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CALC A shaft is drilled from the surface to the center of

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 85P Chapter 13

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 85P

CALC A shaft is drilled from the surface to the center of the earth (see Fig. 13.24). As in Example 13.10 (Section 13.6), make the unrealistic assumption that the density of the earth is uniform. With this approximation, the gravitational force on an object with mass m, that is inside the earth at a distance r from the center, has magnitude (as shown in Example 13.10) and points toward the center of the earth. (a) Derive an expression for the gravitational potential energy U(r) of the object–earth system as a function of the object’s distance from the center of the earth. Take the potential energy to be zero when the object is at the center of the earth. (b) If an object is released in the shaft at the earth’s surface, what speed will it have when it reaches the center of the earth?

Step-by-Step Solution:

Solution 85P Step 1 Gm m r The gravitational force at a point inside the earth, F = E g RE 3 Where, G - universal gravitational constant mE mass of earth m - mass of the object r - Distance of the object from the centre of the earth and r < R E RE Radius of the earth We know that, the gravitational force, F = - dU/dr g Where, U - Gravitational potential Step 2 a) We can write, dU = - F . g r Or, U = F g dr 0 r That is, U = GmEm rdr RE 3 0 r Gm E Taking the constants outside, U = R 3 r dr E 0 Integrating the equation, we get, Gm m 2 U = 3 [ 2 0 to r RE Gm m r 2 U = 1 E 2 3 R E But, we know that, the gravitational potential at the surface of the earth is, U = -Gm m/R 2 E E So, the potential should be continuous at this limit. Therefore, we can write,Ef r = R , Gm m r 2 Gm m 1 E 3 + C = E 2 RE RE Where “C” is a constant. Eut r = R Gm m R 2 Gm m Then, 1 E 3 E + C = E 2 RE RE Or, 1Gm E + C = Gm E 2 RE RE Gm m Gm m That iC = E 1 E RE 2 RE Gm m C = 3 E 2 RE Therefore, the gravitational potential at a point “r” inside the earth is, 3 Gm E 1 Gm E r 2 U = 2 R + 2 3 E RE Rearranging, we get, 1 Gm E r2 U = 2 RE [3 R 2 ] E

Step 3 of 3

Chapter 13, Problem 85P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 85P from 13 chapter was answered, more than 385 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13. This full solution covers the following key subjects: EARTH, center, Object, gravitational, shaft. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 85P from chapter: 13 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. The answer to “CALC A shaft is drilled from the surface to the center of the earth (see Fig. 13.24). As in Example 13.10 (Section 13.6), make the unrealistic assumption that the density of the earth is uniform. With this approximation, the gravitational force on an object with mass m, that is inside the earth at a distance r from the center, has magnitude (as shown in Example 13.10) and points toward the center of the earth. (a) Derive an expression for the gravitational potential energy U(r) of the object–earth system as a function of the object’s distance from the center of the earth. Take the potential energy to be zero when the object is at the center of the earth. (b) If an object is released in the shaft at the earth’s surface, what speed will it have when it reaches the center of the earth?” is broken down into a number of easy to follow steps, and 144 words.

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