Does the earths spinning motion cause an object to weigh less at the
equator than at the poles due to centrifugal force?

Yes. There are two  latitude effects:


The first is that the rotation of the Earth imposes an additional
acceleration on the body that opposes gravitational acceleration.

The second reason is the Earth's equatorial bulge (itself also caused
by centrifugal force), which causes objects at the equator to be
further from the planet's centre than objects at the poles.

The combined result of these two effects is that g is 0.052 m/sē more,
hence the force due to gravity of an object is 0.5% more, at the poles
than at the equator.
http://en.wikipedia.org/wiki/Gee

These two effects, latutide and elevation are described by equation:

total gravity = 978032(1 + .0053 sin2(q ) - .000024 sin4(q )) - .3086 h + a, 

where q is the latitude, h is the sea level elevation (in meters) and
a is the anomaly (in milligals). Total gravity will be given in
milliGals.
http://www.uwlax.edu/faculty/skala/grav2les.htm

Other effects, tides e.g. exists (as learned commenters pointed out) and are
described here:

" Myths about Gravity and Tides
There are tides everywhere on Earth, including not just oceans and ... "
www.jal.cc.il.us/~mikolajsawicki/tides_new2.pd

The field is 
Geodesy  also called geodetics, scientific discipline that deals with
the measurement and representation of the earth, its gravitational
field and geodynamic phenomena (polar motion, earth tides, and crustal
motion) in three-dimensional time varying space.
http://en.wikipedia.org/wiki/Geodesy

Yes. Although it's not really accurate to say it's due to centrifugal
force at the equator, but rather because of a lack of centripetal
force.

   In the same way, objects on the equator weigh a minute amount less
at midnight than at noon.

   Now of course, so far we're only talking of a single abstract
object on an abstract earth. Let's add a layer of realism and include
the atmosphere. The fluid air will tend to rush towards the night side
of Earth, increasing the air pressure there. The increased pressure
will bear downwards on the earlier considered object, actually making
it a little heavier.

In the same way, objects on the equator weigh a minute amount less
at midnight than at noon.

That's not right is it? At midnight the sun is below your feet and
pulling you towards the earths surface a little harder. At noon it's
pulling you up.

And the moon has a bigger effect on your weight than the sun would.

As far as wind rushing around on to the night side i think that's
probibly due more to pressure differences in hot and cool gasses than
anything. I have never actuialy heard or read anything that supports
that phenomanon. Have further reading?

Hi egon,

Yes, at midnight you are accelerating towards the Sun, against the
Earth; towards the Sun away from Earth at noon. And the Moon would
make a contribution too. These would be extra layers of realism, such
as I metioned before in regards to air pressure. As far as Earth & the
"object" are concerned, the object is pressed harder against Earth due
to their mutual interaction at noon than a midnight.

   And the actual sum total behavior of the atmosphere does include
the pressure differentials due to temperature gradients. But there's
still a component to the behavior due to centrifugal effects.
Factoring in the thermal contributions is yet another layer of
realism.

   By the way, Egon is my favorite Ghostbusters character. He rocks!

One does weigh about 0.5% less at the equator than at the poles.  The
main cause of this is the centrifugal pseudoforce due to the Earth's
rotation.  This accounts for about 2/3 of the effect.  The remaining
1/3 of the effect is due to the fact that the Earth isn't a perfect
sphere;  it's slightly flattened, and the radius at the poles is less
than that at the equator by about 21.5 km.  This means that someone
standing on the equator is 21.5 km further from the center of mass of
the Earth, and feels a correspondingly smaller gravitational
acceleration than someone standing at one of the poles.

The so-called equatorial bulge is, itself, caused by the rotation of
the Earth, so in terms of primary causes, the entire weight-change
effect is really due to the rotational motion of the Earth.  See
<http://en.wikipedia.org/wiki/Equatorial_bulge>

As others have pointed out, there are other things that can affect the
net acceleration one experiences at the Earth's surface:  bouyancy
effects due to atmospheric pressure changes; gravitational
accelerations due to the sun, moon, other planets, etc.; elevation
changes due to local topography (as opposed to the equatorial bulge);
local and regional variations in the density of the rocks that make up
the Earth, etc.

"In the same way, objects on the equator weigh a minute amount less
at midnight than at noon."

