14.8
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…
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.
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Q1: Why is apparent weight different from true weight at the equator?
At the equator, objects move in a circle due to Earth's rotation, requiring centripetal acceleration toward the rotational axis. The spring scale reading (apparent weight) equals true weight minus the centripetal force needed for this circular motion. At the poles, centripetal force is zero, so apparent weight equals true weight.
Q2: How does Earth's rotation affect the centripetal force on an object?
Centripetal force is proportional to Earth's radius and the square of its angular speed. Since Earth's angular speed is relatively small, the centripetal force is also small. At the equator, this force accounts for only a 0.34% difference between true and apparent weight, making it a minor correction.
Q3: What forces act on an object suspended from a scale at the equator?
Three forces act on the object: gravitational force directed toward Earth's center, tension force in the spring directed away from the center, and centripetal force directed toward the rotation axis. The tension force (apparent weight) balances the gravitational force minus the centripetal force requirement.
Q4: How fast would Earth need to rotate for objects at the equator to feel weightless?
For objects at the equator to feel weightless, the centripetal force would need to equal the gravitational force. Calculations show Earth's orbital period would need to be only 84 minutes instead of 24 hours. This demonstrates how slowly Earth actually rotates relative to what would be required for weightlessness.
Q5: Why do objects at the poles experience no centripetal force effect?
At the poles, objects do not move in a circle around Earth's rotational axis; they remain stationary relative to the axis. Since there is no circular motion, no centripetal acceleration is needed. Therefore, apparent weight equals true weight at the poles.
Q6: What is the relationship between tension force and apparent weight on a scale?
The tension force in the spring scale equals the apparent weight of the object. This reading represents the normal force preventing the object from falling through the scale. At the equator, this tension is less than the gravitational force because some gravitational force provides the necessary centripetal acceleration.
Q7: How does variation in acceleration due to gravity relate to Earth's rotation?
Earth's rotation causes variation in acceleration due to gravity near the earth s surface by reducing the effective gravitational acceleration at the equator. The centripetal acceleration required for circular motion at the equator reduces the net downward acceleration, making objects appear lighter on a scale compared to the poles.