Amith2006 said:
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.
<br />
\begin{align}<br />
a_r & = a_u - a_s \\<br />
& = (\Omega + \omega_r)^2 R - \Omega^2 R \\<br />
& = \Omega^2 R + 2 \Omega \omega_r R + \omega_r^2 R - \Omega^2 R \\<br />
& = 2 \Omega \omega_r R + \omega_r^2 R \\<br />
& = 2 \Omega u + u^2 / R \\<br />
\end{align}<br />
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" 
