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The carth has a radius of \(6.38 \times 10^{6} \mathrm{m}\) and turns on its axis once every \(23,9 \mathrm{h}\) . (a) What is the tangential specd (in \(\mathrm{m} / \mathrm{s} )\) of a person living in Ecuador, a country that lies on the equator? (b) At what latitude (i.e., the angle \(\theta\) in the drawing) is the tangential speed one-third that of a person living in Ecuador?

Short Answer

Expert verified
(a) 466.5 m/s; (b) 70.5 degrees latitude.

Step by step solution

01

Understand the problem

We are given the Earth's radius and its rotation period. We need to calculate the tangential speed of a person living on the equator and then find at what latitude the speed is one-third of the equator's speed.
02

Calculate the Earth's circumference at the equator

The formula for the circumference, which is the distance traveled in one complete rotation, is given by \[ C = 2 \pi r \]where \( r = 6.38 \times 10^{6} \text{ m} \). Thus, \[ C = 2 \times \pi \times 6.38 \times 10^{6} \text{ m} \].
03

Calculate tangential speed at the equator

The tangential speed \( v \) at the equator can be calculated using the circumference and the period of rotation. Given, the rotation period is \( 23.9 \text{ hours} \), which should be in seconds:\[ T = 23.9 \times 3600 \text{ s} \]. The tangential speed is:\[ v = \frac{C}{T} \].
04

Calculate tangential speed at latitude \( \theta \)

At latitude \( \theta \), the radius of the circular path is \( R \cos \theta \), where \( R \) is the Earth's radius. The tangential speed is then:\[ v_\theta = \frac{2 \pi (R \cos \theta)}{T} \].
05

Equate tangential speeds for given conditions

We need \( v_\theta = \frac{1}{3}v \). So:\[ \frac{2 \pi (R \cos \theta)}{T} = \frac{1}{3} \cdot \frac{2 \pi R}{T} \].Solving, we find:\[ \cos \theta = \frac{1}{3} \].
06

Calculate latitude \( \theta \)

From \( \cos \theta = \frac{1}{3} \), solve for \( \theta \):\[ \theta = \cos^{-1}(\frac{1}{3}) \].
07

Solve and calculate numerical values

1. Solve the equations numerically: - Earth's circumference: \( C = 2 \pi \times 6.38 \times 10^{6} \approx 4.011 \times 10^{7} \text{ m} \). - Convert period to seconds: \( T = 23.9 \times 3600 \approx 86040 \text{ s} \). - Tangential speed at equator: \( v \approx \frac{4.011 \times 10^{7}}{86040} \approx 466.5 \text{ m/s} \).2. Solve \( \theta = \cos^{-1}(\frac{1}{3}) \) to find \( \theta \approx 70.5^\circ \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Earth's Rotation
The Earth rotates around its axis once approximately every 24 hours. This rotation is responsible for the phenomenon of day and night. A complete rotation means that every point on the surface of the Earth travels through a circular path.
Understanding the Earth's rotation is crucial when calculating tangential speed, which is the linear speed of any object located on the Earth's surface.
* The Earth's equatorial circumference is about 40,070 kilometers or, in this exercise, roughly 40,110,000 meters as calculated from the given radius of 6.38 million meters.
* The rotation period is used to determine how fast a point on the Earth's surface is moving in meters per second.
  • This rotation period was given as 23.9 hours, which must be converted into seconds since speed is typically expressed in meters per second.
The tangential speed varies with latitude because the distance from the axis of rotation to any point on the surface decreases as you move away from the equator to the poles. At the equator, where this distance is greatest, the tangential speed is maximized.
Latitude Calculation
Latitude plays a significant role in determining the tangential speed of a point on Earth's surface. Latitude measures the angle between a point on Earth and the equator.
In this exercise, calculating the latitude where the tangential speed is one-third of that at the equator involves understanding how latitude affects radius and, consequently, speed.
* At any given latitude \( \theta \), the effective radius of rotation is \( R \cos \theta \). This adjustment accounts for the fact that as you move north or south of the equator, the circle of rotation becomes smaller.
* To solve for \( \theta \) when given a specific speed condition, we use the equation \( v_\theta = \frac{1}{3}v \), leading to \( \cos \theta = \frac{1}{3} \).
By solving \( \theta = \cos^{-1}(\frac{1}{3}) \), we determine the latitude to be approximately 70.5 degrees. This latitude is significant as it shows how rotation speed decreases as you move from the equator.
Centripetal Motion
Centripetal motion governs how objects move in circles, such as points on Earth due to its rotation. A fundamental aspect of centripetal motion is that there is a constant force acting towards the center of the circle, keeping the object in motion.
In the context of Earth's rotation, this centripetal force is generated by gravity, pulling objects towards Earth’s center.
* Tangential speed and centripetal force are deeply interlinked; the required centripetal force needs to increase with tangential speed.
* The formula for centripetal force is \( F_c = \frac{mv^2}{r} \), where \( F_c \) is the centripetal force, \( m \) is mass, \( v \) is tangential speed, and \( r \) is the radius of the circular path.
  • As noted in the exercise, the equator has the highest tangential speed due to the larger radius, thus requiring a larger centripetal force to maintain circular motion.
  • At higher latitudes, the reduced radial distance decreases the speed and the centripetal force needed.
Understanding this balance is essential for grasping how objects move smoothly without flying off their path, thanks to Earth's gravitational pull acting as the necessary centripetal force.

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Most popular questions from this chapter

A quarterback throws a pass that is a perfect spiral. In other words, the football does not wobble, but spins smoothly about an axis passing through cach end of the ball. Suppose the ball spins at \(7,7 \mathrm{rev} / \mathrm{s}\). In addition, the ball is thrown with a linear speed of \(19 \mathrm{~m} / \mathrm{s}\) at an angle of \(55^{\circ}\) with respect to the ground. If the ball is caught at the same height at which it left the quarterback's hand, how many revolutions has the ball made while in the air?

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A car is traveling with a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. The wheels have a radius of \(0.300 \mathrm{~m} .\) If the car speeds up with a linear acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{~s},\) find the angular displacement of each wheel during this period.

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