Apparent Weight and the Earth’s Rotation

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Apparent Weight and the Earth’s Rotation

Nächstes Video14.9: Variation in Acceleration due to Gravity near the Earth’s Surface

Consider an object of mass m suspended from a spring scale attached to a rigid support at the Earth's equator. The scale reading is the object's apparent weight mg', equal to the magnitude of tension force Ft in the spring.

If the Earth was a non-rotating body, the tension force would be equal to the gravitational force Fg acting on the object, keeping it in equilibrium.

Recall that the magnitude of the gravitational force is the true weight mg of the object.

However, since the Earth rotates about an axis passing through its poles, the object at the equator describes a circle around the Earth's rotational axis. Therefore, it has a centripetal acceleration directed towards the Earth's center.

Hence, the forces acting on the object are the gravitational force directed towards the Earth's center, the tension force in the spring directed away from the Earth's center, and the centripetal force directed towards the axis of rotation.

Therefore, the apparent weight equals the true weight minus the centripetal force. At the poles, the centripetal force on the object is zero, implying that mg' equals mg.

Apparent Weight and the Earth’s Rotation

Since all objects on the Earth's surface move through a circle every 24 hours, there must be a net centripetal force on each object, directed towards the center of that circle. The points of the north and south poles are the only exception to this rule.

For an object on the Earth's equator, the net centripetal force that accounts for its rotation is the Earth's pull towards its center, or the weight minus the normal force that prevents it from piercing into the Earth's surface. This force, called the apparent weight, is thus the object's weight minus the centripetal force.

The centripetal force is proportional to the Earth's radius and the square of the Earth's angular speed. The angular speed is relatively small; hence, the centripetal force, which is also the difference between the true and apparent weights, is itself quite small. At the equator, an object's apparent weight is only 0.34% less than its true weight, a small correction.

We can further understand this minor correction by considering how quickly the Earth would have to rotate such that all objects on the equator would feel weightless owing to the Earth's rotation. Calculations imply that the orbital period of the Earth, which is 24 hours, would have to be only 84 minutes instead.

This text is adapted from Openstax, University Physics Volume 1, Section 13.2: Gravitation Near Earth's Surface.