Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 11, Problem 74P

(a)

To determine

To Calculate:The magnitude of gravitational force due to M1 and M2 at x=3a .

(a)

Expert Solution
Check Mark

Answer to Problem 74P

  |F|=Gma2[M19+M24.84]

Explanation of Solution

Given data:

Radius of M1 is R2=2a

  R1=a

Center has shifted, x=0.8a

Formula Used:

Gravitational field:

  g=Gm1r2r^

Where, G is the gravitational constant, m is the mass and r is the distance of the point where field is calculated.

The gravitational force is a force of attraction between any two bodies having mass and separated by a distance d . Mathematically, the gravitational force can be represented by:

  F=Mg

  F=GmMr2r^

Where, G is the gravitational constant, m and M are the masses and r is the distance between them.

Calculation:

Physics for Scientists and Engineers, Chapter 11, Problem 74P , additional homework tip  1

The magnitude of force due to M1 at r=3 a

  |F1|=GM1m9a2

Again, due to M2 at r=3a0.8a=2.2a

Force, |F2|=GM2m(2.2a)2=GM2m4.84a2

Hence, net force

  |F|=|F1|+|F2|

At r=3a

  |F|=Gma2[M19+M24.84]

Conclusion:

Thus, the magnitude of gravitational force due to M1 and M2 at x=3a is |F|=Gma2[M19+M24.84] .

(b)

To determine

The magnitude of gravitational force due to M1 and M2 at x=1.9a .

(b)

Expert Solution
Check Mark

Answer to Problem 74P

  |F|=GM2m1.21a2

Explanation of Solution

Given data:

Radius of M1 is R2=2a

  R1=a

Center has shifted, x=0.8a

Formula Used:

Gravitational field:

  g=Gm1r2r^

Where, G is the gravitational constant, m is the mass and r is the distance of the point where field is calculated.

The gravitational force is a force of attraction between any two bodies having mass and separated by a distance d . Mathematically, the gravitational force can be represented by:

  F=Mg

  F=GmMr2r^

Where, G is the gravitational constant, m and M are the masses and r is the distance between them.

Calculation:

Physics for Scientists and Engineers, Chapter 11, Problem 74P , additional homework tip  2

The value of |F1| , due to M1 at r=1.9 a <(R2=2a)

Hence, |F1|=0

The value of |F2| ; due to M2 at r=1.9a0.8a=1.1a

  |F2|=GM2m(1.1a)2

Hence, net force |F|=|F2|+|F1|

  |F|=GM2m1.21a2

Conclusion:

Thus, the magnitude of gravitational force due to M1 and M2 at x=1.9a is |F|=GM2m1.21a2 .

(c)

To determine

The magnitude of gravitational force due to M1 and M2 at x=0.9a .

(c)

Expert Solution
Check Mark

Answer to Problem 74P

  |F|=0

Explanation of Solution

Given data:

Radius of M1 is R2=2a

  R1=a

Center has shifted, x=0.8a

Formula Used:

Gravitational field:

  g=Gm1r2r^

Where, G is the gravitational constant, m is the mass and r is the distance of the point where field is calculated.

The gravitational force is a force of attraction between any two bodies having mass and separated by a distance d . Mathematically, the gravitational force can be represented by:

  F=Mg

  F=GmMr2r^

Where, G is the gravitational constant, m and M are the masses and r is the distance between them.

Calculation:

Physics for Scientists and Engineers, Chapter 11, Problem 74P , additional homework tip  3

The value of |F2| , due to M2 , at r=0.9a0.8a=0.1a<<R2

  |F2|=0

Again, the value of |F1| , due to M1 , at r=0.9a<R1

So, |F1|=0

Hence, net force

  |F|=|F2|+|F1|=0+0

  |F|=0

Conclusion:

Thus, the magnitude of gravitational force due to M1 and M2 at x=0.9a is |F|=0 .

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Chapter 11 Solutions

Physics for Scientists and Engineers

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