Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
Question
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Chapter 13, Problem 76AP

(a)

To determine

To determine: The magnitude of the relative acceleration as a function of m .

(a)

Expert Solution
Check Mark

Answer to Problem 76AP

Answer: The magnitude of the relative acceleration as a function of m is 2.77(1+m5.98×1024kg)m/s2 .

Explanation of Solution

Explanation:

Given information:

A object of mass m is distance from the Earth’s center is 1.20×107m .

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 13, Problem 76AP

Figure I

Formula to calculate the relative acceleration is,

arel=a1a2 (I)

a1 is the acceleration of the object having mass m .

a2 is the acceleration of the Earth.

Formula to calculate the gravitational force exerted by the object on the Earth is,

F12=GMEmr2 (II)

ME is the mass of the Earth.

m is the mass of the object.

r is the distance of object having mass m from the Earth center.

By Newton’s law the force exerted by the object is,

F12=ma1 (III)

From equation (II) and equation (III) is,

ma1=GMEmr2a1=GMEr2

The forces F12 and F21 both are gravitational force equal in magnitude and opposite in nature.

F12=F21

F21 is the force exerted by Earth on the object having mass m .

Substitute GMEmr2 for F12 .

GMEmr2=F21F21=GMEmr2 (IV)

By Newton’s law the force exerted by the Earth is,

F21=MEa2 (V)

From equation (IV) and equation (V) is,

MEa2=GMEmr2a2=Gmr2

Substitute Gmr2 for a2 and GMEr2 for a1 in equation (I).

arel=GMEr2(Gmr2)

Substitute 5.972×1024kg for ME , 6.67×1011Nm2/kg2 for G and 1.20×107m for r to find arel .

arel=6.67×1011Nm2/kg2×5.972×1024kg(1.20×107m)2(6.67×1011Nm2/kg2×m(1.20×107m)2)=2.77(1+m5.98×1024kg)m/s2 (VI)

Conclusion:

Therefore, the magnitude of the relative acceleration as a function of m is 2.77(1+m5.98×1024kg)m/s2 .

(b)

To determine

To determine: The magnitude of the relative acceleration for m=5.00kg .

(b)

Expert Solution
Check Mark

Answer to Problem 76AP

Answer: The magnitude of the relative acceleration for m=5.00kg is 2.77m/s2 .

Explanation of Solution

Explanation:

Given information:

A object of mass m is distance from the Earth’s center is 1.20×107m .

From equation (VI) the relative acceleration is,

arel=2.77(1+m5.98×1024kg)m/s2

Substitute 5.00kg for m to find arel .

arel=2.77(1+5.00kg5.98×1024kg)m/s2=2.77m/s2

Conclusion:

Therefore, the magnitude of the relative acceleration for m=5.00kg is 2.77m/s2 .

(c)

To determine

To determine: The magnitude of the relative acceleration for m=2000kg .

(c)

Expert Solution
Check Mark

Answer to Problem 76AP

Answer: The magnitude of the relative acceleration for m=2000kg is 2.77m/s2 .

Explanation of Solution

Explanation:

Given information:

A object of mass m is distance from the Earth’s center is 1.20×107m .

From equation (VI) the relative acceleration is,

arel=2.77(1+m5.98×1024kg)m/s2

Substitute 5.00kg for m to find arel .

arel=2.77(1+2000kg5.98×1024kg)m/s2=2.77m/s2

Conclusion:

Therefore, the magnitude of the relative acceleration for m=2000kg is 2.77m/s2 .

(d)

To determine

To determine: The magnitude of the relative acceleration for m=2.00×1024kg .

(d)

Expert Solution
Check Mark

Answer to Problem 76AP

Answer: The magnitude of the relative acceleration for m=2.00×1024kg is 3.7m/s2 .

Explanation of Solution

Explanation:

Given information:

A object of mass m is distance from the Earth’s center is 1.20×107m .

From equation (VI) the relative acceleration is,

arel=2.77(1+m5.98×1024kg)m/s2

Substitute 5.00kg for m to find arel .

arel=2.77(1+2.00×1024kg5.98×1024kg)m/s2=3.7m/s2

Conclusion:

Therefore, the magnitude of the relative acceleration for m=2.00×1024kg is 3.7m/s2 .

(e)

To determine

To determine: The pattern of variation of relative acceleration with m .

(e)

Expert Solution
Check Mark

Answer to Problem 76AP

Answer: The relative acceleration is directly proportional to the mass m .

Explanation of Solution

Explanation:

Given information:

A object of mass m is distance from the Earth’s center is 1.20×107m .

From equation (VI) the relative acceleration is,

arel=2.77(1+m5.98×1024kg)m/s2

This is the linear equation and shows the relative acceleration is directly proportional to the object having mass m . As m increases to become comparable to the mass of the Earth, the acceleration increases and can become arbitrarily large.

Conclusion:

Therefore, the relative acceleration is directly proportional to the object having mass m .

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Chapter 13 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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