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The Rocket Equation 
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#1
Jan2809, 04:36 PM

P: 7

In deriving the rocket equation, there is one part I don't understand. The velocity of exhaust with respect to the body is assumed to be constant, where:
v(exhaust wrt body)=v(exhaust wrt inertial)v(body wrt inertial) So assuming a constant mass flow rate, the rocket propellant exerts a constant force on the rocket and hence in space, uniform acceleration. But how can an observer on the accelerating rocket observe the rocket propellant being ejected with a constant velocity? 


#2
Jan2809, 06:31 PM

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#3
Jan2809, 07:14 PM

P: 7

Oops. So assuming a constant mass flow rate the propellant exerts a constant force on the rocket so the rocket's acceleration increases as follows:
a(t)=F[1/m(t)] where F is a constant (until fuel runs out) where m(t) is the mass of the rocket at time t. m(t)=m(initial)bt where b is a constant (mass flow rate) Hence v_rocket(t)=Fln(m(t))/b assuming v(0)=0 But why would an observer in the rocket observe a constant propellant velocity? 


#4
Jan2809, 07:26 PM

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The Rocket Equation



#5
Jan2809, 07:36 PM

P: 7

Oh, yeah. The engine is on the back of the rocket, so it accelerates with the rocket, so if an engine ejects exhaust at a velocity v, then this is what is observed from the rocket's point of view.
Now I feel somewhat embarrassed, but at least the rocket equation makes sense now. 


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