What is the Coriolis force at the equator?

[Amith2006]

I was going through the concepts of Coriolis force but got stuck at the point that the Coriolis force is zero at the equator. Consider a person moving along the equator with a velocity v. Mathematically,
Coriolis force= -2m(w x v)
In this case though the latitude is zero, the angular velocity vector and the tangential velocity vector are perpendicular. This gives a non zero Coriolis force! I know something is wrong with my reasoning but can't figure it out. The only way out seems that that the 2 vectors are not perpendicular. Something to do with angle between 2 lines in 3-D space? Thanx in advance.

 

[D H]

It depends on what you mean by the term "Coriolis force". Physicists now use $F_c = -2m \boldsymbol{\omega}\times \mathbf v$. Coriolis' paper preceded the development of the cross product by about half a century. Coriolis used a different, and almost certainly clunkier, notation.

Atmospheric and oceanic scientists consider the Coriolis force to be the local horizontal component of what physicists call the Coriolis force. At the equator, $\boldsymbol{\omega}\times \mathbf v$ is directed purely upward. It has zero horizontal component. So to an atmospheric scientist, the Coriolis force is zero at the equator because they use a different definition of Coriolis force than the definition used by physicists.

 

[Cleonis]

Mathematically,
Coriolis force= -2m(w x v)
In this case though the latitude is zero, the angular velocity vector and the tangential velocity vector are perpendicular. This gives a non zero Coriolis force!

This non-zero factor gives rise to the Eötvös effect. (The effect was named after him because he was the first to identify it.)

Let me give an example.
Let an airship be suspended at some point above the equator. (Airships are also known as zeppelins.) The airship is stationary with respect to the rotating Earth, and it is trimmed for neutral buoyancy.

At the equator the velocity of circumnavigating the Earth's axis at the rate of one revolution per day is about 465 m/s. The radial distance is 6378 kilometer, and the required centripetal force is about 0.034 Newton per kilogram of mass

The amount of suspension force required is the mass of the airship (multiplied with the gravitational acceleration), minus those 0.034 Newtons/kg. In other words: any object co-rotating with the Earth at the equator has its measured 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 0.0375 Newtons/kg. Cruising at 25 m/s due West the total velocity is 465 - 25 = 440 m/s, requiring about 0.0305 Newtons/kg. So if the airship is neutrally buoyant while cruising due east, it will not be neutrally buoyant anymore after a U-turn; the airship must be retrimmed.

Note the contrasting case: on a non-rotating planet, making the same U-turn would not result in a change of buoyancy.


The following derivation illustrates the physical nature of the Eötvös effect:
$$a_u$$ is the total required centripetal acceleration when moving along the surface of the Earth.
$$a_s$$ is the required centripetal acceleration when stationary with respect to the Earth.
$$\Omega$$ is the angular velocity of the Earth: one revolution per [[Sidereal day]].
$$\omega_r$$ is the angular velocity of the airship relative to the angular velocity of the Earth.
$$(\Omega + \omega_r)$$ is the total angular velocity of the airship.
$$\omega_r * R = u$$ is the airship's velocity (velocity relative to the Earth).
$$R$$ is the Earth's radius.

$$\begin{align} a_r & = a_u - a_s \\ & = (\Omega + \omega_r)^2 R - \Omega^2 R \\ & = \Omega^2 R + 2 \Omega \omega_r R + \omega_r^2 R - \Omega^2 R \\ & = 2 \Omega \omega_r R + \omega_r^2 R \\ & = 2 \Omega u + u^2 / R \\ \end{align}$$


The second term $$u^2 / R$$ is very small, as it includes division by R, the Earth's radius.
The first term $$2 \Omega u$$ equals the Coriolis term.

What is valid for airships is of course also valid for air mass itself. Air mass that has a velocity relative to he Earth is subject to the Eötvös effect.

Up until recently there was no pressing need for meteorologists and oceanographers to take the Eötvös effect into account in the models that are used in the big numbercrunching computers. But by now the models have been pushed to a level where they do need to include the Eötvös effect.

