Thursday, January 14, 2010
Suppose that a certain instant of time t the velocity of the rocket is v and its rest mass m. Suppose that the rocket ejects gas of rest mass k units/sec. Consider another instant of time t + dt, where dt is an infinitesimally small interval of time. The rocket has ejected kdt amount of rest mass. Suppose that the rocket's velocity has increased by an amount dv during this interval. Since there is no external force on the rocket, the momentum of the (ejected gas + rocket) system at t must be equal to that at t+dt.
The momentum of the rocket at time t is given by . The momentum of the rocket at t+dt is given as . The gas ejected by the rocket has a of mass and moves at a non-relativistic velocity u relative to the rocket in the opposite direction to the rocket. In other words, its velocity as seen by an observer on Earth will be v-u in the direction of the rocket (this is shown in the diagram). Hence, the momentum of the ejected gas will be .
Applying conservation of momentum at t and t+dt we have,
We can now use Taylor series expansion and retain only the first order terms in Eqn (1) to obtain,
Now we know that the mass of the rocket reduces at a rate k, in other words -kdt = dm. Hence, (2) can be rewritten as,
Posted by Krishna Kant Chintalapudi at 8:01 AM