Eötvös effect

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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,

Ω is the rotation rate of the Earth
u is the velocity in latitudinal direction (east-west)
φ is the latitude where the measurements are taken.
v is the velocity in longitudinal direction (north-south)
R is the radius of the Earth

Contents

[edit] Physical explanation

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

[edit] Motion along the equator

To focus on the buoyancy, the example of airships will be discussed. An airship with a mass of 10 tons (10,000 kg), with a cruising velocity of 25 meters per second (90 km/h, 55 mi/h)

To calculate what it takes for the airship 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 6378 kilometer (the Earth's equatorial radius), at 465 m/s, is about 0.034 newton per kilogram of mass. For the 10 ton airship that amounts to about 340 newtons. The amount of buoyancy force required is the mass of the airship (multiplied with the gravitational acceleration), minus those 340 newtons. 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 375 newtons. Cruising at 25 m/s due west the total velocity is 465 - 25 = 440 m/s, requiring about 305 newtons. So if the airship is neutrally buoyant while cruising due East, it will not be neutrally buoyant anymore after a U-turn; after the U-turn, the weight of the 10,000 kilogram airship has increased by about 7 kilograms; the airship will have to be re-trimmed. On the other hand: on a non-rotating planet, making the same U-turn would not result in a need to retrim the airship.

[edit] 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 airship between being stationary with respect to the Earth and having a velocity with respect to the Earth.

Notation:

au is the total centripetal acceleration when moving along the surface of the Earth.
as is the centripetal acceleration when stationary with respect to the Earth.
Ω is the angular velocity of the Earth: one revolution per Sidereal day.
ωr is the angular velocity of the airship relative to the angular velocity of the Earth.
(Ω + ωr) is the total angular velocity of the airship.
ωr * R = u is the airship's velocity (velocity relative to the Earth).
R is the Earth's radius.
ar = au − as
ar = (Ω + ωr)2R − Ω2R
a_r = \Omega^2 R + 2 \Omega \omega_r R + \omega_r^2 R - \Omega^2 R
a_r = 2 \Omega \omega_r R + \omega_r^2 R
ar = 2Ωu + u2 / 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}

The first term in the above formula is the vertical component of the Coriolis effect caused by the earth's rotation. (The Coriolis effect is directed perpendicular to the earth's axis of rotation, hence the cosine.) It gives a positive contribution when moving eastwards, and a negative one when moving to the west.

The second term represents the required centripetal acceleration for the airship to follow the curvature of the earth. It is independent of both the earth's rotation and the direction of motion.

[edit] Explanation

The force of gravity and the normal force. The resultant force acts as the required centripetal force.
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, there is an Equatorial bulge. The force of gravity is directed towards the center of the Earth. The normal force is perpendicular to the local surface.

On the poles and on the equator the force of gravity and the normal force are exactly in opposite direction. At every other latitude the two are not exactly opposite, so there is a resultant force, that acts towards the Earth's axis. 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. Whenever the shape of the solid Earth is not entirely in equilibrium with its rate of rotation, then the shear stress deforms the solid Earth over a period of millions of years until the shear stress is resolved.)

Again the example of an airship is convenient for discussing the forces that are at work. When the airship 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.

If an airship has an eastward velocity, then the airship is in a sense "speeding". The situation is comparable to a racecar on a banked circuit with an extremely slippery road surface. If the racecar is going too fast then the car will drift wide. For an airship in flight that means a reduction of the weight, compared to the weight when stationary with respect to the Earth.

If the airship has a westward velocity then the situation is like that of a racecar on a banked circuit going too slow: on a slippery surface the car will slump down. For an airship that means an increase of the weight.

The first term of the Eötvös effect is proportional to the component of the required centripetal force perpendicular to the local Earth surface, and is thus described by a cosine law: the closer to the Equator, the stronger the effect.

[edit] Motion along 60 degrees latitude

The Eötvös effect for an object moving eastward along 60 degrees latitude. The object tends to move away from the Earth's axis.
The Eötvös effect for an object moving eastward along 60 degrees latitude. The object tends to move away from the Earth's axis.
The Eötvös effect for an object moving westward along 60 degrees latitude. The object tends to be pulled towards the Earth's axis.
The Eötvös effect for an object moving westward along 60 degrees latitude. The object tends to be pulled towards the Earth's axis.

An object located at 60 degrees latitude, co-moving with the Earth, is following a circular trajectory, with a radius of about 3190 kilometer, and a velocity of about 233 m/s.

That circular trajectory requires a centripetal force of about 0.017 newton for every kilogram of mass; 170 newtons for a 10,000 kilogram airship. To calculate the Eötvös effect at 60 degrees latitude, the component that is perpendicular to the local surface (the local vertical) is taken, which at 60 degrees latitude is half the total force. Hence, at 60 degrees latitude, any object co-moving with the Earth has its weight reduced by about 0.08 percent, thanks to the Earth's rotation.

When the airship is cruising at 25 m/s towards the east the total velocity becomes 233 + 25 = 258, which requires a centripetal force of about 208 newtons; local vertical component about 104 newtons. Cruising at 25 m/s towards the west the total velocity becomes 233 - 25 = 208 m/s, which requires a centripetal force of about 135 newtons; local vertical component about 68 newtons. Hence after making a U-turn the airship will have to be retrimmed, because of the 4 kilogram difference in required buoyancy force.

The diagrams also show the component in the direction parallel to the local surface. In meteorology and in oceanography, it is customary to refer to the effects of the component parallel to the local surface as the Coriolis effect.

[edit] 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.
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