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Chapter 11: Problem 39

Assume the Earth to be a sphere of uniform density. (a) Calculate its rotational kinetic energy. (b) Suppose that this energy could be harnessed for our use. For how long could the Earth supply \(1.00 \mathrm{~kW}\) of power to each of the \(6.17 \times 10^{9}\) persons on the Earth?

Short Answer

Expert verified
The rotational kinetic energy of the Earth is 2.56x10^29 J. This energy could be harnessed to supply 1.00 kW of power to each of the 6.17 billion people on Earth for approximately 131 days.

Step by step solution

01

Calculate the Earth's rotational kinetic energy

First, we need to calculate the rotational kinetic energy of the Earth. The rotational kinetic energy (\( K \)) is given as \[ K = 0.5 \times I \times \omega^2 \], where \( I \) is the moment of inertia and \( \omega \) is the angular speed. The moment of inertia for a sphere of uniform density is \[I = 0.4 \times M \times R^2 \], and the angular speed of the Earth is \[ \omega = \frac{2\pi}{24 \times 60 \times 60} \text{ rad/s} \], as it makes one rotation in one day, which is converted here to seconds. The mass of the Earth is \( M = 5.97 \times 10^{24} \) kg, and its radius is \( R = 6.38 \times 10^{6} \) m. Substitute these values in the formula to get \( K \).
02

Calculate the total power

Next, we need to calculate the total power that needs to be supplied for every person on Earth. This will be \( P = 1.00 \, \mathrm{kW} \times 6.17 \times 10^{9} \) (as there are 6.17 billion people). Convert the power from kilowatt to watt (since 1 kW = 1000 W).
03

Find the time

Finally, we need to solve for the time \( t \) for which the rotational kinetic energy of the Earth can provide power to each person. We know that power is the rate at which energy is used, so \( P = \frac{E}{t} \). Rearranging, we find \( t = \frac{E}{P} \), where \( E \) is the total energy (i.e., the rotational kinetic energy of the Earth) and \( P \) is the total power. Substitute \( E \) and \( P \) into the equation to get \( t \).

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Most popular questions from this chapter

Delivery trucks that operate by making use of energy stored in a rotating flywheel have been used in Europe. The trucks are charged by using an electric motor to get the flywheel up to its top speed of \(624 \mathrm{rad} / \mathrm{s}\). One such flywheel is a solid, homogeneous cylinder with a mass of \(512 \mathrm{~kg}\) and a radius of \(97.6 \mathrm{~cm} .\) (a) What is the kinetic energy of the flywheel after charging? ( \(b\) ) If the truck operates with an average power requirement of \(8.13 \mathrm{~kW}\), for how many minutes can it operate between chargings?

The oxygen molecule has a total mass of \(5.30 \times 10^{-26} \mathrm{~kg}\) and a rotational inertia of \(1.94 \times 10^{-46} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about an axis through the center perpendicular to the line joining the atoms. Suppose that such a molecule in a gas has a mean speed of \(500 \mathrm{~m} / \mathrm{s}\) and that its rotational kinetic energy is two- thirds of its translational kinetic energy. Find its average angular velocity.

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A proton (nucleus of the hydrogen atom) is being accelerated in a linear accelerator. In each stage of such an accelerator the proton is accelerated along a straight line at \(3.60 \times 10^{15} \mathrm{~m} / \mathrm{s}^{2}\). If a proton enters such a stage moving initially with a speed of \(2.40 \times 10^{7} \mathrm{~m} / \mathrm{s}\) and the stage is \(3.50 \mathrm{~cm}\) long, compute \((a)\) its speed at the end of the stage and \((b)\) the gain in kinetic energy resulting from the acceleration. The mass of the proton is \(1.67 \times 10^{-27} \mathrm{~kg} .\) Express the energy in electron-volts.
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