ToShow : The element produces a gravitational field at a point x 0 on the x axis in the region x 0 > L is given by d g 0 = − G M L ( x 0 − x ) 2 d x .

Question

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Answer 1

Question

Chapter 11, Problem 68P

(a)

To determine

ToShow: The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

(a)

Expert Solution

Explanation of Solution

Given information :

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used :

Acceleration due to gravity for small mass element ( dm ) can be obtained by:

dg=−Gdmr2

Calculation:

The gravitational field at P due to the element dm is

dg→x=−Gdmr2i^

dm=MLdx

Distance to the point on the axis:

r=x0−x

dg→x={−GML(x0−x)2dx}i^

(b)

To determine

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point x0 due to the rod.

(b)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Calculation:

g→x=−GML∫0Ldx(x0−x)2i^g→x={−GML[1(x0−x)]0L}i^g→x=−GMxo(x0−L)i^

Conclusion:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

(c)

To determine

To Calculate: The gravitational force on a point particle of mass m0 at x0 .

(c)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

Force of gravity

F=mg

Where, m is the mass and gis the acceleration due to gravity.

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculation:

By using the definition of gravitational field and the result from part (b) to express F→g at x=x0 is:

F→g=m0g→xF→g=−GMm0x0(x0−L)i^

Conclusion:

The gravitational force on a point particle of mass m0 at x0 is F→g=−GMm0x0(x0−L)i^

(d)

To determine

To Show:For x0>>L , the field of the rod approximates the field of a point particle of mass M at x=0 .

(d)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculations:

g→x=−GMxo(x0−L)i^

g→x=−GMx02(1−Lx0)i^

For x0>>L , the second term in parentheses is very small.

g→x≈−GMx02i^

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Students have asked these similar questions

A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

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simple diagram to illustrate the setup for each law- coulombs law and biot savart law

Answer 2

Question

Chapter 11, Problem 68P

(a)

To determine

ToShow: The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

(a)

Expert Solution

Explanation of Solution

Given information :

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used :

Acceleration due to gravity for small mass element ( dm ) can be obtained by:

dg=−Gdmr2

Calculation:

The gravitational field at P due to the element dm is

dg→x=−Gdmr2i^

dm=MLdx

Distance to the point on the axis:

r=x0−x

dg→x={−GML(x0−x)2dx}i^

(b)

To determine

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point x0 due to the rod.

(b)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Calculation:

g→x=−GML∫0Ldx(x0−x)2i^g→x={−GML[1(x0−x)]0L}i^g→x=−GMxo(x0−L)i^

Conclusion:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

(c)

To determine

To Calculate: The gravitational force on a point particle of mass m0 at x0 .

(c)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

Force of gravity

F=mg

Where, m is the mass and gis the acceleration due to gravity.

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculation:

By using the definition of gravitational field and the result from part (b) to express F→g at x=x0 is:

F→g=m0g→xF→g=−GMm0x0(x0−L)i^

Conclusion:

The gravitational force on a point particle of mass m0 at x0 is F→g=−GMm0x0(x0−L)i^

(d)

To determine

To Show:For x0>>L , the field of the rod approximates the field of a point particle of mass M at x=0 .

(d)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculations:

g→x=−GMxo(x0−L)i^

g→x=−GMx02(1−Lx0)i^

For x0>>L , the second term in parentheses is very small.

g→x≈−GMx02i^

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Answer 3

Question

Chapter 11, Problem 68P

(a)

To determine

ToShow: The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

(a)

Expert Solution

Explanation of Solution

Given information :

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used :

Acceleration due to gravity for small mass element ( dm ) can be obtained by:

dg=−Gdmr2

Calculation:

The gravitational field at P due to the element dm is

dg→x=−Gdmr2i^

dm=MLdx

Distance to the point on the axis:

r=x0−x

dg→x={−GML(x0−x)2dx}i^

(b)

To determine

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point x0 due to the rod.

(b)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Calculation:

g→x=−GML∫0Ldx(x0−x)2i^g→x={−GML[1(x0−x)]0L}i^g→x=−GMxo(x0−L)i^

Conclusion:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

(c)

To determine

To Calculate: The gravitational force on a point particle of mass m0 at x0 .

(c)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

Force of gravity

F=mg

Where, m is the mass and gis the acceleration due to gravity.

