Question

Assume that the standard kilogram would weigh exactly \(9.80 \mathrm{~N}\) at sea level on the equator if the Earth did not rotate. Then take into account the fact that the Earth does rotate, so that this object moves in a circle of radius \(6370 \mathrm{~km}\) (the Earth's radius) in one day. (a) Determine the centripetal force needed to keep the standard kilogram moving in its circular path. ( \(b\) ) Find the force exerted by the standard kilogram on a spring balance from which it is suspended at the equator (its apparent weight).

Short answer

The centripetal force needed to keep the standard kilogram moving in a circular path is calculated in step 1. The apparent weight of the standard kilogram on a spring balance at the equator is calculated in step 3. Note: Both results are in Newtons (N).

Step-by-step solution

Calculation of Centripetal Force

First we convert 1 day to seconds (24 hours/day × 60 min/hour × 60 sec/min = 86400 sec). Now we need to find the velocity (v), where \(v= \frac{2πr}{T}\), with \( r = 6370 km = 6.37 × 10^6 m \) (radius of Earth), \( T = 86400 sec \) (time for 1 rotation). Then we plug the values into the formula for centripetal force \( F_c= \frac{mv^2}{r} \), where \( m = 1 kg \) (mass of the object).

Calculation of Gravitational Force

Next, we calculate the gravitational force \( F_g \) exerted by the Earth on the 1kg mass using the formula: \( F_g=mg \), where \( g = 9.80 m/s^2 \) (acceleration due to gravity). This gives us the actual weight of the standard kilogram.

Calculation of Apparent weight

Finally, the apparent weight of the standard kilogram is the difference between the gravitational force and the centripetal force. So, we subtract the centripetal force obtained in step 1 from the gravitational force obtained in step 2.

Key concepts

Centripetal Acceleration

Imagine being inside a car that’s taking a sharp turn. You feel yourself pushed outward, away from the curve. This is a result of a force due to centripetal acceleration. Centripetal acceleration is necessary for any object moving in a circle, and it always points toward the center of the circle. Its magnitude is given by the equation \( a_c = \frac{v^2}{r} \), where \( v \) is the velocity of the object and \( r \) is the radius of the circular path.

In our textbook exercise, the Earth’s rotation provides the centripetal acceleration needed to keep objects on the equator moving in a circular path. Since the velocity is based on one complete rotation of Earth in one day, and the path is the Earth's circumference, we can calculate centripetal acceleration and the corresponding force.

Gravitational Acceleration

Gravitational acceleration, also known around Earth as \( g \), is the acceleration that the Earth imparts to objects due to gravity. It is approximately \( 9.80 \text{ m/s}^2 \) at sea level but can vary slightly with altitude and location. Despite common belief, gravitational acceleration does not change based on an object's mass.

For our exercise, we're considering the pull of gravity on a standard kilogram mass at sea level. It determines how much the standard kilogram would weigh without any other influences. The gravitational force on the mass is the product of this acceleration and the mass of the object (in our case, 1 kg), leading to the standard force of gravity, commonly referred to as weight, before adjustments for rotation are made.

Apparent Weight

Apparent weight might be confusing; it's different from the actual weight because it's the force felt by the object due to gravity after other factors are accounted for—in this case, the Earth's rotation. When an object like the standard kilogram is in circular motion due to the Earth's rotation, a part of the gravitational force is 'used up' to provide the necessary centripetal force to keep the object moving in its path.

Thus, the apparent weight is actually the normal force exerted on the object, which becomes less than the gravitational force due to this 'use' of force for centripetal motion. Our exercise's solution subtracts the centripetal force from the gravitational force to find the apparent weight of the kilogram on a scale at the equator.

Circular Motion

The concept of circular motion describes the movement of an object along a circular path. It can be the orbiting of satellites, the spinning of a merry-go-round, or the rotation of Earth itself. This motion always involves a force that directs towards the center of the circle, which we call the centripetal force.

In our problem, the Earth's rotation provides a real-world example of circular motion on a grand scale, exerting a centripetal force on objects at the equator. This force reduces the object's apparent weight as observed in our step-by-step solution. The interplay of gravitational and centripetal forces is a fascinating demonstration of how celestial mechanics can have down-to-earth (pun intended!) implications, like slightly altering the way things weigh at different points on Earth.