Problem 23 - University Physics with Modern Physics

Question

The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) What is the radial acceleration of an object at the earth's equator? Give your answer in m/s\(^2\) and as a fraction of \(g\). (b) If \(a_{rad}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

Step-by-Step Solution

Answer

(a) 0.034 m/s², approximately 0.0035g. (b) Approximately 1.41 hours.

Step 1: Convert Radius to Meters

The Earth's radius is given as 6380 km. To convert this to meters, multiply by 1000 (since 1 km = 1000 m). \[ R = 6380 \, \text{km} \times 1000 \, \frac{\text{m}}{\text{km}} = 6.38 \times 10^6 \, \text{m} \]

Step 2: Calculate Angular Velocity

The Earth completes one full rotation in 24 hours. To calculate the angular velocity, use the formula:\[ \omega = \frac{2\pi}{T} \]where \( T = 24 \times 3600 \, \text{seconds}\) (since 1 hour = 3600 seconds).\[ \omega = \frac{2\pi}{86400 \, \text{s}} \approx 7.27 \times 10^{-5} \, \text{rad/s} \]

Step 3: Calculate Radial Acceleration

The radial (or centripetal) acceleration can be calculated using:\[ a_{\text{rad}} = R\omega^2 \]Substitute \( R = 6.38 \times 10^6 \, \text{m} \) and \( \omega = 7.27 \times 10^{-5} \, \text{rad/s} \):\[ a_{\text{rad}} = 6.38 \times 10^6 \, \text{m} \times (7.27 \times 10^{-5} \, \text{rad/s})^2 \approx 0.034 \, \text{m/s}^2 \]

Step 4: Express Radial Acceleration as a Fraction of g

The acceleration due to gravity \( g \) is approximately \( 9.81 \, \text{m/s}^2 \). The fraction is:\[ \frac{a_{\text{rad}}}{g} = \frac{0.034 \, \text{m/s}^2}{9.81 \, \text{m/s}^2} \approx 0.0035 \]

Step 5: Determine Rotation Period for Objects to Fly Off

For objects to fly off, the radial acceleration needs to equal \( g \). Thus, set \( R\omega^2 = g \) and solve for \( \omega \):\[ \omega^2 = \frac{g}{R} \Rightarrow \omega = \sqrt{\frac{9.81}{6.38 \times 10^6}} \approx 0.00124 \, \text{rad/s} \]We can find the period \( T \) from \( \omega \):\[ \omega = \frac{2\pi}{T} \Rightarrow T = \frac{2\pi}{\omega} \approx \frac{2\pi}{0.00124} \approx 5065 \, \text{seconds} \approx 1.41 \, \text{hours} \]

Key Concepts

Radial Acceleration

Radial acceleration is a type of acceleration experienced by an object moving in a circular path. This acceleration points towards the center of the circle and is responsible for keeping the object in its curved path. Imagine you are swinging a ball tied to a string; the tension in the string causes the ball to constantly accelerate towards the center. In physics, this is known as radial or centripetal acceleration.

To calculate radial acceleration (\( a_{rad} \)), we use the formula:\[ a_{rad} = R\omega^2 \]where \( R \) is the radius of the circular path and \( \omega \) is the angular velocity. For objects on the Earth's equator, this means they experience radial acceleration due to Earth's rotation. This radial acceleration is small compared to the gravitational acceleration, \( g \), of \( 9.81 \, \text{m/s}^2 \).

Centripetal Force

Centripetal force is the force that keeps an object moving in a circular path. Without it, objects would move in a straight line due to inertia. This force is always directed towards the center of the circle.

For example, in the context of Earth's rotation, the centripetal force is provided by gravity. Gravity pulls objects towards Earth's center, allowing them to maintain their curved, circular motion. The magnitude of this force can be calculated using the formula:\[ F_{c} = ma_{rad} \]where \( m \) is the mass of the object and \( a_{rad} \) is the radial acceleration. It's important to note that while gravity acts as the centripetal force keeping us grounded, the sensation or effect of this force is less noticeable due to the Earth's vast size.

Angular Velocity

Angular velocity is a measure of how quickly an object rotates or revolves around a central point. It describes the rate of rotation and is typically measured in radians per second (rad/s).

• For Earth, completing one revolution every 24 hours means that its angular velocity is very small yet significant in its own nature.


• The formula to find angular velocity \( \omega \) is:\[ \omega = \frac{2\pi}{T} \]where \( T \) is the time taken for one complete rotation (here, 86400 seconds for the Earth). Calculating this using the given numbers results in approximately \( 7.27 \times 10^{-5} \, \text{rad/s} \).

This measure is crucial, as it directly influences the radial acceleration experienced on Earth's surface.

Earth Rotation

Earth's rotation is a fascinating aspect of our planet's motion. It not only determines the day and night cycle but also exerts visible effects on tides and climate. As Earth spins on its axis, every point on the Equator moves in a circular path with a velocity determined by the rotation.

The Earth's rotation has several implications:

• Objects on the Earth's surface are slightly "lighter" due to the centrifugal force generated by the spin acting against gravity.


• The periodic time for one rotation, or a day, is 24 hours, but theoretically, if the Earth spun much faster, this could drastically change life, leading to objects potentially being flung off into space.

Through mathematical calculations, we understand that for objects to overcome Earth's gravity and move into space due to centrifugal force, the rotation would need to be much faster — with a period reducing to approximately 1.41 hours.