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Escape Velocity. According to Newton’s law of gravitation,

Fundamentals of Differential Equations | 8th Edition | ISBN: 9780321747730 | Authors: R. Kent Nagle, Edward B. Saff, Arthur David Snider ISBN: 9780321747730 43

Solution for problem 25E Chapter 3.4

Fundamentals of Differential Equations | 8th Edition

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Fundamentals of Differential Equations | 8th Edition | ISBN: 9780321747730 | Authors: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Fundamentals of Differential Equations | 8th Edition

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Problem 25E

Problem 25E

Problem

Escape Velocity. According to Newton’s law of gravitation, the attractive force between two objects varies inversely as the square of the distances between them. That is,Fg = GM1M2/r2 where M1 and M2 are the masses of the objects, r is the distance between them (center to center), Fg is the attractive force, and G is the constant of proportionality. Consider a projectile of constant mass m being fired vertically from Earth (see Figure 3.12). Let t represent time and v the velocity of the projectile.  Figure 3.12 Projectile escaping from Earth

(a) Show that the motion of the projectile, under Earth’s gravitational force, is governed by the Equation  where r is the distance between the projectile and the center of Earth, R is the radius of Earth, M is the mass of Earth, and g = GM/R2

(b) Use the fact that dr/dt = v to obtain

(c) If the projectile leaves Earth’s surface with velocity v0, show that

(d) Use the result of part (c) to show that the velocity of the projectile remains positive if and only if The velocity  is called the escape velocity of Earth.

(e) If g = 9.81 m/sec2 and R = 6370 km for Earth, what is Earth’s escape velocity?

(f ) If the acceleration due to gravity for the Moon is gm = g/6 and the radius of the Moon is Rm = 1738 km, what is the escape velocity of the Moon?

Step-by-Step Solution:
Step 1 of 3

Solution:-

Step1

(a)We have to show that the motion of the projectile.Earth’s gravitational force, is  where r is the distance between the projectile and the center of Earth, R is the radius of Earth, M is the mass of Earth, and g = .

We have

According to Newton’s law of gravitation, the attractive force between two objects varies inversely as the square of the distances between them.

 That is,

where  and  are the masses of the objects, r is the distance between them (center to center),  is the attractive force, and G is the constant of proportionality.

    (since,

Therefore, where r is the distance between the projectile and the center of Earth, R is the radius of Earth, M is the mass of Earth.

Step2

(b) Use the fact that dr/dt = v to obtain .

Therefore,

Step3

(c) If the projectile leaves Earth’s surface with velocity v0,  we have to show that

If the projectile leaves Earth’s surface with velocity

Integrate both side we get

Step4

--------(1)

When r=R, v=

Put value of C in (1) we get

Therefore,

Step5

(d) we have to use the result of part (c) to show that the velocity of the projectile remains positive if and only if The velocity  is called the escape velocity of Earth.

Result of part(c) is


Therefore, This shows that the velocity of the projectile remains positive.

Step6

(e) If g = 9.81 m/sec2 and R = 6370 km for Earth, what is Earth’s escape velocity?

 Earth’s escape velocity ,  is

    =

   =

   =

=

=11178.30

=11.18

Therefore, Earth’s escape velocity is .

Step7

(f ) If the acceleration due to gravity for the Moon is gm = g/6 and the radius of the Moon is Rm = 1738 km, what is the escape velocity of the Moon?

 The escape velocity of the Moon ,  is

 =

     =

    =

=

=

=75.39

=

=2383.83

= 2.38 km/sec

Therefore,  The escape velocity of the Moon is 2.38 km/sec.

Step 2 of 3

Chapter 3.4, Problem 25E is Solved
Step 3 of 3

Textbook: Fundamentals of Differential Equations
Edition: 8
Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider
ISBN: 9780321747730

This full solution covers the following key subjects: EARTH, velocity, projectile, escape, show. This expansive textbook survival guide covers 67 chapters, and 2118 solutions. This textbook survival guide was created for the textbook: Fundamentals of Differential Equations , edition: 8. The full step-by-step solution to problem: 25E from chapter: 3.4 was answered by , our top Calculus solution expert on 07/11/17, 04:37AM. The answer to “Escape Velocity. According to Newton’s law of gravitation, the attractive force between two objects varies inversely as the square of the distances between them. That is,Fg = GM1M2/r2 where M1 and M2 are the masses of the objects, r is the distance between them (center to center), Fg is the attractive force, and G is the constant of proportionality. Consider a projectile of constant mass m being fired vertically from Earth (see Figure 3.12). Let t represent time and v the velocity of the projectile. Figure 3.12 Projectile escaping from Earth(a) Show that the motion of the projectile, under Earth’s gravitational force, is governed by the Equation where r is the distance between the projectile and the center of Earth, R is the radius of Earth, M is the mass of Earth, and g = GM/R2(b) Use the fact that dr/dt = v to obtain (c) If the projectile leaves Earth’s surface with velocity v0, show that (d) Use the result of part (c) to show that the velocity of the projectile remains positive if and only if The velocity is called the escape velocity of Earth.(e) If g = 9.81 m/sec2 and R = 6370 km for Earth, what is Earth’s escape velocity?(f ) If the acceleration due to gravity for the Moon is gm = g/6 and the radius of the Moon is Rm = 1738 km, what is the escape velocity of the Moon?” is broken down into a number of easy to follow steps, and 237 words. Fundamentals of Differential Equations was written by and is associated to the ISBN: 9780321747730. Since the solution to 25E from 3.4 chapter was answered, more than 285 students have viewed the full step-by-step answer.

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