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# Momentum: Rocket Firing in Space

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A rocket, which is in deep space and initially at rest relative to an inertial reference frame, has a mass of 265*10^105 kg, of which 200*10^5 kg is fuel. The engine is fired for 250s, during which fuel is consumed at the rate of 480 kg/s. The speed of the exhaust product relative to the rocket is 3.23 km/s.
a) What is the rocket's thrust?
b) What is the mass of the rocket after the 250s firing?
c) What is the speed of the rocket after the 250s firing?

##### Solution Summary

The change in momentum of the rocket body is dP(rocket) = d(Mv)= dM *v + M * dv. This equation says that the rocket changes momentum due to its loss in mass (which is just the mass of gas expelled) times its initial velocity plus the mass of the rocket times its change in velocity. To get the momentum of the expelled gas, we need to know its velocity.

##### Solution Preview

At initial time t, the rocket is moving with velocity v and has a mass M. Thus the momentum of the system is just Mv and will always remain so since momentum is conserved for the system. After a small amount of time dt, the rocket will have expelled a mass dm of gas and the rocket body will have gained a small amount of speed v + dv. The change in momentum of the rocket body is

dP(rocket) = d(Mv)= dM *v + M * dv ...(1)

This equation says that the rocket changes momentum due to its loss in mass (which is just the mass of gas expelled) times its initial velocity plus the mass ...

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