See also:
http://www.nioz.nl/public/annual_report/2008/gerkema.pdf"

 

[Amith2006]

It depends on what you mean by the term "Coriolis force". Physicists now use $F_c = -2m \boldsymbol{\omega}\times \mathbf v$. Coriolis' paper preceded the development of the cross product by about half a century. Coriolis used a different, and almost certainly clunkier, notation.

Atmospheric and oceanic scientists consider the Coriolis force to be the local horizontal component of what physicists call the Coriolis force. At the equator, $\boldsymbol{\omega}\times \mathbf v$ is directed purely upward. It has zero horizontal component. So to an atmospheric scientist, the Coriolis force is zero at the equator because they use a different definition of Coriolis force than the definition used by physicists.

Great! Got your point buddy. The mathematical part is clear. Now, I have a conceptual hitch on Coriolis force. Consider 2 people sitting diametrically opposite on a merry-go-round. While it is rotating, one of them throws a ball to the other. From an inertial frame of reference, it is clear that the ball takes a straight path. My question is- though it is straight path, as the thrower has a tangential component of velocity at the moment of release, it should fall at a point which is displaced a bit from the intended point in space. Please correct me if I am wrong.

 

[Cleonis]

Consider 2 people sitting diametrically opposite on a merry-go-round. While it is rotating, one of them throws a ball to the other. From an inertial frame of reference, it is clear that the ball takes a straight path. My question is- though it is straight path, as the thrower has a tangential component of velocity at the moment of release, it should fall at a point which is displaced a bit from the intended point in space. Please correct me if I am wrong.

What is the goal of your question? Is your goal better understanding of meteorology?

Let me explain why I ask: in meteorology and oceanography the velocities involved are comparatively small. Let's say one revolution of that merry-go-round takes 10 seconds. If you would throw a ball at a velocity that scales to typical wind velocities on Earth, then it takes that ball something like an hour to make it to the other side of the merry-go-round.
The point is: the tangential velocity of co-moving with the merry-go-round is many times larger than the velocity relative to the merry-go-round platform (if you use a velocity that scales to typical wind velocity).

So the example of a merry-go-round is particularly unsuitable as a model for what matters in Meteorology and Oceanography.

 

[Amith2006]

What is the goal of your question? Is your goal better understanding of meteorology?

Let me explain why I ask: in meteorology and oceanography the velocities involved are comparatively small. Let's say one revolution of that merry-go-round takes 10 seconds. If you would throw a ball at a velocity that scales to typical wind velocities on Earth, then it takes that ball something like an hour to make it to the other side of the merry-go-round.
The point is: the tangential velocity of co-moving with the merry-go-round is many times larger than the velocity relative to the merry-go-round platform (if you use a velocity that scales to typical wind velocity).

So the example of a merry-go-round is particularly unsuitable as a model for what matters in Meteorology and Oceanography.

Cleonis,
Are you into Meteorology or Oceanography? Just curious! Let me tell you that Meteorology is not really my area. I just felt that such details are to be known by a physics graduate irrespective of their specialization. I understand that the merry-go-round model is unsuitable in meteorology, but considering just the merry-go-round where the throw velocity is comparable to the tangential velocity, there will be a perceivable displacement, isn't it?

 

[Cleonis]

Cleonis,
Are you into Meteorology or Oceanography? Just curious!

I'm interested in the rotation-of-Earth effect that is taken into account in the fluid dynamics of the Earth.

considering just the merry-go-round where the throw velocity is comparable to the tangential velocity, there will be a perceivable displacement, isn't it?

Obviously there will be a displacement, yes.
Another example of a rotating platform is a wheelshaped spacestation that is rotating to pull G's. (I rather like the simplification of a gravity-free environment.)
There's a Java applet on my website called http://www.cleonis.nl/physics/ejs/spacestation_vertical_throw_simulation.php" , for exploring that example. Tangential velocity and radial velocity can be set up in a range of values.

 

[Amith2006]

Thanx guys especially Cleonis for the time and effort taken in clarifying my doubts.