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Chapter 7: Problem 12

The earth's radius is \(6.37 \times 10^{\circ} \mathrm{m} ;\) it rotates once every 24 hours. a. What is the earth's angular speed? b. Viewed from a point above the north pole, is the angular velocity positive or negative? c. What is the speed of a point on the equator? d. What is the speed of a point on the earth's surface halfway between the equator and the pole? (Hint: What is the radius of the circle in which the point moves?)

Short Answer

Expert verified
a. The Earth's angular speed is approximately \(7.27 \times 10^{-5}\) rad/s. b. The angular velocity is positive. c. The speed of a point on the equator is approximately \(465.1 m/s.\) d. The speed of a point at halfway between the equator and the pole is approximately \(328.98 m/s\)

Step by step solution

01

Calculate the Earth's angular speed

The angular speed \(w\) is given by the formula \(w = 2\pi / T\), where \(T\) is the period. In this case, the Earth rotates once every 24 hours, which is equivalent to \(24 \times 60 \times 60\) seconds. So, calculating \(w\) as \(w = 2\pi / (24 \times 60 \times 60)\) gives the Earth's angular speed.
02

Determine the direction of the Earth's angular velocity

Angular velocity, like velocity, has a direction. The convention is to take counterclockwise motion as positive. From a point above the North Pole, the Earth's rotation appears to be counterclockwise. Therefore, the angular velocity is positive.
03

Calculate the speed of a point on the equator

The speed of a point on the rotating body is given by the formula \(v = r \cdot w\), where \(r\) is the radius and \(w\) is the angular speed. Plug in the values for the Earth's radius (\(6.37 \times 10^{6} m\)) and the calculated angular speed to determine the speed of a point on the equator.
04

Calculate the speed of a point halfway between the equator and the pole

This point effectively moves in a circle with a different radius. The effective radius is the Earth's radius times the cosine of the latitude. For a point halfway between the equator and the pole, this latitude is 45 degrees. Therefore, the effective radius \(r_{eff}= r \times cos(45掳)\). The speed of the point is then \(v = r_{eff} \times w\). Plug in the values for the effective radius and the calculated angular speed to get the speed.

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how quickly an object rotates around a central point. In the context of Earth's rotation, it is used to describe how fast the Earth spins around its axis. This can be calculated using the formula \(w = \frac{2\pi}{T}\), where \(w\) is the angular velocity, and \(T\) is the period of rotation.
For Earth, which completes one full rotation in 24 hours (or \(24 \times 60 \times 60\) seconds), the angular velocity can be calculated as \(w = \frac{2\pi}{86400}\). This tells us how fast all points on the Earth move in the rotational circle.
Understanding angular velocity is crucial in physics as it helps in predicting the speed and direction of rotation.
Earth's Rotation
The rotation of the Earth is a fundamental phenomenon that results in the day and night cycle. Earth rotates about its axis from west to east, which is why we observe the sun rising in the east and setting in the west. From an observational point above the North Pole, this rotation appears counterclockwise.
This movement has implications on the angular velocity, making it positive when viewed from this perspective. The convention is that counterclockwise rotations are deemed positive, and clockwise are negative.
This rotating motion influences not only the daily cycle but also impacts global weather patterns and ocean currents. Understanding Earth's rotation is crucial for geosciences and astronomy.
Equator Speed
The equator is the imaginary line around the middle of the Earth, equidistant from both poles. Due to Earth's rotation, the speed of a point on the equator is of great interest. To compute this linear speed, the formula \(v = r \cdot w\) is used, where \(v\) is the linear speed, \(r\) is the Earth's radius, and \(w\) is the angular velocity.
Given the Earth's radius is \(6.37 \times 10^6\) meters and using the previously calculated angular velocity, the equator's speed can be calculated. This helps in understanding Earth's dynamics and is crucial for activities like satellite deployment and navigation.
This speed is essentially the longest distance a point on Earth covers due to rotation in a single day.
Latitude
Latitude has a significant effect on rotational speed because it dictates the size of the circle a point on Earth travels. Points at different latitudes rotate along different-sized circles. For example, at the equator, the circle is largest, while nearer to the poles, the circle is smaller.
When calculating the speed of a point halfway between the equator and the North Pole, the concept of effective radius is applied. This effective radius is calculated using the formula \(r_{eff} = r \times \cos(\text{latitude})\). For a location at 45 degrees latitude, this means \(r_{eff} = 6.37 \times 10^6 \cdot \cos(45掳)\).
Understanding the effect of latitude helps in a variety of fields, from aviation to meteorology, as it influences the Earth's atmospheric phenomena.

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Most popular questions from this chapter

To throw a discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about \(1.8 \mathrm{m}\). If the thrower takes \(1.0 \mathrm{s}\) to complete one revolution, starting from rest, what will be the speed of the discus at release?

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