This discusses some basics about Earth. This issure is minor, all in all.

Polar Versus Equatorial Weight Suppose we have a certified mass of 1,000 pounds. Demonstrate analytically that the mass will weigh 1,000 pounds force at the North Pole but only 996 pounds force at the equator?

♦   Newton's Second Law will be applied. In each case the system is the thousand pound mass. The forces acting will be gravity and the support force of the spring of the scale. Our inertial reference has the center of Earth as the origin. This means we assume the center of Earth moves in a straight line at a constant speed). The vector triad, e_r(t), e_θ(t) and K will be used. Imagine the mass to rest on a spring scale at the North Pole and later at the Equator. The sketches depict these measurements of weight:

At the North pole the position vector of the mass initiates at the center of Earth and extends the distance, r, through the bottom of the scale to the mass. The mass, in constant contact with the scale, rotates or pivots as Earth rotates beneath it.

A precise answer is not needed; let the radius of Earth equal 3960 miles and the surface acceleration of gravity equal 32.2 ft/s²).

For the weighing at the equator, the mass and scale supporting and the place where it rests are rotating with Earth. The position vector of the mass and scale rotates about Earth's center. So we differentiate position correctly as follows:

Now enter there results into Newton's Second Law:

A gravity force will act on the mass on the spring scale whether at the North Pole or equator. At the North Pole the mass has no component of momentum relative to the Earth center. The K-axis is directed vertical from the weing place at either pole. The mass and scale just pivot or rotate about the axis vertically through them. Locations on Earth's equator are not inertial. The mass there has momentum; to weigh a mass there will incur a small error.