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculation:

By using the definition of gravitational field and the result from part (b) to express F→g at x=x0 is:

F→g=m0g→xF→g=−GMm0x0(x0−L)i^

Conclusion:

The gravitational force on a point particle of mass m0 at x0 is F→g=−GMm0x0(x0−L)i^

(d)

To determine

To Show:For x0>>L , the field of the rod approximates the field of a point particle of mass M at x=0 .

(d)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculations:

g→x=−GMxo(x0−L)i^

g→x=−GMx02(1−Lx0)i^

For x0>>L , the second term in parentheses is very small.

g→x≈−GMx02i^

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Answer 4

Question

Chapter 11, Problem 68P

(a)

To determine

ToShow: The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

(a)

Expert Solution

Explanation of Solution

Given information :

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used :

Acceleration due to gravity for small mass element ( dm ) can be obtained by:

dg=−Gdmr2

Calculation:

The gravitational field at P due to the element dm is

dg→x=−Gdmr2i^

dm=MLdx

Distance to the point on the axis:

r=x0−x

dg→x={−GML(x0−x)2dx}i^

(b)

To determine

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point x0 due to the rod.

(b)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Calculation:

g→x=−GML∫0Ldx(x0−x)2i^g→x={−GML[1(x0−x)]0L}i^g→x=−GMxo(x0−L)i^

Conclusion:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

(c)

To determine

To Calculate: The gravitational force on a point particle of mass m0 at x0 .

(c)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

Force of gravity

F=mg

Where, m is the mass and gis the acceleration due to gravity.

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculation:

By using the definition of gravitational field and the result from part (b) to express F→g at x=x0 is:

F→g=m0g→xF→g=−GMm0x0(x0−L)i^

Conclusion:

The gravitational force on a point particle of mass m0 at x0 is F→g=−GMm0x0(x0−L)i^

(d)

To determine

To Show:For x0>>L , the field of the rod approximates the field of a point particle of mass M at x=0 .

(d)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculations:

g→x=−GMxo(x0−L)i^

g→x=−GMx02(1−Lx0)i^

For x0>>L , the second term in parentheses is very small.

g→x≈−GMx02i^

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 11, Problem 68P

(a)

To determine

ToShow: The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

(a)

Expert Solution

Explanation of Solution

Given information :

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used :

Acceleration due to gravity for small mass element ( dm ) can be obtained by:

dg=−Gdmr2

Calculation:

The gravitational field at P due to the element dm is

dg→x=−Gdmr2i^

dm=MLdx

Distance to the point on the axis:

r=x0−x

dg→x={−GML(x0−x)2dx}i^

(b)

To determine

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point x0 due to the rod.

(b)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Calculation:

g→x=−GML∫0Ldx(x0−x)2i^g→x={−GML[1(x0−x)]0L}i^g→x=−GMxo(x0−L)i^

Conclusion:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

(c)

To determine

To Calculate: The gravitational force on a point particle of mass m0 at x0 .

(c)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

Force of gravity

F=mg

Where, m is the mass and gis the acceleration due to gravity.

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculation:

By using the definition of gravitational field and the result from part (b) to express F→g at x=x0 is:

F→g=m0g→xF→g=−GMm0x0(x0−L)i^

Conclusion:

The gravitational force on a point particle of mass m0 at x0 is F→g=−GMm0x0(x0−L)i^

(d)

To determine

To Show:For x0>>L , the field of the rod approximates the field of a point particle of mass M at x=0 .

(d)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculations:

g→x=−GMxo(x0−L)i^

g→x=−GMx02(1−Lx0)i^

For x0>>L , the second term in parentheses is very small.

g→x≈−GMx02i^

Answer 6

Chapter 11, Problem 68P

(a)

To determine

ToShow: The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

(a)

Expert Solution

Explanation of Solution

Given information :

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used :

Acceleration due to gravity for small mass element ( dm ) can be obtained by:

dg=−Gdmr2

Calculation:

The gravitational field at P due to the element dm is

dg→x=−Gdmr2i^

dm=MLdx

Distance to the point on the axis:

r=x0−x

dg→x={−GML(x0−x)2dx}i^

Answer 7

Chapter 11, Problem 68P

(a)

To determine

ToShow: The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Answer 8

(a)

Expert Solution

Explanation of Solution

Given information :

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used :

Acceleration due to gravity for small mass element ( dm ) can be obtained by:

dg=−Gdmr2

Calculation:

The gravitational field at P due to the element dm is

dg→x=−Gdmr2i^

dm=MLdx

Distance to the point on the axis:

r=x0−x

dg→x={−GML(x0−x)2dx}i^

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point x0 due to the rod.

