/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 The Earth rotates ahout its axis... [FREE SOLUTION] | 91Ó°ÊÓ

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The Earth rotates ahout its axis once in one sidereal day of 23 h 56 min, Calculate the centripetal acceleration of a point located on the equator. Calculate the centripetal mccelcration of a point located at a latitude of \(45^{\circ}\).

Short Answer

Expert verified
Centripetal acceleration at the equator is approximately 0.034 m/s²; at 45° latitude, it is about 0.024 m/s².

Step by step solution

01

Determine the Rotation Period in Seconds

First, convert the sidereal day, which is 23 hours and 56 minutes, into seconds. Since 1 hour has 3600 seconds and 1 minute has 60 seconds, calculate:\[23 \text{ hours} = 23 \times 3600 = 82800 \text{ seconds} \]\[56 \text{ minutes} = 56 \times 60 = 3360 \text{ seconds} \]Thus, the sidereal day in seconds is:\[82800 + 3360 = 86160 \text{ seconds}\]
02

Calculate Angular Velocity of Earth

Angular velocity \( \omega \) is the change in angular displacement per unit time. Since Earth rotates once per sidereal day:\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{86160} \text{ rad/s} \]
03

Calculate Centripetal Acceleration at the Equator

The formula for centripetal acceleration \( a_c \) is:\[ a_c = \omega^2 r \]where \( r \) is the radius of the Earth (approximately 6371 km or 6.371 x 10^6 meters):\[ a_c = \left(\frac{2\pi}{86160}\right)^2 \times 6.371 \times 10^6 \]Calculate to find the centripetal acceleration at the equator.
04

Calculate Effective Radius at 45° Latitude

At any latitude \( \theta \), the effective radius \( r_{eff} \) of rotation is given by:\[ r_{eff} = r \cos(\theta) \]For 45° latitude:\[ r_{eff} = 6.371 \times 10^6 \times \cos(45^{\circ}) \]
05

Calculate Centripetal Acceleration at 45° Latitude

Using the formula for centripetal acceleration again with the effective radius:\[ a_{c(45)} = \omega^2 r_{eff} \]Substitute the values from Step 4 and Step 2 to calculate this acceleration.

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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 spins around its axis, which is an imaginary line passing from the North Pole to the South Pole. This rotation is not something we feel directly because it is a smooth and constant movement. However, it significantly affects everything on Earth, contributing to the cycle of day and night as well as affecting weather patterns. A full rotation around the axis takes approximately 24 hours to complete, but to be more precise with astronomical observations, this period is measured as a sidereal day, which lasts 23 hours, 56 minutes, and 4 seconds.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves around a specific point or axis. For the Earth, this refers to how quickly it spins around its own axis. It is calculated using the formula:
  • \( \omega = \frac{2\pi}{T} \)
  • where \( \omega \) is the angular velocity and \( T \) is the rotation period (or the period of one full rotation).
  • For Earth, with a sidereal day of 86160 seconds, this calculation gives the rate at which Earth completes its rotation.
Angular velocity is crucial in determining how forces like centripetal force act upon objects on Earth's surface.
Latitude Effect on Acceleration
Centripetal acceleration is the acceleration that causes an object moving in a circular path to change its motion so that it continues along the circular path. At different latitudes, the effective radius for this motion changes, altering the centripetal acceleration.- At the equator, the radius is maximum and is equal to Earth’s radius. - As you move toward the poles, the effective radius of rotation decreases because you have to consider the latitude. - This is described by the formula: \[ r_{eff} = r \cos(\theta) \] - Here, \( \theta \) is the latitude, and \( r \) is Earth's radius. - Thus, at 45° latitude, for example, the effective radius becomes \( r_{eff} = r \cos(45^{\circ}) \). This means that objects at the equator experience the highest centripetal acceleration, which gradually decreases as you move toward the latitudes, impacting the weight and motion of objects at different latitudes.
Sidereal Day
A sidereal day is the time it takes for the Earth to complete one rotation relative to the stars, not just the Sun. This is about 4 minutes shorter than a solar day, which forms the basis of our 24-hour day system. - Conversion of the 23 hours and 56 minutes into seconds is crucial for precision in calculations. - With one sidereal day being 86160 seconds long, this understanding is vital in computations involving Earth's rotation and movements. This precise measurement is often used in astronomy and physics because it reflects the true rotation period of the Earth, unaffected by its orbit around the Sun.
Effective Radius Calculation
In circular motion, like the Earth's rotation, the radius is the distance from the center of the circle to the path around it. The effective radius varies with latitude because the Earth is not a perfect sphere. - At the Equator, the radius equals the Earth's radius (approximately 6371 km). - At other latitudes, the radius is adjusted based on the cosine of the latitude angle: \[ r_{eff} = r \cos(\theta) \] - This adjustment accounts for the curved path followed by points on Earth's surface. - Thus, a point at 45° latitude will have a reduced effective radius due to \( \cos(45^{\circ}) \). Calculating the effective radius is essential for understanding centripetal forces and their effect on motion at different geographical locations.

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