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The Rocket Equation |
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| Jan28-09, 04:36 PM | #1 |
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The Rocket Equation
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? |
| Jan28-09, 06:31 PM | #2 |
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| Jan28-09, 07:14 PM | #3 |
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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? |
| Jan28-09, 07:26 PM | #4 |
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The Rocket Equation |
| Jan28-09, 07:36 PM | #5 |
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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. |
| Jan28-09, 07:50 PM | #6 |
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