## Rocket Propulsion

Rockets provide a wonderful example of

Momentum Conservation.As momentum in one direction is given to the rocket's exhaust gases, momentum in the other direction is given to the rocket itself.First, think of two masses connected by a lightweight (massless!) compressed spring. When the two spring apart, conservation of momentum tells us the Center of Mass remains where it was (or

movingas it was).P

_{Tot,i}= p_{1i}+ p_{2i}= 0 + 0 = 0P

_{Tot,f}= p_{1f}+ p_{2f}= P_{Tot,i}= 0p

_{1f}+ p_{2f}= - m_{1}v_{1f}+ m_{2}v_{2f}= 0m

_{1}v_{1f}= - m_{2}v_{2f}A rocket does much the same thing. Momentum is given to the exhaust gases in one direction so the rocket itself must gain momentum in the other direction. Instead of sending out a single chunk of mass, the rocket fuel is burned

continuously. "Continuously" is a code word meaning "use calculus here".The

resultsarev _{f}= v_{i}+ v_{exh}ln [ M_{i}/ M_{f}]and

Thrust = v

_{exh}[ dM / dt ]Now, for the details, . . .

Consider a rocket initially moving with velocity

vand having mass M + dm. The rocket engine burns rocket fuel which leave with an exhaust speed of v_{e}. During a short amount of time dt, the rocket engine burns an amount of fuel |dm|.(The

signof dm is important and is often the cause of some concern and confusion).The initial momentum of the rocket and fuel is

P

_{init}= (m + |dm|)(v)P

_{init}= m v + |dm| vAfter the amount of fuel |dm| is burned the rocket increases its velocity to

v+ dvand its mass decreases to m so the final momentum of the rocket and fuel isP

_{final}= (m)(v + dv) + (|dm|)(v - v_{e})P

_{final}= m v + m dv + |dm| v - |dm| v_{e}We expect the total momentum of "the system" to remain constant,

P

_{final}= P_{init}m v + m dv + |dm| v - |dm| v

_{e}= m v + |dm| v

m v+ m dv +|dm| v- |dm| v_{e}=m v+|dm| vm dv - |dm| v

_{e}= 0The mass is

decreasing so the change in mass dm is intrinsicly negative, dm < 0.m dv + dm v

_{e}= 0m dv = - v

_{e}dmdv = - v

_{e}[ dm/m ]Now we need to integrate this; remember v

_{e}, the exhaust speed, is aconstant.Remember, M

_{i}> M_{f}so the natural log will be positive. Of course, we would expect that since we also expect v_{f}> v_{i}.

Center of MassSummaryReturn to ToC, Ch9, Linear Momentum(c) Doug Davis, 2001; all rights reserved