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A hammer with mass \(m\) is dropped from rest from a height \(h\) above the earth's surface. This height is not necessarily small compared with the radius \(R_{\mathrm{E}}\) of the earth. If you ignore air resistance, derive an expression for the speed \(v\) of the hammer when it reaches the surface of the earth. Your expression should involve \(h, R_{\mathrm{E}},\) and \(m_{\mathrm{E}},\) the mass of the earth.

Short Answer

Expert verified
The speed \( v \) is \[ v = \sqrt{2 G m_{\text{E}} \left( \frac{1}{R_{\text{E}}} - \frac{1}{R_{\text{E}} + h} \right)} \].

Step by step solution

01

Understand the Concepts

Before solving, let's recall that energy conservation can be used to find the speed of an object dropped from a height. The gravitational potential energy at the initial height is converted into kinetic energy just before reaching the ground.
02

Write the Energy Conservation Equation

The energy conservation principle states that the initial potential energy equals the final kinetic energy. The potential energy at height \( h \) is \( U_i = -\frac{G m m_{\text{E}}}{R_{\text{E}} + h} \) and the kinetic energy at Earth’s surface is \( K_f = \frac{1}{2} m v^2 \).
03

Calculate the Initial and Final Potential Energies

The initial potential energy \( U_i \) when the hammer is at height \( h \): \ \[ U_i = -\frac{G m m_{\text{E}}}{R_{\text{E}} + h} \]\The final potential energy \( U_f \) when the hammer reaches the Earth's surface \( h=0 \): \ \[ U_f = -\frac{G m m_{\text{E}}}{R_{\text{E}}} \]
04

Set Up the Conservation of Energy Equation

Equate initial total energy to final total energy: \ \[ -\frac{G m m_{\text{E}}}{R_{\text{E}} + h} = \frac{1}{2} m v^2 - \frac{G m m_{\text{E}}}{R_{\text{E}}} \]
05

Simplify and Solve for v

Rearrange the equation to solve for \( v \): \ \[ \frac{1}{2} m v^2 = \frac{G m m_{\text{E}}}{R_{\text{E}}} - \frac{G m m_{\text{E}}}{R_{\text{E}} + h} \]\Simplify by eliminating \( m \): \ \[ \frac{1}{2} v^2 = G m_{\text{E}} \left( \frac{1}{R_{\text{E}}} - \frac{1}{R_{\text{E}} + h} \right) \]\Finally, solve for \( v \): \ \[ v = \sqrt{2 G m_{\text{E}} \left( \frac{1}{R_{\text{E}}} - \frac{1}{R_{\text{E}} + h} \right)} \]
06

Finalize the Expression

The expression for the speed \( v \) of the hammer when it reaches the surface of the Earth is: \ \[ v = \sqrt{2 G m_{\text{E}} \left( \frac{1}{R_{\text{E}}} - \frac{1}{R_{\text{E}} + h} \right)} \] \ This equation incorporates the gravitational parameter, the heights involved, and the Earth's radius.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy (U_i) is the energy an object possesses due to its position in a gravitational field. This concept is crucial when an object, like a hammer, is dropped from a height. The energy is stored because of the gravitational attraction between the object and Earth.
The formula for gravitational potential energy at a height \( h \) above the Earth's surface is given by:\[U_i = -\frac{G m m_{\text{E}}}{R_{\text{E}} + h}\]where \( G \) is the gravitational constant, \( m \) is the mass of the hammer, \( m_{\text{E}} \) is the mass of Earth, and \( R_{\text{E}} \) is the radius of Earth.
When the hammer is at height \( h \), it has maximum potential energy. As it falls, this potential energy is converted into kinetic energy as it approaches the Earth. Understanding this energy transformation is key to applying the principle of energy conservation.
Kinetic Energy
Kinetic energy (K) is the energy an object has due to its motion. When an object is in motion, like a hammer falling to the ground, this energy is in play. For falling objects, kinetic energy increases as potential energy decreases.
The formula for kinetic energy is:\[K_f = \frac{1}{2} m v^2\]where \( m \) is the mass of the object, and \( v \) is its velocity.
In the context of a hammer falling from a height, just before it hits the ground, all initial potential energy has been converted to kinetic energy. Understanding this shift is essential for applying energy conservation principles. As the hammer speeds up, its kinetic energy increases, illustrating the inverse relationship between kinetic and potential energy.
Gravitational Constant
The gravitational constant (G) is a fundamental constant in nature, representing the strength of gravity. It is used in Newton's law of universal gravitation and is crucial in calculating gravitational forces and energy.
In mathematical terms, \( G \) is used in equations such as:
  • The gravitational force equation \( F = \frac{G m_1 m_2}{r^2} \)
  • The potential energy equation \( U = -\frac{G m m_{\text{E}}}{R_{\text{E}} + h} \)
The value of \( G \) is approximately \( 6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2 \). This constant is pivotal in calculating how gravitational forces affect the movement of objects, such as a hammer in free fall.
Understanding \( G \) helps us predict how objects interact in a gravitational field, like how fast a hammer will fall to Earth.
Free Fall Motion
Free fall motion occurs when gravity is the only force acting on an object. In the absence of air resistance, this motion is characterized by a constant acceleration towards the Earth.
With free fall, an object starts at rest and accelerates due to gravity (\( g \)). However, when discussing larger distances or heights like in our scenario, we must consider more precise gravitational influences, including variations at different heights.
Key points about free fall include:
  • Acceleration due to gravity is approximately \( 9.81 \text{m/s}^2 \) near the Earth's surface.
  • In vacuum conditions, all objects fall with the same acceleration irrespective of their mass.
  • The velocity of a freely falling object continuously increases as it falls.
Considering these concepts with our hammer, free fall motion describes how it speeds up as it nears the Earth, transforming potential energy into kinetic energy smoothly and predictably.

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Most popular questions from this chapter

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be \(15.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the center and \(2.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the surface. What is the acceleration due to gravity at the surface of this planet?

The point masses \(m\) and 2\(m\) lie along the \(x\) -axis, with \(m\) at the origin and 2\(m\) at \(x=L .\) A third point mass \(M\) is moved along the \(x\) -axis. (a) At what point is the net gravitational force on \(M\) due to the other two masses equal to zero? (b) Sketch the \(x\) -component of the net force on \(M\) due to \(m\) and \(2 m,\) taking quantities to the right as positive. Include the regions \(x < 0, 0 < x < L,\) and \(x > L .\) Be especially careful to show the behavior of the graph on either side of \(x=0\) and \(x=L .\)

Rhea, one of Saturn's moons, has a radius of 765 \(\mathrm{km}\) and an acceleration due to gravity of 0.278 \(\mathrm{m} / \mathrm{s}^{2}\) at its surface. Calculate its mass and average density.

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Planet Vulcan. Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to \(\frac{2}{3}\) of the average orbit radius of Mercury. What would be the orbital period of such a planet? (Such a planet was once postulated, in part to explain the precession of Mercury's orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury's precession has been explained by general relativity.)

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