read said:
Q'= -m'(V_ex-V)^2/2
T'=((mV^2)/2)'=mVV'
We get different answers for these two expressions. The way I worked out these two, is this way: Suppose our rocket at one instant has mass m, moving at speed v, then a short time later, the mass and velocity of our ship have both changed by some small amounts, to v+ \Delta v and m+ \Delta m So by conservation of mass, we will need to have a small amount of propellant ejected, with mass equal to - \Delta m and we define it to be moving at speed v_{ex}+v (i.e. the exhaust velocity is the relative velocity of the propellant with respect to the rocket). Therefore, using conservation of momentum, (i.e. total momentum before = total momentum after), we get the equation:
mv = (m+ \Delta m)(v+ \Delta v) - \Delta m(v_{ex} +v)
Now, cancelling some terms, and dividing by the small time interval \Delta t, we get:
m \frac{\Delta v}{\Delta t} + \frac{\Delta m \Delta v}{\Delta t} = v_{ex} \frac{\Delta m}{\Delta t}
Now, take the limit that these small changes are infinitesimally small changes, then the second term on the left hand side disappears (because it is much smaller than the other terms), so we get left with:
m \frac{dv}{dt} = v_{ex} \frac{dm}{dt}
hooray! Now, it is possible to find things like the rate of change of the KE of the rocket and the rate of change of total KE, by using a very similar line of reasoning as above. Right, so to find out what will be the small change in KE of the rocket, we simply use the KE of rocket after, minus the KE of the rocket before, so we get:
\Delta T = \frac{1}{2} (m+ \Delta m)(v+ \Delta v )^2 - \frac{1}{2}mv^2
And straight away, I am going to ignore any terms which have a product of more than one 'delta', since these terms are going to be much smaller, when we take our limit of very small changes, so therefore, we get:
\Delta T = mv \Delta v + \frac{1}{2} v^2 \Delta m
So, now, again dividing by \Delta t and taking the limit of very small change, we get:
\frac{dT}{dt}=mv \frac{dv}{dt} + \frac{1}{2} v^2 \frac{dm}{dt}
And now we simply use the relation \dot{m} v_{ex} = m \dot{v} with the first term of our equation for T, to get the nicer expression:
\frac{dT}{dt} = \frac{dm}{dt} v (v_{ex} + \frac{1}{2} v )
hooray. And now, to find the change in KE of the system, we use the KE of the rocket after, and plus the KE of the propellant, and minus the KE of the rocket before (since we want to find the total change in KE of the system). So we get:
\Delta Q = \Delta T - \frac{1}{2} \Delta m (v_{ex} + v)^2
Note: we are adding the KE of the propellant, but since the mass of the propellant is - \Delta m, we get a minus sign. So again, discarding any terms where there is a product of delta's, we have:
\Delta Q = mv \Delta v - v_{ex} v \Delta{m} - \frac{1}{2} v_{ex}^2 \Delta m
So, divide by delta t and take limit of small changes:
\dot{Q} = mv \dot{v} - \dot{m} v_{ex} v - \frac{1}{2} v_{ex}^2 \dot{m}
But luckily, using our equation \dot{m} v_{ex} = m \dot{v} the first two terms on the right-hand side cancel, so we end up with:
\dot{Q} = - \frac{1}{2} \dot{m} v_{ex}^2
