9781118230725.pdf

Integrating leads to

in which Miis the initial mass of the rocket and Mfits final mass. Evaluating the
integrals then gives

(second rocket equation) (9-88)

for the increase in the speed of the rocket during the change in mass from Mito
Mf. (The symbol “ln” in Eq. 9-88 means the natural logarithm.) We see here the
advantage of multistage rockets, in which Mfis reduced by discarding successive
stages when their fuel is depleted. An ideal rocket would reach its destination
with only its payload remaining.

vfvivrel ln

Mi

Mf



vf

vi

dvvrel

Mf

Mi

dM

M

,

REVIEW & SUMMARY 243

rocket’s mass. However,Mdecreases and aincreases as fuel

is consumed. Because we want the initial value of ahere, we

must use the intial value Miof the mass.

Calculation:We find

(Answer)

To be launched from Earth’s surface, a rocket must have

an initial acceleration greater than. That is, it

must be greater than the gravitational acceleration at the

surface. Put another way, the thrust Tof the rocket engine

must exceed the initial gravitational force on the rocket,

which here has the magnitude Mig, which gives us

(850 kg)(9.8 m/s^2 )8330 N.

Because the acceleration or thrust requirement is not met

(hereT6400 N), our rocket could not be launched from

Earth’s surface by itself; it would require another, more

powerful, rocket.

g9.8 m/s^2

a

T

Mi



6440 N

850 kg

7.6 m/s^2.

Sample Problem 9.09 Rocket engine, thrust, acceleration

In all previous examples in this chapter, the mass of a system
is constant (fixed as a certain number). Here is an example of
a system (a rocket) that is losing mass. A rocket whose initial
massMiis 850 kg consumes fuel at the rate The
speedvrelof the exhaust gases relative to the rocket engine is
2800 m/s. What thrust does the rocket engine provide?

KEY IDEA

Thrust Tis equal to the product of the fuel consumption
rateRand the relative speed vrelat which exhaust gases are
expelled, as given by Eq. 9-87.

Calculation: Here we find

(Answer)

(b) What is the initial acceleration of the rocket?

KEY IDEA

We can relate the thrust Tof a rocket to the magnitude aof
the resulting acceleration with TMa, where M is the

6440 N6400 N.

TRvrel(2.3 kg/s)(2800 m/s)

R2.3 kg/s.

Additional examples, video, and practice available at WileyPLUS

Center of Mass The center of massof a system of nparticles is

defined to be the point whose coordinates are given by

(9-5)

or (9-8)

whereMis the total mass of the system.

:rcom

1

M 

n

i 1

mi:ri,

xcom

1

M 

n

i 1

mixi, ycom

1

M

n

i 1

miyi, zcom

1

M 

n

i 1

mizi,

Review & Summary

Newton’s Second Law for a System of Particles The

motion of the center of mass of any system of particles is governed

byNewton’s second law for a system of particles,which is

. (9-14)

Here F is the net force of all the externalforces acting on the sys-

:

net

F

:

netMa

:

com

tem,Mis the total mass of the system, and is the acceleration

of the system’s center of mass.

a:com