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Unformatted text preview: Paul A. Tlplér ' i. 'PhYSiCS II andenginm for scientists Fumfh Edition
_ Volume I
. Mania: .
IDmililatinns and Warren rm. FREEMAN} mmmtwomu 'Pmusmns Rocket Propulsion Rocket Propulsion is a striking exampie of the conservation of momentum in
action. The mathematical description of rocket propulsion can became quite
complex because the mass of the tocket changes continuously as it bums fuel
and expels exhaust gas. The easiest approach is to compute the change in the
momentum of the total system (including the athaust gas) for some time in
terval and use Newton’s law in the form Pen 2 dP/dt, where Fexl is the net
force acting on the rocket _ Consider a rocket moving with speed '0 relative to the earth (Figpre SAME)
If the fuel is bmned at a constant rate, R = Idm/dfﬁ, the rocket's umss at time t is m = mu — Rt . — 335
where m0 is the initial'inass of the rocket The momentum of the syﬁem at
time t is P; = me At a later time t + ﬁt, the rocket has expelled gas of mass R ﬁt If the gas is
exhausted at a speed ﬂex relative to the rocket, the Veiecity of the gas Ieletive to
the earth is v — Hg. The rocket then has a mass m —R A! and is moving at a speed I} + AI: (Figure 8—45) IAMI Figure 8115 242 CHAPTER 3 Systems of Particles and Conservation of Momentum L The momentum of the system at t + Mia
Pf = (m — Rotﬂv + do) + Rare; " Hex)
=mo+ movwoRot ~RA£Av+vRM— uﬂRM
mmv+ onf HexRﬁt where we have dropped the term R or do, which is the product of two vory
small quantities, and therefore negligible compared with the others. The
change in momentum is AP=Pf—Pi=on—uﬂRot
and "E  orE  u R 3 Be At a: 3" . . '
As At approaches. zero, Ari/At approaches the derivative do/dt, which '15 the .' 
acceleration. For a rocket moving upward near the surface of the earth, 1:“ = emg. Setting dP/dt = Pm = —mg gives 'us the rocket equation: do '
or; = Ruex + PM = Rem  mg 33?
Rocket equation
or
do Re Re ' I' " The quantity Ratio: is the force exerted on the rocket by the exhausting fuel.
This is Called the thrust? ﬁlm “at . 8‘39 Definfrisnmﬂostet thrust Equation 338 is solved by integrating both sides with respect to time. For a
rocket starting at rest at t = U, the result is v : —uex m(WTF) ._ gt 340 as can be veriﬁed by taking the time derivative of o. The payload of a rocket
is the final mass, on, after all the fuel has been burned. The burn time tb is
given by m; = ”In  Rtb, or m “m
thr: UR f 841 Thus. a rocket starting at rest with mass mg, and payload of mf, attains a final
speed as
waging—3a _ 342 Has! speed of roster assuming the acceleration of gravity to be constant. ...
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 Spring '08
 WHITE
 Physics

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