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Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about \(1.1 \times 10^{20} \mathrm{J}\)

Short Answer

Expert verified
The earth's rotational kinetic energy is approximately \(2.57 \times 10^{29} \text{ J}\), while its translational kinetic energy around the sun is about \(2.68 \times 10^{33} \text{ J}\).

Step by step solution

01

Identify the Formulas

For rotational kinetic energy, we use \( KE_{rot} = \frac{1}{2} I \omega^2 \) where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a uniform sphere, \( I = \frac{2}{5}mR^2 \) where \( m \) is the mass and \( R \) is the radius. For translational kinetic energy in circular motion, we use \( KE_{trans} = \frac{1}{2} mv^2 \), where \( v \) is the linear velocity.
02

Calculate Rotational Kinetic Energy

We know: \( m = 5.972 \times 10^{24} \text{ kg} \), \( R = 6.371 \times 10^6 \text{ m} \), and the earth rotates once a day. Thus, \( \omega = \frac{2 \pi}{86400} \) \( \text{rad/s} \). Using \( I = \frac{2}{5}mR^2 \), calculate \( I \), then find \( KE_{rot} = \frac{1}{2} I \omega^2 \).
03

Calculate Translational Kinetic Energy

The earth travels around the sun with an orbital period of 1 year and an approximate distance of \( 1.496 \times 10^{11} \) m. The orbital velocity is \( v = \frac{2 \pi R_{orbit}}{T} \), where \( T = 3.156 \times 10^7 \) s. Substitute this \( v \) into \( KE_{trans} = \frac{1}{2} mv^2 \) and calculate.
04

Compare with US Annual Energy Consumption

Compare both the computed kinetic energy values of the earth with the annual energy consumption of the US, which is \( 1.1 \times 10^{20} \text{ J} \).

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Key Concepts

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

Rotational Kinetic Energy
The concept of rotational kinetic energy is crucial when considering the movement of objects that rotate around an axis. Rotational kinetic energy of an object is given by the formula:
  • \( KE_{rot} = \frac{1}{2} I \omega^2 \)
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
For the Earth, this energy arises from its rotation around its axis. Imagine spinning a basketball; the energy needed to keep it spinning is like Earth's rotational kinetic energy.
In the calculation of Earth's rotational kinetic energy, the Earth is treated as a uniform sphere. Using the moment of inertia specifically for such a solid sphere allows us to use the formula:
\( I = \frac{2}{5}mR^2 \). This incorporates both the mass \( m \) and the radius \( R \) of Earth.
By inserting the known values for Earth's mass and radius, and its angular velocity, we can compute its rotational energy.
Translational Kinetic Energy
Translational kinetic energy pertains to the movement of an object through space along a path, such as Earth's orbit around the Sun. This energy can be computed with the formula:
  • \( KE_{trans} = \frac{1}{2} mv^2 \)
This type of kinetic energy is analogous to the energy of a car moving straight down a road. Similarly, Earth follows a nearly circular path around the Sun.
In calculating Earth's translational kinetic energy, we need the orbital velocity, \( v \), which can be figured out by using the formula:
  • \( v = \frac{2 \pi R_{orbit}}{T} \)
whereby \( R_{orbit} \) is the average radius of Earth's orbit, and \( T \) is the orbital period (one year).
Substituting these values into the translational kinetic energy equation enables us to understand the energy Earth exhibits due to its journey around the Sun.
Moment of Inertia
The moment of inertia is a measure of how difficult it is to change an object's rotation. In simple terms, it's the rotational equivalent of mass in linear motion. The formula for a uniform sphere, like Earth, is:
  • \( I = \frac{2}{5} m R^2 \)
where \( m \) is the mass and \( R \) is the radius. This signifies the distribution of mass in rotation.
Larger moment of inertia means more energy required to change the spinning motion.
This concept becomes very important when calculating rotational kinetic energy. In effects, it dictates how mass and radius impact rotational motion. For Earth, these values are crucial in figuring out the energy involved with its rotation.
Angular Velocity
Angular velocity defines the speed of rotation around an axis and is essential in calculating rotational kinetic energy. It is expressed as:
  • \( \omega = \frac{2\pi}{T} \)
where \( T \) is the rotational period. In Earth's case, one full rotation around its axis occurs in approximately 24 hours.
Angular velocity is measured in radians per second, showcasing how fast the Earth spins.
This rotational speed is fundamental for computing the rotational kinetic energy. With the angular velocity, we can accurately determine how this spinning contributes to the total kinetic energy of the Earth.

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Most popular questions from this chapter

Two identical wheels are moving on horizontal surfaces. The center of mass of each has the same linear speed. However, one wheel is rolling, while the other is sliding on a frictionless surface without rolling. Each wheel then encounters an incline plane. One continues to roll up the incline, while the other continues to slide up. Eventually they come to a momentary halt, because the gravitational force slows them down. Each wheel is a disk of mass \(2.0 \mathrm{kg} .\) On the horizontal surfaces the center of mass of each wheel moves with a linear speed of \(6.0 \mathrm{m} / \mathrm{s}\). (a) What is the total kinetic energy of each wheel? (b) Determine the maximum height reached by each wheel as it moves up the incline.

A rod is lying on the top of a table. One end of the rod is hinged to the table so that the rod can rotate freely on the tabletop. Two forces, both parallel to the tabletop, act on the rod at the same place. One force is directed perpendicular to the rod and has a magnitude of \(38.0 \mathrm{N}\). The second force has a magnitude of \(55.0 \mathrm{N}\) and is directed at an angle \(\theta\) with respect to the rod. If the sum of the torques due to the two forces is zero, what must be the angle \(\theta ?\)

Conceptual Example 13 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of \(125 \mathrm{kg} \cdot \mathrm{m}^{2} .\) When one person is standing on the carousel at a distance of \(1.50 \mathrm{m}\) from the center, the carousel has an angular velocity of \(0.600 \mathrm{rad} / \mathrm{s} .\) However, as this person moves inward to a point located \(0.750 \mathrm{m}\) from the center, the angular velocity increases to \(0.800 \mathrm{rad} / \mathrm{s} .\) What is the person's mass?

As seen from above, a playground carousel is rotating counter-clockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. The carousel has a radius of \(1.50 \mathrm{m},\) an initial angular speed of \(3.14 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(125 \mathrm{kg} \cdot \mathrm{m}^{2} .\) The mass of the person is \(40.0 \mathrm{kg} .\) Find the final angular speed of the carousel after the person climbs aboard.

A small 0.500-kg object moves on a frictionless horizontal table in a circular path of radius \(1.00 \mathrm{m}\). The angular speed is \(6.28 \mathrm{rad} / \mathrm{s}\). The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than \(105 \mathrm{N}\), what is the radius of the smallest possible circle on which the object can move?

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