Ok, in the previous posts I mistakenly took "u" to be the initial velocity, since I'm use to a certain convention where "u" is the initial velocity, but in this case it is the exhaust velocity, sorry
Well, I think I'm beginning to understand a bit what's going on here.
The gain in kinetic energy due to the rocket emitting a small amount of matter
only and
not from added source, is given by:
phyzmatix said:
\frac{1}{2}M_0u^2(1-e^{-v/u})
(I still don't completely understand why this is the equation and why it is only determined by the exhaust velocity and not other velocity components)
Now you can calculate the
total kinetic energy gained by the rocket as follows (this
total kinetic energy gained is due to the rocket emitting small amounts of matter
plus extra supplied kinetic energy from other sources.)
Consider the problem consisting of three different components:
1.) The exhaust cloud
2.) The fuel inside the rocket (this is the part of the total mass of the rocket which will be decreasing)
3.) The shell of the rocket (without fuel) (this is the part of the total mass of the rocket which will stay the same)
Each of these components have a mass and a velocity.
1.) For the shell of the rocket:
Initial mass: m_{r}
Initial velocity: v_{0}
Final mass: m_{r}
Final velocity: v_{0}+\Delta v
2.) For the fuel inside the rocket:
Initial mass: m_{f}
Initial velocity: v_{0}
Final mass: m_{f}-\Delta m
Final velocity: v_{0}+\Delta v
3.) For the exhaust cloud:
Initial mass: 0
Final mass: \Delta m
Final velocity: v_{0}+\Delta v-u
It can be seen that the initial and final velocity of the fuel and the rocket is the same, since they are moving together and the final velocity of the exhaust cloud is again an inertial frame, thus it has a component due to moving with the rocket ( v_{0}+\Delta v ) minus the component at which it is emitting from the rocket, since it is emitted in the opposite direction to which the rocket is moving, to accelerate the rocket ( u )
The initial kinetic energy is thus:
T_{0}=\frac{1}{2}m_{r}v^{2}_{0}+\frac{1}{2}m_{f}v^{2}_{0}
which is just due to the shell of the rocket and the fuel, since the exhaust cloud does not exist yet.
The final kinetic energy is thus:
T_{F}=\frac{1}{2}m_{r}\left(v_{0}+\Delta v\right)^{2}+\frac{1}{2}\left(m_{f}-\Delta m\right)\left(v_{0}+\Delta v\right)^{2}+\frac{1}{2}\Delta m\left(v_{0}+\Delta v-u\right)^{2}
T_{F}=\frac{1}{2}m_{r}\left(v^{2}_{0}+2v_{0}\Delta v+\Delta v^{2}\right)+\frac{1}{2}\left(m_{f}-\Delta m\right)\left(v^{2}_{0}+2v_{0}\Delta v+\Delta v^{2}\right)+\frac{1}{2}\Delta m\left(v^{2}_{0}+2v_{0}\Delta v-2v_{0}u+\Delta v^{2}-2\Delta vu+u^{2}\right)
T_{F}=\frac{1}{2}m_{r}v^{2}_{0}+m_{r}v_{0}\Delta v+\frac{1}{2}m_{r}\Delta v^{2}+\frac{1}{2}m_{f}v^{2}_{0}+m_{f}v_{0}\Delta v+\frac{1}{2}m_{f}\Delta v^{2}-\frac{1}{2}\Delta mv^{2}_{0}-\Delta mv_{0}\Delta v-\frac{1}{2}\Delta m\Delta v^{2}
+\frac{1}{2}\Delta mv^{2}_{0}+\Delta mv_{0}\Delta v-\Delta mv_{0}u+\frac{1}{2}\Delta m\Delta v^{2}-\Delta m\Delta vu+\frac{1}{2}\Delta mu^{2}
T_{F}=\frac{1}{2}m_{r}v^{2}_{0}+m_{r}v_{0}\Delta v+\frac{1}{2}m_{r}\Delta v^{2}+\frac{1}{2}m_{f}v^{2}_{0}+m_{f}v_{0}\Delta v+\frac{1}{2}m_{f}\Delta v^{2}-\Delta mv_{0}u-\Delta m\Delta vu+\frac{1}{2}\Delta mu^{2}
Thus the gain in kinetic energy is equal to the final kinetic energy minus the initial kinetic energy:
T_{F}-T_{0}=m_{r}v_{0}\Delta v+\frac{1}{2}m_{r}\Delta v^{2}+m_{f}v_{0}\Delta v+\frac{1}{2}m_{f}\Delta v^{2}-\Delta mv_{0}u-\Delta m\Delta vu+\frac{1}{2}\Delta mu^{2}
Now I know you can group terms together to reduce it more, but I didn't do it, because I don't really know which terms to group together to get the best result, because if I'm right you have subtract \frac{1}{2}M_0u^2(1-e^{-v/u}) from the above answer to get the energy that needs to be supplied by other sources, right?
Plus if the rocket is "held fixed on a test bed" does it imply that it is stationary and have a constant velocity of 0 or that it has a velocity, but can't accelerate?
