It's a fine source.

Quote:
Originally Posted by

**Hiro**
I know what he meant, and he was wrong, as are you. The rotation has NOTHING to do with the gravity. Ever.

You can't even construct the virtual centripetal force vector because the solid surface is under the person standing, not on the outside radius as it is in a centrifuge. There simply is no force trying to throw things off the surface of the Earth because the earth is spinning, so there is no reduction in gravity because of it.

Differences in gravity between Poles and equator are due to the shape of the Earth and differences in local composition density contributions.

Let me give you a question to consider. I assume that you would agree that on the equator, one is moving (relative to the center of mass,

*M*, of the earth) in a circular path, of radius equal to the equatorial radius of the earth,

*R*, and a period of one day. In which case, would you agree that your motion includes a radial acceleration,

*a*, (normal to your instantaneous velocity vector) given by

*a* = -

*ω*²

*R* , [1]

and that to maintain that acceleration (circular motion) requires a centripetal force,

*F*, given by

*F* = -

*mω*²

*R*, [2]

where

*m* is your mass and

*ω* your angular velocity?

Hopefully you would agree, because that is simple Newtonian physics. That centripetal force is, of course, provided by a fraction of your gravitational attraction to the earth, and your apparent weight,

*mg* (the residual reaction between you and the surface you are standing on), is reduced by that amount. Or more rigorously, if

*G* is the gravitational constant and

*g* is the acceleration due to gravity in the local coordinate system (your initial acceleration if you jump off a building), for mechanical equilibrium between yourself and the surface of the earth:

*mg* =

*GMm*/

*R*² -

*mω*²

*R* , [3]

While there actually is no such thing as a centrifugal force - it is just Newton's 3rd law reaction to a centripetal force - there is indeed a centripetal force acting in this system.

So, I assume that you are drawing the distinction between pure gravitational attraction and actual acceleration due to gravity in a dynamic system. If the discussion is static gravitational attraction then you are correct, but when discussing actual acceleration due to gravity (i.e. the acceleration of an object in free fall in the local coordinate system), the angular velocity,

*ω*, of the coordinate system has the effect described, leading to an acceleration given by dividing [3] though by your mass,

*m*,

*g* =

*GM*/

*r*² -

*ω*²

*r* . [4]

The second term on the RHS is the effect of the earth's rotation. Does that explicit distinction reconcile the two views?