutrally buoyant while cruising due East, it will not be neutrally buoyant anymore after a U-turn; the airship will have to be re-trimmed.

On the other hand: on a non-rotating planet making the same U-turns would not result in a need to retrim the airship.

Derivation of the formula for simplified case

Derivation of the formula for motion along the Equator.

A convenient coordinate system in this situation is the inertial coordinate system that is co-moving with the center of mass of the Earth. Then the following is valid: objects that are at rest on the surface of the Earth, co-rotating with the Earth, are circling the Earth's axis, so they are in centripetal acceleration with respect to that inertial coordinate system.

What is sought is the difference in centripetal acceleration of the zeppelin between being stationary with respect to the Earth and having a velocity with respect to the Earth.

Notation:

$\omega _{r}$ is the angular velocity of the cargolifter relative to the angular velocity of the Earth.

$\Omega$ is the angular velocity of the Earth: one revolution per Sidereal day.

(Hence the total angular velocity of the cargolifter is $\Omega +\omega _{r}$)

$a_{s}$ is the centripetal acceleration when stationary with respect to the Earth.

$a_{u}$ is the total centripetal acceleration when moving along the surface of the Earth.

$(\omega _{r})*R=u$ is the cargolifter's velocity (velocity relative to the Earth)

$R$ is the Earth's radius

$a_{r}=a_{u}-a_{s}=(\Omega +\omega _{r})^{2}R-\Omega ^{2}R$

$a_{r}=((\Omega +\omega _{r})^{2}-\Omega ^{2})R$

$a_{r}=(\Omega ^{2}+2\Omega \omega _{r}+\omega _{r}^{2}-\Omega ^{2})R$

$a_{r}=2\Omega \omega _{r}R+\omega _{r}^{2}R$

$a_{r}=2\Omega u+\omega _{r}^{2}R$

It can readily be seen that in the case of motion along the equator the formula for any latitude simplifies into the formula above.

$a_{r}=2\Omega u\cos \phi +{\frac {u^{2}+v^{2}}{R}}$

Explanation of the cosine in the first term

The force of gravity and the normal force. The resultant force acts as the required centripetal force.

The mathematical derivation for the Eötvös effect for motion along the Equator explains the factor 2 in the first term of the Eötvös correction formula.

Because of its rotation, the Earth is not spherical in shape, it is an oblate spheroid. The force of gravity is directed towards the center of the Earth. The normal force is perpendicular to the local surface. The forces represented in the diagram account for the motion with respect to the inertial frame of reference that is co-moving with the center of mass of the Earth.

The diagram applies for objects that have only a moderate velocity relative to the Earth. (If a rocket sled would speed along the equator, at such a velocity that it approaches orbital velocity, then the surface of the Earth would not have to exert a normal force.)

Except on the poles and on the Equator the force of gravity and the normal force are not exactly in opposite direction, so there is a resultant force, that acts in centripetal direction. At every latitude there is precisely the amount of centripetal force that is necessary to maintain an even thickness of the atmospheric layer. (The solid Earth is ductile. If the shape of the solid Earth would not be close to an oblate spheroid, then the shear stress would deform the solid Earth over a period of millions of years until the shear stress is resolved.)

Again the example of a zeppelin is convenient for discussing the forces that are at work. When the cargolifter has a velocity relative to the Earth in latitudinal direction then the weight of the cargolifter is not the same as when the cargolifter is stationary with respect to the Earth.

When the cargolifter has a velocity relative to the Earth in latitudinal direction then it will tend to veer of course. When moving eastward it will tend to veer to the Equator, when moving westward the poleward force will tend to pull it to the nearest pole.

The tendency to veer of course is described by a sine law: the closer to the Equator the weaker the effect in a direction that is parallel to the local Earth surface. The Eötvös effect is the component perpendicular to the local Earth surface, and is thus described by a cosine law: the closer to the Equator the stronger the effect.

Reference

• The Coriolis effect PDF-file. 870 KB 17 pages. A general discussion by the meteorologist Anders Persson of various aspects of geophysics, covering the Coriolis effect as it is taken into account in Meteorology and Oceanography, the Eötvös effect, the Foucault pendulum, and Taylor columns.