(b)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Calculation:

g→x=−GML∫0Ldx(x0−x)2i^g→x={−GML[1(x0−x)]0L}i^g→x=−GMxo(x0−L)i^

Conclusion:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Answer 13

(b)

To determine

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point x0 due to the rod.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Calculation:

g→x=−GML∫0Ldx(x0−x)2i^g→x={−GML[1(x0−x)]0L}i^g→x=−GMxo(x0−L)i^

Conclusion:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

To Calculate: The gravitational force on a point particle of mass m0 at x0 .

(c)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

Force of gravity

F=mg

Where, m is the mass and gis the acceleration due to gravity.

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculation:

By using the definition of gravitational field and the result from part (b) to express F→g at x=x0 is:

F→g=m0g→xF→g=−GMm0x0(x0−L)i^

Conclusion:

The gravitational force on a point particle of mass m0 at x0 is F→g=−GMm0x0(x0−L)i^

Answer 19

(c)

To determine

To Calculate: The gravitational force on a point particle of mass m0 at x0 .

Answer 20

(c)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

Force of gravity

F=mg

Where, m is the mass and gis the acceleration due to gravity.

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculation:

By using the definition of gravitational field and the result from part (b) to express F→g at x=x0 is:

F→g=m0g→xF→g=−GMm0x0(x0−L)i^

Conclusion:

The gravitational force on a point particle of mass m0 at x0 is F→g=−GMm0x0(x0−L)i^

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

To determine

To Show:For x0>>L , the field of the rod approximates the field of a point particle of mass M at x=0 .

(d)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculations:

g→x=−GMxo(x0−L)i^

g→x=−GMx02(1−Lx0)i^

For x0>>L , the second term in parentheses is very small.

g→x≈−GMx02i^

Answer 25

(d)

To determine

To Show:For x0>>L , the field of the rod approximates the field of a point particle of mass M at x=0 .

Answer 26

(d)

Expert Solution

Explanation of Solution

Given information:

Mass of the rod =M

Length of the rod =L

Length of an element of the rod =dx

Mass of an element of the rod =dm

Formula used:

The total gravitational field at the point x0 due to the rod is: g→x=−GMxo(x0−L)i^

Calculations:

g→x=−GMxo(x0−L)i^

g→x=−GMx02(1−Lx0)i^

For x0>>L , the second term in parentheses is very small.

g→x≈−GMx02i^

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Ch. 11 - Prob. 1P Ch. 11 - Prob. 2P Ch. 11 - Prob. 3P Ch. 11 - Prob. 4P Ch. 11 - Prob. 5P Ch. 11 - Prob. 6P Ch. 11 - Prob. 7P Ch. 11 - Prob. 8P Ch. 11 - Prob. 9P Ch. 11 - Prob. 10P

ToShow : The element produces a gravitational field at a point x 0 on the x axis in the region x 0 > L is given by d g 0 = − G M L ( x 0 − x ) 2 d x .

Concept explainers

Videos

ToShow: The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Explanation of Solution

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point x0 due to the rod.

Explanation of Solution

To Calculate: The gravitational force on a point particle of mass m0 at x0 .

Explanation of Solution

To Show:For x0>>L , the field of the rod approximates the field of a point particle of mass M at x=0 .

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

ToShow : The element produces a gravitational field at a point x 0 on the x axis in the region x 0 &gt; L is given by d g 0 = − G M L ( x 0 − x ) 2 d x .

Concept explainers

Videos

ToShow: The element produces a gravitational field at a point x0 on the x axis in the region x0>L is given by dg0=−GML(x0−x)2dx .

Explanation of Solution

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point x0 due to the rod.

Explanation of Solution

To Calculate: The gravitational force on a point particle of mass m0 at x0 .

Explanation of Solution

To Show:For x0>>L , the field of the rod approximates the field of a point particle of mass M at x=0 .

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

ToShow : The element produces a gravitational field at a point x 0 on the x axis in the region x 0 > L is given by d g 0 = − G M L ( x 0 − x ) 2 d x .