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The Earth turns with a very high velocity, around its own axis and around the Sun. So why can't we feel that it's turning, but we can still feel earthquake.

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I guess it's the same way you can't "feel" that you are driving 100KM/h in a car, you only "feel" acceleration or deceleration. – David Freitas Jul 20 '11 at 10:37
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@David Freitas: That's a pseudo-explanation and a terrible analogy. – Qmechanic♦ Jul 20 '11 at 18:39
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@Qmechanic Agree. – lamwaiman1988 Jul 21 '11 at 1:08
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@David Freitas: You are under acceleration when the earth is turning, though... – Christian Mann Jul 21 '11 at 19:16
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@Pieter Müller: i) When one drives on a road there are all kinds of vibrations; ii) if the car is on the rotating Earth, one would have to assume that the road is an initial frame, which is the very assumption that OP is questioning in the first place; iii) or if we imagine that the "car" is really a spaceship in empty space, then David Freitas is comparing the fact one cannot feel the velocity of the spaceship, due to Galilean invariance of inertial frames, with the unrelated fact that one cannot feel the centrifugal acceleration on the surface of Earth, which is in an accelerated frame. – Qmechanic♦ Nov 5 '12 at 16:31
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Dan's answer is essentially good, but miss one effect : the Coriolis effect. You can imagine a planet spinning much more rapidly than the earth, but at a constant angular speed. On that quickly rotating planet, the explanation of Dan would still stand, but as soon as on moves, we would feel a lateral Coriolis force.

The Coriolis acceleration is $2\vec{\Omega}\times\vec v$, where $\vec{\Omega}$ is the (vectorial) angular frequency of the planet's rotation and $\vec v$ the speed of the object moving. For an object moving at the speed of sound (340 m/s) near the Earth's pole, where the effect is maximum, the Coriolis acceleration is $$2\frac{2\pi}{24\times60\times60}\times 340 \simeq \frac{12\times 340}{24\times 3600}\sim \frac1{20} = 5\times10^{-2} \mathrm{m}\cdot\mathrm{s}^{-2}.$$ This corresponds to an acceleration which is half a percent of the gravity acceleration, for a situation which is already quite far from everyday life.

This small effect can accumulate over long distance and can have visible effects, notably at meteorological scales. In some sense, we feel the Earth rotation when we feel the dominant wind direction in our region. The parameter characterizing the intensity of the Coriolis effect for a phenomenon is the Rossby number, which is big if the Coriolis effect is negligible. If the phenomenon you analyse has a typical speed $v$, occur over a distance $L$, the Rosby number is essentialy proportional to the ration of the rotation period (24 h in our case) over the time $v/L$ it takes to go over the typical distance.

For meteorological depressions, the wind take several days to go over the thousands of kilometres they span, and the Coriolis effect has an important effect. To really feel the effect in everyday's life, one would need to be on a planet with a day of a few seconds, like the Little Prince's lamplighter's planet ! Of course, if you don't live on a rapidly rotating asteroid, you can see the effect on a carousel.

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Another way to feel (err... actually to see) the rotation of the Earth is with a Foucault pendulum. It's behavior can also be explained in terms of Coriolis forces. – Edgar Bonet Jul 20 '11 at 19:57
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Because the rotation of the earth is very smooth and doesn't change, the centripetal acceleration we feel is very nearly constant. This means that the (small) centrifugal force from the rotation gets added to gravity to make up the "background force" we don't notice.

Earthquakes are not at all smooth and the accelerations involved are large and change direction a lot. This makes it easy to feel them.

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The rate of change in acceleration is sometimes called “jerk”. It can be used to quantify how much passengers on a vehicle are shaken. – Edgar Bonet Jul 20 '11 at 20:05
I thought that centrifugal forces are fictitious ..... therefore, I am not willing to accept your answer. Bah humbug. – ldog Feb 2 '12 at 2:15
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There's always this standard reply to your objection. – Dan Feb 2 '12 at 4:17
   
@ldog: Centrifugal forces are not fictitious. They arise when the frame of reference you consider is not inertial (it accelerates in some way). Because it only appears in some frames of reference, and because inertial frames are sometimes seen as the "proper" way to do things, it is sometimes said that it isn't a "real force". Well, it's a measurable quantity, and it has units of force. What else is it? – Dan Feb 2 '12 at 4:24
@ldog: Furthermore, in general relativity, gravity is just a kinematic effect caused by the observer not being in an inertial reference frame. – Dan Feb 10 '12 at 20:11

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