Why doesnt everything fly away from the spinning earth, why doesnt
centrifical force through us out??

I think the easiest way to explain gravity is like this:

Imagine having four people, spread out, each holding the corner of a
blanket nearly taut.  Then, have someone place a basketball in the
center.  The basketball makes a "dent" in the blanket.  If you were to
place a ping pong ball now along any edge, it will fall into the
basketball in the center.  Thus, the dent created by the basketball
causes ping pong balls to roll toward it -- in a way it's attracting
it.

Gravity in space acts like this, but in three dimensions.  The earth
(or the Sun or any other massy body for that matter) makes a dent in
the fabric of space and draws other objects toward it.  The closer
another object is to it, the stronger the pull is and the faster
objects fall toward it (like the ping pong ball that rolls faster and
faster toward the basketball as it gets closer).

In the case of us not flying out into space from the Earth, the
centripetal force of the gravity (pulling us in) is equal to or
stronger than any rotational force with which the Earth's rotation is
thrusting us.

As for centrifugal force, it doesn't really ever exist.  It's sort of
an imaginary force, and physics books used to teach people the
centrifugal force drew objects in rotation out.  In fact, it's just
the opposite.  The force drawing an object out is actually tangent to
the rotation (an object generally wants to keep going in the same
direction it's currently in -- a marble going around a track, if it
encounters a hole in the track, is going to keep going in the same
direction in which it last travelled).  The centripetal force is what
draws the object in (e.g. a string attached to an object you're
spinning around, or in this case, gravity pulling you into the Earth).
 Centrifugal force is the imaginary opposite force supposedly
counteracting the centripetal force.

      --<-
     /    \
    |      | A
     \    /
      ->--

Thus, an object at point A rotating counter-clockwise around the
center would experience the following:

^  (up)    inertial/tangential force
<- (left)  centripetal force pulling in the body
-> (right) imaginary centrifugal force counteracting the centripetal force


--Joey

6378100 m = radius of earth
465.1 m/s = velocity of earth's rotation at the equator. 

So, a 70 kg person at the equator would feel
70kg*(465.1m/s^2)/6378100m or 2.374 Newtons less, or about half a
pound less force from gravity than standing at the north pole. So it
does have an effect!

Also, I just did this for fun, the centripetal force we experience (as
Joey says, this is not an actual force, but a tendency for an object
in motion to continue in a straight line) from the earth orbiting the
sun is:
velocity of earth around sun = 29 951 m/s
radius of rotation = 149 598 000 000 m
So a 70 kg person feels 70/149598000000*29951^2, or 0.4198 N, or .094
lb less of gravity from rotation at night, and .094 lb more during the
day. Wow, I didn't think it was that much.

purkinje--

you were right to be surprised.  The sun calculation is not correct. 
The earth and you are both in free fall toward the sun, so the sun
does not pull toward or away from the earth.  There are very slight
tidal forces from the sun, meaning you weigh a little more at dawn and
dusk than at noon and midnight, but tidal forces are orders of
magnitude smaller than what you have calculated.

...does not pull *you* toward or away...

that's a roger, racecar.  I think what purkinje perhaps meant to say,
ahem ahem, is that you feel 0.9 lb less force at the poles than at the
equator...

How can you say that my numbers are orders of magnitude off when you
don't even give an alternate calculation? However, you are on the
right track, I realized-- we do not feel ANY centripetal force from
the sun since we are in free fall. Any 'tidal forces' would only be
from an elliptical orbit. However, you still feel .5 lbs less between
the equator and the poles, since you are not free falling, but are
resisting the force of gravity on the earth's surface.

That's exactly right, purk.

You are not in free fall toward the earth, but you are toward the sun,
and there's the difference.

Since you would like to see a calculation of tidal forces, here it is:

The difference in force felt at dawn/dusk versus noon/midnight is:

F = 3GmMr/D^3

where G is the universal gravitational constant (6.67e-11 m^3/kg/s^2), 
m is your mass (70 kg), 
M is the mass of the sun (1.99e30 kg),
r is the radius of the earth (6.38e6 m), and
D is the earth-sun distance (1.50e11 m).
 
Plugging in, your wight is .000053 N less at noon/midnight than at dawn/dusk.
That's .0000118 lb less, and is indeed about 4 orders of magnitude
different from .4 N.

Also, the tidal force does not depend on the orbit being elliptical. 
Every massive object exerts a tidal force on every other object of
finite size.  It's basically a 'stretching' force that arises from the
fact that the massive object pulls harder on whichever side of the
object being stretched is closest.

Cool, thanks for clarifying the details!

In the pub last night, I was flying and the earth was spinning and I
got thrown out. Does this help to answer your question?

Best

The radius of the Earth is approximately
R = 6378.1 kilometers

The Earth makes one rotation every day

In one day you will travel the farthest at the equator. That distance
is the circumference of a circle
d = 2 pi R = 40074.7842 kilometers

so the velocity at the equator is 
v = d/T = (2 pi R)/day = 463.828521 m / s

Centripetal acceleration is given by
v^2/r = 0.0337305619 m / s^2

The gravitational acceleration at the surface of the earth is approximately
g = 9.8 m/s^2

The ratio of the maximum centripital acceleration to the acceleration
due to gravity is
v^2/r / g = 0.00344189407

So the centripital acceleration is less than 1% of gravitational acceleration.

For them to be equal a day would be 1.40802379 hours long. So we
aren't rotating nearly fast enough to fly off.