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Okey, I'm thinking about a problem like this:

Suppose we walk along the equator, let's choose the east directon (for example). Let our walking speed be $v$, then from the coordinate system that follows earths rotation, we expect a coriolisacceleration to act upon us.

We know that the coriolisacceleration can be calculated as $2 \vec{\omega} \times \vec{r}$, but choosing our rotating coordinate system such that $x$ is in the direction of east, and $z$ radially out towards us from the center of earth, then, since the rotationvector of earth points in the positive $y$ direction, the cross product shows us that the coriolisacceleration points in the $z$ - direction, but how can this be the case. Shouldn't we expect that it's pointing upwards (y - direction), since we would experience a higher angular velocity by walking along the equator?

Thanks.

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2 Answers 2

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Walking along the equator, the Coriolis force pushes neither left nor right, but instead up or down (from the walker's perspective), which is exactly the $z$ direction you defined.

You can make sense of this like so: If you were to walk straight ahead along the equator (let's stick with east as in your example), you are orbiting the center of rotation faster than someone standing still. This increases your centrifugal force, which you can interpret as an apparent radial force outwards that appears in addition to the centrifugal force you would experience if you were standing still.

If you were to walk due west, you would see the opposite effect: you orbit slower, making the centrifugal force weaker, which you can also frame as an apparent force pointing radially inward, countering part of the centrifugal force you would expect if you were standing still.

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  • $\begingroup$ Did you reverse the two directions by mistake? Earth spins from West to East. $\endgroup$
    – RC_23
    Apr 21, 2022 at 21:42
  • $\begingroup$ @RC_23 Indeed I did, thanks for pointing it out. Corrected now. $\endgroup$
    – noah
    Apr 21, 2022 at 21:49
  • $\begingroup$ @noah Thank you for your answer. I'm now thinking, what if we walked north from the equator? Wouldn't this mean that the angular velocity of earth and the direction of our walking speed are parallell, meaning we wouldn't get a coriolis effect. Is this expected? I'm thinking that it should be, since now we are walking "perpendicular" in a sense to the rotation of earth, meaning we won't experience that additional centrifugal force you were talking about? $\endgroup$
    – Tanamas
    Apr 22, 2022 at 17:42
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    $\begingroup$ @Tanamas Yes that's right, if you walk north or south at the equator, there is (right at the equator) no Coriolis force, because you are not changing your angular velocity, nor the distance from the rotational axis. Mathematically that's easy to see as the cross product of two parallel vectors is zero. $\endgroup$
    – noah
    Apr 22, 2022 at 19:58
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In geophysics the rotation-of-Earth-effect that you are describing is called the Eötvös effect.

The wikipedia article was initiated by me (and has remained largely unchanged since.)

In geophysics the expression 'Coriolis effect' is by convention used for rotation-of-Earth-effect for motion component parallel to the local surface. That is: in geophysics by convention Coriolis effect on motion is treated as if the objects in motion are confined to moving parallel to the local surface.

In geophysics distinction between Coriolis effect and Eötvös effect is by convention; the underlying physics is the same.


The following is copied from the wikipedia article (that I wrote)

The most common design for a gravimeter for field work is a spring-based design; a spring that suspends an internal weight. The suspending force provided by the spring counteracts the gravitational force. A well-manufactured spring has the property that the amount of force that the spring exerts is proportional to the extension of the spring from its equilibrium position (Hooke's law). The stronger the effective gravity at a particular location, the more the spring is extended.

For the calculations it will be assumed that the internal weight has a mass of ten kilograms (10 kg; 10,000 g). It will be assumed that for surveying a method of transportation is used that gives good speed while moving very smoothly: an airship. Let the cruising velocity of the airship be 25 metres per second (90 km/h; 56 mph).

To calculate what it takes for the internal weight of a gravimeter to be neutrally suspended when it is stationary with respect to the Earth, the Earth's rotation must be taken into account. At the equator, the velocity of Earth's surface is about 465 metres per second (1,674 km/h; 1,040 mph). The amount of centripetal force required to cause an object to move along a circular path with a radius of 6378 kilometres (the Earth's equatorial radius), at 465 m/s, is about 0.034 newtons per kilogram of mass. For a 10,000-gram internal weight, that amounts to about 0.34 newtons. The amount of suspension force required is the mass of the internal weight (multiplied by the acceleration of gravity) minus those 0.34 newtons. In other words: any object co-rotating with the Earth at the equator has its measured weight reduced by 0.34 percent, as a consequence of 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 0.375 newtons. Cruising at 25 m/s due West the total velocity is 465 - 25 = 440 m/s, requiring about 0.305 newtons. This gives a difference of about 0.07 Newtons (0.375 - 0.305).

So if the internal weight is neutrally suspended while cruising due east, it will not be neutrally suspended anymore after a U-turn. After the U-turn, the weight of the 10,000 gram internal weight has increased by about 7 grams; the spring of the gravimeter must extend some more to accommodate the larger weight. On the other hand: on a non-rotating planet, making the same U-turn would not result in a change of gravimetric reading.

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