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Chapter 3: Problem 25

The earth has a radius of \(6380 \mathrm{~km}\) and turns around once on its axis in \(24 \mathrm{~h}\). (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of \(g .\) (b) If \(a_{\mathrm{rad}}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We'll see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

Short Answer

Expert verified
The radial acceleration of an object at the Earth's equator is approximately \(0.034 \mathrm{m/s}^2\) or \(0.0035g\). If the Earth rotated about its axis every \(1.41\) hours or less, the radial acceleration at the equator would be greater than gravity and objects would fly off into space.

Step by step solution

01

Determine the speed at the Earth's equator

Firstly, we will calculate the speed, \(v\), of an object at the Earth's equator. The speed of an object moving in a circular path is calculated by the formula \(v = \frac{d}{t}\), where \(d\) is the distance travelled and \(t\) is the time taken to travel that distance. Since the object is moving along the Earth's equator, the distance is the circumference of the Earth, \(\mathrm{C} = 2\pi r\), with \(r = 6380\mathrm{~km} = 6.38 \times 10^6\mathrm{~m}\), and the time is 24 hours = \(86400\mathrm{~s}\). Thus \(v\approx 465.1\mathrm{~m/s}\).
02

Compute the radial acceleration at the Earth's equator

Secondly, we calculate the radial acceleration of an object at the Earth's equator using the formula \(a_{\mathrm{rad}} = \frac{v^2}{r}\). Substituting \(v = 465.1\mathrm{~m/s}\) from the previous step and \(r = 6.38 \times 10^6\mathrm{~m}\) we get \(a_{\mathrm{rad}} \approx 0.034 \mathrm{m/s}^2\). The acceleration can also be computed as a fraction of the gravitational acceleration, \(g = 9.8 \mathrm{m/s}^2\), which gives \(\frac{a_{\mathrm{rad}}}{g} \approx 0.0035\).
03

Calculate the period for radial acceleration to be greater than gravity

Finally, if the radial acceleration is greater than the gravitational acceleration, then objects would fly off the surface into space. Solving for the period in the formula for radial acceleration, \(a_{\mathrm{rad}} = \frac{4 \pi^2 r}{T^2}\), with \(a_{\mathrm{rad}} = g\), we get \(T \approx 1.41 \mathrm{~hours}\). This means if the Earth rotates about its axis every 1.41 hours, the radial acceleration at the equator would be equal to the gravitational acceleration, and any faster would cause objects to fly off into space.

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Key Concepts

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

Uniform Circular Motion
When an object follows a circular path at a constant speed, it is said to be in uniform circular motion. Despite the speed being constant, the direction of the velocity changes at every point along the path. Therefore, the object is accelerating because acceleration isn't just about changes in speed, but also changes in the direction of motion.

An example of uniform circular motion is the motion of the Earth as it rotates on its axis. An object on the Earth's equator travels a considerable distance due to the large circumference of the Earth, yet it takes a full 24 hours to complete one rotation, thereby maintaining a constant speed throughout.
Centripetal Acceleration
The acceleration that keeps an object moving in a circular path and directs it towards the center of the rotation is known as centripetal acceleration. The term 'centripetal' comes from Latin words meaning 'center seeking'. As related to the exercise, you previously learned that the formula for centripetal, or radial, acceleration is given by \(a_{\text{rad}} = \frac{v^2}{r}\), where \(v\) is the velocity of the object and \(r\) is the radius of the circular path.

In the case of Earth's rotation, the centripetal acceleration at the equator can be calculated using the Earth's radius and the velocity of an object at the equator. It is important to understand that this acceleration is what keeps the Earth's atmosphere and everything on the surface from flying off into space, as long as it does not exceed the force of gravity.
Earth's Rotation
The Earth's rotation refers to the planet spinning on its axis. This rotation is responsible for the cycle of day and night as different parts of the Earth face the Sun or turn away from it. The speed and path of the Earth's rotation also contribute to the centripetal force felt by objects on its surface. As demonstrated in the exercise, if the Earth were to rotate much more quickly, with a period less than the calculated 1.41 hours, objects could potentially be flung off due to a centripetal acceleration greater than the gravitational pull on those objects.

Understanding the effects of Earth's rotation on the environment and on human perception includes acknowledging forces such as the Coriolis effect, which affects weather patterns and ocean currents. However, under normal circumstances with a 24-hour rotation period, objects remain safely secured on the planet's surface.

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Most popular questions from this chapter

When a train's velocity is \(12.0 \mathrm{~m} / \mathrm{s}\) eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined \(30.0^{\circ}\) to the vertical on the windows of the train. (a) What is the horizontal component of a drop's velocity with respect to the earth? With respect to the train? (b) What is the magnitude of the velocity of the raindrop with respect to the earth? With respect to the train?

A large number of seeds are observed, and their initial launch angles are recorded. The range of projection angles is found to be \(-51^{\circ}\) to \(75^{\circ},\) with a mean of \(31^{\circ} .\) Approximately \(65 \%\) of the seeds are launched between \(6^{\circ}\) and \(56^{\circ} .\) (See W. J. Garrison et al., "Ballistic seed projection in two herbaceous species," Amer. J. Bot., Sept. 2000 , \(87: 9,1257-64 .)\) Which of these hypotheses is best supported by the data? Seeds are preferentially launched (a) at angles that maximize the height they travel above the plant; (b) at angles below the horizontal in order to drive the seeds into the ground with more force; (c) at angles that maximize the horizontal distance the seeds travel from the plant; (d) at angles that minimize the time the seeds spend exposed to the air.

A dog in an open field is at rest under a tree at time \(t=0\) and then runs with acceleration \(\vec{a}(t)=\left(0.400 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\imath}-\left(0.180 \mathrm{~m} / \mathrm{s}^{3}\right) t \hat{\jmath}\) How far is the dog from the tree \(8.00 \mathrm{~s}\) after it starts to run?

At \(t=0\) a rock is projected from ground level with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(53.0^{\circ}\) above the horizontal. Neglect air resistance. At what two times \(t\) is the rock \(5.00 \mathrm{~m}\) above the ground? At each of these two times, what are the horizontal and vertical components of the velocity of the rock? Let \(v_{0 x}\) and \(v_{0 y}\) be in the positive \(x\) - and \(y\) -directions, respectively.

A jet plane is flying at a constant altitude. At time \(t_{1}=0,\) it has components of velocity \(v_{x}=90 \mathrm{~m} / \mathrm{s}, v_{y}=110 \mathrm{~m} / \mathrm{s} .\) At time \(t_{2}=30.0 \mathrm{~s}\) the components are \(v_{x}=-170 \mathrm{~m} / \mathrm{s}, v_{y}=40 \mathrm{~m} / \mathrm{s}\). (a) Sketch the velocity vectors at \(t_{1}\) and \(t_{2} .\) How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.
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