Question

of the a) (5 p) Interpret the rocket equation dv(t)M(t)=-udM(t) [EQ.1) within the framework law of momentum conservation, written in a closed system; here M(t) is the rocket mass, at time t, whereas dM(t) is by definition, dM(t)=M(t+dt)-M(t); dM(t)=|dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)=v(t+dt)-v(t), i.e. the increase in the velocity of the rocket through the period of time dt; u is the relative velocity of the exhaust gas, with respect to the rocket.

Answer 1

Solution

The rocket equation is nothing but conservation of linear momentum stated in a particular case of closed system of rocket and expelled gas.

Dolbron The of a closed lonear givers rocket equanian Luck) M(A) - - UdMCA) is nothing but conservation linear momentum systems. Momentum conservatias Stakes that case of a closed systear no external forces 3 the net momentum of the system must be conserved. Here consides the rocket and the exhaust one combined cosed systems since there are no external forces acting on the combined system, ů when we If MCA) is the mass of crocket and it changed its velocity from vef) to Nettde) by expelling and amount 09 gas dM(A) = -(M Clado) - MCE)) Conservation of momentum a) Initial momentury of system Final momentum of syptens. case of roocket change in momentens of od gas, such that total change in momenten is zero making momentens consequed. Mct) (NCE +df) - VCR)) = (M(t)-M (ftat)) or in gas ystems, change in momentum of rocket = Mct dut) DMCA U which is the roclect equation.