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Chapter 5: Problem 13

Does the Earth do any work on the Moon as the Moon moves in its orbit?

Short Answer

Expert verified
Answer: No, the Earth does not do any work on the Moon as it moves in its orbit. This is because the gravitational force vector and the Moon's displacement vector are always perpendicular to each other, resulting in zero work done.

Step by step solution

01

Understand the concept of work

Work is done by a force when it displaces an object in the direction of the force. Mathematically, work (W) is calculated as the dot product of force (F) and displacement (d): W = F ⋅ d = |F| |d| cos(θ), where θ is the angle between the force vector and the displacement vector.
02

Determine the gravitational force between Earth and Moon

The gravitational force between the Earth and the Moon can be determined using Newton's law of universal gravitation: F = G * (m_E * m_M) / r^2, where F is the gravitational force, G is the gravitational constant (6.67430 x 10^{-11} m^3 kg^{-1} s^{-2}), m_E is the mass of the Earth (5.972 × 10^{24} kg), m_M is the mass of the Moon (7.342 × 10^{22} kg), and r is the average distance between the Earth and the Moon (3.844 × 10^8 m).
03

Analyze the direction of the gravitational force and the Moon's orbit

The gravitational force between the Earth and the Moon always acts along the line joining their centers. The direction of this force is radially inward, pointing towards the center of the Earth. The Moon's orbit is approximately circular, and its motion is perpendicular to the line joining the Earth and the Moon. Therefore, the displacement vector of the Moon is always perpendicular to the gravitational force vector acting on it.
04

Calculate the angle between the gravitational force and the Moon's displacement

Since the gravitational force is always directed towards the center of the Earth and the Moon's displacement is perpendicular to it, the angle between the force vector and the displacement vector (θ) is 90°.
05

Calculate the work done by the Earth on the Moon

Now we can calculate the work done (W) by the gravitational force on the Moon using the formula: W = F ⋅ d = |F| |d| cos(θ), Considering the angle between the force vector and the displacement vector is 90° and cos(90°) = 0, the work done becomes: W = |F| |d| cos(90°) = 0
06

Conclusion

Based on our calculations, the Earth does not do any work on the Moon as it moves in its orbit. This is because the gravitational force vector and the displacement vector are always perpendicular to each other (θ = 90°), resulting in zero work done.

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Most popular questions from this chapter

A car, of mass \(m,\) traveling at a speed \(v_{1}\) can brake to a stop within a distance \(d\). If the car speeds up by a factor of \(2, v_{2}=2 v_{1},\) by what factor is its stopping distance increased, assuming that the braking force \(F\) is approximately independent of the car's speed?

A horse draws a sled horizontally across a snowcovered field. The coefficient of friction between the sled and the snow is \(0.195,\) and the mass of the sled, including the load, is \(202.3 \mathrm{~kg}\). If the horse moves the sled at a constant speed of \(1.785 \mathrm{~m} / \mathrm{s}\), what is the power needed to accomplish this?

A \(1500-\mathrm{kg}\) car accelerates from 0 to \(25 \mathrm{~m} / \mathrm{s}\) in \(7.0 \mathrm{~s}\) What is the average power delivered by the engine \((1 \mathrm{hp}=746 \mathrm{~W}) ?\) a) \(60 \mathrm{hp}\) c) \(80 \mathrm{hp}\) e) \(180 \mathrm{hp}\) b) \(70 \mathrm{hp}\) d) \(90 \mathrm{hp}\)

A horse draws a sled horizontally on snow at constant speed. The horse can produce a power of \(1.060 \mathrm{hp} .\) The coefficient of friction between the sled and the snow is \(0.115,\) and the mass of the sled, including the load, is \(204.7 \mathrm{~kg}\). What is the speed with which the sled moves across the snow?

Eight books, each \(4.6 \mathrm{~cm}\) thick and of mass \(1.8 \mathrm{~kg}\), lie on a flat table. How much work is required to stack them on top of one another? a) 141 J c) 230 e) 14 b) 23 J d) 0.81
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