fooly said:
If that is so, what causes the weight of a person to be lighter at the equator?
Short answer to your question: It's the Earth's rotation. There's a direct effect and an indirect one.
First, we need a working definition of weight. In some physics texts you'll see weight defined as the product of gravitational acceleration and mass. In a few, you'll see weight defined as the sum of all real forces except gravity. Some call this latter definition "apparent weight". A few call it "scale weight" because this is what a spring scale measures. It's this scale weight that you are asking about. If you are standing still on the ground, your scale weight is the normal force the ground exerts on you (plus a tiny bit from buoyancy; I'll ignore that).
Why not use weight as mass times gravitational acceleration? The answer is that don't feel gravitational acceleration. From the perspective of general relativity, gravitation is a fictitious force; we don't feel fictitious forces. They aren't "real". From the perspective of Newtonian mechanics, gravitation is a real force, but we don't feel it because it affects every cell equally. Gravitation alone doesn't induce any stresses or strains in our bodies. Or in an accelerometer, for that matter. An accelerometer at rest on the surface of the Earth will register an acceleration of about 1g upward because accelerometers don't sense gravitational acceleration.
Back to the problem at hand, I'll look at things from the perspective of an Earth-fixed frame. This is a frame rotating with the Earth. In this frame, a person standing still on the surface of the Earth has zero velocity. The person isn't moving, so the normal force, gravitational force, and fictitious centrifugal force must add to zero. As noted above, it's the normal force that we feel as weight. Even if the gravitational force was the same at the equator and the poles, this outward centrifugal force will mean that the normal force at the equator is smaller than it is at the poles.
Now look at things from the perspective of an inertial frame. Here our person standing still with respect to the (rotating) surface Earth is no longer standing still. The person instead is undergoing uniform circular motion, one revolution per sidereal day about the Earth's rotation axis. The net force acting on the person is exactly that needed to result in this circular motion. In this frame, there are only two forces acting on the person, the gravitational force exerted the Earth as a whole and the normal force exerted by the Earth's surface. Do the math and the normal force from this perspective will be exactly the same as the normal force calculated from the perspective of the rotating frame. Real forces are the same in all reference frames. It's only the fictitious forces that are frame dependent. Once again, even if gravitational force is the same at the poles and the equator, the normal force will be smaller at the equator compared to at the poles.
Bottom line: Regardless of perspective, the Earth's rotation is directly responsible for a reduction in the normal force.
The Earth's rotation is also indirectly responsible for an additional decrease in the normal force. This direct cause assumes that gravitational acceleration is uniform over the surface of the Earth. This is not the case. The Earth's rotation makes the Earth take on the shape of an oblate spheroid. There's a bulge at the equator. The poles are closer to the center of the Earth than is the equator. This in turn means that gravitational acceleration is greater at the poles than at the equator.
