Apparent gravity versus latitude* The acceleration due to...

Apparent gravity versus latitude* The acceleration due to gravity measured in an earthbound coordinate system is denoted by $g .$ However, because of the Earth's rotation, $g$ differs from the true acceleration due to gravity, $g_{0}$. Assuming that the Earth is perfectly spherical, with radius $R_{e}$ and angular velocity $\Omega_{e}$, find $g$ as a function of latitude $\lambda$. (Assuming the Earth to be spherical is actually not justified $-$ the contribution to the variation of $g$ with latitude due to the polar flattening is comparable to the effect calculated here.)

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Key Concepts

True Gravitational Acceleration

This is the acceleration due to gravity as dictated by Newton’s law of universal gravitation. It represents the gravitational pull exerted by Earth on objects and would be the only acceleration experienced in a perfectly non-rotating, spherical Earth scenario.

Centrifugal Acceleration

Due to the Earth's rotation, an outward centrifugal acceleration acts on objects away from the axis of rotation. This fictitious force reduces the net effective acceleration on a body by partially counteracting true gravity. Its magnitude depends on the rate of rotation and the distance from the rotation axis.

Apparent Gravity

Apparent gravity is the effective acceleration that an object experiences when both true gravitational acceleration and the opposing centrifugal acceleration are taken into account. It is this net acceleration that is measured by scales and experienced as weight on the surface of the rotating Earth.

Latitude Dependence

The magnitude of the centrifugal acceleration varies with latitude since the distance from an object to Earth's axis of rotation changes with latitude. At the equator, where this distance is maximal, the centrifugal effect is strongest, while at the poles, it vanishes. This variation leads to a latitude-dependent effective gravitational acceleration experienced on Earth.

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