That statement is false.  Objects on the equator weigh (very) slightly
more at dawn and dusk than at noon and midnight, because of the tidal
force from the sun.  However, the tidal force from the moon is
stronger.  Objects weigh more when the moon is near the horizon, and
less when it's overhead or on the other side of the earth.  Tidal
forces from the sun and moon are tiny (more than 10,000 times smaller)
than the centrifugal force generated by the rotation of the earth.  As
hfshaw pointed out, the bulge at the equator is also an important
factor.  There is some relevant discussion at
http://answers.google.com/answers/threadview?id=194122

When you go from the equator to the pole, your weight changes by about
as much as it does when you drink a glass of water.  The change in
your weight caused by the moon and sun are comparable to the change in
your weight when you trim your fingernails.

"The fluid air will tend to rush towards the night side
of Earth, increasing the air pressure there."

That statement is false too.  There is no tendency for air to go to
the 'night side'.  There are tides in the atmosphere as well as the
ocean, but there are 2 'high tides' a day.  Like the ocean and the
earth itself, the atmosphere tends to bulge both toward AND AWAY from
the moon (or sun) and the be thinner in a direction 90 degrees away
from that of the massive body.

rracecarr,

   Your points are correct factually. But in principle what I've said
is also correct. There has to be a centrifugal component to the
overall weight of an object at noon vs midnight, even if that
component isn't predominant. There also must be a centrifugal
component to the overall behavior of Earth's fluid atmosphere, even if
it isn't predominant. These are additional layers of realism such as
what I cited. My point was only to show how the OP's orginal idea can
be extended beyond the two body problem of Earth/object, to include
other elements of the world which in fact all make contributions to
the sum of the circumstances.

Hi qed100,

I'm not sure I agree.  I don't understand why there has to be a
difference, even a tiny one, between the weight of an object at noon
vs midnight.  You don't have to worry about the fact that the earth is
revolving around the sun.  Both the earth and the object on the earth
are in free fall toward the sun, so the only thing that causes
differences in the weight of the object are are changes in the
strength and direction of the gravitational field of the sun as you
move the object to different locations on the earth.  The changes are
called tidal forces, and they are identical to quite high precision on
the sides of the earth facing toward and away from the sun.  Right?

OK. Let me clear up a few things that were in the comments.

First, let's not get all concerned about the use of the term
"centrifugal force". Physicists use it all the time, we all know what
it means, and it is a perfectly legitimate concept.

Second, I realize that the point of the comment about the day and
night sides was just to say that things are more complicated, which is
certainly true, but let's see what the real effect is. To a high
degree of approximation the weight on the night side and on the day
side due to the effects of solar gravity and the earth's acceleration
should be the same, as race car has observed. Let's put ourselves in
the reference frame of the earth as it orbits the sun. At the center
of the earth, the gravitational acceleration from the sun exactly
cancels the centrifugal accelaration due to the earth's orbit, as it
must. On the night side, you are farther from the sun, so gravity is
weaker and the centrifugal force is stronger, thus you are a little
bit lighter. On the day side, the gravitational force is stronger and
the centrifugal force is weaker so, again, you are a little bit
lighter. As race car could have told us, this is one way to understand
why the solar tidal bulge is on both the day side and the night side
of the earth. Since the diameter of the earth is tiny compared to the
distance to the sun, it is a very accurate approximation to take the
first term in the Taylor expansion and thus the small increment in the
force is the same on either side. If you want to go to the next term
in the Taylor series, you note that the gravity is going like 1/r^2,
while the centrifugal force is going like r, so the sum of the forces
has positive curvature. Thus the change in the gravitational force is
ever so slightly greater than the change in the centrifugal force on
the side closer to the sun, so you are ever so slightly lighter on the
day side than on the night side, in contradiction to what QED said.

Third, a higher atmospheric pressure will not make someone heavier.
Quite the contrary, the higher the density of the air, the more
bouyant force you will experience. Thus a higher pressure will make
you lighter, not heavier, as any talkative helium balloon will tell
you.

"Third, a higher atmospheric pressure will not make someone heavier.
Quite the contrary, the higher the density of the air, the more
bouyant force you will experience. Thus a higher pressure will make
you lighter, not heavier, as any talkative helium balloon will tell
you."

   Yes, you're right.

Hey rracecarr,

   After considering it I think you may be right about the noon vs midnight thing.