Question

The period of rotation of the Earth on its axis is \(24 \mathrm{~h}\). At this angular velocity, the centripetal acceleration at the surface of the Earth is small compared with the acceleration due to gravity. What would Earth's period of rotation have to be for the magnitude of the centripetal acceleration at its surface at the Equator to be equal to the magnitude of the acceleration due to gravity? (With this period of rotation, you could levitate just above the Earth's surface!) a) \(0.043 \mathrm{~h}\) d) \(1.41 \mathrm{~h}\) b) \(0.340 \mathrm{~h}\) e) \(3.89 \mathrm{~h}\) c) \(0.841 \mathrm{~h}\) f) \(12.0 \mathrm{~h}\)

Short answer

Answer: d) 1.41 h

Step-by-step solution

1. Express centripetal acceleration in terms of angular velocity

The centripetal acceleration, \(a_c\), can be expressed in terms of angular velocity, \(\omega\), and Earth's radius \(R\) as: \(a_c = \omega^2 R\).

2. Express angular velocity in terms of the period of rotation

The angular velocity is related to the period of rotation, \(T\), as \(\omega = \frac{2\pi}{T}\).

3. Set centripetal acceleration equal to gravity acceleration

We are given that the magnitude of centripetal acceleration is equal to the acceleration due to gravity, \(g\). Therefore, we have the equation: \(\omega^2 R = g\).

4. Substitute the expression of angular velocity

From step 2, plug in the value of \(\omega\) in terms of \(T\): \(\left(\frac{2\pi}{T}\right)^2 R = g\).

5. Solve for the period of rotation, \(T\)

Now, solve the equation for \(T\). Rearrange the equation as follows: \(T^2 = \left(\frac{4\pi^2 R}{g}\right) = \frac{4\pi^2(6.37\times10^6 \mathrm{~m})}{9.81 \mathrm{~m/s^2}}\) \(T^2 = 1.616 \times 10^7 \mathrm{~s}^2\) \(T = \sqrt{1.616 \times 10^7\mathrm{~s}^2} = 4022 \mathrm{~s}\) Now convert this value to hours by dividing by 3600 (since there are 3600 seconds in an hour): \(T = \frac{4022}{3600} = 1.117 \mathrm{~h}\).

6. Choose the closest given option

Looking at the given options, the closest value to our calculated period of rotation \(T\) is \(1.41 \mathrm{~h}\). Thus, the correct choice is option d) \(1.41 \mathrm{~h}\).

Key concepts

Centripetal Acceleration

When an object moves in a circle, it experiences an inward force that keeps it moving along the curved path. This force causes what is known as centripetal acceleration, which always points towards the center of the circle. For an object moving at a constant speed in a circular path, the centripetal acceleration is given by the formula:
\( a_c = \frac{v^2}{r} \),
where \(v\) is the linear velocity of the object and \(r\) is the radius of the circle. Another way to express centripetal acceleration is in terms of angular velocity,
\( a_c = \omega^2 r \),
where \(\omega\) is the angular velocity. On Earth, at the equator, the centripetal acceleration is considerably less than the gravity's pull, which allows us to stay firmly on the ground rather than being flung into space.

Angular Velocity

Angular velocity is the rate at which an object rotates or revolves around a central point. In other words, it's how fast the angle is changing as something spins around an axis. For Earth or any astronomical body, it is what determines the period of rotation - the time it takes for one complete spin.
Mathematically, angular velocity is represented as:
\( \omega = \frac{2\pi}{T} \),
where \(T\) is the object's period of rotation. Angular velocity helps us understand not just how quickly Earth rotates, but also what happens to the forces experienced on the rotating body - such as the centripetal acceleration we feel due to Earth's rotation.

Acceleration Due to Gravity

The acceleration due to gravity is the acceleration that Earth imparts to objects on or near its surface. In most introductory physics contexts, the acceleration due to gravity is denoted by \(g\), with a standard average value of approximately
\(9.81 \mathrm{m/s^2}\)
on Earth's surface. However, it can vary slightly based on your location on Earth, primarily due to the planet's rotation and shape. We compare other forces to this acceleration to understand the experiences of objects on Earth. For example, when we say centripetal acceleration must equal gravitational acceleration for levitation at the Earth's surface, we equate centripetal force to the weight of an object to find when this levitation scenario would occur.

Period of Rotation Calculation

Calculating the period of rotation involves using the various physical formulas that relate angular velocity, centripetal acceleration, and gravity. Given that centripetal and gravitational forces balance out exactly when an object is in 'levitation', we can set up an equation
\( \omega^2 R = g \),
where \(R\) is the radius of Earth. Using the formula for angular velocity \( \omega\) in terms of period \(T\), we get
\( \left(\frac{2\pi}{T}\right)^2 R = g \).
From here, you solve for \(T\) to find the period at which this balance occurs. For real-world applications, this calculation is purely hypothetical because Earth cannot physically rotate fast enough for its surface gravity to be entirely negated by centripetal acceleration without disintegrating. However, it's a thought-provoking way to understand the interplay of rotations, forces, and accelerations in celestial mechanics.