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Chapter 13: Problem 65

Falling Hammer. 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 of the hammer when it reaches the Earth's surface is given by: \[ v = \sqrt{2 G m_E \left( \frac{1}{R_E} - \frac{1}{R_E + h} \right)} \].

Step by step solution

01

Gravitational Potential Energy at Height

When a hammer is dropped from a height \( h \), it possesses gravitational potential energy given by \[ U_i = -\frac{G m m_E}{R_E + h} \]where \( G \) is the gravitational constant, \( m \) is the mass of the hammer, \( m_E \) is the mass of the Earth, and \( R_E \) is the Earth's radius.
02

Gravitational Potential Energy at Earth's Surface

When the hammer reaches the Earth's surface, the gravitational potential energy is \[ U_f = -\frac{G m m_E}{R_E} \].
03

Conservation of Mechanical Energy

The total mechanical energy is conserved, meaning initial potential energy plus initial kinetic energy is equal to final potential energy plus final kinetic energy. Since the hammer starts from rest: \[ K_i = 0 \] Thus, \[ U_i + K_i = U_f + K_f \].
04

Express Kinetic Energy at Earth's Surface

Substituting the expressions for the potential energies and the initial kinetic energy, we have:\[ -\frac{G m m_E}{R_E + h} = -\frac{G m m_E}{R_E} + \frac{1}{2} m v^2 \], where \( v \) is the speed of the hammer at the Earth's surface.
05

Solve for Final Speed

Rearrange the above expression to solve for \( v^2 \):\[ \frac{1}{2} m v^2 = \frac{G m m_E}{R_E} - \frac{G m m_E}{R_E + h} \].Factor out the common term:\[ \frac{1}{2} m v^2 = G m m_E \left( \frac{1}{R_E} - \frac{1}{R_E + h} \right) \]. Solving for \( v \):\[ v^2 = 2 G m_E \left( \frac{1}{R_E} - \frac{1}{R_E + h} \right) \]. \[ v = \sqrt{2 G m_E \left( \frac{1}{R_E} - \frac{1}{R_E + h} \right)} \].

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Key Concepts

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

Conservation of Mechanical Energy
In physics, the Conservation of Mechanical Energy principle is crucial when analyzing systems like a hammer falling to Earth. It states that the total mechanical energy of an isolated system remains constant if only conservative forces, like gravity, do work. This means that the sum of the kinetic energy and potential energy of a system is always the same, as long as there are no non-conservative forces at play, such as air resistance.
In the case of a falling hammer, it starts with a certain amount of gravitational potential energy due to its height above Earth, and this energy is converted into kinetic energy as it falls. Mathematically, this principle is expressed as:
  • Initial potential energy + initial kinetic energy = final potential energy + final kinetic energy
  • If the hammer starts from rest, its initial kinetic energy is zero, simplifying the equation.
Understanding this principle helps in deriving expressions for the speed of the hammer as it hits the ground, using known quantities like height and mass.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the hammer falls towards Earth, its gravitational potential energy is transformed into kinetic energy, causing it to accelerate. The formula for kinetic energy is given by:
  • Kinetic energy (\(K\)) = \(\frac{1}{2}mv^2\)
  • \(m\) represents the mass, and \(v\) is the velocity of the object.
As the hammer approaches the ground, its velocity increases, resulting in an increase in kinetic energy. Initially, the kinetic energy is zero because the hammer is at rest. As energy is conserved, the increase in kinetic energy equals the decrease in potential energy. By understanding kinetic energy, we can calculate the hammer's speed upon impact. This concept is vital for predicting motion in physics.
Gravitational Constant
The gravitational constant, denoted by \(G\), is a fundamental constant in physics that appears in the formula for gravitational force and potential energy. It describes the strength of gravity's pull in our universe and is approximately \(6.674 \times 10^{-11} \) m鲁 kg鈦宦� s鈦宦�. This constant helps us calculate how massive objects interact with one another due to gravity.
When determining the gravitational potential energy of the hammer at a certain height, \(G\) is used in the formula:
  • Gravitational potential energy = \(-\frac{G m m_E}{R_E + h}\)
  • \(m_E\) is Earth's mass, and \(m\) is the hammer's mass.
Incorporating \(G\) into calculations ensures precise computation of the forces and energies at play when objects fall due to gravity. Its value remains constant across a variety of situations, making it an indispensable part of solving gravitational problems.
Earth's Radius
Earth's radius, symbolized as \(R_E\), is an important factor in determining gravitational potential energy. The average radius of Earth is approximately 6,371 kilometers or about 3,959 miles. When calculating gravitational potential energy, the distance from the center of the Earth is crucial, as gravitational effects decrease with distance. For objects above the surface, this distance is \(R_E + h\), where \(h\) is the height above Earth's surface.
In gravitational equations, like the one for potential energy:
  • Initial potential energy = \(-\frac{G m m_E}{R_E + h}\)
  • Final potential energy = \(-\frac{G m m_E}{R_E}\)
Understanding the role of Earth's radius helps us apply conservation laws accurately by determining how gravity acts over various distances. It's a crucial datum for making precise predictions about the behavior of falling objects.

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Most popular questions from this chapter

An \(8.00-\mathrm{kg}\) point mass and a 15.0 -kg point mass are held in place 50.0 \(\mathrm{cm}\) apart. A particle of mass \(m\) is released from a point between the two masses 20.0 \(\mathrm{cm}\) from the \(8.00-\mathrm{kg}\) mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

A thin, uniform rod has length \(L\) and mass \(M . \mathrm{A}\) small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. E13.32). (a) Calculate the gravitational potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when \(x\) is much larger than \(L .\) (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix B.) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4\()\) . Show that your answer reduces to the expected result when \(x\) is much larger than \(L .\)

Binary Star-Equal Masses. Two identical stars with mass \(M\) orbit around their center of mass. Each orbit is circular and has radius \(R,\) so that the two stars are always on opposite sides of the circle. (a) Find the gravitational force of one star on the other. (b) Find the orbital speed of each star and the period of the orbit. (c) How much energy would be required to separate the two stars to infinity?

Deimos, a moon of Mars, is about 12 \(\mathrm{km}\) in diameter with mass \(2.0 \times 10^{15} \mathrm{kg} .\) Suppose you are stranded alone on Deimos and want to play a one-person game of baseball. You would be the pitcher, and you would be the batter! (a) With what speed would you have to throw a baseball so that it would go into a circular orbit just above the surface and return to you so you could hit it? Do you think you could actually throw it at this speed? (b) How long (in hours) after throwing the ball should you be ready to hit it? Would this be an action-packed baseball game?

In March \(2006,\) two small satellites were discovered orbiting Pluto, one at a distance of \(48,000 \mathrm{km}\) and the other at \(64,000 \mathrm{km} .\) Pluto already was known to have a large satellite Charon, orbiting at \(19,600 \mathrm{km}\) with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.
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