Problem 83 - Precalculus Mathematics for Calculus

Question

The earth rotates about its axis once every 23 h 56 min 4 s, and the radius of the earth is 3960 mi. Find the linear speed of a point on the equator in \(\mathrm{mi} / \mathrm{h}\).

Step-by-Step Solution

Answer

The linear speed is approximately 1038 mi/h.

Step 1: Calculate the Earth's Rotation Period in Hours

First, convert each time component (hours, minutes, seconds) of Earth's rotation period to hours. The period is given as 23 hours, 56 minutes, and 4 seconds. Convert the minutes to hours: 56 minutes = 56/60 hours = 0.9333... hours. Convert the seconds to hours: 4 seconds = 4/3600 hours = 0.0011... hours. Add them all together: 23 + 0.9333 + 0.0011 = 23.9344 hours.

Step 2: Find the Circumference of the Earth at the Equator

Use the formula for the circumference of a circle, which is the diameter times pi: \( C = 2 \pi r \), where \( r \) is the radius of the earth. Substitute the radius of Earth at the equator \( r = 3960 \text{ mi} \). So, the circumference \( C \) is \( 2 \times \pi \times 3960 \). This computes to approximately 24877.1 mi.

Step 3: Calculate the Linear Speed

The linear speed \( v \) of a point on the equator can be found by dividing the circumference by the rotation period: \( v = \frac{C}{T} \), where \( C = 24877.1 \text{ mi} \) and \( T = 23.9344 \text{ hours} \). By substituting these values, we have \( v = \frac{24877.1}{23.9344} \approx 1038 \text{ mi/h} \).

Key Concepts

Earth Rotation

The Earth rotates around its axis, completing one full turn approximately every 24 hours. However, to be exact, the rotational period is 23 hours, 56 minutes, and 4 seconds. This precision is important for scientific calculations. Understanding Earth's rotation helps explain why we experience day and night.
Calculating exact values involves converting time units. By converting hours, minutes, and seconds into a single unit (hours), calculation becomes more straightforward. This simplifies the calculation of linear speed, as we are comparing all values in the same unit. For instance, Earth's rotational period translates into roughly 23.9344 hours when converted precisely. This is fundamental when determining phenomena such as linear speed at the equator.

Circumference Calculation

To understand linear speed at the equator, one must first calculate the Earth's circumference. The Earth can be modeled as a sphere with a radius of around 3960 miles at the equator.
The formula for the circumference of a circle is important here. It is given by \[ C = 2 \pi r \], where \( r \) represents the radius. Substituting \( r = 3960 \text{ mi} \) results in\[ C = 2 \times \pi \times 3960 \approx 24877.1 \text{ mi}. \]
This value is crucial for calculating linear speed, as it represents the distance a point on the equator travels in one full rotation of the Earth. Knowing this distance allows us to determine how fast a point moves as the Earth rotates.

Unit Conversion

Converting units accurately is essential when solving real-world problems. In this exercise, we begin with Earth's rotational period expressed in hours, minutes, and seconds, and we aim to convert it all into hours.
Simplifying various units into a single type allows calculations to be uniform and accurate. For example, to convert 56 minutes into hours, divide by 60, resulting in 0.9333 hours. Similarly, converting 4 seconds into hours by dividing by 3600 yields 0.0011 hours.

• 56 minutes = \( \frac{56}{60} \approx 0.9333 \) hours
• 4 seconds = \( \frac{4}{3600} \approx 0.0011 \) hours

The total is calculated by adding these components together with the full hours to get 23.9344 hours.
This meticulous conversion ensures accurate computation of linear speed, enabling consistent and reliable results for various scientific applications.