Question
Book Icon
Chapter 11, Problem 65P

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 65P

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 .

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term, Chapter 11, Problem 65P

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 65P

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 65P

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 65P

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 65P

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 .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A solid copper sphere of mass M and radius R has a cavity of radius ½ R. Inside the cavity a particle of mass m placed a distance d > R from the center of the sphere along the line connecting the centers of the sphere and the cavity. Find the gravitational force on m.
(a) Imagine that a space probe could be fired as a projectile from the Earth's surface with an initial speed of 5.78 × 10* m/s relative to the Sun. What would its speed be when it is very far from the Earth (in m/s)? Ignore atmospheric friction, the effects of other planets, and the rotation of the Earth. (Consider the mass of the Sun in your calculations.) 38107.8 m/s (b) What If? The speed provided in part (a) is very difficult to achieve technologically. Often, Jupiter is used as a "gravitational slingshot" to increase the speed of a probe to the escape speed from the solar system, which is 1.85 × 10“ m/s from a point on Jupiter's orbit around the Sun (if Jupiter is not nearby). If the probe is launched from the Earth's surface at a speed of 4.10 x 104 m/s relative to the Sun, what is the increase in speed needed from the gravitational slingshot at Jupiter for the space probe to escape the solar system (in m/s)? (Assume that the Earth and the point on Jupiter's orbit lie along the…
(a) Imagine that a space probe could be fired as a projectile from the Earth's surface with an initial speed of 5.34 x 104 m/s relative to the Sun. What would its speed be when it is very far from the Earth (in m/s)? Ignore atmospheric friction, the effects of other planets, and the rotation of the Earth. (Consider the mass of the Sun in your calculations.) m/s (b) What If? The speed provided in part (a) is very difficult to achieve technologically. Often, Jupiter is used as a "gravitational slingshot" to increase the speed of a probe to the escape speed from the solar system, which is 1.85 x 104 m/s from a point on Jupiter's orbit around the Sun (if Jupiter is not nearby). If the probe is launched from the Earth's surface at a speed of 4.10 x 104 m/s relative to the Sun, what is the increase in speed needed from the gravitational slingshot at Jupiter for the space probe to escape the solar system (in m/s)? (Assume that the Earth and the point on Jupiter's orbit lie along the same…

Chapter 11 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

Ch. 11 - Prob. 7OQCh. 11 - Prob. 8OQCh. 11 - Prob. 9OQCh. 11 - Rank the following quantities of energy from...Ch. 11 - Prob. 11OQCh. 11 - Prob. 12OQCh. 11 - Prob. 13OQCh. 11 - Prob. 14OQCh. 11 - Prob. 1CQCh. 11 - Prob. 2CQCh. 11 - Prob. 3CQCh. 11 - Prob. 4CQCh. 11 - Prob. 5CQCh. 11 - Prob. 6CQCh. 11 - Prob. 7CQCh. 11 - Prob. 8CQCh. 11 - In his 1798 experiment, Cavendish was said to have...Ch. 11 - Prob. 1PCh. 11 - Prob. 2PCh. 11 - A 200-kg object and a 500-kg object are separated...Ch. 11 - Prob. 4PCh. 11 - Prob. 5PCh. 11 - Prob. 6PCh. 11 - Prob. 7PCh. 11 - Prob. 8PCh. 11 - Prob. 9PCh. 11 - Prob. 10PCh. 11 - A spacecraft in the shape of a long cylinder has a...Ch. 11 - (a) Compute the vector gravitational field at a...Ch. 11 - Prob. 13PCh. 11 - Two planets X and Y travel counterclockwise in...Ch. 11 - Prob. 15PCh. 11 - Prob. 16PCh. 11 - Prob. 17PCh. 11 - Prob. 18PCh. 11 - Plasketts binary system consists of two stars that...Ch. 11 - As thermonuclear fusion proceeds in its core, the...Ch. 11 - Comet Halley (Fig. P11.21) approaches the Sun to...Ch. 11 - Prob. 22PCh. 11 - Prob. 23PCh. 11 - Prob. 24PCh. 11 - Prob. 25PCh. 11 - A space probe is fired as a projectile from the...Ch. 11 - Prob. 27PCh. 11 - Prob. 28PCh. 11 - Prob. 29PCh. 11 - Prob. 30PCh. 11 - Prob. 31PCh. 11 - Prob. 32PCh. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Prob. 36PCh. 11 - Prob. 37PCh. 11 - Prob. 38PCh. 11 - Prob. 39PCh. 11 - Prob. 40PCh. 11 - Prob. 41PCh. 11 - Prob. 42PCh. 11 - Prob. 43PCh. 11 - Prob. 44PCh. 11 - Prob. 45PCh. 11 - Prob. 46PCh. 11 - Let gM represent the difference in the...Ch. 11 - Prob. 48PCh. 11 - Prob. 49PCh. 11 - Two stars of masses M and m, separated by a...Ch. 11 - Prob. 51PCh. 11 - Prob. 52PCh. 11 - Prob. 53PCh. 11 - Prob. 54PCh. 11 - Prob. 55PCh. 11 - Prob. 56PCh. 11 - Prob. 57PCh. 11 - Prob. 58PCh. 11 - Prob. 59PCh. 11 - Prob. 60PCh. 11 - Prob. 61PCh. 11 - Prob. 62PCh. 11 - Prob. 63PCh. 11 - Prob. 64PCh. 11 - Prob. 65P
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill