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242 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Figure 9-22(a) An accelerating rocket of
massMat time t, as seen from an inertial
reference frame. (b) The same but at time
tdt. The exhaust products released dur-
ing interval dtare shown.

forces acting on it. For this one-dimensional motion, let Mbe the mass of the

rocket and vits velocity at an arbitrary time t(see Fig. 9-22a).

Figure 9-22bshows how things stand a time interval dtlater. The rocket now

has velocity vdvand mass MdM, where the change in mass dMis a negative

quantity.The exhaust products released by the rocket during interval dthave

massdMand velocity Urelative to our inertial reference frame.

Conserve Momentum.Our system consists of the rocket and the exhaust

products released during interval dt. The system is closed and isolated, so the lin-

ear momentum of the system must be conserved during dt; that is,

PiPf, (9-82)

where the subscripts iandfindicate the values at the beginning and end of time

intervaldt.We can rewrite Eq. 9-82 as

MvdM U(MdM)(vdv), (9-83)

where the first term on the right is the linear momentum of the exhaust products

released during interval dtand the second term is the linear momentum of the

rocket at the end of interval dt.

Use Relative Speed.We can simplify Eq. 9-83 by using the relative speed vrelbe-

tween the rocket and the exhaust products, which is related to the velocities relative to

the frame with

.

In symbols, this means

(vdv)vrelU,

or Uvdvvrel. (9-84)

Substituting this result for Uinto Eq. 9-83 yields, with a little algebra,

dM vrelM dv. (9-85)

Dividing each side by dtgives us

(9-86)

We replace dM/dt(the rate at which the rocket loses mass) by R, where Ris the

(positive) mass rate of fuel consumption, and we recognize that dv/dtis the accel-

eration of the rocket. With these changes, Eq. 9-86 becomes

RvrelMa (first rocket equation). (9-87)

Equation 9-87 holds for the values at any given instant.

Note the left side of Eq. 9-87 has the dimensions of force (kg/s m/s

kg m/s^2 N) and depends only on design characteristics of the rocket engine —

namely, the rate Rat which it consumes fuel mass and the speed vrelwith which that

mass is ejected relative to the rocket. We call this term Rvrelthethrustof the rocket

engine and represent it with T. Newton’s second law emerges if we write Eq. 9-87 as

TMa, in which ais the acceleration of the rocket at the time that its mass is M.

Finding the Velocity

How will the velocity of a rocket change as it consumes its fuel? From Eq. 9-85

we have

dvvrel

dM

M

.



dM

dt

vrelM

dv

dt

.



velocity of rocket

relative to frame

velocity of rocket

relative to products

velocity of products

relative to frame 

x

M v

System boundary

(a)

x

M+dM v+ dv

System boundary

(b)

• dM
  U

The ejection of mass from

the rocket's rear increases

the rocket's speed.