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?

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.