Physics for Scientists and Engineers, Volume 1

9th Edition

ISBN: 9781133954156

Author: Raymond A. Serway

Publisher: CENGAGE L

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Textbook Question

Chapter 21, Problem 21.72AP

Review, (a) H it has enough kinetic energy, a molecule at the surface of the Earth can “escape the Earth’s gravitation” in the sense that it can continue to move away from the Earth forever as discussed in Section 13.6. Using the principle of conservation of energy, show that the minimum kinetic energy needed for “escape” is m₀gR_E where m₀ is the mass of the molecule, g is the free-fall acceleration at the surface, and R_E is the radius of the Earth, (b) Calculate the temperature for which the minimum escape kinetic energy is ten times the average kinetic energy of an oxygen molecule.

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Chapter 21, Problem 21.71AP

Chapter 21, Problem 21.73AP

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(a) Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and place it in a circular low earth orbit—that is, an orbit whose altitude above the earth’s surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE = 6370 km.) Ignore the kinetic energy that the spacecraft has on the ground due to the earth’s rotation. (b) Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. Ignore the gravitational effects of the sun, the moon, and the other planets. (c) Justify the statement “In terms of energy, low earth orbit is halfway to the edge of the universe.”

The escape velocity from a massive object is the speed needed to reach an infinite distance from it and have just slowed to a stop, that is, to have just enough kinetic energy to climb out of the gravitational potential well and have none left. You can find the escape velocity by equating the total kinetic and gravitational potential energy to zero E = = muesc - GmM/r=0 Vesc = √2GM/r where G is Newton's constant of gravitation, M is the mass of the object from which the escape is happening, and r is its radius. This is physics you have seen in the first part of the course, and you should be able to use it to find an escape velocity from any planet or satellite. For the Earth, for example the escape velocity is about 11.2 km/s, and for the Moon it is 2.38 km/s. A very important point about escape velocity: it does not depend on what is escaping. A spaceship or a molecule must have this velocity or more away from the center of the planet to be free of its gravity, 1. In the atmosphere of…

(a) State Newton's law of gravitation in word form. (b) Use the law to show that that the escape velocity for a planet is given by the expression v = 2go R. You can assume that the gravitational potential per unit -GM where mass U, at the surface of the planet, is given by the expression U = R M and R are the mass and radius of the planet. (c) Assuming that the Earth is a uniform sphere of radius 6.4 x 10 m and mass 6.0 x 1024 kg, calculate: (i) the gravitational potential at the surface of the Earth and (ii) the gravitational potential at a point 6.0 x 105 m above the surface of the Earth. (d) A 5 kg mass is to be delivered to a point in space where the gravitational effect on the mass is negligible. Calculate: (i) the escape velocity for the mass. (iii) the work done in taking the mass to a point 6.0 x 105 m above the surface of the Earth. (iii) the work needed to deliver the mass to the point where the gravitational potential is zero.

Chapter 21 Solutions

Physics for Scientists and Engineers, Volume 1

Ch. 21 - Two containers hold an ideal gas at the same...Ch. 21 - (i) How does the internal energy of an ideal gas...Ch. 21 - Prob. 21.3QQ Ch. 21 - Prob. 21.4QQ Ch. 21 - Cylinder A contains oxygen (O2) gas, and cylinder...Ch. 21 - An ideal gas is maintained at constant pressure....Ch. 21 - Prob. 21.3OQ Ch. 21 - A helium-filled latex balloon initially at room...Ch. 21 - Prob. 21.5OQ Ch. 21 - Prob. 21.6OQ

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Solution Summary: The author explains that the minimum kinetic energy needed for escape is m_0gR

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Chapter 21 Solutions