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(a) What is the magnitude of the centripetal acceleration of an object on Earth's equator due to the rotation of Earth? (b) What would Earth's rotation period have to be for objects on the equator to have a centripetal acceleration of magnitude \(9.8 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
Centripetal acceleration: 0.034 m/s². Required rotation period: 84.6 minutes.

Step by step solution

01

Understand the Problem

We're tasked with finding two things: (a) the magnitude of the centripetal acceleration at Earth's equator and (b) the required rotation period of Earth for this acceleration to be equal to gravitational acceleration (\(9.8 \mathrm{~m/s}^2\)).

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Key Concepts

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

Earth's Rotation
The concept of Earth's rotation is fundamental to understanding centripetal acceleration. Earth rotates about its axis, completing one full rotation approximately every 24 hours. This rotation causes different points on Earth to travel in circular paths. When an object travels in a circle, it experiences centripetal acceleration, which acts towards the center of the circle.
For someone standing on the equator, the path is the largest circle possible due to the Earth’s spherical shape. As a result, the speed is higher at the equator compared to other latitudes, maximizing the centripetal force.
Understanding this, you can see why the magnitude of centripetal acceleration is dependent on both the speed of the rotation and the radius of the path, which is approximately the radius of the Earth at the equator. The formula for centripetal acceleration, \( a_c = \frac{v^2}{r} \), highlights the need for both speed \(v\) and radius \(r\).
Equator
The equator is an essential reference point in understanding centripetal acceleration during Earth's rotation. It is an imaginary line circling the Earth, equally distant from both poles, and dividing the Earth into the Northern and Southern Hemispheres.
Due to Earth's spherical shape, the equator is the widest circle of latitude, meaning it's also the point on the Earth's surface farthest from the poles. As such, it has the longest distance to travel in a rotation, and hence experiences the greatest linear speed during Earth's rotation.
This heightened speed results in the maximum centripetal acceleration at the equator compared to other latitudes. This is why, when calculating centripetal acceleration, the artifact of a rotating Earth is most pronounced here. For calculations involving centripetal acceleration at the equator, the radius \(r\) is effectively the radius of the Earth, about 6,371 kilometers.
Gravitational Acceleration
Gravitational acceleration is a key concept when looking at the forces acting on objects on Earth. It is the acceleration due to Earth's gravitational pull, typically measured as \(9.8 \mathrm{~m/s}^2\). This is the force that gives weight to physical objects and causes them to fall towards the ground when dropped.
In considering centripetal acceleration, the concept plays a critical role especially in part (b) of the exercise. We explored what would happen if this centripetal acceleration achieved the magnitude of gravitational acceleration.
If Earth's rotation were fast enough for the centripetal acceleration at the equator to equal gravitational acceleration, objects would effectively become "weightless." This means the inward force required to keep them in a circular path would be the same as the force pulling them to Earth’s surface. For this scenario to occur, the period of rotation would need to decrease significantly, to a much shorter period than 24 hours. This intriguing concept highlights the balance and interplay between gravitational and centripetal forces.

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Most popular questions from this chapter

You are kidnapped by political-science majors (who are upset because you told them political science is not a real science). Although blindfolded, you can tell the speed of their car (by the whine of the engine), the time of travel (by mentally counting off seconds), and the direction of travel (by turns along the rectangular street system). From these clues, you know that you are taken along the following course: \(50 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(4.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(30 \mathrm{~s}\). At that point, (a) how far are you from your starting point, and (b) in what direction relative to your initial direction of travel are you?

A ball is thrown horizontally from a height of \(20 \mathrm{~m}\) and hits the ground with a speed that is three times its initial speed. What is the initial speed?

An electron having an initial horizontal velocity of magnitude \(1.00 \times 10^{9} \mathrm{~cm} / \mathrm{s}\) travels into the region between two horizontal metal plates that are electrically charged. In that region, the electron travels a horizontal distance of \(2.00 \mathrm{~cm}\) and has a constant downward acceleration of magnitude \(1.00 \times 10^{17} \mathrm{~cm} / \mathrm{s}^{2}\) due to the charged plates. Find (a) the time the electron takes to travel the \(2.00 \mathrm{~cm},(\mathrm{~b})\) the vertical distance it travels during that time, and the magnitudes of its (c) horizontal and (d) vertical velocity components as it emerges from the region.

You are to throw a ball with a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) at a target that is height \(h=5.00 \mathrm{~m}\) above the level at which you release the ball (Fig. 4-58). You want the ball's velocity to be horizontal at the instant it reaches the target. (a) At what angle \(\theta\) above the horizontal must you throw the ball? (b) What is the horizontal distance from the release point to the target? (c) What is the speed of the ball just as it reaches the target?

You throw a ball toward a wall at speed \(25.0 \mathrm{~m} / \mathrm{s}\) and at angle \(\theta_{0}=40.0^{\circ}\) above the horizontal (Fig. 4-35). The wall is distance \(d=\) \(22.0 \mathrm{~m}\) from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?

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