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# supplement8 - Paul A Tlplér i'PhYSiCS II andenginm for...

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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 8-115 242 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+ on-f 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 3-3? 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 3-38 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 3-40 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- 8-41 Thus. a rocket starting at rest with mass mg, and payload of mf, attains a final speed as waging—3a _ 3-42 Has! speed of roster assuming the acceleration of gravity to be constant. ...
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